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How do I find the extraneous solution of sqrt(x+4)=x-2?
Mar 8, 2018
$x = 0 \text{ is an extraneous solution}$
Explanation:
$\text{square both sides to 'undo' the radical}$
${\left(\sqrt{x + 4}\right)}^{2} = {\left(x - 2\right)}^{2}$
$\Rightarrow x + 4 = {x}^{2} - 4 x + 4$
$\text{rearrange into standard form}$
$\Rightarrow {x}^{2} - 5 x = 0 \leftarrow \textcolor{b l u e}{\text{in standard form}}$
$\Rightarrow x \left(x - 5\right) = 0$
$\Rightarrow x = 0 \text{ or } x = 5$
$\textcolor{b l u e}{\text{As a check}}$
$\text{Substitute these values into the original equation}$
$x = 0 \to \sqrt{4} = 2 \text{ and } 0 - 2 = - 2$
$2 \ne - 2 \Rightarrow x = 0 \textcolor{red}{\text{ is an extraneous solution}}$
$x = 5 \to \sqrt{9} = 3 \text{ and "5-2=3larr" True}$
$\Rightarrow x = 5 \textcolor{red}{\text{ is the solution}}$ |
Alright, so you want to do some LaTeX-ing? OK, so maybe you have to do some LaTeX-ing. I know. I feel your pain. But honestly, it's good for you! Like brussels sprouts! No, bear with me on this. It's not as bad as it seems. For some reason people really love to wax philosophic about LaTeX and can write big complicated websites on it. When I learned it, I was like you. All we want are the facts, ma'am. I'll try to cut through some of the cruft. Here's the stuff you really need to know: WYSIWYG vs WYSIWYW Compiling by hand LaTeX editors (for whimps, don't do this! I won't be able to help you much) Shameless referencing, especially when you've got a template to work off of (it's a good way to learn; see how someone else did things) Symbol reference sheets The important stuff: Lists: \begin{itemize} or \begin{enumerate} and \end{itemize} or \end{enumerate} Each new item has an \item tag Enumerate is the numbered list. Math Mode ($x$ or or eqnarray) Equation Arrays: \begin{eqnarray*} \\ \end{eqnarray*} The * suppresses line numbers Need line breaks after every line Use & to line up separate lines Parenthesis: the \left( \right) trick. Proof Trees (for this class) For the future: on writing papers, articles, Bibtex, etc. See, that wasn't so bad, was it?
Allons-y! Back to CIS160 Page |
# square root of 4489 by division method
Minimum number of more plants to make the number of rows and number of columns same. For a P.T. MEDIUM. He wants to plant these in such a way that the number of rows and the number of columns remain same. 3. Join now. A gardener has 1000 plants. Join now. Ask your question. In a right-angle triangle if two sides are given then third side can be calculated using the Pythagoras theorem. Answer. The remainder obtained is $$53$$. In each row, the number of children is 22. Also, find the square root of the perfect square so obtained: 5. Find the Square Root of the Following Number by Division Method. Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. The square root of $$3136$$ is calculated as follows. Find the square root of each of the following numbers by division method. (i) 2304 (ii) 4489 (iii) In this, first group the numbers into two starting from unit place, these groups are called periods. Find the number of digits in the square root of each of the following numbers (without any calculation). A. (vii) 5776 (viii) 7921 (ix) 576 (x) 1024 (xi) 3136 (xii) 900. Hence number to be added to $$1000$$ to make it perfect square, .\begin{align} &= {32^2} - 1000\\ &= 1024 - 1000\\ &= 24 \end{align}, Thus, the required number of plants $$=24$$. Find the number of digits in the square root of each of the following numbers (without any calculation): (i) 64 (ii) 144 (iii) 4489 (iv) 27225 (v) 390625. Hence, the square root of 2304 is 48. 4489. Therefore 41 must be subtracted from 825 to get a perfect square. Let us learn here how to find the square root of numbers which … \)This shows that, The required perfect square is $$6412 + 149 = 6561$$. If number of rows and number of columns are equal then number of plants has to be a perfect square. satapathyaradhana46 satapathyaradhana46 Here is your answer in the attachment hope it helps you . 1. For example, the square root of 16 is 4, because 16 is a perfect square of 4, such as: 4 2 = 16 and √16 = 4. Join now. Find the square roots of each of the following numbers by Division method: (i) 2304 (ii) 4489 (iii) 3481 (iv) 529 (v) 3249 (vi) 1369. Ask your question. Hence, the length of the side of a square is 21 m. Hence, the gardener requires 24 more plants. Finding square root by division method This is followed by finding the Square Roots of Decimal, the topic is explained in 6 steps. Add your answer and earn points. Therefore, perfect square can be obtained by subtracting $$53$$ from the given number $$1989$$. Also find the square root of the perfect square so obtained. A gardener has $$1000$$ plants. Square root of $$252$$ is calculated as follows. What must be subtracted from the numbers so as to get perfect square, It is evident that square of $$20$$ is less than $$402$$ by $$2$$. Find the square roots of each of the following numbers by Division method: (i) 2304 (ii) 4489 (iii) 3481 (iv) 529 (v) 3249 (vi) 1369, (vii) 5776 (viii) 7921 (ix) 576 (x) 1024 (xi) 3136 (xii) 900. \begin{align}&= 500 - 16\\&= 484\end{align}, Number of children left out in PT drill arrangement $$= 16$$, Instant doubt clearing with Cuemath Advanced Math Program. The square of $$57$$ is less than $$3250$$ by $$1$$. Here, we get remainder 53. Find the minimum number of plants he needs more for this. Therefore, required perfect square $$= 1989 − 53 = 1936$$, Square root of $$3250$$ can be calculated by long division method as follows, The remainder obtained is $$1$$. The square root of $$3249$$ is calculated as follows. To ask any doubt in Math download Doubtnut: https://goo.gl/s0kUoe Question: Find the square root of each of the following numbers by Division method. We get a remainder 0 and quotient as 5 7 using division method. (i) $$2304$$ (ii) $$4489$$ … 1. What must be added to the numbers so as to get perfect square. \begin{align}&= {23^2} - 525\\&= 529 - 525\\&= 4\end{align}, The required perfect square is $$525 + 4 = 529$$. Therefore, required perfect square $$= 825 − 41 = 784$$. Therefore, number of digits in square root =, Therefore, the number of digits in square root =. (iii) 3481. Join now. The square root of $$900$$ is calculated as follows. ... Find the square root of the following number by Division method. Here, we have to find the number which should be subtracted from total number of children to make it a perfect square. The remainder is $$41$$.it shows that the square of $$28$$ is less than $$825$$ by $$41$$. 2. Log in. \begin{align}&= {16^2} - 252\\&= 256 - 252\\&= 4\end{align}, The required perfect square is $$252 + 4 = 256$$. Therefore. Join now. Log in. \begin{align}\sqrt {4489} = 67\end{align}. The square root of $$5776$$ is calculated as follows. (i) If $$AB = 6 \,\rm{ cm}$$, $$BC = 8\, \rm{cm}$$, find $$AC$$, (ii) If $$AC = 13\,\rm{cm},$$ $$BC = 5 \,\rm{cm}$$, find $$AB$$. Find the square root of the following decimal numbers: (i) 2.56 (ii) 7.29 (iii) 51.84 (iv) 42.25 (v) 31.36. The square root of $$4489$$ is calculated as follows. Ans. Here, we get remainder 2. In a right triangle $$\rm{}ABC$$, $$\rm{}∠B = 90°$$. Find the least number which must be added to each of the following numbers so as to get a perfect square. If number of rows and number of column are equal then number of palant has to be a prefect square. B. Square root of $$2.56$$ is calculated as:-, Square root of $$51.84$$ is calculated as, Square root of $$42.25$$ is calculated as follows, Square root of $$31.36$$ is calculated as follows. Therefore 53 must be subtracted from 1989 to get a perfect square. Therefore, perfect square can be obtained by subtracting $$16$$ from the given number. Reasoning: When a number is large, even the method of trading the square root by prime factorization becomes lengthy and difficult, so the division method is used. drill, they have to stand in such a manner that the number of rows is equal to the number of columns. Here, we get remainder 31. Find the least number which must be added to each of the following numbers so as to get a perfect square. C. 73. Add your answer and earn points. Since remainder is zero and number of digits and left in the given number. After that, the concept of Estimating Square Roots is also explained. The remainder is $$16$$. $$AC = 13\;\rm{cm}$$ , $$BC = 5\;\rm{cm}$$ , $$AB$$=? Area of the square = side of a square x side of a square, \begin{align}441\;\rm{m^2} = \text{(side of a square)}^2\end{align}, $${\text{Side of a square}} = \;\sqrt {441}= 21\;{\rm{m}}$$. (ii) 4489. Answered Square root of 0.4489 by long division method 2 See answers karankumar8461 is waiting for your help. Find the square root of the following number by Division method… Find the least number which must be subtracted from each of the following numbers so as to get a perfect square. $$AB = 6\;\rm{cm}$$ $$BC = 8\;\rm{cm}$$ $$AC=$$ ? $${80^2} < 6412$$The remainder is \(12. Here, we get remainder 1. Find the square root of 4489 by division method - 25901342 1. New questions in Math. How many children would be left out in this arrangement. 1. This chapter contains a total of 4 exercises. |
# Shortest hex dumping program
### Challenge
Create a console program to display each byte of a file.
### Winning
Since this is , fewest bytes wins.
### Rules
• Program must be a console application, meaning that it will be ran from some sort of command-line interpreter;
• Every byte must be uppercase hexadecimal, separated by a space, and it must be 2 digits; (put number 0 before it if it has 1 digit)
• File must be read using IO or alternative, and not hard-coded;
• File path must be specified as a command-line argument or a user prompt (like STDIN);
### Example
test.txt (ends with LF)
Hello World!
"$(gc$args -ra|% *ay|%{'{0:X2}'-f+$_})" Try it online! -5 bytes thanks to mazzy • thanks for ./.input.tio. 40 bytes with CRLF preserved. – mazzy Jul 10 '19 at 5:29 • I learned this from the Kevin Cruijssen solution. – Andrei Odegov Jul 10 '19 at 8:41 # Java 11, 156 154 bytes import java.nio.file.*;interface M{static void main(String[]a)throws Exception{for(int b:Files.readAllBytes(Path.of(a[0])))System.out.printf("%02X ",b);}} -2 bytes thanks to @Holger. Try it online by using ./.input.tio as argument file-path, which will have a given input as file-content. Explanation: import java.nio.file.*; // Required import for Files and Paths interface M{ // Class static void main(String[]a) // Mandatory main method throws Exception{ // With mandatory thrown clause for the readAllBytes builtin a[0] // Get the first argument Path.of( ) // Get the file using that argument as path Files.readAllBytes( ) // Get all bytes from this file for(int b: ) // Loop over each of them: System.out.printf( // And print the current byte "%02X ",b);}} // As uppercase hexadecimal with leading 0 // and trailing space as delimiter • What's the rationale behind using interface instead of class? – JakeDot Jul 10 '19 at 10:06 • @JakeDot main is required to be public, interface methods are always public, interface is shorter than class + public. – Grimmy Jul 10 '19 at 11:00 • With Java 11, you can use Path.of instead of Paths.get – Holger Jul 10 '19 at 12:02 • @Holger Thanks! :) – Kevin Cruijssen Jul 10 '19 at 12:27 • @Grimy since Java 9, interface methods are not always public, but they are public unless explicitly declared private. – Holger Jul 10 '19 at 12:30 # PHP, 6059 54 bytes <?=wordwrap(bin2hex(implode(file($argv[1]))),2,' ',1);
• -1 byte thanks to manassehkatz
• -5 bytes thanks to Blackhole
Try it online!
• Should be able to drop the trailing ?> and save 2 bytes, or if that doesn't work then replace ?> with a semicolon and save 1 byte. – manassehkatz-Moving 2 Codidact Jul 10 '19 at 4:34
• Use implode(file($x)) instead of file_get_contents($x) (-4 bytes). – Blackhole Jul 10 '19 at 13:45
• And wordwrap(), with 1 as the last parameter, is one byte shorter than chunk_split(). – Blackhole Jul 10 '19 at 13:54
# Perl 5 (-aF//), 23 bytes
printf"%02X ",ord for@F
TIO
# APL (Dyalog Unicode), 16 bytes
Anonymous tacit prefix function. Returns (and implicitly prints, if the value isn't otherwise consumed) a two-row matrix with the top 4 bits represented as a decimal number 0–15 in the top row and the bottom 4 bits similarly represented in the bottom row. That is, the matrix has as many columns as the file has bytes.
16 16⊤83 ¯1∘⎕MAP
Try it online!
⎕MAP map the argument filename to an array
∘ with parameters:
¯1 the entire length of the file
83 read as 8-bit integers
16 16⊤ convert (anti-base) to 2-position hexadecimal
• @facepalm42 It very much is in hexadecimal. E.g. H is 72, which is 4×16¹+8×16⁰ or [4,8]₁₆. Hence the first column in the example reads [4,8]. – Adám Jul 9 '19 at 10:28
• Oh, I completely forgot! Sorry. – facepalm42 Jul 9 '19 at 10:29
## Python 3, 59 bytes
-11 bytes thanks to Mostly Harmless!
-8 bytes thanks to James K Polk!
-24 bytes thanks to Blue!
print(' '.join('%02X'%ord(i)for i in open(input()).read()))
Try it online!
This is pretty straightforward; it opens a filename given as input on STDIN, reads it, converts each character to its ASCII value, converts each number to hex, strips off the "0x" that precedes hexademical values in Python, pads the value with a zero if necessary, then joins the values together with spaces.
• Can save a few bytes with '%02X'%ord(i) instead of slicing the output of hex – Mostly Harmless Jul 9 '19 at 23:22
• @MostlyHarmless Done! -11 bytes. Thanks! – mprogrammer Jul 10 '19 at 0:37
• how about '%02X' instead of '%02x', and get rid of .upper() – President James K. Polk Jul 10 '19 at 0:50
• You can save the bytes from the import sys by using raw_input() as the filename instead; rules allow user prompting. – Blue Jul 10 '19 at 12:11
• @Blue Thanks! And it's even shorter in Python 3, where you can just do input() – mprogrammer Jul 10 '19 at 15:32
# Bash, 33 23 bytes
...with a lot of help:
-3 thanks to manatwork
-4 thanks to spuck
-3 thanks to Nahuel Fouilleul
echo xxd -c1 -p -u $1 Try it online! Note that the TIO link above uses input - we can write files locally, so this shows it working as a program taking a file path. • Minor reductions: xxd -u -p$1|fold -2|tr \\n \ . – manatwork Jul 9 '19 at 13:07
• Thanks, any idea how to get the \n and \ to work in the 'this' link version? EDIT: I added another escape character. – Jonathan Allan Jul 9 '19 at 13:25
• If I understand you correctly, you just want to change from double quotes to single quotes: Try it online! – manatwork Jul 9 '19 at 13:41
• Awesome thank you! – Jonathan Allan Jul 9 '19 at 13:51
• xxd -c1 -p -u $1|tr \\n \ – spuck Jul 10 '19 at 4:35 # Kotlin, 13012710493 92 bytes fun main(a:Array<String>){java.io.File(a[0]).readBytes().forEach{print("%02X ".format(it))}} Try it online! Edit: -11 bytes thanks to @ChrisParton Edit: Working TIO Edit: -1 byte thanks to @KevinCruijssen • Could you ditch the import and reference File as java.io.File instead? – Chris Parton Jul 10 '19 at 5:43 • @ChrisParton right you are, thanks! – Quinn Jul 10 '19 at 12:59 • Here a working TIO. You can use ./.input.tio as file-path argument, and it will use the STDIN as file-content. :) – Kevin Cruijssen Jul 10 '19 at 18:36 • @KevinCruijssen thanks! just updated answer – Quinn Jul 10 '19 at 18:40 • I don't know Kotlin, but the TIO still works if I remove the space at a:Array, so I think you can save a byte. – Kevin Cruijssen Jul 10 '19 at 19:51 # Dart, 140 134 bytes import'dart:io';main(a){print(new File(a[0]).readAsBytesSync().map((n)=>n.toRadixString(16).toUpperCase().padLeft(2,'0')).join(' '));} Try it online! -6 bytes because I forgot to reduce variable names • +1 for dart. Such an underrated language. – vasilescur Jul 9 '19 at 14:26 • Hard to golf with, since it's basically JS without the very lax type system – Elcan Jul 9 '19 at 14:27 # Haskell, 145 143 bytes import System.Environment import Text.Printf import Data.ByteString main=getArgs>>=Data.ByteString.readFile.(!!0)>>=mapM_(printf"%02X ").unpack • A little bit shorter: import Data.ByteString plus main=getArgs>>=Data.ByteString.readFile.(!!0)>>=mapM_(printf"%02X ").unpack. – nimi Jul 9 '19 at 18:07 # Rust, 141 bytes (contributed version) use std::{io::*,fs::*,env::*};fn main(){for x in File::open(args().nth(1).unwrap()).unwrap().bytes(){print!("{:02X} ",x.unwrap())}println!()} # Rust, 151 bytes (original version) fn main(){std::io::Read::bytes(std::fs::File::open(std::env::args().nth(1).unwrap()).unwrap()).map(|x|print!("{:02X} ",x.unwrap())).count();println!()} • -10 bytes: TIO – Herman L Jul 10 '19 at 16:04 ## bash+Stax, 6+4+1=11 bytes This is complete theory craft at this point. You can't actually run this. If everything works according to its spec this would work, but not everything does yet. The bash script is ]<$1
and the stax program must be compiled and saved to ] is
╛↕ßú┼_
Set your character set to ISO 8859-1 (Windows-1252 won't work here) and go
Unpacked and explained
_ push all input as a single array
F run the rest of the program for each element of the array
|H write the hex of the byte to standard output
| write a space to standard output
# Emojicode, 186 162 bytes
📦files🏠🏁🍇🔂b🍺📇🐇📄🆕🔡👂🏼❗️❗️🍇👄📫🍪🔪🔡🔢b❗️➕256 16❗️1 2❗️🔤 🔤🍪❗️❗️🍉🍉
Try it online here.
Ungolfed:
📦 files 🏠 💭 Import the files package into the default namespace
🏁 🍇 💭 Main code block
🔂 b 💭 For each b in ...
🍺 💭 (ignoring IO errors)
📇 🐇 📄 💭 ... the byte representation of the file ...
🆕 🔡 👂🏼 💭 ... read from user input:
❗️ ❗️ 🍇
👄 💭 Print ...
📫 💭 ... in upper case (numbers in bases > 10 are in lower case) ...
🍪 💭 ... the concatenation of:
🔪 🔡 🔢 b ❗️ ➕ 256 💭 b + 256 (this gives the leading zero in case the hex representation of b is a single digit) ...
16 💭 ... represented in hexadecimal ...
❗️
1 2 💭 ... without the leading one,
❗️
🔤 🔤 💭 ... and a space
🍪
❗️❗️
🍉
🍉
# Perl 6, 45 bytes
@*ARGS[0].IO.slurp(:bin).list.fmt('%02X').say
Try it online!
• @*ARGS[0] is the first command-line argument.
• .IO turns that (presumed) filename into an IO::Path object.
• .slurp(:bin) reads the entire file into a Buf buffer of bytes. (Without the :bin the file contents would be returned as a Unicode string.)
• .list returns a list of the byte values from the buffer.
• .fmt('%02X') is a List method that formats the elements of the list using the given format string, then joins them with spaces. (Convenient!)
• .say prints that string.
• Based on the Python answer, a TIO Link is in fact quite possible. – Draco18s no longer trusts SE Jul 9 '19 at 21:35
• Some rearranging can remove the .list for 41 bytes – Jo King Jul 13 '19 at 4:01
# Node.js, 118 bytes
console.log([...require("fs").readFileSync(process.argv[2])].map(y=>(y<16?0:"")+y.toString(16).toUpperCase()).join )
What the result looks like:
Btw the content of test.txt in the example is as follows:
做乜嘢要輸出大楷姐，搞到要加番toUpperCase()去轉番，咁就13byte啦。
(Why on earth is upper-case output necessary. I had to add the conversion with toUpperCase(), and that cost 13 bytes.)
# D, 98 Bytes
import std;void main(string[]s){File(s[1]).byChunk(9).joiner.each!(a=>writef("%02X ",a.to!byte));}
Try it Online!
# Python 3, 75 bytes
Mostly a copy of Maxwell's python 2 answer.
import sys
print(' '.join('%02X'%b for b in open(sys.argv[1],'rb').read()))
• you mean probably sys.argv[1]. with sys.argv[0] this script works more like a quine ;-) – anion Jul 10 '19 at 13:22
• @anion: oops, haha, fixing ... – President James K. Polk Jul 10 '19 at 13:32
# Racket, 144 bytes
This submission does output a trailing space, and no trailing newline. Let me know if this is considered a loophole :)
(command-line #:args(f)(for([b(call-with-input-file f port->bytes)])(printf"~a "(string-upcase(~r b #:base 16 #:min-width 2 #:pad-string"0")))))
## Cleaned up
(command-line #:args (f)
(for ([b (call-with-input-file f port->bytes)])
(printf "~a "
(string-upcase
(~r b #:base 16 #:min-width 2 #:pad-string "0")))))
# Forth (gforth), 71 bytes
: f slurp-file hex 0 do dup c@ 0 <# # # #> type space 1+ loop ;
1 arg f
Try it online!
TIO has 3 arg in the last line because TIO passes "-e bye" to the command line parser before passing in the code
### Code Explanation
: f \ start a function definition
slurp-file \ open the file indicated by the string on top of the stack,
\ then put its contents in a new string on top of the stack
hex \ set the interpreter to base 16
0 do \ loop from 0 to file-length - 1 (inclusive)
dup c@ \ get the character value from the address on top of the stack
0 <# # # #> \ convert to a double-length number then convert to a string of length 2
type \ output the created string
space \ output a space
loop \ end the loop
; \ end the word definition
1 arg f \ get the filename from the first command-line argument and call the function
## Javascript, 155 bytes
for(b=WScript,a=new ActiveXObject("Scripting.FileSystemObject").OpenTextFile(b.Arguments(0));;b.echo(('0'+a.read(1).charCodeAt(0).toString(16)).slice(-2)))
## VBScript, 143 bytes
set a=CreateObject("Scripting.FileSystemObject").OpenTextFile(WScript.Arguments(0)):while 1 WScript.echo(right("0"+Hex(Asc(a.read(1))),2)):wend
# Wolfram Language (Mathematica), 94 89 bytes
Print@ToUpperCase@StringRiffle@IntegerString[BinaryReadList@Last@$ScriptCommandLine,16,2] Try it online! The code is quite self-explanatory because of the long command names. It should be read mostly from right to left: $ScriptCommandLine is a list of {scriptname, commandlinearg1, commandlinearg2, ...}
Last@... extracts the last command-line argument
IntegerString[...,16,2] converts each byte to a 2-digit hex string (lowercase)
StringRiffle@... converts this list of strings into a single string with spaces
ToUpperCase@... converts the string to uppercase
Print@... prints the result to stdout
# Gema, 45 characters
?=@fill-right{00;@radix{10;16;@char-int{?}}}
Sample run:
bash-5.0$gema '?=@fill-right{00;@radix{10;16;@char-int{?}}} ' <<< 'Hello World!' 48 65 6C 6C 6F 20 57 6F 72 6C 64 21 0A Try it online! # Pyth, 12 bytes jdcr1.Hjb'w2 Try it online! Takes input as user prompt (no way to access command-line arguments AFAIK). jd # join on spaces c 2 # chop into pieces of length 2 r1 # convert to uppercase .H # convert to hex string, interpreting as base 256 (*) jb # join on newlines ' # read file as list of lines w # input() (*) I'm not 100% sure if this is intended, but one base 256 digit (as in, one character), will always convert into exactly 2 hex digits, eliminating the need to pad with zeroes. ### C#.NET Framework 4.7.2 - 235213203191175 140 bytes Try it online! using System.IO;class P{static void Main(string[]a){foreach(var b in File.ReadAllBytes(a[0])){System.Console.Write(b.ToString("X2")+" ");}}} using System; using System.IO; namespace hexdump { class Program { static void Main(string[] args) { // Read the bytes of the file byte[] bytes = File.ReadAllBytes(args[0]); // Loop through all the bytes and show them foreach (byte b in bytes) { // Show the byte converted to hexadecimal Console.Write(b.ToString("X2") + " "); } } } } • I think the following will save some bytes (now 181 I think): using System.IO;class P{static void Main(string[] a){if(a.Length>0 && File.Exists(a[0])){foreach(var b in File.ReadAllBytes(a[0])){System.Console.Write($"{b.ToString("X2")} ");}}}} – PmanAce Jul 17 '19 at 15:05
• @PmanAce If you remove some of the whitespace, it gets down to 175. – facepalm42 Jul 18 '19 at 5:58
# 05AB1E, 18 bytes
IvyÇh2j' 0.:' Jvy?
Try it online!
Explanation:
IvyÇh2j' 0.:' Jvy?
Iv Loop through each character in input
y Push current character
Ç ASCII value |
# THE MILLIMETER- AND SUBMILLIMETER-WAVE SPECTRUM OF METHYL CARBAMATE $(H_{2}NCOOCH_{3})$
Please use this identifier to cite or link to this item: http://hdl.handle.net/1811/20565
Files Size Format View
2002-TB-11.jpg 207.1Kb JPEG image
Title: THE MILLIMETER- AND SUBMILLIMETER-WAVE SPECTRUM OF METHYL CARBAMATE $(H_{2}NCOOCH_{3})$ Creators: Winnewisser, M.; De Lucia, Frank C.; Herbst, E.; Sastry, K. V. L. N. Issue Date: 2002 Publisher: Ohio State University Abstract: Although most organic molecules found in interstellar clouds are unsaturated in nature, saturated and near-saturated molecules are detected in so-called hot-core'' sources near the sites of high-mass star formation. One molecule searched for in such sources is glycine, the simplest amino acid and the amino derivative of acetic acid, but the results of this search have been ambiguous. Because there is more methyl formate than acetic acid in hot-core sources, it is reasonable to search for an amino derivative of methyl formate, the internal rotor known as methyl carbamate. The lowest energy, or syn, con-former of this species has heretofore only been studied at frequencies through 40 GHz. Using our fast-scan spectrometer (FASSST''), we have measured and analyzed many new lines of the rotational-torsional spectrum of methyl carbamate in the ground torsional state of the syn conformer. Our new laboratory measurements should make it possible to search for this species in space. Description: Author Institution: Department of Physics, The Ohio State University; Department of Physics, University of New Brunswick URI: http://hdl.handle.net/1811/20565 Other Identifiers: 2002-TB-11 |
Data models
Models are formal Pythons objects telling the mapper how to map DynamoDB data to regular Python and vice versa.
Bare minimal model
A bare minimal model with only a hash_key needs only to define a __table__ and a hash_key.
from dynamodb_mapper.model import DynamoDBModel
class MyModel(DynamoDBModel):
__table__ = u"..."
__hash_key__ = u"key"
__schema__ = {
u"key": int,
#...
}
The model can then be instanciated and used like any other Python class.
>>> data = MyModel()
>>> data.key = u"foo/bar"
Initial values can even be specified directly in the constructor. Otherwise, unless :ref:defaults are provided <using-default-values>, all fields are set to None
>>> data = MyModel(key=u"foo/bar")
>>> repr(data.key)
"u'foo/bar'"
While this is not stricly speaking related the mapper itself, it seems important to clarify this point as this is a key feature of Amazon's DynamoDB.
Amazon's DynamoDB has support for 1 or 2 keys per objects. They must be specified at table creation time and can not be altered. Neither renamed nor added or removed. It is not even possible to change their values whithout deleting and re-inserting the object in the table.
The first key is mandatory. It is called the hash_key. The hash_key is to access data and controls its replications among database partitions. To take advantage of all the provisioned R/W throughput, keys should be as random as possible. For more informations about hash_key, please see Amazon's developer guide
The second key is optional. It is called the range_key. The range_key is used to logically group data with a given hash_key. :ref:More informations below <range-key>.
Data access relying either on the hash_key or both the hash_key and the range_key is fast and cheap. All other options are very expensive.
We intend to add migration tools to Dynamodb-mapper in a later revision but do not expect miracles in this area.
This is why correctly modeling your data is crucial with DynamoDB.
Creating the table
Unlike other NoSQL engines like MongoDB, tables must be created and managed explicitely. At the moment, dynamodb-mapper abstracts only the initial table creation. Other lifecycle managment operations may be done directly via Boto.
To create the table, use :py:meth:~.ConnectionBorg.create_table with the model class as first argument. When calling this method, you must specify how much throughput you want to provision for this table. Throughput is mesured as the number of atomic KB requested or sent per second. For more information, please see Amazon's official documentation.
from dynamodb_mapper.model import DynamoDBModel, ConnectionBorg
conn = ConnectionBorg()
Important note: Unlike most databases, table creation may take up to 1 minute. during this time, the table is not usable. Also, you can not have more than 10 tables in CREATING or DELETING state any given time for your whole Amazon account. This is an Amazon's DynamoDB limitation.
• /etc/boto.cfg
• ~/.boto
• or AWS_ACCESS_KEY_ID and AWS_SECRET_ACCESS_KEY environment variables
If none of these places defines them or if you want to overload them, please use :py:meth:~.ConnectionBorg.set_credentials before calling create_table.
For more informations on the connection manager, pease see :py:class:~.ConnectionBorg
Region
To change the AWS region from the the default us-east-1, use :py:meth:~.ConnectionBorg.set_region before any method that creates a connection. The region defaults to RegionInfo:us-east-1.
You can list the currently available regions like this:
>>> import boto.dynamodb
>>> boto.dynamodb.regions()
[RegionInfo:us-east-1, RegionInfo:us-west-1, RegionInfo:us-west-2,
RegionInfo:ap-northeast-1, RegionInfo:ap-southeast-1, RegionInfo:eu-west-1]
Namespacing the models
This is more an advice, than a feature. In DynamoDB, each customer is allocated a single database. It is highly recommended to namespace your tables with a name of the form <application>-<env>-<model>.
Deep schema definition and validation with Onctuous
Onctous (http://pypi.python.org/pypi/onctuous) has been integrated into DynamoDB-Mapper as part of 1.8.0 release cycle.
Before writing any validator relying on Onctuous, there is a crucial point to take into account. Validators are run when loading from DynamoDB and when saving to DynamoDB. save stores the output of the validators while reading functions feeds the validators with raw DynamoDB values that is to say, the serialized output of the validators.
Hence, validators must be accept both serialized and already de-serialized input. As of Onctuous 0.5.2, Coerce can safely do that as it checks the type before attempting anything.
To sum up, schema entries of the form
• base type (int, unicode, float, dict, list, ...) works seamlessly.
• datetime type: same special behavior as before
• [validators] and {'keyname': validators} are automatically (de-)serialized
• callable validators (All, Range, ...) MUST accept both serialized and de-serialized input
Here is a basic schema example using deep validation:
from dynamodb_mapper.model import DynamoDBModel
from onctuous.validators import Match, Length, All, Coerce
from datetime import datetime
class Article(DynamoDBModel):
__table__ = "Article"
__hash_key__ = "slug"
__schema__ = {
# Regex validation. Input and output are unicode so no coercion problem
"slug": Match("^[a-z0-9-]+\$"),
# Regular title and body definition
"title": unicode,
"body": unicode,
# Special case for dates. Not that you would have to handle
# (de-)serialization yourself if you wanted to apply condition
"published_date": datetime,
# list of tags. I force unicode as an example even though it is not
# strictly speaking needed here
"tags": [All(Coerce(unicode), Length(min=3, max=15))],
}
Using auto-incrementing index
For those comming from SQL-like world or even MongoDB with its UUIDs, adding an ID field or using the default one has become automatic but these environement are not limited to 2 indexes. Moreover, DynamoDB has no built-in support for it. Nonetheless, Dynamodb-mapper implements this feature at a higher level while. For more technical background on the :ref:internal implementation <auto-increment-internals>.
If the field value is left to its default value of 0, a new hash_key will automatically be generated when saving. Otherwise, the item is inserted at the specified hash_key.
Before using this feature, make sure you really need it. In most cases another field can be used in place. A good hint is "which field would I have marked UNIQUE in SQL ?".
• for users, email or login field shoud do it.
• for blogposts, permalink could to it too.
• for orders, datetime is a good choice.
In some applications, you need a combination of 2 fields to be unique. You may then consider using one as the hash_key and the other as the range_key or, if the range_key is needed for another purpose, combine try combining them.
At Ludia, this is a feature we do not use anymore in our games at the time of writing.
So, when to use it ? Some applications still need a ticket like approach and dates could be confusing for the end user. The best example for this is a bugtracking system.
Use case: Bugtracking System
from dynamodb_mapper.model import DynamoDBModel, autoincrement_int
class Ticket(DynamoDBModel):
__table__ = u"bugtracker-dev-ticket"
__hash_key__ = u"ticket_number"
__schema__ = {
u"ticket_number": autoincrement_int,
u"title": unicode,
u"description": unicode,
u"tags": set, # target, version, priority, ..., order does not matter
u"comments": list, # probably not the best because of the 64KB limitation...
#...
}
# Create a new ticket and auto-generate an ID
ticket = Ticket()
ticket.title = u"Chuck Norris is the reason why Waldo hides"
ticket.tags = set([u'priority:critical', u'version:yesterday'])
ticket.description = u"Ludia needs to create a new social game to help people all around the world find him again. Where is Waldo?"
ticket.save()
print ticket.ticket_number # A new id has been generated
# Create a new ticket and force the ID
ticket = Ticket()
ticket.ticket_number = 42
ticket.save() # create or replace item #42
print ticket.ticket_number # id has not changed
To prevent accidental data overwrite when saving to an arbitrary location, please see the detailed presentation of :ref:saving.
Please note that hash_key=-1 is currently reserved and nothing can be stored at this index.
You can not use autoincrement_int and a range_key at the same time. In the bug tracker example above, it also means that tickets number are distributed on the application scope, not on a per project scope.
This feature is only part of Dynamodb-mapper. When using another mapper or direct data access, you might corrupt the counter. Please see the reference documentation for implementation details and technical limitations.
Using a range_key
Models may define a second key index called range_key. While hash_key only allows dict like access, range_key allows to group multiple items under a single hash_key and to further filter them.
For example, let's say you have a customer and want to track all it's orders. The naive/SQL-like implementation would be:
from dynamodb_mapper.model import DynamoDBModel, autoincrement_int
class Customer(DynamoDBModel):
__table__ = u"myapp-dev-customers"
__schema__ = {
u"order_ids": set,
#...
}
class Order(DynamoDBModel):
__table__ = u"myapp-dev-orders"
__hash_key__ = u"order_id"
__schema__ = {
u"order_id": autoincrement_int,
#...
}
# Get all orders for customer "John Doe"
customer = Customer(u"John Doe")
order_generator = Order.get_batch(customer.order_ids)
But this approach has many drawbacks.
• It is expensive:
• An update to generate a new autoinc ID
• An insertion for the new order item
• An update to add the new order id to the customer
• It is risky:
• Items are limited to 64KB but the order_ids set has no growth limit
• To get all orders from a giver customer, you need to read the customer first
and use a :py:meth:~.DynamoDBModel.get_batch request
As a first enhancement and to spare a request, you can use datetime instead of autoincrement_int for the key order_id but with the power of range keys, you could to get all orders in a single request:
from dynamodb_mapper.model import DynamoDBModel
from datetime import datetime
class Customer(DynamoDBModel):
__table__ = u"myapp-dev-customers"
__schema__ = {
#u"orders": set, => This field is not needed anymore
#...
}
class Order(DynamoDBModel):
__table__ = u"myapp-dev-orders"
__range_key__ = u"order_id"
__schema__ = {
u"order_id": datetime,
#...
}
# Get all orders for customer "John Doe"
Order.query(u"John Doe")
Not only is this approach better, it is also much more powerful. We could easily limit the result count, sort them in reverse order or filter them by creation date if needed. For more background on the querying system, please see the :ref:accessing data <accessing-data> section of this manual.
Default values
When instanciating a model, all fields are initialised to "neutral" values. For containers (dict, set, list, ...) it is the empty container, for unicode, it's the empty string, for numbers, 0...
It is also possible to specify the values taken by the fields when instanciating either with a __defaults__ dict or directly in __init__. The former applies to all new instances while the later is obviously on a per instance basis and has a higher precedence.
__defaults__ is a {u'keyname':default_value}. __init__ syntax follows the same logic: Model(keyname=default_value, ...).
default_value can either be a scalar value or a callable with no argument returning a scalar value. The value must be of type matching the schema definition, otherwise, a TypeError exception is raised.
Example:
from dynamodb_mapper.model import DynamoDBModel, utc_tz
from datetime import datetime
# define a model with defaults
class PlayerStrength(DynamoDBModel):
__table__ = u"player_strength"
__hash_key__ = u"player_id"
__schema__ = {
u"player_id": int,
u"strength": unicode,
u"last_update": datetime,
}
__defaults__ = {
u"strength": u'weak', # scalar default value
u"last_update": lambda: datetime.now(utc_tz), # callable default value
}
>>> player = PlayerStrength(strength=u"chuck norris") # overload one of the defaults
>>> print player.strength
chuck norris
>>> print player.lastUpdate
2012-12-21 13:37:00.00000 |
# CPA and CMA Review Questions
Park Co. uses the equity method to account for its January 1, 2011,purchase of Tun Inc.'s common stock. On January 1,2011, the fair value of Tun's FIFO inventory and land exceeded their carrying amounts. How do these excesses of the fair values over carrying amounts affect Park's reported equity in...
Choice "b" is correct. Park would record the additional COGS associated with the undervalued beginninginventory by debiting Investment Income and crediting the Investment in Tun account. Since thedifference between book...
## Related Questions in Accounting Concepts and Principles
• ### CPA and CMA Review Questions
(Solved) March 14, 2012
When the equity method is used to account for investment in common stock , which of the following affect(s ) the investor's reported investment income? A Change in Fair Value of Investee's Common Stock / Cash Dividends from Investee A) Yes ...
When the equity method is used to account for investment in common stock, which of the following affect(s) the investor's reported investment income? A Change in
• ### Financial Accounting
(Solved) October 17, 2014
the assignment is attached in file.
• ### Accounting Concepts written/math
(Solved) August 15, 2011
If there is a blank space a number must go there unless it specifies a 0 can be placed there. The drop down boxes have the selections that can be used, please try not to make up your own label. It makes it hard to align it with the correct anwer.
• ### Jaynes Inc. acquired all of Aaron Co's common stock on January 1, 2010, by using 11,000 Shares of $1... (Solved) November 05, 2014 Jaynes Inc. acquired all of Aaron Co.s common stock on January 1 , 2010, by issuing 11,000 shares of$ 1 par value common stock . Jaynes shares had a $17 per share fair value. On that date, Aaron reported a net book value of$120,000. However, its equipment (with a five-year remaining life
Dear Student, The work has been done perfectly . As per the requirement the question has been solved and the answer is done in Microsoft excel .The solution has been done after analizing...
• ### CPA and CMA Review Questions
(Solved) March 13, 2012
The following information was extracted from Gil Co.'s December 31, 2011, balance sheet: Noncurrent Assets: Long-term investments in marketable equity securities (at fair value) $96,450 Stockholders' equity: Accumulated other comprehensive income** ($25,000) **Includes a net unrealized |
Most students studying science and technologies need to know differential equations and some common techniques to solve them, exactly or numerically. Differential equations can be considered as an application of calculus and is a foundation for many subjects, including physics, engineering, probability (stochastic differential equations, for example) & statistics, economics, and finance.
Most students who get a good foundation in calculus will have no difficulties in learning ordinary and partial differential equations, but for students who missed some basics, we can help you to fill in some gaps.
A quick example for those who know nothing about differential equations: the easiest differential equation is $$\frac{d}{dx}f(x)=0$$ or $$y'=0$$. The solution is the set of all real or complex numbers, depends on your needs. This is because, for any constant function $$y=c,$$ its derivative is zero. But solve equations like $$y''+y'+y=0,$$ needs a bit more efforts than just doing one integration.
Visit the Math E-learning site or the front page to contact us. |
# Pfaffian problem
The problem of describing the integral manifolds of maximal dimension for a Pfaffian system of Pfaffian equations
$$\tag{* } \theta ^ \alpha = 0 ,\ \ \alpha = 1 \dots q ,$$
given by a collection of $q$ differential $1$- forms which are linearly independent at each point in a certain domain $M \subset \mathbf R ^ {n}$( or on a certain manifold). A submanifold $N \subset M$ is called an integral manifold of the system (*) if the restrictions of the forms $\theta ^ \alpha$ to $N$ are identically zero. The problem was posed by J. Pfaff (1814).
From a geometric point of view the system (*) determines an $( n - q )$- dimensional distribution (a Pfaffian structure) on $M$, that is, a field
$$x \mapsto P _ {x} = \ \{ {y \in \mathbf R ^ {n} } : {\theta _ {x} ^ \alpha ( y) = 0 } \} ,\ \ x \in M ,$$
of $( n - q )$- dimensional subspaces, and the Pfaffian problem consists of describing the submanifolds of maximum possible dimension tangent to this field. The importance of the Pfaffian problem lies in the fact that the integration of an arbitrary partial differential equation can be reduced to a Pfaffian problem. For example, the integration of a first-order equation
$$F \left ( x ^ {i} , u , \frac{\partial u }{\partial x ^ {i} } \right ) = 0$$
reduces to the Pfaffian problem for the Pfaffian equation $\theta = d u - p _ {i} d x ^ {i} = 0$ on the submanifold (generally speaking with singularities) of the space $\mathbf R ^ {2n+} 1$ defined by the equation
$$F ( x ^ {i} , u , p _ {i} ) = 0 .$$
A completely-integrable Pfaffian system (and also a single Pfaffian equation of constant class) can be locally reduced to a simple canonical form. In these cases the solution of the Pfaffian problem reduces to the solution of ordinary differential equations. In the general case (in the class of smooth functions) the Pfaffian problem has not yet been solved (1989). The Pfaffian problem was solved by E. Cartan in the analytic case in his theory of systems in involution (cf. Involutional system). The formulation of the basic theorem of Cartan is based on the concept of a regular integral element. A $k$- dimensional subspace $E _ {k}$ of the tangent space $T _ {x} M$ is called a $k$- dimensional integral element of the system (*) if
$$\theta ^ \alpha ( E _ {k} ) = 0 ,\ \ d \theta ^ \alpha ( E _ {k} \wedge E _ {k} ) = 0 ,\ \alpha = 1 \dots q .$$
The subspace $S ( E _ {k} )$ of the cotangent space $T _ {r} ^ {*} M$ generated by the $1$- forms $\theta ^ \alpha \mid _ {x}$, $( v \llcorner d \theta ^ \alpha ) \mid _ {x}$, where $v \in E _ {k}$ and $\llcorner$ is the operation of interior multiplication (contraction), is called the polar system of the integral element $E _ {k}$. The integral element $E _ {k}$ is regular if there exists a flag $E _ {k} \supset {} \dots \supset E _ {1} \supset 0$ for which
$$\mathop{\rm dim} E _ {i} = i ,\ \ \mathop{\rm dim} S ( E _ {i} ) = {\max \mathop{\rm dim} } S ( E _ {i} ^ \prime ) ,$$
where the maximum is taken over all $i$- dimensional integral elements $E _ {i} ^ \prime$ containing $E _ {i-} 1$. Cartan's theorem asserts the following: Let $N$ be a $k$- dimensional integral manifold of a Pfaffian system with analytic coefficients and let, for a certain $x \in N$, the tangent space $T _ {x} N$ be a regular integral element. Then for any integral element $E _ {k+} 1 \supset T _ {x} N$ of dimension $k + 1$ there exists in a certain neighbourhood of the point $x$ an integral manifold $\widetilde{N}$, locally containing $N$, for which $E _ {k+} 1 = T _ {x} \widetilde{N}$. Cartan's theorem has been generalized to arbitrary differential systems given by ideals in the algebra of differential forms on a manifold (the Cartan–Kähler theorem).
#### References
[1] E. Cartan, "Sur la théorie des systèmes en involution et ses applications à la relativité" Bull. Soc. Math. France , 59 (1931) pp. 88–118 MR1504975 Zbl 0002.26401 Zbl 57.0551.02 [2] E. Cartan, "Leçons sur les invariants intégraux" , Hermann (1922) MR0355764 Zbl 48.0538.02 [3] P.K. Rashevskii, "Geometric theory of partial differential equations" , Moscow-Leningrad (1947) (In Russian) [4] S. Sternberg, "Lectures on differential geometry" , Prentice-Hall (1964) MR0193578 Zbl 0129.13102 [5] P.A. Griffiths, "Exterior differential systems and the calculus of variations" , Birkhäuser (1983) MR0684663 Zbl 0512.49003
## Pfaffian problems and partial differential equations.
Let
$$\tag{a1 } F _ {h} \left ( x ^ {i} , u ^ {j} , \frac{\partial ^ \alpha }{\partial x ^ \alpha } u ^ {k} \right ) = 0 ,$$
$$h = 1 \dots p,\ i = 1 \dots n,\ j = 1 \dots m,$$
$$\alpha = ( a _ {1} \dots a _ {n} ) ,$$
$$| \alpha | = a _ {1} + \dots + a _ {n} \leq r,\ a _ {i} \in \{ 0, 1, . . . \} ,$$
be a system $p$ partial differential equations for $m$ functions in $n$ variables of order $\leq r$. Introduce the variables
$$p ^ {\alpha , k } ,\ \ 1 \leq | \alpha | \leq r,\ \ k = 1 \dots m .$$
Replacing the equations (a1) with the equations
$$\tag{a2 } \widetilde{F} {} _ {h} ( x ^ {i} , u ^ {j} , p ^ {\alpha ,k } ) = 0$$
and adding to this the Pfaffian system
$$\tag{a3 } dp ^ {\alpha , k } - \sum _ { i= } 1 ^ { n } p ^ {\alpha ( i), k } d x ^ {i} = 0 ,\ \ 0 \leq | \alpha | \leq r- 1 ,$$
where $p ^ {0,k} = u ^ {k}$ and $\alpha ^ {(} i) = ( a _ {1} \dots a _ {i-} 1 , a _ {i} + 1 , a _ {i+} 1 \dots a _ {n} )$ if $\alpha = ( a _ {1} \dots a _ {n} )$ for $i = 1 \dots n$, one finds a system (a2)–(a3) of equations which are equivalent to equations (a1) in a suitable sense. Thus, if (locally) (a2) defines a subvariety $M$ in $( x ^ {i} , u ^ {j} , p ^ {\alpha , k } )$- space, then a solution of the Pfaffian problem (a3) on $M$ defines a solution of (a1) in the sense that the projection onto $\mathbf R ^ {n} \times \mathbf R ^ {m}$( or $\mathbf C ^ {n} \times \mathbf C ^ {m}$ as the case may be) gives the graph of a solution of (a1).
For instance, in the case of a single second-order equation
$$F \left ( x ^ {1} , x ^ {2} , u ,\ \frac{\partial u }{\partial x ^ {1} } , \frac{\partial u }{\partial x ^ {2} } ,\ \frac{\partial ^ {2} u }{\partial x ^ {1} \partial x ^ {1} } ,\ \frac{\partial ^ {2} u }{\partial x ^ {1} \partial x ^ {2} } ,\ \frac{\partial ^ {2} u }{\partial x ^ {2} \partial x ^ {2} } \right ) = 0$$
one has for (a2) and (a3), respectively,
$$\tag{a2\prime } \widetilde{F} ( x ^ {1} , x ^ {2} , u , p ^ {1} , p ^ {2} ,\ p ^ {11} , p ^ {12} , p ^ {22} ) = 0,$$
$$\tag{a3\prime } \left . \begin{array}{c} du = p ^ {1} dx ^ {1} + p ^ {2} dx ^ {2} , \\ dp ^ {1} = p ^ {11} dx ^ {1} + p ^ {12} dx ^ {2} , \\ dp ^ {2} = p ^ {12} dx ^ {1} + p ^ {22} dx ^ {2} \end{array} \right \} .$$
The main equations are (a2); the remaining equations (a3) express that the solutions of (a2) of interest are $r$- jets (cf. Jet and Partial differential equations on a manifold) of functions $\mathbf R ^ {n} \rightarrow \mathbf R ^ {m}$. This leads to the idea of a system of partial differential equations on a manifold of order $r$ as being determined by a set of functions on the $r$- th jet bundle; cf. Partial differential equations on a manifold for more details.
In the setting of equations like (a2), (a3) the following generalization of Frobenius' theorem on complete integrability is of interest. Let $\omega ^ {1} , \dots , \omega ^ {r}$ be a set of differential forms on a manifold $M$ and $f ^ { 1 } , \dots , f ^ { s }$ a set of functions on $M$. Let $m \in M$ be such that $f ^ { i } ( m) = 0$, $i = 1 \dots s$. Suppose that
i) $d \omega ^ {i}$ and $df ^ { j }$ are in the ideal of differential forms generated by $\omega ^ {1} , \dots , \omega ^ {r} ; f ^ { 1 } \dots f ^ { s }$;
ii) the $\omega ^ {i}$ are linearly independent at $m$.
(Recall that the linearly independent $1$- forms $\omega ^ {1} \dots \omega ^ {r}$ form an involutive system if $d \omega ^ {i}$ is in the ideal generated by the $\omega ^ {i}$, cf. Involutive distribution.) Then there is a unique germ of a submanifold $N$ at $m$ of dimension $n$, $r+ n = \mathop{\rm dim} M$, such that the differential forms $\omega ^ {i}$ and functions $f ^ { j }$ restricted to $N$ are zero. Further if $x ^ {1} \dots x ^ {n}$ are functions on $M$ near $m$ such that $\omega ^ {1} \dots \omega ^ {r} , dx ^ {1} \dots dx ^ {n}$ are linearly independent at $m$, then the $x ^ {1} \dots x ^ {n}$ give a coordinate chart of $N$ near $m$.
## Cartan–Kähler theorem for differential systems defined by ideals.
Let $\theta ^ {a} = 0$, $a = 1 \dots q$, be a Pfaffian system on $M$ and let $N$ be an integral manifold of this system. Then obviously the $d \theta ^ {a}$ and $\theta ^ {a} \wedge \omega$, where $\omega$ is any differential form on $M$, are also zero on $N$. Thus all the elements of the differential ideal generated by $\theta ^ {1} \dots \theta ^ {q}$ in the differential algebra of exterior differential forms $F( M)$( cf. Differential form; Differential ring) are zero on $N$. This leads to the idea of a differential system (of equations) on $M$ as being defined by such an ideal. From now on let $M$ be a real analytic manifold. Let ${\mathcal F} ( M)$ be the associated sheaf to $F( M)$, i.e. ${\mathcal F} ( M)$ is the sheaf of germs of rings of differential forms on $M$. Let ${\mathcal O} ( M)$ be the sheaf of analytic functions on $M$ and let ${\mathcal F} _ {p} ( M)$ be the ${\mathcal O} ( M)$- module of $p$- forms on $M$. A differential system on $M$ is a graded differential subsheaf ${\mathcal G}$ of ideals of ${\mathcal F} ( M) = {\mathcal F}$, i.e. ${\mathcal F} {\mathcal G} = {\mathcal G} = {\mathcal G} {\mathcal F}$( the ideal property), ${\mathcal G}$ is generated by the ${\mathcal G} _ {p} = {\mathcal F} _ {p} \cap {\mathcal G}$( the graded property) and $d {\mathcal G} \subset {\mathcal G}$( the differential property). A $p$- dimensional integral manifold for ${\mathcal G}$ is a submanifold $N$ of $M$ on which ${\mathcal G}$ is zero. For each $m \in M$ let $\mathop{\rm Gr} _ {p} ( m)$ be the Grassmann manifold of $p$- dimensional subspaces of the tangent space $T _ {m} M$. The union of the $\mathop{\rm Gr} _ {p} ( m)$ for $m \in M$ has a natural structure of a real-analytic manifold and the projection $\mathop{\rm Gr} _ {p} ( m) \ni E _ {p} \rightarrow m$ then defines a locally trivial fibre bundle $\mathop{\rm Gr} _ {p} ( M) \rightarrow M$. An element $E _ {p} \in \mathop{\rm Gr} _ {p} ( m)$ is called a contact element at $m$. Such an element is an integral element of ${\mathcal G} _ {p}$ if $\omega ( E _ {p} ) = 0$ for all $\omega \in {\mathcal G} _ {p}$; it is an integral element of a differential system ${\mathcal G}$ if for all $E _ {q} \subset E _ {p}$, $0 \leq q \leq p$, $E _ {q}$ is an integral element of ${\mathcal G} _ {p}$. An integral element of dimension zero (i.e. a point of $M$) is an integral point (which is simply a solution of the equations $f( m) = 0$ for the functions $f \in {\mathcal G} _ {0}$). The polar element of an integral element $E _ {p}$ for ${\mathcal G}$ is the element $P( E _ {p} ) \supset E _ {p}$ consisting of all vectors $v \in T _ {m} M$ such that the span of $v , E _ {p}$ is an integral element of ${\mathcal G}$. Let $z ^ {i _ {1} \dots i _ {p} }$, $1 \leq i _ {1} < \dots < i _ {p} \leq n$, be the Grassmann coordinates of $E _ {p}$( cf. Exterior algebra; these are only defined up to a common scalar multiple). Now associate to ${\mathcal G} _ {p}$ the sheaf ${\mathcal G} _ {p} ^ {0}$ of ${\mathcal O} ( M)$- modules in ${\mathcal O} ( \mathop{\rm Gr} _ {p} ( M))$ consisting of all the functions $\sum _ {i \leq i _ {1} < \dots < i _ {p} \leq n } a _ {i _ {1} \dots i _ {p} } z ^ {i _ {1} \dots i _ {p} }$ for all $p$- forms $\sum _ {1 \leq i _ {1} < \dots < i _ {p} \leq n } a _ {i _ {1} \dots i _ {p} } dx ^ {i _ {1} } \wedge \dots \wedge dx ^ {i _ {p} } \in {\mathcal G} _ {p}$. Let ${\mathcal I} ( {\mathcal G} _ {p} )$ be the set of integral elements of ${\mathcal G} _ {p}$( so that ${\mathcal I} ( {\mathcal G} _ {p} )$ is a certain subset of the Grassmann bundle $\mathop{\rm Gr} _ {p} ( M)$). The element $E _ {p}$ is called a regular integral element if ${\mathcal G} _ {p} ^ {0}$ is a regular local equation for ${\mathcal I} ( {\mathcal G} _ {p} )$ at $E _ {p}$ and $\mathop{\rm dim} ( P( E _ {p} ))$ is constant near $E _ {p}$ on ${\mathcal I} ( {\mathcal G} _ {p} )$. Recall that a subsheaf ${\mathcal A} \subset {\mathcal O} ( X)$, where $X$ is a manifold, is a regular local equation for (its set of zeros) $N \subset X$ at $m \in N \subset X$ if locally around $m$ there exist sections $s _ {1} \dots s _ {t} \in \Gamma ( U, {\mathcal O} ( X))$ such that the $ds _ {1} \dots ds _ {t}$ are linearly independent on $U$ and $m ^ \prime \in N \cap U$ if and only if $s _ {1} ( m ^ \prime ) = \dots = s _ {t} ( m ^ \prime ) = 0$.
The first Cartan–Kähler existence theorem is now as follows. Let $N$ be a $p$- dimensional integral manifold of $G$ which defines a regular element $T _ {m} N \subset T _ {m} M$ at $m \in N \subset M$. Suppose that there is a submanifold $M ^ \prime$ of $M$ containing $N$ and of dimension $n+ p+ 1 - \mathop{\rm dim} P( T _ {m} N)$ such that $\mathop{\rm dim} ( T _ {m} M ^ \prime \cap P( T _ {m} N)) = p+ 1$. Then locally around $m$ there exists a unique integral manifold $N ^ \prime$ of dimension $p+ 1$ contained in $M ^ \prime$.
If $\mathop{\rm dim} P( T _ {m} N) = p+ 1$, the only possible choice (locally) for $M ^ \prime$ is $M$ itself, and there is a unique integral manifold of dimension $p+ 1$ extending $N$. If $\mathop{\rm dim} P( T _ {m} N) = p+ 2$ there is "one arbitrary function worth" freedom in choosing $M ^ \prime$ and one re-encounters the phenomenon that the solution of a partial differential equation may depend on arbitrary functions (such as $u _ {x} = u _ {t}$ with as solutions any function of the form $\phi ( x+ t)$). The second Cartan–Kähler existence theorem, which is obtained by repeated application of the first, details the dependence on initial conditions and arbitrary functions.
An immediate corollary of the first Cartan–Kähler existence theorem is as follows. Suppose one is given an integral element $E _ {p+} 1$ of dimension $p+ 1$ of the differential system ${\mathcal G}$ at $m \in M$ which contains a regular integral element $E _ {p}$. Then there exists (locally) an integral manifold $N$ of dimension $p+ 1$ such that $T _ {m} N = E _ {p+} 1$.
#### References
[a1] P. Libermann, C.-M. Marle, "Symplectic geometry and analytical mechanics" , Reidel (1987) pp. Chapt. V, Appendix 3 (Translated from French) MR0882548 Zbl 0643.53002 [a2] E. Cartan, "Les systèmes différentielles extérieurs et leur applications géométriques" , Hermann (1945) [a3] E. Cartan, "Sur l'intégration des systèmes d'équations aux différentielles totales" Ann. Sci. Ec. Norm. Sup. , 18 (1901) pp. 241–311 Zbl 32.0351.04 [a4] E. Kähler, "Einführung in die Theorie der Systeme von Differentialgleichungen" , Teubner (1934) Zbl 0011.16103 Zbl 60.0401.08 [a5] M. Kuranishi, "Lectures on exterior differential systems" , Tata Inst. (1962) [a6] J. Dieudonné, "Eléments d'analyse" , 4 , Gauthier-Villars (1977) pp. Chapt. XVIII, Sect. 13 MR0467780
How to Cite This Entry:
Pfaffian problem. Encyclopedia of Mathematics. URL: http://encyclopediaofmath.org/index.php?title=Pfaffian_problem&oldid=48174
This article was adapted from an original article by D.V. Alekseevskii (originator), which appeared in Encyclopedia of Mathematics - ISBN 1402006098. See original article |
# The Chi-squared Distribution¶
The chi-squared distribution arises in statistics. If $$Y_i$$ are $$n$$ independent Gaussian random variates with unit variance then the sum-of-squares,
$X_i = \sum_i Y_i^2$
has a chi-squared distribution with $$n$$ degrees of freedom.
gsl_ran_chisq(nu)
This function returns a random variate from chi-squared distribution with nu degrees of freedom. The distribution function is,
$p(x) dx = {1 \over 2 \Gamma(\nu/2) } (x/2)^{\nu/2 - 1} \exp(-x/2) dx$
for $$x \geq 0$$.
gsl_ran_chisq_pdf(x, nu)
This function computes the probability density $$p(x)$$ at $$x$$ for a chi-squared distribution with nu degrees of freedom, using the formula given above.
gsl_ran_chisq_P(x, nu)
gsl_ran_chisq_Q(x, nu)
gsl_ran_chisq_Pinv(P, nu)
gsl_ran_chisq_Qinv(Q, nu)
These functions compute the cumulative distribution functions $$P(x), Q(x)$$ and their inverses for the chi-squared distribution with nu degrees of freedom. |
Free Shipping On Orders Over $49. Details Talk to a Mustang Enthusiast 1-877-887-1105 M-F 8:30A-11P, Sat-Sun 8:30A-9P # CDC Interior Quarter Window Covers - Charcoal (05-09 All)$0.00
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## Features & Details
• Show Winning Look.
• Matches OEM Interior.
• Sold as a Set of Two.
• Fits All 05-09 Mustangs.
Complete Your Interior. When you're adding the finishing touches to your interior, a set of CDC interior quarter window louvers gives your ride's interior a distinctive look that is sure to impress.
Perfect with Exterior Louvers. Exterior quarter window louvers give your 'Stang a classic muscle-car look on the outside, but they're not so great looking from the inside. Exercise a little attention to detail by adding a set of blackout covers and enhance your interior's appearance.
Matches OEM. These high quality inserts are designed to match your charcoal OEM interior perfectly for a seamless look.
Easy Installation. Adding a set of CDC Interior Quarter Window Louvers to your ride is as simple as cleaning the glass with the included alcohol wipe, peeling the backing off the 3M tape, and applying the inserts. Get the blue ribbon look you want in minutes.
Fitment: 2005 2006 2007 2008 2009 Details
MPN# 112040
CA Residents: WARNING: Cancer and Reproductive Harm - www.P65Warnings.ca.gov
What's in the Box
Hardware included:
Installation Info
Installation Time
(approx) 15 Minutes
Difficulty Level:
Simple installation for anyone
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## Customer Photos (22) Submit Your Photo
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##### 3 Top Rated Questions
What does it look like from the outside
• The cloth material is haphazardly cut and mounted to a cheap looking piece of white plastic. They look terrible even through the narrow louver slits.
My interior is not black, are there any that aren't black? And I'm planning on buying these with some louvers but will it look bad through the slits of the louvers?
• Currently, these are only available in the charcoal color.
I know that is says 05-09 but I was wondering if they would fit in the 2010 cause the windows do not look all that different in size..?
• No unfortunately, they don't fit properly on the 2010+ Mustangs.
## Customer Reviews (100+) Write a Review
Overall Rating 4.6
out of 5 stars
Installation Time:Less than 1 Hour
Installation Cost:0-\$100
• Fit 4.742268 4.7
• Appearance 4.597938 4.6
• Bang For Your Buck 4.40625 4.4
• 1/5
July 14, 2019
Poor dissign
Easy to install but had to cut to fit. Fit in perfect after they where cut to fit.
Boltz
Year: 2014 Submodel: GT
• 5/5
June 08, 2019
Almost perfect
If the back side facing the window was black and if they used black tape they would have been a simple 2 minute installation but had to buy can of flat black paint and spray them
Mypony07
Year: 2007 Submodel: V6
• 5/5
Installation Time: Less than 1 Hour
May 21, 2019
These covers fit perfectly and look factory. Highly reccomended for finishing the interior look of your mustang!
AaronT
Year: 2007 Submodel: GT500
• 5/5
Installation Time: Less than 1 Hour
April 13, 2019
No Tape Needed
Got these window covers in about a day.They do come with 2 sided tape. No need they fit right in.Just like a glove.However I do recommend window louvers as you will be able to see the tape.
JasonB
Year: 2009 Submodel: V6
• 3/5
Installation Time: Less than 1 Hour
April 03, 2019
Gave u\it two for a reason....
This product, though, it is made well, seems to not have had anyone on the design or testing board. The overall pics shown here, both in Cust. Pics and the Products pics are accurate. However, none show the flip side. (I have posted one that does.) The reverse side is exposed. This allows the material that is used as the "base" to show. Either a person in the RandD dept. just fell flat on their face or they did no testing with it at all. The side that will end up facing the outside is WHITE. The point of the rear blackouts is to actually blackout the rear quarters. Why make the baseboard of the item in WHITE. On top of that, the double-sided foam tape they use is also WHITE and instead of running the tape around the outer edges of the product, they stuck them right down the middle of it. (Two strips) I really think they dropped the ball on this part. When looking from the outside, you can see something has been placed in the window because the color is so bright it comes through the factory tint. In my case, I used a permanent black marker and colored in the white and removed the double-sided tape and lined it up better along the edges instead of down the middle. (See my pic in Cust. pics for a before shot.)
Scorpius
Year: 2008 Submodel: GT
• 5/5
Installation Time: Less than 1 Hour
March 31, 2019
Great product
Easy install and perfectly matched If you want the more coupe style and want no rear windows this will be perfect for you
Eric09
Year: 2008 Submodel: GT
• 5/5
Installation Time: Less than 1 Hour
March 09, 2019
Much better look
Just what I needed. Highly recommend didn't even have to use the tape.
EricH
Year: 2006 Submodel: GT
• 5/5
Installation Time: Less than 1 Hour
March 08, 2019
Easy
Nicely made easy install. Took minutes. Shipment arrived on time
Mustang
Year: 2006 Submodel: GT
• 5/5
Installation Time: Less than 1 Hour
March 07, 2019
CDC interior quarter window covers
The interior quarter window covers is the final fit to my Shelby quarter window covers. It gives my interior that professional finish. This is a must for anyone looking to give their car a quality appearance.
FrankJ
Year: 2008 Submodel: GT500
• 5/5
Installation Time: Less than 1 Hour
February 03, 2019
They fit perfectly
I was tired of seeing the backside of my louvers even through my tint, they fit perfectly painted the backside black to hide the white backside of them. Didn't even need the 3m tape they fit snug without it and give the interior a great custom look...took less than 10 minutes to install, but recommend painting the backside to anyone who buys a set
Bryan
Year: 2007 Submodel: V6
• 5/5
Installation Time: Less than 1 Hour
February 03, 2019
Great product, stylish
Great addition to interior. Matches existing interior, I deleted my qtr window.
ShaneP
Year: 2006 Submodel: GT
• 5/5
Installation Time: Less than 1 Hour
January 04, 2019
Look great
These look and fit perfectly. Very simple to install and will stay even without using the provided 3M tape.
KeithC
Year: 2005 Submodel: GT
• 5/5
Installation Time: Less than 1 Hour
December 19, 2018
Pretty solid
They’re easy to install and add a bit of privacy in the car
MasonM
Year: 2006 Submodel: GT
• 5/5
Installation Time: Less than 1 Hour
December 07, 2018
Covers
Easy to install fits great and looks,great only thing is it would be better if they came in color variation love them
Candelario
Year: 2009 Submodel: V6
• 3/5
Installation Time: Less than 1 Hour
October 29, 2018
Window covers
Would've given a 5 just based on inside appearance and fit, since they look good, and are snug without having to use the tape strips. However, cost is too high for what you get, and also covered the back of it with black construction paper, to hide the tape and white finish from being seen on the outside.
TomK
Year: 2007 Submodel: GT500
• 5/5
Installation Time: Less than 1 Hour
October 06, 2018
Interior quarter window covers 2008 GT 500
Fit was perfect, didn't need the tape strips. Looks great.
KeithB
Year: 2008 Submodel: GT500
• 3/5
Installation Time: Less than 1 Hour
October 04, 2018
Excellent idea :)
1.It is an excellent idea to cover the window if you have a louver on it (You wouldnt want to see the underside of the louver). 2.The price is too much for the material that was used. The underside is white and for the meticulous few, you could see the white through the louver. I painted it black. 3.The fit is bang on. I didnt have to use the 3m tape. It sits perfectly flush against the window. 4. Overall, it looks great but should have been cheaper :)
EricS
Year: 2005 Submodel: V6
• 5/5
Installation Time: Less than 1 Hour
October 03, 2018
Perfect Fit
Great fit. Didn't need the tape. Just wish it was leather.
JamesH
Year: 2005 Submodel: V6
• 5/5
Installation Time: Less than 1 Hour
October 02, 2018
Fit and finish is perfect, As for the back side, a quick shot of flat black spray paint solves the issue of the ugly window facing surface. I used flat black so that there would be no reflection. And with the side vents on, they look great. No need to use the tape either so painted over those too.
terry
Year: 2008 Submodel: GT
• 2/5
Installation Time: Less than 1 Hour
September 28, 2018
Look terrible from the outside.
They look great from the inside, but lousy from the outside. Unless you have the exterior black out overlay for the window, they look like terrible. Shows noticeably through the louvers I bought. Should have also bought the outside black out overlay. Too late now. Looking at the inside of the louvers is better than seeing these show through on the outside. Really would be easy to just make them covered with material on both sides, because they fit snug and really don't need the adhesive.
PeteP
Year: 2008 Submodel: Bullitt |
## anonymous one year ago Three consecutive integers have a sum of 87 .. What are the integers ?!
The options are endless...Any answer choices?
2. anonymous
No...
3. freckles
Let n-1 be an integer. The next integer is n. The next integer after that would be n+1. so just solve the equation: (n-1)+n+(n+1)=87
What he said xD Any questions?
5. anonymous
I get 2n^3=86 then what?
6. freckles
oh mine how did you get that
Did you tell her to divide 86 into 3 first THEN do this?
8. freckles
how many n's do you see here n, n , n?
9. anonymous
3..
10. freckles
n+n+n=3n
11. freckles
-1+1=0
12. freckles
$(n-1)+n+(n+1)=87 \\(n+n+n)+(-1+1)=87 \\ 3n+0=87 \\ 3n=87$
13. freckles
n+n+n=3n just like 4+4+4=3(4) or 12
14. anonymous
15. freckles
3n=87 gives n=29 n-1=? n+1=?
16. anonymous
28,29,30
17. freckles
put n to be 29 these questions I just asked you are equivalent to: n-1=29-1=28 n+1=29+1=30 yes 28+29+30=87
18. freckles
yep we had to find the number before and after 29 since we used (n-1) and n and (n+1) to be are integer where n we found to be 29
19. anonymous
Thank you .. Can u stick around and help me with some other problems?
20. freckles
I can probably help with one more
21. anonymous
Write 7/3 as a mixed number
22. anonymous
How do I start?
23. freckles
how many times does 3 go into 7? or how many 3's are in 7?
24. anonymous
2.3
25. freckles
7=3+3+1 how many 3's do you see?
26. anonymous
2
27. freckles
right so far we have: $\frac{7}{3}=2 \frac{?}{3}$
28. freckles
the question mark is the remainder
29. anonymous
1
30. freckles
7=3+3+1 right that 1 right there was left over
31. freckles
so the question mark is 1
32. anonymous
Thank you!
33. freckles
np |
# 2d Lattice Unit Cell
2D periodic charge sheets: electrostatics Sohrab Ismail-Beigi November 12, 2013 1 Motivation These notes deal with the electrostatics of periodic 2D charge distributions as well as verti-cally stacked 2D distributions which have the same in-plane periodicity but which may have di erent charge distributions inside their unit cells. Amitava has 3 jobs listed on their profile. analysis of the effect of mesh connectivity, unit cell aspect ratio and mesh density is conducted. Goto Contents: In this example an orthorhombic lattice is simulated. Primitive unit cells contain one lattice point only. There are two ways to specify the atomic positions. Introduction to the theory of lattice dynamics M. The crystal basis is defined by. In the Wigner–Seitz cell, the lattice point is at the center of the cell, and for most Bravais lattices, the shape is not a parallelogram or parallelepiped. 1 x 2 = 2 à 4 F. Write the conventional symbol for each of these rotations. c, Schematic of the primarily 1D lamellar assemblies of gold triangular nanoprisms. is able to build the whole lattice without gaps/overlaps. Plots of the Wigner-Seitz cell and Brillouin Zone including high symmetry directions labeled. Select any point that OD passes through - say P. these crystal systems all have α=β=γ=90°. Read the portion of your textbook that pertains to lattice planes, Miller indices, and X-ray diffraction. In this work, we assume a¼20. The random lattice styles are 1d, 2d, and 3d lattices with a1 = 1 0 0 and a2 = 0 1 0 and a3 = 0 0 1. Energy ħω; momentum ħq •Concept of the phonon density of states •Einstein and Debye models for lattice heat capacity. Primitive cells, Wigner-Seitz cells, and 2D lattices 4P70, Solid State Physics Chris Wiebe. The Bravais lattice is the same as the lattice formed by all the. com and I'll get right on it!. primitive cell. Newton’s second law of motion applied to each of the masses can be expressed as m u¨ n n 1 = −(k 1+ k 2) u 1 + k 1u n 2 + k 2u n −1 2, m u¨ n 2 = −(k 1+ k 2) u n 2 + k 1u n 1 + k 2u n +1 1, (1) where u n j ≡ u n(j. The Brillouin zone is the WS cell in the reciprocal lattice. Such a periodic replication of the deformed unit cell renders the entire graphene lattice under a macroscopically homogeneous. ¾Primitive cell (P): only contain one lattice point. View Amitava Banerjee’s profile on LinkedIn, the world's largest professional community. Unit Cells: A Three-Dimensional Graph. lattice constant of the sc lattice and R is the radius of the sphere. ¾A lattice can be described in terms of unit cell and lattice. Then the reciprocal lattice can be generated using primitive vectors 123 2π b=×a V a, 23 2 1 π =×aa V b, 312 2π =×aa V) b, (2. The lattice parameters define the size and shape of the unit cell. Draw the Wigner-Seitz cells for a 2D hexagonal lattice. determining which 2D Bravais lattice best ts the experimental data from an image being processed with CIP. ! From now on, we will call these distinct lattice types Bravais lattices. 4284um and the hole diameter to 0. 4 %, packing efficiency of bcc is 68 %. S1, we present the optimized atomic structure of ReS 2. Unit cell: Building block, repeat in a regular way. unit cell. Here there are 14 lattice types (or Bravais lattices). To obtain the band structure of the considered 2D PhC waveguide, the PWE method has been employed [ 1 , 5 ]. The random lattice styles are 1d, 2d, and 3d lattices with a1 = 1 0 0 and a2 = 0 1 0 and a3 = 0 0 1. We are now going to verify band structure of 2D hexagonal lattice as reported in reference [1]. Unit Cell, Primitive Cell and Wigner-Seitz Cell. Example: 2D hexagonal lattice. g, the lattice formed by the Aatoms shown by dashed lines) is triangular with a Bravais lattice spacing 2 × sin60 × a= √ 3a, where ais the spacing between neighboring atoms. Repetition of the unit cell generates the entire crystal. The lattice parameters define the size and shape of the unit cell. Point symmetry elements. (See Figure 5a in Kittel on page 7). Consider a square lattice with lattice spacing 1, and suppose the density of electrons = 3:14159 electrons per unit cell. Novoselov, and A. See also: blender-chemicals by Patrick Fuller. Also, the choice of a nonprimitive (centered) unit cell adds lattice points in direct space on the left but subtracts them in frequency space on the right. Each of the various lattice parameters are designated by the letters a, b, and c. bxc is the volume of the parallelepiped of sides a, b and c, which is the volume of the unit cell in the real space. py, specifying the unit cell atoms and their positions. Electron microscopy in molecular cell biology II unit cell Fourier transform 2D lattice transforms. F, B, E) ane also urit cell Tota Count Name Bhavls Parametess Lattice. The patterns that appear when filaments either overfill or under-fill the lattice are reminiscent of those observed in other physical systems involving 2d lattices. Remember, since the unit cell is a cube, all of the sides are the same length (represented by "a"), and all angles are 90°. The Bravais lattice are the distinct lattice types which when repeated can fill the whole space. Point symmetry elements. This relationship is tricky to think of three-dimensional space. It gives 14 3D Bravais lattice. Crystal lattices as gratings • Crystals can be described as three-dimensional lattices • A lattice point at the origin of the unit cell is reproduced by three translations a, b and c and three angles α, βand γ, (lattice or unit cell parameters) • An infinite number of lattice planes run through the lattice points and their. The cross-section of the unit cell of this alternative profile is. lattice points, and these are associated with a set of basis vectors. Handout on Bravais Lattices Illustrations of the Isometric, Hexagonal, and Tetragonal Bravais lattices and unit cells Illustrations of the Orthorhombic, Monoclinic, and Triclinic Bravais lattices and unit cells. 2 A conventional (cubic) unit cell of zincblende, ZnS, is shown below. Note that the primitive cells of the centered lattice is not the unit cell commonly drawn. Point coordinates for unit cell center are a/2, b/2, c/2 ½½½ Point coordinates for unit cell corner are 111 Translation: integer multiple of lattice constants identical position in another unit cell z x y a b c 000 111 y z x 2c b b. FCC 1st Brillouin zone: 270. ¾Primitive cell (P): only contain one lattice point. For three dimensions all distinct lattices may be described by a few types of centering. Wigner-Seitz cell: primitive cell with lattice point at its center enclosed region is W-S cell for 2D hexagonal lattice d. CONSISTENT ASYMPTOTIC HOMOGENIZATION METHOD FOR LATTICE STRUCTURES BASED ON THE VIRTUAL POWER PRINCIPLE A Thesis Presented to the Graduate School of Clemson University In Partial Ful llment of the Requirements for the Degree Master of Science Mechanical Engineering by David Bracho December 2016 Accepted by: Dr. A unit cell of the lattice is mapped into the simulation box (scaled and rotated), so that it now has (perhaps) a modified size and orientation. crystal symmetry. Modeling of crack propagation in 2D brittle finite lattice structures assisted by additive manufacturing Yan Wu, Li Yang Department of Industrial Engineering, University of Louisville, Louisville, KY 40292 Abstract The failure characteristics of lattice structures are of significant importance in various. !The unit cell is defined in terms of the lengths of the edges of the parallelepiped ( a, b, c) and the three angles between these. Example, for the 2D lattice above: • The volume (3D), area (2D), or length (1D) of a WS primitive cell can be given in terms of the primitive vectors, and is independent of the choice of the primitive vectors a1a2 3 a1. arrangement. This unit cell will reduce its energy by fluctuating between the states zero, up, and down. A 2D plane strain lattice approach was regarded to discretize the continuum domain of the unit cell based on Voronoi tessellation. Parameters. Translation of lattice Rotation lattice. All of this structural information is used to build up a larger system by translation. LatticeData [{" type ", id}, …] gives data for the lattice of the specified type with identifier id. So sometimes it makes sense not to use a primitive unit cell but one which fits better to the symmetry of the problem. py, specifying the unit cell atoms and their positions. Figure 2 Graphite unit cell. The density of Eu is 5. Packing efficiency of the simple cubic lattice is 52. Unit cell is Primitive (1 lattice point) but contains TWO atoms in the Basis. It can be seen that bð1Þ ¼ b 1, b ð2Þ ¼ b 2, b ð3Þ ¼ b 1, and b ð4Þ ¼ b 2. A NOVEL NONLOCAL LATTICE PARTICLE FRAMEWORK FOR MODELING OF SOLIDS. In addition need to specify positions of nearest neighbors in order to have them connected with a bond. There is an infinite number of choices for primitive unit cell. Goto Contents: In this example an orthorhombic lattice is simulated. Create a two dimensional lattice in blender. 0) [source] ¶ Sets the overall structure. FCC 1st Brillouin zone: 270. The choice of unit cell is not unique. A Bravais lattice has the following properties: * All of the points in the lattice can be accessed by properly chosen primitive translation vectors * The parallelepiped formed by the primitive trans. In a unit cell, every constituent particle( atom, molecule or ion ) has a specific and fixed position called lattice site. Figure 5: 2D Lattice Packing Fraction First, consider the packing fraction for the 2D square Bravais lattice shown in Figure 5(a). Draw your own lattice planes This simulation generates images of lattice planes. The crystal basis is defined by. The vectors that span the primitive unit cell are usually chosen to go from (0,0,0) to the nearest equivalent point on the lattice. This is of course only true for crystalline phases, where, as mentioned, a systematic long-range order exists. On a macroscale, Yan et al. A lattice is a regular array of points in space. (Here, the unit cell contains one star, so Z = 1)!The unit cell is the basic repeat unit from which the entire crystal can be built. This topic briefly explains about the number of atoms in a unit cell and the types of unit cell. All of this structural information is used to build up a larger system by translation. Then the reciprocal lattice can be generated using primitive vectors 123 2π b=×a V a, 23 2 1 π =×aa V b, 312 2π =×aa V) b, (2. It is best thought of as a face-centered cubic array of anions with an interpenetrating fcc cation lattice (or vice-versa). You form this cell by taking all the perpendicular bisectrices or lines connecting a lattice point to its neighboring lattice points. causes a lattice distortion from the ideal octahedral layered structure. calculations for two-dimensional (2D) materials are discussed together with possible solutions. In vitro cultured cells tightly adhere on 3D plasmonic nanoelectrodes precisely in the plasmonic hot spots, making this kind of investigation possible. Lattice¶ class Lattice (a1, a2=None, a3=None) ¶ Unit cell of a Bravais lattice, the basic building block of a tight-binding model. Scattering by a Unit Cell amplitudescatteredby single electron amplitudescatteredby atoms in unit cell F hkl h a d AC MCN d h h c 00 G 2 1 2 00 sinT O a hx a h x AC AB MCN AC AB RBS S O G I S O G I S G O O 2 2 2 / 3 1 3 1 3 1 c c c c For atoms A & C For atoms A & B hu a hx S S I 2 2 3c1c If atom B position: u x/a For 3D: I 2S hu kv lw phase. The lattice points in a cubic unit cell can be described in terms of a three-dimensional graph. The conventional primitive unit cell has the shortest and most nearly equal lattice vectors bounding it. Interactive examples 2D crystal builder Planes and HKL's Finite size effect. The unit cell lengths are identical, the angel between the base vectors is 75 degrees. In polycrystalline graphite, the bond between the individual layers is broken, so their orientation is random. primitive cell. Because all three cell-edge lengths are the same in a cubic unit cell, it doesn't matter what orientation is used for the a, b, and c axes. volume (float) – New volume of the unit cell in A^3. The patterns that appear when filaments either overfill or under-fill the lattice are reminiscent of those observed in other physical systems involving 2d lattices. Close-packed structures can be designated by specifying the total number of layers in the hexagonal unit cell followed by the letter H, R or C to indicate the lattice type. This is an online tool to visualise a plane associated with a specific set of miller indices. Figure 1: (a) 2D PhC waveguide, (b) the unit cell of the 2D PhC waveguide, and (c) the element of the unit cell. PRIMITIVE CELLS •The unit cell is a volume that can fill all of space, without gaps, when translated by lattice vectors •The choice of unit cell is also non-unique, so it is convenient to have a standard for assigning them. scale_lattice (volume) [source] ¶ Performs a scaling of the lattice vectors so that length proportions and angles are preserved. The following pictorial shows the thought process. There are two ways to specify the atomic positions. In this expression, R is a lattice vector between a pair of unit cells: R =ua +vb+wc; u,v, and w are integers and the dot product k R. the unit cell may have certain angles constrained to be 90° indicating the presence of an axis (see next page). To set up a periodic structure in FHI-aims, all three lattice vectors as well as the atomic positions in the unit cell must be speci ed. • Lattice means a three-dimensional array of point coinciding with atom positions. 1) the full arrangement of lattice points can be reproduced. Three types of cubic unit cells are covered in this course. Two-Dimensional Space Groups The five basic lattice types There are 17 space groups in the plane, but their unit cells fall into one of five basic shapes as follows:. Move OD so it passes through the origin of the unit cell. The trigonal (or rhombohedral) lattice has three edges of equal length and three equal angles ($eq 90^\circ$). Viciu| AC II | Symmetry in 2D. Introduction to Number of Atoms in a Unit Cell. it is extending in 3D is perpendicular to the 2D plane Unit Cell a b c α β γ from CHEM F101 at Indian Institute of Technology, Roorkee. primitive unit cells have the same area. within the First Brillouin zone. Since there are one lattice sites per sc cubic cell, the density should be 0. In this work, we assume a¼20. Counting Lattice points and Atoms in Unit Cell. This relationship is tricky to think of three-dimensional space. LatticeData [patt] gives a list of all named lattices that match the string pattern patt. • The simple cubic cell (primitive cubic) is the simplest unit cell and has structural particles centered only at its corners. crystal symmetry. All of this structural information is used to build up a larger system by translation. Consider a square lattice in the xyplane with lattice constant a. Parallelipiped. The unit cell in this lattice is a parallelepiped whose base is a rhombus, as shown by the broken lines in Fig. The unit cell in 2D is a hollowed square with a side 278. The crystal system of the reciprocal lattice is the same as the direct lattice (for example, cubic remains cubic), but the Bravais lattice may be different (e. unit cell synonyms, unit cell pronunciation, unit cell translation, English dictionary definition of unit cell. Our program is compatible with data collected directly on polycrystalline thin films, the structure of which critically impacts maroscopic. If you add that point, then you just create a simple square lattice with half the lattice parameter. So this is the first Brillouin zone of the fcc based crystals like diamond and blend crystals. The unit cell • The atoms in a crystal are in a regular repeating pattern called the crystalline lattice. Normal Modes of a 1D Lattice By: Albert Liu The simplest case when examining a crystal structure is the approximation that the positive ions (or multi-atom bases) remain stationary at their Bravais lattice positions R. The lattice parameters define the size and shape of the unit cell. primitive cells. For a Bravais lattice, the primitive lattice vectors span the smallest possible volume, and the resulting unit cell is called the primitive unit cell. Although the main translation periods are chosen arbitrarily,theunitcell volumestill remainsthesame foranychoiceoftheunitvectors. 373 R3 3 The volume of the BCC unit cell is Vunit cell = a 3 where a is the lattice constant. In this article, we develop further the approach proposed by ( 3 ) and ( 11 ). The parameter a represents the total length of the unit cell, b is thickness of the ribs, t is the width of the slit and h is the out-plane thickness of the unit cell (not indicated in the. Example, for the 2D lattice above: • The volume (3D), area (2D), or length (1D) of a WS primitive cell can be given in terms of the primitive vectors, and is independent of the choice of the primitive vectors a1a2 3 a1. The method of was used in to consider lattices with triangular unit cell structure, and for square cell lattices in. 1/8 x 8 + 1 x 1 = 2 atoms/unit cell. At the intracellular level, the MABM approach employs a system of ordinary differential equations to describe quantitatively specific intracellular molecular pathways that determine phenotypic switches among cells (e. Diffraction in 2D Effect of Basis Positions of diffracted spots are identical, but intensity distribution is. In the simplest crystals the structural unit is a single atom, as in copper, silver, gold, iron, aluminium, and the alkali metals. 2D Crystallography Bravais lattices in 2D are called Bravais nets Unit cells in 2D are called unit meshes There are just 5 symmetrically different Bravais nets in 2D The centered rectangular net is the only non-primitive net. determining which 2D Bravais lattice best ts the experimental data from an image being processed with CIP. Crystal lattice is the depiction of three dimensional arrangements of constituent particles (atoms, molecules, ions) of crystalline solids as points. The area of the primitive cell in the direct lattice space is A= a. CBSE Class 12th Chemistry: The smallest group of atoms or molecules, whose repetition at regular intervals in three dimensions produces a crystal, is called as a Unit cell. There is an infinite number of choices for primitive unit cell. The hexagonal closest packed (hcp) has a coordination number of 12 and contains 6 atoms per unit cell. •Lattice vibrations: acoustic and optical branches In three-dimensional lattice with s atoms per unit cell there are 3s phonon branches: 3 acoustic, 3s - 3 optical •Phonon - the quantum of lattice vibration. unit cell volume V0 = a 1[a2, a3]. The density of Eu is 5. The unit cell is described by three edge lengths a = b = c = 2r (r is the host atom radius), and the angles between the edges, alpha = beta = gamma = 90 degrees. Types of unit cell: Primitive cubic unit cell. LatticeData [{" type ", id}, …] gives data for the lattice of the specified type with identifier id. the unit cell • Unit cell includes all of the required points on the lattice needed to repeat the lattice in an infinite array • Arbitrary definition of unit cell, but following rules are good to follow: –Edges of unit cell should match with symmetry of lattice –Edges of unit cell should also be related by the symmetry of lattice. Accordingly, a point located on the corner of a cube is shared by 8 cubes and would count with \$\frac{1}{8. Example: 2D honeycomb net can be represented by translation of two adjacent atoms that form a unit cell for this 2D crystalline structure Example of 3D crystalline structure:. The unit cell can be seen as a three dimension structure containing one or more atoms. In this paper, a new hybrid lattice particle modeling (HLPM) scheme is proposed that considers different particle interaction schemes not only limited to the nearest particles but also extended to the particles in the second nearest neighborhood, and different mesh structures with triangular or rectangular unit cell. Close-packed structures can be designated by specifying the total number of layers in the hexagonal unit cell followed by the letter H, R or C to indicate the lattice type. As examples, we examine scanning tunneling microscope images. Space Lattice, Unit Cell, Basis, Motif,. The considerable. 2D Lattices Two basis vectors generate the lattice There are three lattice parameters which describe this lattice One angle: o Two distances: a, b = 90 in the current example o b a Four (4) Unit Cell shapes in 2D can be used for 5 lattices as follows: Square (a = b, o = 90) Rectangle (a, b, o = 90) 120 Rhombus (a = b, o = 120) Parallelogram. There is an algorithm for constricting the reciprocal lattice from the direct lattice. •In general, the repeating volume (area in 2D) in a crystal is known as the unit cell. 1 Unit cells Any region of space that contains only one lattice point and can be translated by lattice vectors R to fill. The method of was used in to consider lattices with triangular unit cell structure, and for square cell lattices in. All of this structural information is used to build up a larger system by translation. The Wigner-Seitz cell of the reciprocal lattice is called the first Brillouin zone (FBZ). lattice Three possible arrangements yielding c(2x2) structures. This means a1 = 1 0 0 and a2 = 0 1 0. d , Experimental (blue) and simulated (red) SAXS patterns for nanoprisms (95nm edge length, 7nm thickness) assembled into a. Percentage of spaces filled by the particles in the unit cell is known as the packing fraction of the unit cell. Crystallography (II) – Unit Cell, continuation. Conventional Mott Insulators - Background. ‘visible’ domain walls, as there is one unit cell which cannot have a dipole aligned inx direction. (2012) have fabricated macroscopic single gyroid lattices (unit cell sizes between 2 and 8 mm of relative density of 15 %) from stainless steel powder using laser sintering technique. Visualize the resulting structures (e. Example, for the 2D lattice above:. Posts about Lattice written by dhruvbhate. Each ion is 6-coordinate and has a local octahedral geometry. Figure 5: 2D Lattice Packing Fraction First, consider the packing fraction for the 2D square Bravais lattice shown in Figure 5(a). Castro Neto, F. Parameters of a unit cell: A unit cell is characterized by six parameters. Primitive unit cells contain one lattice point only. In two dimensions, there are five Bravais lattices. It's like arranging balls of equal radii on the floor touching eachother, forming a sq (the balls placed on the vertices of the sq of side length twice the radius of the sphere). Physics 123B: Mid-Term February 7, 2013 1. , using jmol). It is generally the 2D projection looking down the [001]/z-axis of the unit cell. Two common choices are the parallelepiped and the Wigner-Seitz cell. Introduction to the theory of lattice dynamics M. In the triclinic lattice, all edges and angles are unequal. Different lattices can be created by modifying the atoms dictionary in lattice. Notice that once we begin stacking the unit cells, we never change the orientation of any subsequent unit cells as they stack. PRIMITIVE CELLS •The unit cell is a volume that can fill all of space, without gaps, when translated by lattice vectors •The choice of unit cell is also non-unique, so it is convenient to have a standard for assigning them. This result is also valid in 3 dimensions 3. The primitive rectangular lattice can also be described by a centered rhombic unit cell, while the centered rectangular lattice can also be described by a primitive rhombic unit cell. py, specifying the unit cell atoms and their positions. The smallest rectangular unit cell includes 2 unit cells of the hexagonal lattice Lattice constant a ax=a ay=a*sqrt(3)/2 Set mesh step size in y-direction Dx=ax/n Dy=ay/m where n,m are integers. The valley formation. It can be seen that bð1Þ ¼ b 1, b ð2Þ ¼ b 2, b ð3Þ ¼ b 1, and b ð4Þ ¼ b 2. The Brillouin Zone is the unit cell in reciprocal space, defined in analogy to the Wigner-Seitz cell in real space Definition of reciprocal Lattice and Brillouin Zone 3D case: given the unit cell vectors a 1, a 2, a 3 of the direct lattice, the reciprocal vector b 1, b 2, b 3 is defined by: b 1 2S a 2 u a 2 a 1 a 2 u a 3 3. This tool also features visualisation of plane in specific lattice structures like Simple Cubic, Body Centered or Face Centered cubic lattice structure. The area for the square unit cell is a2 while the area of the circle is ˇa2=4. •Neither primitive basis vectors nor the primitive unit cells are unique!. Unit cell. The basic cell for this periodicity is called the first Brillouin zone. We review the theory of lattice dynamics, starting from a simple model with two atoms in the unit cell and generalising to the standard formalism used by the scientific community. All of this structural information is used to build up a larger system by translation. An equivalent definition of the primitive unit cell is a cell with one lattice point per cell (each lattice point in the figures above belong to. First we need to nd the reciprocal lattice and rst Brillouin zone. A sq lattice has 1 basis atom at the lower-left corner of the square. 2 shows examples of possible unit cells in a 2D crystal lattice in which the lattice points are symbolized by the open circles, showing some possible unit cells. The average size hand specimen is made of more than1023 atoms, so the lattice of a single crystal is spatially extensive. In both cases, it is the underlying space symmetry of the lattice points (or W, Fe, Cr, Rb,…atoms) that allows for the centerings with the consequence that the conventional unit cells contain two lattice points (or W atoms in case of tungsten), rather than one lattice point (or one W atom) that represents the primitive sub-unit cell. These striated 2D crystals cannot be described by a typical unit cells of 1-2 Å for crystals, but rather long range. 3 Bravais lattices in 2D Bravais lattices, named after the French physicist Auguste Bravais (1811-1863), define all the. the unit cell • Unit cell includes all of the required points on the lattice needed to repeat the lattice in an infinite array • Arbitrary definition of unit cell, but following rules are good to follow: –Edges of unit cell should match with symmetry of lattice –Edges of unit cell should also be related by the symmetry of lattice. The unit cell is the smallest group of atoms, ions or molecules that, when repeated at regular intervals in three dimensions, will produce the lattice of a crystal system. The maps below represent the signal for each unit in the labeled experiment’s dimension. Specify the crystal lattice type for the unit cell, that is, the crystallographic parameters a, b, c (cell lengths) and , , (cell angles). to_unit_cell (bool) – Whether new sites are transformed to unit cell. 1B), it is. This unit assembly is called the `basis’. lattice elements in one unit cell. figuration, in which a skyrmion is obtained in each unit cell of a lattice. There are two orthorhombic Bravais lattices in two dimensions: Primitive rectangular and centered rectangular. These are defined by how you can rotate the cell contents (and get the same cell back), and if there are any mirror planes within the cell. Thus the WS unit cell is not appropriate when one deals with graphene transport problem, which is explained below. In the simplest crystals the structural unit is a single atom, as in copper, silver, gold, iron, aluminium, and the alkali metals. The demonstration shows how to express the direction OD in the 3-D lattice shown. Our program is compatible with data collected directly on polycrystalline thin films, the structure of which critically impacts maroscopic. primitive cells. The Brillouin Zone is the unit cell in reciprocal space, defined in analogy to the Wigner-Seitz cell in real space Definition of reciprocal Lattice and Brillouin Zone 3D case: given the unit cell vectors a 1, a 2, a 3 of the direct lattice, the reciprocal vector b 1, b 2, b 3 is defined by: b 1 2S a 2 u a 2 a 1 a 2 u a 3 3. The Body-Centred Cubic Lattice! The primitive cell of the BCC lattice is defined by the translation vectors: a 1 a 2 a 3 x y z a 1 = ‰ a (x + y - z) a 2 = ‰ a (-x+y + z) a 3 = ‰ a (x - y + z) a where x, y, and z are the Cartesian unit vectors. Our interest to hexagonal 2D structures has been inspired by graphene stud-ies, studies of the excitonic spectra of bacteriorhodpsin [3,4], as well as by the possibilities to realize such 2D or quasi-2D structures using modern technolo-gies [5,6]. [email protected] make_snapshot()). This is of course only true for crystalline phases, where, as mentioned, a systematic long-range order exists. py, specifying the unit cell atoms and their positions. 13 Angstroms, and 5. Different lattices can be created by modifying the atoms dictionary in lattice. A unit cell contains two masses m 1 and m 2 of equal value m and two springs of constants k 1 and k 2. In a second step one constructs the perpendicular bisectors of the connecting lines. Unit Cell, Primitive Cell and Wigner-Seitz Cell. Figure 1: (a) 2D PhC waveguide, (b) the unit cell of the 2D PhC waveguide, and (c) the element of the unit cell. Novoselov, and A. that is larger than the. So sometimes it makes sense not to use a primitive unit cell but one which fits better to the symmetry of the problem. These translation vectors connect the lattice pt at the origin to the points at the body centres (and make a rhombohedron). Lattice unit cells (unit lattices) are parameterizable, analyzable, patternable and manufacturable to support the design. These 14 lattice types can cover all possible Bravais lattices. Bravais lattices in 2 dimensions In each of 0-dimensional and 1-dimensional space there is just one type of Bravais lattice. Inserting this volume into (4), we obtain for continuum 2lt 2. Lattices can be specified by standard names such as "FaceCenteredCubic" and "CoxeterTodd". This tool also features visualisation of plane in specific lattice structures like Simple Cubic, Body Centered or Face Centered cubic lattice structure. Lattice¶ class Lattice (a1, a2=None, a3=None) ¶ Unit cell of a Bravais lattice, the basic building block of a tight-binding model. Also, the choice of a nonprimitive (centered) unit cell adds lattice points in direct space on the left but subtracts them in frequency space on the right. Unit cells can be used to build the entire lattice. primitive cell. In 2D, a non-primitive unit cell has one additional lattice point exactly centered within it and is called a body-centered non-primitive unit cell. These are defined by how you can rotate the cell contents (and get the same cell back), and if there are any mirror planes within the cell. The unit cell, depicted in red, contains a complete circle. in a clearer way. A calcium fluoride unit cell, like that shown in Figure 17, is also an FCC unit cell, but in this case, the cations are located on the lattice points; equivalent calcium ions are located on the lattice points of an FCC lattice. Move OD so it passes through the origin of the unit cell. The upper right structure actually appearing in graphite, stacked layers of graphene. unit cell. In reality, we have to deal with finite sizes. within the First Brillouin zone. The unit cell is that unique part of the crystal structure such that when translated along parallel lines, generates the entire crystal. The details of this difference will be reported elsewhere (13). In the simplest crystals the structural unit is a single atom, as in copper, silver, gold, iron, aluminium, and the alkali metals. edu) Most inorganic materials we usually deal with are crystalline, meaning that they are periodic at the atomic scale. The 2D lattice shown here has a primitive unit cell containing two points (a black one and a blue one). The maps below represent the signal for each unit in the labeled experiment’s dimension. 6 Reproduced acceleration response for the last mass in the nite pe-riodic lattice with M=2, 5, and 10 unit cells. These translation vectors connect the lattice pt at the origin to the points at the body centres (and make a rhombohedron). the unit cell • Unit cell includes all of the required points on the lattice needed to repeat the lattice in an infinite array • Arbitrary definition of unit cell, but following rules are good to follow: –Edges of unit cell should match with symmetry of lattice –Edges of unit cell should also be related by the symmetry of lattice. 2 g cm-3, calculate the number of atoms present in 200 g of element. 1 Definition Lattices are in appearance very similar to open cell foams but differ in that lattice member deformation is stretch-dominated, as opposed to bending*. Draw the Wigner-Seitz cells for a 2D hexagonal lattice. Bravais lattices in 2 dimensions In each of 0-dimensional and 1-dimensional space there is just one type of Bravais lattice. a r Starting with a point the lattice translation vector (basis vector) can generate the lattice Note: Basis vector should not be confused with the basis ( the motif. The Plan view generally displays a 2×2 array of unit cells. For example, the (100) spot vanishes for the fcc lattice due to the extra face-centered atom at (u,v,w) = (½, 0, ½). Atoms at the corner of the 2D unit cell contribute only 1/ 4 to unit cell count. CBSE Class 12 Chemistry Notes: The Solid State. Let the unit cell length be given by a. Crystal lattice is the depiction of three dimensional arrangements of constituent particles (atoms, molecules, ions) of crystalline solids as points. The vertical unit cell length is half the horizontal unit cell length. 1) the full arrangement of lattice points can be reproduced. The crystal basis is defined by. In the dispersion relation, or the band diagram, of the structure (Fig. In the case of a rectangular two dimensional lattice the unit cell is the rectangle, whose sides are the vectors a1 and a2. !A primitive unit cell contains only one lattice point. Dove∗ Department of Earth Sciences, University of Cambridge, Downing Street, Cambridge CB1 8BL, UK Abstract. (unit cell size ≈ 400 nm) that function as multi-focal lens, interference filter and wave guide. This means a1 = 1 0 0 and a2 = 0 1 0. is the volume associated with one lattice point. Therefore, unit cells are the repeating units of crystal lattices. 4284um and the hole diameter to 0. The lattice can therefore be generated by three unit vectors, a 1, a 2 and a 3 and a set of integers k, l and m so that each lattice point, identified by a vector r, can be obtained from:. Sketch the Bravais lattice, identify the basis, and de ne the primitive unit cell for a 2D CuO. 2D Lattices Two basis vectors generate the lattice There are three lattice parameters which describe this lattice One angle: o Two distances: a, b = 90 in the current example o b a Four (4) Unit Cell shapes in 2D can be used for 5 lattices as follows: Square (a = b, o = 90) Rectangle (a, b, o = 90) 120 Rhombus (a = b, o = 120) Parallelogram. A primitive cell is a unit cell that contains exactly one lattice point. |
# How to generate bold face numbers in LaTeX as they appear in Microsoft Word or PowerPoint?
In MS Word and PowerPoint you have a set of these bold faced numbers written like
More specifically
Does anyone know if you can generate these using LaTeX? It does not seem to be the output of either mathbf, mathfrak, mathscr
Thanks!
The right command in LaTeX is \mathbb, but the default fonts only have blackboard bold letter, not blackboard bold numbers.
The easiest solution is to use a Opentype Math font that has blackboard bold numbers; but this requires XeTeX or LuaTeX engine. For example, in ConTeXt you can use Cambria Math font as follows:
\setupbodyfont[cambria]
\starttext
$\Bbb 1234567890$
\stoptext
In LaTeX, I believe that one can use fontspec package to load opentype fonts.
If you just need boldface 1, you can use dsfont package:
\documentclass{article}
\usepackage{dsfont}
\begin{document}
$\mathds{1}$
\end{document} |
# Math Help - Improper integral
1. ## Improper integral
I need to find the Improper integral of this thing, how do I do that?
2. Originally Posted by asi123
I need to find the Improper integral of this thing, how do I do that?
RonL
3. Originally Posted by CaptainBlack
RonL
But it isn't divergent as both integrals are convergent.
RonL |
# Definition of topological space
The definition of a topological space is a set with a collection of subsets (the topology) satisfying various conditions. A metric topology is given as the set of open subsets with respect to the metric. But if I take an arbitrary topology for a metric space, will this set coincide with the metric topology?
I'm trying to justify why we call the elements of a topology "open". If my above question is true, then at least in a metric space, the set of open sets is equivalent to the topology of the metric space. So am I right in thinking that when we remove the metric, we are generalising this equivalence by defining the open sets as those that satisfy the conditions of a topology?
• I think your question is related to when a topology on a space is metrizable (I.E you can define a metric function that generates the same topology). In general this requires some properties on the topological space itself. – DBS Feb 4 '16 at 14:36
• The metrizability depend on the topology ! – Surb Feb 4 '16 at 14:37
• What do you mean by "arbitrary topology for a metric space"? Do you mean an arbitrary topology on the points of the metric space? If so, then there is no reason that it will necessarily coincide with the metric topology. The collection of all open subsets in the 'metric topology' will satisfy the general definition for a topology, but there are topologies which are non-metrizable. Thus the notion of a topological space is a generalization of metric space. – Apollo Feb 4 '16 at 14:39
• Ok so if the open sets wrt the metric are not equivalent to the open sets in a topology (of a metric space), why are the sets in a topology called open?? I get this is a generalisation but there seems to be no correspondence, despite the same name being used. My problem may be more due to the language more than anything else – Ted Jh Feb 4 '16 at 14:48
Suppose you've got a set, $X$. If you equip $X$ with a metric $d$, now the pair $(X,d)$ is a metric space.
This metric generates a topology on $X$. You consider the collection $\mathcal{B}$ of sets of the form $$B(x,r):= \{y \in X:d(x,y) < r \}.$$
Now the collection $\mathcal{B}$ is not a topology in and of itself, but lets you build one by taking arbitrary unions and finite intersections of sets from $\mathcal{B}$. The resulting collection $\mathcal{T}$ satisfies the conditions needed for a collection to be considered a topology on $X$. This is exactly the metric topology on $X$ with respect to $d$. So in this setting, a subset of $X$ will be open (is in $\mathcal{T}$) if and only if it is open with respect to the metric (that you can fit a ball of some radius around each point in the set).
But you could just as easily muster up a collection of subsets $\mathcal{T}'$ of $X$ that don't spawn from a metric on $X$, yet still satisfy the conditions to be a topology. The open sets in $(X,d, \mathcal{T})$ will be in general very different from those in $(X, \mathcal{T}')$.
You can build a topology from scratch by declaring, "I want these sets to be open." Then you do what's necessary to make sure everything satisfies the conditions to have a topology. This is the concept of a basis for a topology. So in this setting the idea of "open" in topological spaces is really just an abstraction of the "open" you're used to in metric spaces.
Except for the trivial case of a metric space with only one element, there is always at least one topology on a metric space that is not the same as the metric topology, namely the discrete topology in which only the empty set and the whole space are open. An example of a very interesting topology on $\Bbb{R}$ that is not the metric topology is the lower limit topology.
As the answer to your first question is "no", you may want to rethink your second question.
• So in the lower limit topology the "open" sets are the half-open intervals? I just find that a bizarre use of language, probably the cause of my confusion. Why not call it anything else, since it's hardly ever going to correspond to the intuitive idea of open! – Ted Jh Feb 4 '16 at 15:10
• I don't know what your intuitive idea of open is, so it's hard to comment on that. – Rob Arthan Feb 4 '16 at 20:49
• Just a set that's open wrt a metric. Whereas I wouldn't think to call an arbitrary set that happens to satisfy the topology axioms open. But I'm more happy with the idea now, thanks! – Ted Jh Feb 4 '16 at 21:20 |
# American Institute of Mathematical Sciences
October 2018, 14(4): 1565-1577. doi: 10.3934/jimo.2018021
## Frequency $H_{2}/H_{∞}$ optimizing control for isolated microgrid based on IPSO algorithm
Key Lab of Industrial Computer Control Engineering of Hebei Province, College of Electric Engineering, Yanshan University, Qinhuangdao 066004, China
* Corresponding author: Zhong-Qiang Wu
Received February 2017 Revised August 2017 Published January 2018
Fund Project: The first author is supported by the Hebei Natural Science(F2016203006)
Affected by the fluctuation of wind and load, large frequency change will occur in independently islanded wind-diesel complementary microgrid. In order to suppress disturbance and ensure the normal operation of microgrid, a $H_{2}/H_{∞}$ controller optimized by improved particle swarm algorithm is designed to control the frequency of microgrid. $H_{2}/H_{∞}$ hybrid control can well balance the robustness and the performance of system. Particle swarm algorithm is improved. Adaptive method is used to adjust the inertia weight, and cloud fuzzy deduction is used to determine the learning factor. Improved particle swarm algorithm can solve the problem of local extremum, so the global optimal goal can be achieved. It is used to optimize $H_{2}/H_{∞}$ controller, so as to overcome the conservative property of solution by linear matrix inequality and improve the adaptive ability of controller. Simulation results show that with a $H_{2}/H_{∞}$ controller optimized by improved particle swarm algorithm, the frequency fluctuations caused by the wind and load is decreased, and the safety and stable operation of microgrid is guaranteed.
Citation: Zhong-Qiang Wu, Xi-Bo Zhao. Frequency $H_{2}/H_{∞}$ optimizing control for isolated microgrid based on IPSO algorithm. Journal of Industrial & Management Optimization, 2018, 14 (4) : 1565-1577. doi: 10.3934/jimo.2018021
##### References:
show all references
##### References:
Independent wind-diesel microgrid
The dynamic model of microgrid
Flow chart of PSO algorithm
Cloud membership function of $D(i, {g}_{best})$
Cloud membership functions of $c_{1}$, $c_{2}$
Load and maximum power output of wind in microgrid
Output power of diesel generator
Frequency deviation of microgrid (with $H_{2}/H_{\infty}$ control based on LMI)
Frequency deviation of microgrid (with $H_{2}/H_{\infty}$ control based on PSO algorithm)
Frequency deviation of microgrid (with $H_{2}/H_{\infty}$ control based on IPSO algorithm)
The fuzzy rules of $c_{1}$ and $c_{2}$
Rules $D(i, {g}_{best})$ $c_{1}$ $c_{2}$ 1 Near Small Big 2 Middle Middle Middle 3 Far Big Small
Rules $D(i, {g}_{best})$ $c_{1}$ $c_{2}$ 1 Near Small Big 2 Middle Middle Middle 3 Far Big Small
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2018 Impact Factor: 1.025 |
# Review “An Efficiency Comparison of Document Preparation Systems Used in Academic Research and Development”
Today someone email of of the mailing list that I subscribed about a article wrote by Markus Knauff and Jelica Nejasmic with the title “An Efficiency Comparison of Document Preparation Systems Used in Academic Research and Development” and published at PLOS ONE. This is a open access journal so if you want to read the article before read this post just click on the first link and enjoy the article.
## How you can compare apples and oranges?
At the introduction we have
Microsoft Word is based on a principle called “What you see is what you get” (WYSIWYG), which means that the user immediately sees the document on the screen as it will appear on the printed page. LaTeX, in contrast, embodies the principle of “What you get is what you mean” (WYGIWYM), which implies that the document is not directly displayed on the screen and changes, such as format settings, are not immediately visible.
The authors are saying that Microsoft Word and LaTeX are different starting from their designs. So how do you want to compare it?
## Novices vs Experts
About the division of the participants in groups we have
The participants were divided into 4 groups with 10 participants in each group: Word novices, Word experts, LaTeX novices, and LaTeX experts. Participants were classified as “novices” if they had less than 500 hours of experience with the respective program and “experts” if they had more than 1000 hours of experience with the respective program.
This classification of novices and experts is too much simplistic. With this separation will not be hard to find a Word expert that doesn’t know the difference of add a figure as embed element or as a link.
Note
Figures included as a link are automatically updated when another application change it. LaTeX only support this type of image inclusion so for a far comparison should be requested that Word users use only this type of figure inclusion.
Note
In the same way that with this criterion is easy to find Word expert that doesn’t know basic features of Word is also easy to find LaTeX expert that doesn’t know basic features of LaTeX.
## The continuous text
Note
You will find the text used here.
The continuous text looks fair to me.
From the result, I still have questions of how much the LaTeX users know of the IDE they use.
## The table text
Note
You will find the table used here.
The table has multi columns and multi lines that, in LaTeX, request the use of the packages multicols and multirow, respectively. Reproduce this table will be almost impossible for any LaTeX novice user and any LaTeX expert will probably use an third party tool, like Calc2LaTeX, to get this done.
The result is, more or less, what I expected. Word and LaTeX users had the same amount of orthographic and grammatical mistakes but the amount of formatting erros and typos is high among LaTeX users (probably because the requested table isn’t easy to be reproduce in LaTeX). I don’t know how the amount of written text was calculate so I will not make any comment.
## The equation text
Note
You will find the equations used here.
Except for the equation numbering doesn’t start at one this should be easy for any Word or LaTeX expert. I expect that LaTeX novice have some issues with the last equation since it request the alignment of equations and the use of \nonumber to get ride of the equation numbering in the first two lines.
Note
In LaTeX, the last equation should be type as:
\begin{align}
\frac{\partial}{\partial \phi_c(j)} \frac{1}{a(\phi_c)} &= \frac{-1}{(a
(\phi_c))^2} \frac{\partial}{\partial \phi_c(j)} a(\phi_c) \nonumber \\
&= \frac{-1}{(a(\phi_c))^2} \sum_{r \in \mathcal{R}} \exp(C(r)/c) x_r(j))
\nonumber \\
&= \frac{-1}{a(\phi_c)} p_c(x(j))
\end{align}
If you like you could replace ( with \left( and ) with \right).
As expect in this case LaTeX has a small advantage.
## Usability questionnaire
This is the part that I’m little happy about since it confirms what many of us “know”:
• “Word users rated their respective software as less efficient than LaTeX users”,
• “LaTeX users rated the learnability of their respective software as poorer than Word users”.
## Discussion
The author first said that
“Another characteristic of our study is that it is practically impossible to evaluate LaTeX without also evaluating the used editors. In fact, our research measured the efficiency of Word against LaTeX in combination with some editor interfaces.”
and next that
“our results suggest that LaTeX reduces the user’s productivity and results in more orthographical, grammatical, and formatting errors, more typos, and less written text than Microsoft Word over the same duration of time.”
From my experience, the fact that more orthographical and grammatical erros among LaTeX users is due the fact that many LaTeX IDE don’t have the spell check feature enable out of the box and many users don’t configure it. I would like to see how Word users performs with Word spell check disable.
## Others reviews
You may also like to read others reviews: |
Simplifying complex fractions is a process that can range from easy to difficult based on how many terms are present in the numerator and denominator, whether any of the terms are variables, and, if so, the complexity of the variable terms. For example, the product of the two complex numbers (3+2i)*(5-3i) works as follows: First. Like. Abs[A]^2 But when I try. The product of the first terms is 3*5=15. 18 0 obj<> Outer. A complex rational expression is -6*-i=6i, i^2=-1. endobj For example, to simplify the sum of (a+bi) and (c+di), first identify that a and c are the real number portions, and add them together. Mar 18, 2008 #1 Okay, I think I did this right, but I'm going to post it here just to make sure. 2 0 obj<> ±16 (12 + 8i) 4i 1 2i ± 4i + (2 + 10i) 13 ± 4i (15 ± 3i) ± !±49 The product of the outer terms is 3*(-3i). Hot Network Questions Which software is good with generally contracted basis sets? Simplify the complex number and write it in standard form. 25 0 obj<> 6i+-1=-1+6i. Simplifying Expressions with Complex Numbers. For the sample, this would be a*d. Inner. Simplifying a Complex Number. 9. When I take. D�P�!+ (! Simplifying complex expressions. The sum of (a+bi) and (c+di) is written as (a+c) + (b+d)i. Simplifying Complex Numbers. ∴-6i^3+i^2=-1+6i. For the sample expression, this would be b*d. Finally, add all four products together. stream 21 0 obj<> The result for the sample binomial multiplication of (a+b)(c+d) is ac+ad+bc+bd. Easy simplifying complex fractions worksheet. A*Conjugate[A] // FullSimplify The output is as desired. When mathematics was first used the primary purpose was for counting. We know ads can be annoying, but they’re what allow us to make all of wikiHow available for free. endobj Click on pop out icon or print icon to worksheet to print or download. Use basic rules like combining the like terms together, FOIL method, multiplying the top and bottom by the complex conjugate of the denominator to simplify complex-valued expressions. 24 0 obj<> Simplifying Complex Numbers Simplify. Simplifying complex numbers worksheet. The product of the first terms is 3*5=15. You need to apply special rules to simplify these expressions with complex numbers. %���� In this activity, students will match the complex number expressions that are equivalent. endobj By signing up you are agreeing to receive emails according to our privacy policy. Using the numerical example of (3+3i) + (5-2i), the imaginary portions of the two complex numbers are 3i and -2i. The following calculator can be used to simplify ANY expression with complex numbers. Here is a pdf worksheet you can use to practice how to solve negative square roots as well as simplifying numbers using the imaginary unit i. 15 0 obj<> Then multiply the numerator and denominator and simplify as follows: Notice in the second step above, the denominator contains the terms. How do you simplify an expression like 8(i^60)? In the given example, the inner terms are b*c. Last. ±9 ) (12 + 5i) 6 + 3i 3 + 9i! For the sample 15-9i+10i+6, you can add the 15 and 6 together and add the -9i and the 10i together. Last Updated: March 3, 2020 Let $\omega=e^{i2\pi/2015}$, evaluate $\sum\limits_{k=1}^{2014}\frac{1}{1+\omega^k+\omega^{2k}}$ 0. Russia makes military move with Biden set to take office. $2.50. Include your email address to get a message when this question is answered. EMOJI - Complex Numbers: Simplifying Radicals with i. by . Large polynomial power of a complex number. x���� �1 The product of the last terms is (2i)*(-3i). Some sample complex numbers are 3+2i, 4-i, or 18+5i. (Note: and both can be 0.) @ 19 0 obj<> 14 0 obj<> Abs[A]^2 Abs[B]^2 I am using Mathematica 11.2.0. simplifying-expressions complex… 5.01 Simplifying Imaginary Numbers and Powers of i … For example: 8 - 3i/-2i? You need to use the Least Common Divisor (LCD) Rule. Example 2: to simplify 2 −3i2 +3i. One version has … 22 0 obj<> Please help us continue to provide you with our trusted how-to guides and videos for free by whitelisting wikiHow on your ad blocker. … This article was co-authored by our trained team of editors and researchers who validated it for accuracy and comprehensiveness. This fact is used in simplifying expressions where the denominator of a quotient is complex. 5. endobj Answers match to form a square.Two levels of the puzzle are included. To represent a complex number, we use the algebraic notation, z = a + ib with i ^ 2 = -1 The complex number online calculator, allows to perform many operations on complex numbers. Log in AG Ankit G. Numerade Educator. Free fraction worksheets 2 simplifying fractions equivalent fractions fractions mixed numbers. Indeed, it is always possible to put any complex number into the form , where and are real numbers. Share this: Click … Outer. The product of the two inner terms is 2i*5. To overcome the issue we have created a super cool & handy online tool ie., Simplifying Complex Fractions Calculator. Already have an account? the real parts with real parts and the imaginary parts with imaginary parts). For the algebraic example of (a+bi) plus (c+di), the imaginary portions are b and d. Adding these together algebraically gives the result (b+d)i. Therefore we can split up large exponents like so: 8i^60 = 8 * (i^4)^15 = 8 * (1)^15 = 8. Learn more... A complex number is a number that combines a real portion with an imaginary portion. If you thought fractions were scary before, a complex rational expression will scare you even more! By using this website you agree to our cookie policy. A B Conjugate[A] Conjugate[B] Why mathematica is not showing the output as. 16 0 obj<> Addition / Subtraction - Combine like terms (i.e. From this representation, the magnitude of a complex number is defined as the point on the Cartesian plane where the real and the imaginary parts intersect. �*�F-zmE'�O�JK�Qۨ>�h��~Ct]���:�E�wHw:(�����B��dx+�%�Y;�0��tr���U^f���3�Y�w#>��7J��_��\j��,�(�Gx�U�{�. 3 ways to simplify complex numbers wikihow simplifying foldable i speak math mistakes your finest number please Mar 2008 37 0. Because we want to offer everything that you need in one genuine and also efficient origin, we all current very helpful details on many subject areas plus topics. For the sample, this would be a*c. Outer. By … wikiHow's Content Management Team carefully monitors the work from our editorial staff to ensure that each article is backed by trusted research and meets our high quality standards. 3 0 obj<> These would be the two terms that appear in the middle, which are the second term of the first binomial and the first term of the second binomial. endobj Thanks to all authors for creating a page that has been read 30,556 times. The L in FOIL represents the last terms of each binomial. The O in FOIL tells you to multiply the “outer” terms. Imaginary is the term used for the square root of a negative number, specifically using the notation i=−1{\displaystyle i={\sqrt {-1}}}. By using our site, you agree to our. This product is -6i. x�c� The x-axis represents the real part, with the imaginary part on the y-axis. There are 11 references cited in this article, which can be found at the bottom of the page. Algebra 2 simplifying complex numbers worksheet answers. http://www.purplemath.com/modules/complex.htm, http://www.algebrahelp.com/lessons/simplifying/foilmethod/pg2.htm, https://www.mathsisfun.com/numbers/complex-numbers.html, consider supporting our work with a contribution to wikiHow. Complex Number – any number that can be written in the form + , where and are real numbers. Complex numbers, as any other numbers, can be added, subtracted, multiplied or divided, and then those expressions can be simplified. This calculator simplifies your complex fractions in the blink of an eye & generates the show work for the given inputs. From simplifying complex numbers to division, we have every aspect discussed. These are the first term of the first binomial and the second term of the second binomial. This article has been viewed 30,556 times. The real portion of the simplified complex number will be 8. Simplifying fractions worksheets reducing fraction is one of the very basic concepts the 3rd grade 4th grade and 5th grade children should learn. References. 200+ Algebra Worksheets available here and free to be downloaded! Complex numbers are sometimes represented using the Cartesian plane. As usual we start with demonstrating the technique in a general case, and then give a … First. Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. Please consider making a contribution to wikiHow today. endobj Finding Absolute Values . I am trying to take product of a complex number with its conjugate. For such complex fractions, you need to split the expression into two parts. What if there's only 1 number for the denominator when it comes to complex numbers? wikiHow is where trusted research and expert knowledge come together. Last. stream [�L)� Report. Simplifying Complex Fractions Calculator: Simplifying the fractions can be a little difficult task while calculating the lengthy problems. . Apply the FOIL rule to complex number multiplication. Easy mixed fractions. You multiply with that and it cancels out the 'i' in the denominator. Your support helps wikiHow to create more in-depth illustrated articles and videos and to share our trusted brand of instructional content with millions of people all over the world. If you really can’t stand to see another ad again, then please consider supporting our work with a contribution to wikiHow. -6-15i-8i+20 = 14-23i (Combine terms and simplify). endstream To multiply two complex numbers, set them up as the product of two binomials and apply the FOIL rule. Improper to mixed fractions. The complex symbol notes i. Become adept at adding, subtracting, multiplying, and dividing complex numbers with this free worksheet. 13 0 obj<> Free worksheet pdf and answer key on complex numbers. Applying the numerical example, the sum of (3+3i) + (5-2i) is 8+i. Here, your denominator has a real number of 0 and a complex number of -2. Amid the current public health and economic crises, when the world is shifting dramatically and we are all learning and adapting to changes in daily life, people need wikiHow more than ever. Forums. For this reason, complex fractions are sometimes referred to as "stacked fractions". The complex number calculator is also called an imaginary number calculator. What would the conjugate be for the denominator? 1 0 obj<> Simplifying Complex Numbers Worksheet along with Beneficial Contents. wikiHow's. The union of the set of all imaginary numbers and the set of all real numbers is the set of complex numbers. &s�h}}^^���d���q*�?����f�5po&�O��f ?����r�F9�`��"&&6i#�ef��>���Rd㓋��F�h�[̡���B~��Ƌ�E,�9���6�"��YD�$9_��y�h2[�L�������R�P��\|v��CYv�i�M��2�u�J����. Come to Emaths.net and learn about linear inequalities, course syllabus for intermediate algebra and a … 12 0 obj<> Simplifying A Number Using The Imaginary Unit i – Worksheet. For this, you have to start from solving the denominator first and then move on to solving the simplified fraction. 1 i34 2 i129 3 i146 4 i14 5 i68 6 i97 7 i635 8 i761 9 i25 10 i1294 11 4 i 1 7i. endobj We use cookies to make wikiHow great. So what you're really seeing is (8-3i)/(0-2i). Using actual numbers instead of variables, consider the example of (3+3i) + (5-2i). endstream 11 0 obj<> Simplifying Complex numbers. A complex number, then, is made of a real number and some multiple of i. It is known that i^1 = i = sqrt(-1), i^2 = -1, i^3 = -i, and i^4 = 1. endobj COPMLEX NUMBERS OVERVIEWThis file includes a handwritten and complete page of notes, PLUS a blank student version.Includes:• basic definition of imaginary numbers• examples of simplifying imaginary numbers• examples of adding, subtracting, multiplying, and dividing complex numbers• complex conjugate endobj View 5.01 SImplifying Complex Numbers and Powers of i.pdf from MAT 109 at Nassau Community College. When simplifying complex fractions, there are different ways that you can choose to simplify the problem. The product of the outer terms is 3*(-3i). This product is -9i. (-i)^{6} This product is 10i. B3�y��Þ��|raT���JA��)�Qh&� - Simplifying Imaginary Numbers – Worksheet (Note – All of The Complex Hub’s pdf worksheets are available for download on our Complex Numbers Worksheets page.) What is it? Please consider making a contribution to wikiHow today. E. eraser851. First. Jump To Question Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 … endobj So the conjugate would be 0+2i, or simply 2i. % of people told us that this article helped them. FullSimplify@(A*Conjugate[A]*B*Conjugate[B]) The output is . The F in FOIL means that you multiply the first term of the first binomial by the first term of the second binomial. Pro-Trump rocker claims he's 'destitute' after label cut him. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Simplifying a Complex Number. 17 0 obj<> Every day at wikiHow, we work hard to give you access to instructions and information that will help you live a better life, whether it's keeping you safer, healthier, or improving your well-being. Inner. Simplify the comple… View Full Video. endobj 23 0 obj<> Simplifying Complex Numbers (i) Thread starter eraser851; Start date Mar 18, 2008; Tags complex numbers simplifying; Home. x�}S}PTU��������.Y�ݗY�DH@��gH�T�Ʈ�X}�|��X��L�4�Lq�IIR�p5�\$ؐ�Ҧ����6�[���j�Ι�;����s�RH�(�Ҥ�" endobj There are a surprising number of consequences to the fact that , and one of these is how far one can simplify a complex number. Add these together to get 3+5=8. This product is -9i. The I in FOIL means to multiply the “inner” terms. The result will be 21+i. Pre-University Math Help. Click on pop out icon or print icon to worksheet to print or download. %PDF-1.3 The real portion of the first number is 3, and the real portion of the second complex number is 5. {"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/f\/f6\/Simplify-Complex-Numbers-Step-1.jpg\/v4-460px-Simplify-Complex-Numbers-Step-1.jpg","bigUrl":"\/images\/thumb\/f\/f6\/Simplify-Complex-Numbers-Step-1.jpg\/aid8611950-v4-728px-Simplify-Complex-Numbers-Step-1.jpg","smallWidth":460,"smallHeight":345,"bigWidth":728,"bigHeight":546,"licensing":" |
# [SciPy-user] sine transform prefactor
Lubos Vrbka lists@vrbka....
Sat Jul 26 09:43:13 CDT 2008
hi guys,
since i am not a mathematician, i'd like to ask one question related to
discrete sine transform and its implementation in scipy (sandbox at the
moment).
from definition, the continuous sine transform comes with the constant
prefactor of sqrt(2/pi), i.e.,
FST(f(r)) = f(k) = sqrt(2/pi) int_0^{\infty} f(r) sin(kr) dr
iFST(f(k)) = f(r) = sqrt(2/pi) int_0^{\infty} f(k) sin(kr) dk
how is it in case of a discrete transform? does this sqrt(2/pi) factor
come into play also for discrete transforms, or is it applicable only
for continuous transforms? if it is applicable, is it already included
inside the routines? how is it then distributed among the forward and
reverse transform?
i think the situation here might be complicated by the fact, that the
code in sandbox uses the fast fourier transform for the evaluation of
the sine transform - and this might probably change the constants
involved as well (since the fourier transform should be 'normalized'
with a factor of 1/2pi).
i would be very grateful for any help on this issue. my program seems to
run, but it is dependent on the choice of the constant factors involved
in the transforms. so far i've found the correct combination (it seems
so), but i'd like to make this more rigorous and to really understand _why_.
thanks a lot in advance. best,
--
Lubos _@_"
http://www.lubos.vrbka.net |
# Portal:Mathematics
## The Mathematics Portal
Mathematics is the study of numbers, quantity, space, structure, and change. Mathematicians seek out patterns and formulate new conjectures. Mathematicians resolve the truth or falsity of conjectures by mathematical proofs, which are arguments sufficient to convince other mathematicians of their validity. The research required to solve mathematical problems can take years or even centuries of sustained inquiry. However, mathematical proofs are less formal and painstaking than proofs in mathematical logic. Since the pioneering work of Giuseppe Peano (1858–1932), David Hilbert (1862–1943), and others on axiomatic systems in the late 19th century, it has become customary to view mathematical research as establishing truth by rigorous deduction from appropriately chosen axioms and definitions. When those mathematical structures are good models of real phenomena, then mathematical reasoning often provides insight or predictions.
Through the use of abstraction and logical reasoning, mathematics developed from counting, calculation, measurement, and the systematic study of the shapes and motions of physical objects. Practical mathematics has been a human activity for as far back as written records exist. Rigorous arguments first appeared in Greek mathematics, most notably in Euclid's Elements. Mathematics developed at a relatively slow pace until the Renaissance, when mathematical innovations interacting with new scientific discoveries led to a rapid increase in the rate of mathematical discovery that continues to the present day.
Galileo Galilei (1564–1642) said, "The universe cannot be read until we have learned the language and become familiar with the characters in which it is written. It is written in mathematical language, and the letters are triangles, circles and other geometrical figures, without which means it is humanly impossible to comprehend a single word. Without these, one is wandering about in a dark labyrinth". Carl Friedrich Gauss (1777–1855) referred to mathematics as "the queen of sciences". The mathematician Benjamin Peirce (1809–1880) called the discipline, "the science that draws necessary conclusions". David Hilbert said of it: "We are not speaking here of arbitrariness in any sense. Mathematics is not like a game whose tasks are determined by arbitrarily stipulated rules. Rather, it is a conceptual system possessing internal necessity that can only be so and by no means otherwise." Albert Einstein (1879–1955) stated that "as far as the laws of mathematics refer to reality, they are not certain; and as far as they are certain, they do not refer to reality".
Mathematics is used throughout the world as an essential tool in many fields, including natural science, engineering, medicine, and the social sciences. Applied mathematics, the branch of mathematics concerned with application of mathematical knowledge to other fields, inspires and makes use of new mathematical discoveries and sometimes leads to the development of entirely new mathematical disciplines, such as statistics and game theory. Mathematicians also engage in pure mathematics, or mathematics for its own sake, without having any application in mind. There is no clear line separating pure and applied mathematics, and practical applications for what began as pure mathematics are often discovered.
There are approximately 30,316 mathematics articles in Wikipedia.
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## Selected article
Image credit: User:Melchoir
The real number denoted by the recurring decimal 0.999… is exactly equal to 1. In other words, "0.999…" represents the same number as the symbol "1". Various proofs of this identity have been formulated with varying rigour, preferred development of the real numbers, background assumptions, historical context, and target audience.
The equality has long been taught in textbooks, and in the last few decades, researchers of mathematics education have studied the reception of this equation among students, who often reject the equality. The students' reasoning is typically based on one of a few common erroneous intuitions about the real numbers; for example, a belief that each unique decimal expansion must correspond to a unique number, an expectation that infinitesimal quantities should exist, that arithmetic may be broken, an inability to understand limits or simply the belief that 0.999… should have a last 9. These ideas are false with respect to the real numbers, which can be proven by explicitly constructing the reals from the rational numbers, and such constructions can also prove that 0.999… = 1 directly.
## Selected picture
This spiral diagram represents all ordinal numbers less than $\omega^\omega$. The first (outermost) turn of the spiral represents the finite ordinal numbers, which are the regular counting numbers starting with zero. As the spiral completes its first turn (at the top of the diagram), the ordinal numbers approach infinity, or more precisely $\omega$, the first transfinite ordinal number (identified with the set of all counting numbers, a "countably infinite" set, the cardinality of which corresponds to the first transfinite cardinal number, called $\aleph_0$). The ordinal numbers continue from this point in the second turn of the spiral with $\omega+1$, $\omega+2$, and so forth. (A special ordinal arithmetic is defined to give meaning to these expressions, since the + symbol here does not represent the addition of two real numbers.) Halfway through the second turn of the spiral (at the bottom) the numbers approach $\omega+\omega$, or $\omega\cdot2$. The ordinal numbers continue with $\omega\cdot2+1$ through $\omega\cdot2+\omega=\omega\cdot3$ (three-quarters of the way through the second turn, or at the "9 o'clock" position), then through $\omega\cdot4$, and so forth, up to $\omega\cdot\omega=\omega^2$ at the top. (As with addition, the multiplication and exponentiation operations have definitions that work with transfinite numbers.) As one would expect, the ordinals continue in the third turn of the spiral with $\omega^2+1$ through $\omega^2+\omega$, then through $\omega^2+\omega^2=\omega^2\cdot2$, up to $\omega^2\cdot\omega=\omega^3$ at the top of the third turn. Continuing in this way, the ordinals increase by one power of $\omega$ for each turn of the spiral, approaching $\omega^\omega$ in the middle of the diagram, as the spiral makes a countably infinite number of turns. This process can actually continue through $\omega^{\omega^\omega}$ and $\omega^{\omega^{\omega^\omega}}$, and so on, approaching the first uncountable ordinal number, which (assuming the axiom of choice) corresponds to only the second transfinite cardinal number, $\aleph_1$, the cardinality (according to the continuum hypothesis) of the set of real numbers.
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# Generating random networks
Users can generate random networks. It is, of course, possible to supply your own. The networks should be presented as matrices of 0 and 1. Internally, befwm will check that there are as many rows as there are columns.
## Niche model
Following Williams & Martinez, we have implemented the niche model of food webs. This model represents allometric relationships between preys and predators well, and is therefore well suited to generate random networks.
Random niche model networks can be generated using nichemodel, which takes two arguments: the number of species S, and the desired connectance C:
using BioEnergeticFoodWebs
nichemodel(10, 0.2)
10×10 Array{Int64,2}:
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 1 0 0 0 0 0 0
0 0 0 1 1 1 0 0 0 0
0 0 0 0 0 0 0 0 0 0
0 0 0 1 1 1 1 1 0 0
0 0 0 0 0 0 0 0 0 0
0 0 1 1 1 1 1 1 1 1
Note that there are a number of keyword arguments (optional) that can be supplied: tolerance will give the allowed deviation from the desired connectance, and toltype will indicate whether the error is relative or absolute. |
A propped cantilever beam $\text{XY}$, with an internal hinge at the middle, is carrying a uniformly distributed load of $\text{10 kN/m}$, as shown in the figure.
The vertical reaction at support $\text{X}$ ( in $\text{kN}$, $\textit{in integer}$) is ___________ |
## LATEX-L@LISTSERV.UNI-HEIDELBERG.DE
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The idea with modules is to not define localities. There is already a structure in LaTeX for creating localities, name the environments. In fact, a convenient way to implement environments is by creating a module called "environment", which defines the behavior of environments, including how to define new environments. In fact, one can go one step further and create a module "Environment" which is used to create different styles of environments (say if you want to change that \begin{foo} ... \end{foo} to something else, or perhaps create environments for HTML code, or something). I have put up a link on my home page to the LaTeX/TeX code I once made while exploring these ideas. It is just a copy of what I happened to have in my home directory, so it's a mess. But the stuff is at least available for exploration. Hans Aberg * Email: Hans Aberg * Home Page: * AMS member listing: |
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# DC Circuits Containing Resistors and Capacitors
Module by: OpenStax College. E-mail the author
Summary:
• Explain the importance of the time constant, τ , and calculate the time constant for a given resistance and capacitance.
• Explain why batteries in a flashlight gradually lose power and the light dims over time.
• Describe what happens to a graph of the voltage across a capacitor over time as it charges.
• Explain how a timing circuit works and list some applications.
• Calculate the necessary speed of a strobe flash needed to “stop” the movement of an object over a particular length.
When you use a flash camera, it takes a few seconds to charge the capacitor that powers the flash. The light flash discharges the capacitor in a tiny fraction of a second. Why does charging take longer than discharging? This question and a number of other phenomena that involve charging and discharging capacitors are discussed in this module.
## RC Circuits
An RCRC size 12{ ital "RC"} {} circuit is one containing a resistorRR size 12{R} {} and a capacitor CC size 12{C} {}. The capacitor is an electrical component that stores electric charge.
Figure 1 shows a simple RCRC size 12{ ital "RC"} {} circuit that employs a DC (direct current) voltage source. The capacitor is initially uncharged. As soon as the switch is closed, current flows to and from the initially uncharged capacitor. As charge increases on the capacitor plates, there is increasing opposition to the flow of charge by the repulsion of like charges on each plate.
In terms of voltage, this is because voltage across the capacitor is given by Vc=Q/CVc=Q/C size 12{V rSub { size 8{c} } =Q/C} {}, where QQ size 12{Q} {} is the amount of charge stored on each plate and CC size 12{C} {} is the capacitance. This voltage opposes the battery, growing from zero to the maximum emf when fully charged. The current thus decreases from its initial value of I0=emfRI0=emfR size 12{I rSub { size 8{0} } = { {"emf"} over {R} } } {} to zero as the voltage on the capacitor reaches the same value as the emf. When there is no current, there is no IRIR size 12{ ital "IR"} {} drop, and so the voltage on the capacitor must then equal the emf of the voltage source. This can also be explained with Kirchhoff’s second rule (the loop rule), discussed in Kirchhoff’s Rules, which says that the algebraic sum of changes in potential around any closed loop must be zero.
The initial current is I0=emfRI0=emfR size 12{I rSub { size 8{0} } = { {"emf"} over {R} } } {}, because all of the IRIR size 12{ ital "IR"} {} drop is in the resistance. Therefore, the smaller the resistance, the faster a given capacitor will be charged. Note that the internal resistance of the voltage source is included in RR size 12{R} {}, as are the resistances of the capacitor and the connecting wires. In the flash camera scenario above, when the batteries powering the camera begin to wear out, their internal resistance rises, reducing the current and lengthening the time it takes to get ready for the next flash.
Voltage on the capacitor is initially zero and rises rapidly at first, since the initial current is a maximum. Figure 1(b) shows a graph of capacitor voltage versus time (tt size 12{t} {}) starting when the switch is closed at t=0t=0 size 12{t=0} {}. The voltage approaches emf asymptotically, since the closer it gets to emf the less current flows. The equation for voltage versus time when charging a capacitor CC size 12{C} {} through a resistor RR size 12{R} {}, derived using calculus, is
V=emf(1et/RC) (charging),V=emf(1et/RC) (charging), size 12{V="emf" $$1 - e rSup { size 8{ - t/ ital "RC"} }$$ } {}
(1)
where VV size 12{V} {} is the voltage across the capacitor, emf is equal to the emf of the DC voltage source, and the exponential e = 2.718 … is the base of the natural logarithm. Note that the units of RCRC size 12{ ital "RC"} {} are seconds. We define
τ=RC,τ=RC, size 12{τ= ital "RC"} {}
(2)
where ττ size 12{τ} {} (the Greek letter tau) is called the time constant for an RCRC size 12{ ital "RC"} {} circuit. As noted before, a small resistance RR size 12{R} {} allows the capacitor to charge faster. This is reasonable, since a larger current flows through a smaller resistance. It is also reasonable that the smaller the capacitor CC size 12{C} {}, the less time needed to charge it. Both factors are contained in τ=RCτ=RC size 12{τ= ital "RC"} {}.
More quantitatively, consider what happens when t=τ=RCt=τ=RC size 12{t=τ= ital "RC"} {}. Then the voltage on the capacitor is
V=emf1e1=emf10.368=0.632emf.V=emf1e1=emf10.368=0.632emf. size 12{V="emf" left (1 - e rSup { size 8{ - 1} } right )="emf" left (1 - 0 "." "368" right )=0 "." "632" cdot "emf"} {}
(3)
This means that in the time τ=RCτ=RC size 12{τ= ital "RC"} {}, the voltage rises to 0.632 of its final value. The voltage will rise 0.632 of the remainder in the next time ττ size 12{τ} {}. It is a characteristic of the exponential function that the final value is never reached, but 0.632 of the remainder to that value is achieved in every time, ττ size 12{τ} {}. In just a few multiples of the time constant ττ size 12{τ} {}, then, the final value is very nearly achieved, as the graph in Figure 1(b) illustrates.
## Discharging a Capacitor
Discharging a capacitor through a resistor proceeds in a similar fashion, as Figure 2 illustrates. Initially, the current is I0=V0RI0=V0R size 12{I rSub { size 8{0} } = { {V rSub { size 8{0} } } over {R} } } {}, driven by the initial voltage V0V0 size 12{V rSub { size 8{0} } } {} on the capacitor. As the voltage decreases, the current and hence the rate of discharge decreases, implying another exponential formula for VV size 12{V} {}. Using calculus, the voltage VV size 12{V} {} on a capacitor CC size 12{C} {} being discharged through a resistor RR size 12{R} {} is found to be
V=V0et/RC (discharging).V=V0et/RC (discharging). size 12{V=V"" lSub { size 8{0} } e rSup { size 8{ - t/ ital "RC"} } } {}
(4)
The graph in Figure 2(b) is an example of this exponential decay. Again, the time constant is τ=RCτ=RC size 12{τ= ital "RC"} {}. A small resistance RR size 12{R} {} allows the capacitor to discharge in a small time, since the current is larger. Similarly, a small capacitance requires less time to discharge, since less charge is stored. In the first time interval τ=RCτ=RC size 12{τ= ital "RC"} {} after the switch is closed, the voltage falls to 0.368 of its initial value, since V=V0e1=0.368V0V=V0e1=0.368V0 size 12{V=V rSub { size 8{0} } cdot e rSup { size 8{ - 1} } =0 "." "368"V rSub { size 8{0} } } {}.
During each successive time ττ size 12{τ} {}, the voltage falls to 0.368 of its preceding value. In a few multiples of ττ size 12{τ} {}, the voltage becomes very close to zero, as indicated by the graph in Figure 2(b).
Now we can explain why the flash camera in our scenario takes so much longer to charge than discharge; the resistance while charging is significantly greater than while discharging. The internal resistance of the battery accounts for most of the resistance while charging. As the battery ages, the increasing internal resistance makes the charging process even slower. (You may have noticed this.)
The flash discharge is through a low-resistance ionized gas in the flash tube and proceeds very rapidly. Flash photographs, such as in Figure 3, can capture a brief instant of a rapid motion because the flash can be less than a microsecond in duration. Such flashes can be made extremely intense.
During World War II, nighttime reconnaissance photographs were made from the air with a single flash illuminating more than a square kilometer of enemy territory. The brevity of the flash eliminated blurring due to the surveillance aircraft’s motion. Today, an important use of intense flash lamps is to pump energy into a laser. The short intense flash can rapidly energize a laser and allow it to reemit the energy in another form.
### Example 1: Integrated Concept Problem: Calculating Capacitor Size—Strobe Lights
High-speed flash photography was pioneered by Doc Edgerton in the 1930s, while he was a professor of electrical engineering at MIT. You might have seen examples of his work in the amazing shots of hummingbirds in motion, a drop of milk splattering on a table, or a bullet penetrating an apple (see Figure 3). To stop the motion and capture these pictures, one needs a high-intensity, very short pulsed flash, as mentioned earlier in this module.
Suppose one wished to capture the picture of a bullet (moving at 5.0×102m/s5.0×102m/s) that was passing through an apple. The duration of the flash is related to the RCRC size 12{ ital "RC"} {} time constant, ττ size 12{τ} {}. What size capacitor would one need in the RCRC size 12{ ital "RC"} {} circuit to succeed, if the resistance of the flash tube was 10.0 Ω10.0 Ω size 12{"10" %OMEGA } {}? Assume the apple is a sphere with a diameter of 8.0×10–2m.8.0×10–2m.
Strategy
We begin by identifying the physical principles involved. This example deals with the strobe light, as discussed above. Figure 2 shows the circuit for this probe. The characteristic time ττ size 12{τ} {} of the strobe is given as τ=RCτ=RC size 12{τ= ital "RC"} {}.
Solution
We wish to find CC size 12{C} {}, but we don’t know ττ size 12{τ} {}. We want the flash to be on only while the bullet traverses the apple. So we need to use the kinematic equations that describe the relationship between distance xx size 12{x} {}, velocity vv size 12{v} {}, and time tt size 12{t} {}:
x=vtort=xv.x=vtort=xv. size 12{t= { {x} over {v} } } {}
(5)
The bullet’s velocity is given as 5.0×102m/s5.0×102m/s, and the distance xx size 12{x} {} is 8.0×10–2m.8.0×10–2m. The traverse time, then, is
t=xv=8.0×10–2m5.0×102m/s=1.6×104s.t=xv=8.0×10–2m5.0×102m/s=1.6×104s. size 12{t= { {x} over {v} } = { {0 "." "08"" m"} over {"500 m/s"} } =1 "." 6 times "10" rSup { size 8{ - 4} } " s"} {}
(6)
We set this value for the crossing time tt size 12{t} {} equal to ττ size 12{τ} {}. Therefore,
C=tR=1.6×104s10.0 Ω=16μF.C=tR=1.6×104s10.0 Ω=16μF. size 12{C= { {t} over {R} } = { { left (1 "." 6´"10" rSup { size 8{-4} } right )} over {"10"} } ="16" μF} {}
(7)
(Note: Capacitance CC size 12{C} {} is typically measured in farads, F F , defined as Coulombs per volt. From the equation, we see that CC size 12{C} {} can also be stated in units of seconds per ohm.)
Discussion
The flash interval of 160 μs160 μs size 12{"160" ms} {} (the traverse time of the bullet) is relatively easy to obtain today. Strobe lights have opened up new worlds from science to entertainment. The information from the picture of the apple and bullet was used in the Warren Commission Report on the assassination of President John F. Kennedy in 1963 to confirm that only one bullet was fired.
## RC Circuits for Timing
RCRC size 12{ ital "RC"} {} circuits are commonly used for timing purposes. A mundane example of this is found in the ubiquitous intermittent wiper systems of modern cars. The time between wipes is varied by adjusting the resistance in an RCRC size 12{ ital "RC"} {} circuit. Another example of an RCRC size 12{ ital "RC"} {} circuit is found in novelty jewelry, Halloween costumes, and various toys that have battery-powered flashing lights. (See Figure 4 for a timing circuit.)
A more crucial use of RCRC size 12{ ital "RC"} {} circuits for timing purposes is in the artificial pacemaker, used to control heart rate. The heart rate is normally controlled by electrical signals generated by the sino-atrial (SA) node, which is on the wall of the right atrium chamber. This causes the muscles to contract and pump blood. Sometimes the heart rhythm is abnormal and the heartbeat is too high or too low.
The artificial pacemaker is inserted near the heart to provide electrical signals to the heart when needed with the appropriate time constant. Pacemakers have sensors that detect body motion and breathing to increase the heart rate during exercise to meet the body’s increased needs for blood and oxygen.
### Example 2: Calculating Time: RC Circuit in a Heart Defibrillator
A heart defibrillator is used to resuscitate an accident victim by discharging a capacitor through the trunk of her body. A simplified version of the circuit is seen in Figure 2. (a) What is the time constant if an 8.00-μF8.00-μF size 12{8 "." "00"-mF} {} capacitor is used and the path resistance through her body is 1.00×103Ω1.00×103Ω? (b) If the initial voltage is 10.0 kV, how long does it take to decline to 5.00×102V5.00×102V?
Strategy
Since the resistance and capacitance are given, it is straightforward to multiply them to give the time constant asked for in part (a). To find the time for the voltage to decline to 5.00×102V5.00×102V, we repeatedly multiply the initial voltage by 0.368 until a voltage less than or equal to 5.00×102V5.00×102V is obtained. Each multiplication corresponds to a time of ττ size 12{τ} {} seconds.
Solution for (a)
The time constant ττ size 12{τ} {} is given by the equation τ=RCτ=RC size 12{τ= ital "RC"} {}. Entering the given values for resistance and capacitance (and remembering that units for a farad can be expressed as s/Ωs/Ω size 12{s/ %OMEGA } {}) gives
τ=RC=(1.00× 10 3 Ω)(8.00μF)=8.00ms.τ=RC=(1.00× 10 3 Ω)(8.00μF)=8.00ms. size 12{τ= ital "RC"= $$"1000" %OMEGA$$ $$8 "." "00" μF$$ =8 "." "00"" ms"} {}
(8)
Solution for (b)
In the first 8.00 ms, the voltage (10.0 kV) declines to 0.368 of its initial value. That is:
V=0.368V0= 3.680×103 V at t=8.00ms.V=0.368V0= 3.680×103 V at t=8.00ms. size 12{t=8 "." "00"" ms"} {}
(9)
(Notice that we carry an extra digit for each intermediate calculation.) After another 8.00 ms, we multiply by 0.368 again, and the voltage is
V = 0.368 V = 0.368 3.680 × 10 3 V = 1.354 × 10 3 V at t = 16.0 ms. V = 0.368 V = 0.368 3.680 × 10 3 V = 1.354 × 10 3 V at t = 16.0 ms.
(10)
Similarly, after another 8.00 ms, the voltage is
V ′′ = 0.368 V = ( 0.368 ) ( 1.354 × 10 3 V ) = 498 V at t = 24 .0 ms. V ′′ = 0.368 V = ( 0.368 ) ( 1.354 × 10 3 V ) = 498 V at t = 24 .0 ms.
(11)
Discussion
So after only 24.0 ms, the voltage is down to 498 V, or 4.98% of its original value.Such brief times are useful in heart defibrillation, because the brief but intense current causes a brief but effective contraction of the heart. The actual circuit in a heart defibrillator is slightly more complex than the one in Figure 2, to compensate for magnetic and AC effects that will be covered in Magnetism.
When is the potential difference across a capacitor an emf?
#### Solution
Only when the current being drawn from or put into the capacitor is zero. Capacitors, like batteries, have internal resistance, so their output voltage is not an emf unless current is zero. This is difficult to measure in practice so we refer to a capacitor’s voltage rather than its emf. But the source of potential difference in a capacitor is fundamental and it is an emf.
## PhET Explorations: Circuit Construction Kit (DC only):
An electronics kit in your computer! Build circuits with resistors, light bulbs, batteries, and switches. Take measurements with the realistic ammeter and voltmeter. View the circuit as a schematic diagram, or switch to a life-like view.
## Section Summary
• An RCRC size 12{ ital "RC"} {} circuit is one that has both a resistor and a capacitor.
• The time constant ττ size 12{τ} {} for an RCRC size 12{ ital "RC"} {} circuit is τ=RCτ=RC size 12{τ= ital "RC"} {}.
• When an initially uncharged (V0=0V0=0 size 12{V rSub { size 8{0} } =0} {} at t=0t=0 size 12{t=0} {}) capacitor in series with a resistor is charged by a DC voltage source, the voltage rises, asymptotically approaching the emf of the voltage source; as a function of time,
V=emf(1et/RC) (charging).V=emf(1et/RC) (charging). size 12{V="emf" $$1 - e rSup { size 8{ - t/ ital "RC"} }$$ } {}
(12)
• Within the span of each time constant ττ size 12{τ} {}, the voltage rises by 0.632 of the remaining value, approaching the final voltage asymptotically.
• If a capacitor with an initial voltage V0V0 size 12{V rSub { size 8{0} } } {} is discharged through a resistor starting at t=0t=0 size 12{t=0} {}, then its voltage decreases exponentially as given by
V=V0et/RC (discharging).V=V0et/RC (discharging). size 12{V=V rSub { size 8{0} } e rSup { size 8{ - t/ ital "RC"} } \) } {}
(13)
• In each time constant ττ size 12{τ} {}, the voltage falls by 0.368 of its remaining initial value, approaching zero asymptotically.
## Conceptual questions
### Exercise 1
Regarding the units involved in the relationship τ=RCτ=RC size 12{τ= ital "RC"} {}, verify that the units of resistance times capacitance are time, that is, ΩF=sΩF=s size 12{ %OMEGA cdot F=s} {}.
### Exercise 2
The RCRC size 12{ ital "RC"} {} time constant in heart defibrillation is crucial to limiting the time the current flows. If the capacitance in the defibrillation unit is fixed, how would you manipulate resistance in the circuit to adjust the RCRC size 12{ ital "RC"} {} constant ττ size 12{τ} {}? Would an adjustment of the applied voltage also be needed to ensure that the current delivered has an appropriate value?
### Exercise 3
When making an ECG measurement, it is important to measure voltage variations over small time intervals. The time is limited by the RCRC size 12{ ital "RC"} {} constant of the circuit—it is not possible to measure time variations shorter than RCRC size 12{ ital "RC"} {}. How would you manipulate RR size 12{R} {} and CC size 12{C} {} in the circuit to allow the necessary measurements?
### Exercise 4
Draw two graphs of charge versus time on a capacitor. Draw one for charging an initially uncharged capacitor in series with a resistor, as in the circuit in Figure 1, starting from t=0t=0 size 12{t=0} {}. Draw the other for discharging a capacitor through a resistor, as in the circuit in Figure 2, starting at t=0t=0 size 12{t=0} {}, with an initial charge Q0Q0 size 12{Q rSub { size 8{0} } } {}. Show at least two intervals of ττ size 12{τ} {}.
### Exercise 5
When charging a capacitor, as discussed in conjunction with Figure 1, how long does it take for the voltage on the capacitor to reach emf? Is this a problem?
### Exercise 6
When discharging a capacitor, as discussed in conjunction with Figure 2, how long does it take for the voltage on the capacitor to reach zero? Is this a problem?
### Exercise 7
Referring to Figure 1, draw a graph of potential difference across the resistor versus time, showing at least two intervals of ττ size 12{τ} {}. Also draw a graph of current versus time for this situation.
### Exercise 8
A long, inexpensive extension cord is connected from inside the house to a refrigerator outside. The refrigerator doesn’t run as it should. What might be the problem?
### Exercise 9
In Figure 4, does the graph indicate the time constant is shorter for discharging than for charging? Would you expect ionized gas to have low resistance? How would you adjust RR size 12{R} {} to get a longer time between flashes? Would adjusting RR size 12{R} {} affect the discharge time?
### Exercise 10
An electronic apparatus may have large capacitors at high voltage in the power supply section, presenting a shock hazard even when the apparatus is switched off. A “bleeder resistor” is therefore placed across such a capacitor, as shown schematically in Figure 6, to bleed the charge from it after the apparatus is off. Why must the bleeder resistance be much greater than the effective resistance of the rest of the circuit? How does this affect the time constant for discharging the capacitor?
## Problem Exercises
### Exercise 1
The timing device in an automobile’s intermittent wiper system is based on an RCRC size 12{ ital "RC"} {} time constant and utilizes a 0.500-μF0.500-μF size 12{0 "." "500-"μF} {} capacitor and a variable resistor. Over what range must RR size 12{R} {} be made to vary to achieve time constants from 2.00 to 15.0 s?
#### Solution
range 4 . 00 to 30 . 0 M Ω range 4 . 00 to 30 . 0 M Ω size 12{"range 4" "." "00 to 30" "." "0 M" %OMEGA } {}
### Exercise 2
A heart pacemaker fires 72 times a minute, each time a 25.0-nF capacitor is charged (by a battery in series with a resistor) to 0.632 of its full voltage. What is the value of the resistance?
### Exercise 3
The duration of a photographic flash is related to an RCRC size 12{ ital "RC"} {} time constant, which is 0.100 μs0.100 μs size 12{0 "." "100" μs} {} for a certain camera. (a) If the resistance of the flash lamp is 0.0400Ω0.0400Ω size 12{0 "." "0400" %OMEGA } {} during discharge, what is the size of the capacitor supplying its energy? (b) What is the time constant for charging the capacitor, if the charging resistance is 800800 size 12{"800"" k" %OMEGA } {}?
#### Solution
(a) 2.50 μF2.50 μF size 12{2 "." "50 "mF} {}
(b) 2.00 s
### Exercise 4
A 2.00- and a 7.50-μF7.50-μF size 12{7 "." "50"-mF} {} capacitor can be connected in series or parallel, as can a 25.0- and a 100-kΩ100-kΩ size 12{"100""-k" %OMEGA } {} resistor. Calculate the four RCRC size 12{ ital "RC"} {} time constants possible from connecting the resulting capacitance and resistance in series.
### Exercise 5
After two time constants, what percentage of the final voltage, emf, is on an initially uncharged capacitor CC size 12{C} {}, charged through a resistance RR size 12{R} {}?
86.5%
### Exercise 6
A 500-Ω500-Ω size 12{"500"- %OMEGA } {} resistor, an uncharged 1.50-μF1.50-μF size 12{1 "." "50"-mF} {} capacitor, and a 6.16-V emf are connected in series. (a) What is the initial current? (b) What is the RCRC size 12{ ital "RC"} {} time constant? (c) What is the current after one time constant? (d) What is the voltage on the capacitor after one time constant?
### Exercise 7
A heart defibrillator being used on a patient has an RCRC size 12{ ital "RC"} {} time constant of 10.0 ms due to the resistance of the patient and the capacitance of the defibrillator. (a) If the defibrillator has an 8.00-μF8.00-μF size 12{8 "." "00"-mF} {} capacitance, what is the resistance of the path through the patient? (You may neglect the capacitance of the patient and the resistance of the defibrillator.) (b) If the initial voltage is 12.0 kV, how long does it take to decline to 6.00×102V6.00×102V?
#### Solution
(a) 1.25 kΩ1.25 kΩ size 12{1 "." "25 k" %OMEGA } {}
(b) 30.0 ms
### Exercise 8
An ECG monitor must have an RCRC size 12{ ital "RC"} {} time constant less than 1.00×102μs1.00×102μs size 12{"100" ms} {} to be able to measure variations in voltage over small time intervals. (a) If the resistance of the circuit (due mostly to that of the patient’s chest) is 1.00 kΩ1.00 kΩ size 12{1 "." 00" k" %OMEGA } {}, what is the maximum capacitance of the circuit? (b) Would it be difficult in practice to limit the capacitance to less than the value found in (a)?
### Exercise 9
Figure 7 shows how a bleeder resistor is used to discharge a capacitor after an electronic device is shut off, allowing a person to work on the electronics with less risk of shock. (a) What is the time constant? (b) How long will it take to reduce the voltage on the capacitor to 0.250% (5% of 5%) of its full value once discharge begins? (c) If the capacitor is charged to a voltage V0V0 size 12{V rSub { size 8{0} } } {} through a 100-Ω100-Ω size 12{"100"- %OMEGA } {} resistance, calculate the time it takes to rise to 0.865V00.865V0 size 12{0 "." "865"`V rSub { size 8{0} } } {} (This is about two time constants.)
(a) 20.0 s
(b) 120 s
(c) 16.0 ms
### Exercise 10
Using the exact exponential treatment, find how much time is required to discharge a 250-μF250-μF size 12{"250"-mF} {} capacitor through a 500-Ω500-Ω size 12{"500"- %OMEGA } {} resistor down to 1.00% of its original voltage.
### Exercise 11
Using the exact exponential treatment, find how much time is required to charge an initially uncharged 100-pF capacitor through a 75.0-MΩ75.0-MΩ size 12{"75" "." 0"-M" %OMEGA } {} resistor to 90.0% of its final voltage.
#### Solution
1 . 73 × 10 2 s 1 . 73 × 10 2 s size 12{1 "." "73" times "10" rSup { size 8{ - 2} } " s"} {}
### Exercise 12
Integrated Concepts
If you wish to take a picture of a bullet traveling at 500 m/s, then a very brief flash of light produced by an RCRC size 12{ ital "RC"} {} discharge through a flash tube can limit blurring. Assuming 1.00 mm of motion during one RCRC size 12{ ital "RC"} {} constant is acceptable, and given that the flash is driven by a 600-μF600-μF size 12{"600"-mF} {} capacitor, what is the resistance in the flash tube?
#### Solution
3 . 33 × 10 3 Ω 3 . 33 × 10 3 Ω size 12{3 "." "33"´"10" rSup { size 8{-3} } %OMEGA } {}
### Exercise 13
Integrated Concepts
A flashing lamp in a Christmas earring is based on an RCRC size 12{ ital "RC"} {} discharge of a capacitor through its resistance. The effective duration of the flash is 0.250 s, during which it produces an average 0.500 W from an average 3.00 V. (a) What energy does it dissipate? (b) How much charge moves through the lamp? (c) Find the capacitance. (d) What is the resistance of the lamp?
### Exercise 14
Integrated Concepts
A 160-μF160-μF size 12{"160"-mF} {} capacitor charged to 450 V is discharged through a 31.2-kΩ31.2-kΩ size 12{31 "." "2-k" %OMEGA } {} resistor. (a) Find the time constant. (b) Calculate the temperature increase of the resistor, given that its mass is 2.50 g and its specific heat is 1.67kJkgºC1.67kJkgºC size 12{1 "." "67" { {"kJ"} over {"kg" cdot "deg"C} } } {}, noting that most of the thermal energy is retained in the short time of the discharge. (c) Calculate the new resistance, assuming it is pure carbon. (d) Does this change in resistance seem significant?
#### Solution
(a) 4.99 s
(b) 3.87ºC3.87ºC size 12{3 "." "87"°C} {}
(c) 31.1 kΩ31.1 kΩ size 12{"31" "." "1 k" %OMEGA } {}
(d) No
### Exercise 15
Unreasonable Results
(a) Calculate the capacitance needed to get an RCRC size 12{ ital "RC"} {} time constant of 1.00×103s1.00×103s with a 0.100-Ω0.100-Ω size 12{0 "." "100"- %OMEGA } {} resistor. (b) What is unreasonable about this result? (c) Which assumptions are responsible?
### Exercise 16
Consider a camera’s flash unit. Construct a problem in which you calculate the size of the capacitor that stores energy for the flash lamp. Among the things to be considered are the voltage applied to the capacitor, the energy needed in the flash and the associated charge needed on the capacitor, the resistance of the flash lamp during discharge, and the desired RCRC size 12{ ital "RC"} {} time constant.
### Exercise 17
Consider a rechargeable lithium cell that is to be used to power a camcorder. Construct a problem in which you calculate the internal resistance of the cell during normal operation. Also, calculate the minimum voltage output of a battery charger to be used to recharge your lithium cell. Among the things to be considered are the emf and useful terminal voltage of a lithium cell and the current it should be able to supply to a camcorder.
## Glossary
RC circuit:
a circuit that contains both a resistor and a capacitor
capacitor:
an electrical component used to store energy by separating electric charge on two opposing plates
capacitance:
the maximum amount of electric potential energy that can be stored (or separated) for a given electric potential
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##### Lenses
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# Interactions between phytoplankton and heterotrophic bacteria in freshwater
#### Abstract
The relationship between autotrophic phytoplankton and heterotrophic planktonic bacteria is often called coupling because of a positive correlation between phytoplankton biomass or production and bacterioplankton abundance or production. The objectives of this dissertation are to examine how bacterial-phytoplankton coupling is affected by differences in algal composition, the concentration of dissolved organic carbon (DOC), inorganic nutrient concentration, and intensity of the grazing pressure.^ In this dissertation, I have found that bacterial-phytoplankton relationship is relatively insensitive to the supply of dissolved organic carbon (DOC). The additions of picophytoplankton (Synechococcus) and nanophytoplankton (Scenedesmus) do not significantly affect the coupling between phytoplankton and planktonic bacteria. However, different combinations of inorganic nutrients (N and P) are able to uncouple the bacterial-phytoplankton linkage. I have also designed a useful and simple Coupling Index (CI), which predicts that shifts in phytoplankton and bacterioplankton growth will be unrelated ($\Delta$bact/$\Delta$phyto $\to$ either +$\infty$ or $-\infty$) in systems with inorganic N:P (molar) ratios $<$ about 50. Grazing pressure from large zooplankters such as Daphnia can also unlink the bacterial-phytoplankton coupling by consuming a larger proportion of phytoplankton biomass than bacterial biomass. The altered relationship between phytoplankton and bacteria can also affect the pelagic ecosystem efficiency. The results suggest that different microbial components may have contrasting effects on the presumed enhanced efficiency provided by the microbial loop. I also proposed an automated procedure to measure total dissolved nitrogen (TDN). This procedure recovers 100% of several organic N forms in deionized water, and achieves over 90% recovery for organic-N-spiked natural water samples. This method will be a useful tool for routinely monitoring total dissolved nitrogen in freshwater ecosystems. ^
#### Subject Area
Biology, Ecology|Biology, Microbiology|Biology, Limnology
#### Recommended Citation
Jianhua Le, "Interactions between phytoplankton and heterotrophic bacteria in freshwater" (January 1, 1996). ETD Collection for Fordham University. Paper AAI9628340.
http://fordham.bepress.com/dissertations/AAI9628340
COinS |
Options
Options, which usually occur between brackets ([]) after commands like \addplot, table, or beginnings of environments like \begin{axis} in LaTeX code, are key to most of the functionality of PGFPlots.
The @pgf macro
Use the @pgf {} macro to define options.
PGFPlotsX.@pgfMacro
@pgf { ... }
@pgf some(nested(form({ ... })),
with_multiple_options({ ... }))
Construct Options from comma-delimited key (without value), key = value, key : value, or key => value pairs enclosed in { ... }, anywhere in the expression.
The argument is traversed recursively, allowing { ... } expressions in multiple places.
Multi-word keys need to be either quoted, or written with underscores replacing spaces.
@pgf {
"only marks",
mark_size = "0.6pt",
mark = "o",
color => "black",
}
Another Options can be spliced into one being created using ..., e.g.
theme = @pgf {xmajorgrids, x_grid_style = "white"}
axis_opt = @pgf {theme..., title = "My figure"}
Use {} for empty options that print as [] in LaTeX.
source
For constructors that accept options, they always come first. When omitted, there are assumed to be no options.
julia> c = Coordinates([1, 2, 3], [2, 4, 8]);
julia> p = @pgf PlotInc({ "very thick", "mark" => "halfcircle" }, c);
julia> print_tex(p); # print_tex can be used to preview the generated .tex
coordinates {
(1,2)
(2,4)
(3,8)
}
;
Inside the expression following @pgf, {} expressions can be nested, and can also occur in multiple places.
julia> @pgf a = Axis(
{
"axis background/.style" =
{
top_color = "gray",
bottom_color = "white",
},
ymode = "log"
},
PlotInc(
{
smooth
},
c)
);
which is converted to LaTeX as
julia> print_tex(a)
\begin{axis}[axis background/.style={shade, top color={gray}, bottom color={white}}, ymode={log}]
coordinates {
(1,2)
(2,4)
(3,8)
}
;
\end{axis}
Note
If you use @pgf inside argument lists, make sure you wrap its argument in parentheses, eg
Plot(@pgf({ scatter }), some_table)
Otherwise Julia will also pass the subsequent arguments through @pgf, which results in an error since they are combined into a tuple.
Each option is either a standalone keyword (without value, modifying the plot by itself), or a keyword-value pair. Keywords can be entered
1. as Julia identifiers, which is useful for keywords with no spaces (eg smooth),
2. separated by underscores, which are replaced by spaces (eg only_marks will appear in LaTeX code as only marks),
3. or quoted as strings, eg "very thick".
Values are provided after a =, :, or =>, so the following are equivalent:
1. @pgf { draw = "black" },
2. @pgf { draw : "black" },
3. @pgf { draw => "black" }.
Values should be valid Julia expressions, as they are evaluated, so you cannot use @pgf { draw = black } unless black is assigned to some Julia value in that context.
Note
Keys that contain symbols that in Julia are operators (e.g the key "axis background/.style") have to be entered as strings.
Transformations
In addition to replacing underscores in keys, the following transformations of values are done when the options are written in .tex style:
• A list as a value is written as “comma joined” e.g. [1, 2, 3] -> "1, 2, 3".
• A tuple as a value is written with braces delimiting the elements e.g. (60, 30) -> {60}{30}
Modifying options after an object is created
It is sometimes convenient to set and get options after an object has been created.
You can use getindex, setindex! (ie obj["option"] or obj["option"] = value, respectively), and delete! just like you would for modifiable associative collections (eg a Dict).
julia> c = Coordinates([1, 2, 3], [2, 4, 8]);
julia> p = PlotInc(c);
julia> p["fill"] = "blue";
julia> p["fill"]
"blue"
julia> @pgf p["axis background/.style"] = { shade, top_color = "gray", bottom_color = "white" };
julia> p["axis background/.style"]["top_color"];
julia> p["very thick"] = nothing # Set a value-less options;
julia> delete!(p, "fill");
julia> print_tex(p)
coordinates {
(1,2)
(2,4)
(3,8)
}
;
Working with options
Collections of options are first-class objects: they can be used independently of Plot, Axis, and similar, copied, modified, and merged.
This allows a disciplined approach to working with complex plots: for example, you can create a set of default options for some purpose (eg plots in a research paper, with a style imposed by a journal), and then modify this as needed for individual plots. It is then easy to apply, for example, a “theme” to an axis where the theme is a set of options already saved.
Another use case is creating orthogonal sets of options, eg one for axis annotations and another one for legends, and merging these as necessary.
Extending and combining options
Use ... to splice an option into another one, e.g.
julia> theme = @pgf {xmajorgrids, ymajorgrids};
julia> a = Axis(
@pgf {theme..., title = "Foo"}
);
julia> print_tex(a)
\begin{axis}[xmajorgrids, ymajorgrids, title={Foo}]
\end{axis}
julia> print_tex(theme) # original is not modified
[xmajorgrids, ymajorgrids]
You can also merge sets of options:
julia> O1 = @pgf { color = "red" };
julia> O2 = @pgf { dashed };
julia> O3 = @pgf { no_marks };
julia> print_tex(Plot(merge(O1, O2, O3), Table(1:2, 1:2)))
table[row sep={\\}]
{
\\
1 1 \\
2 2 \\
}
;
Again, the value of original options is unchanged above.
Modifying options
You can modify existing options with push!, append!, and merge!. The first two expect pairs of a string and a value (may be nothing for options like "red"), and are mostly useful when you are generating options using a function. merge! of course accepts options.
julia> opt = @pgf {};
julia> push!(opt, :color => "red", :mark => "x");
julia> append!(opt, [:style => "thick", :mark_options => @pgf { scale = 0.4 }]);
julia> merge!(opt, @pgf { "error bars/y dir=both", "error bars/y explicit" });
julia> print_tex(opt)
[color={red}, mark={x}, style={thick}, mark options={scale={0.4}}, error bars/y dir=both, error bars/y explicit]
All containers with options also support using merge! directly.
julia> a = Axis();
julia> @pgf opts = {xmin = 0, ymax = 1, ybar};
julia> merge!(a, opts);
julia> print_tex(a)
\begin{axis}[xmin={0}, ymax={1}, ybar]
\end{axis}
Empty options
Empty options are not emitted by default, but using in LaTeX code [] can be useful in some cases, eg when combined with global settings \pgfplotsset{every axis plot/.append style={...}}. In order to force printing empty options, it is recommended to use {} in expressions like
@pgf Plot({}, ...)
The PGFPlotsX.Options constructor
PGFPlotsX.OptionsType
Options(args; print_empty)
Options passed to PGFPlots for various structures (table, plot, etc).
Contents emitted in key = value form, or key when value ≡ nothing. Example:
julia> PGFPlotsX.Options(:color => "red", :only_marks => nothing)
[color={red}, only marks]
The constuctor is not exported but part of the API, for use in packages that depend on PGFPlotsX, or code producing complicated plots. It is recommended that the @pgf macro is used in scripts and interactive code.
When print_empty = false (the default), empty options are not printed. Use print_empty = true to force printing a [] in this case.
source |
# [minion-cvs] patches before i dig in to the real patching
Update of /home/minion/cvsroot/doc
In directory moria.seul.org:/home/arma/work/minion/doc
Modified Files:
minion-design.tex
Log Message:
patches before i dig in to the real patching
Index: minion-design.tex
===================================================================
RCS file: /home/minion/cvsroot/doc/minion-design.tex,v
retrieving revision 1.48
retrieving revision 1.49
diff -u -d -r1.48 -r1.49
--- minion-design.tex 8 May 2002 04:38:08 -0000 1.48
+++ minion-design.tex 8 May 2002 05:43:21 -0000 1.49
@@ -142,7 +142,7 @@
communications between MIXes but is unable to observe the reordering
inside each MIX.
-Current research directions on MIX-nets include stop-and-go'' MIX-nets
+Recent research on MIX-nets includes stop-and-go'' MIX-nets
\cite{kesdogan}, distributed flash MIXes'' \cite{flash-mix} and their
weaknesses \cite{desmedt,mitkuro}, and hybrid MIXes \cite{hybrid-mix}.
@@ -289,8 +289,8 @@
Cypherpunk remailer systems, in order to provide a simple extensible
design. We can retain minimal backwards compatibility by remixing''
Type II messages to be Type III messages, thus increasing the anonymity
-set of the Type III network. Type II messages, when travelling between
-remailers capable of understanding both Type II and III, are treated
+set of the Type III network. Type II messages travelling between
+Type III remailers are treated
as plaintext and encrypted to the next remailer in the chain using its
Type III key. The message is sent as a Type III encrypted message, but
it decrypts to reveal the Type II message.
@@ -358,7 +358,8 @@
\subsection{Indistinguishable replies}
-By making forward messages and replies indistinguishable, we prevent an
+By making forward messages and replies indistinguishable even to MIXes,
+we prevent an
adversary from dividing the message anonymity sets into two classes. In
particular, if replies are infrequent relative to forward messages,
an adversary who controls some of the MIXes can more easily trace the
@@ -440,7 +441,7 @@
\end{itemize}
To fulfill the above requirements we use a large-block cipher; that
-is, a cipher that acts as a permution on a block the size of its
+is, a cipher that acts as a permutation on a block the size of its
include LIONESS \cite{BEAR-LIONESS} and SPC \cite{SPC}.
The cryptographic property required is that of a super-pseudo-random
@@ -449,7 +450,7 @@
replays limits the number of oracle queries to 1; this will need
further analysis. In that case the simpler BEAR construction
\cite{BEAR-LIONESS} could be used instead of LIONESS.}
-This ensures that if any bit of
+Thus if any bit of
the encrypted material is changed, the decryption will look like random
bits. An SPRP is also equally secure in the encryption and decryption
directions. See Section \ref{subsec:tagging} for a
@@ -515,7 +516,7 @@
We briefly considered introducing \emph{cover-trash} to frustrate
these tagging attacks; but that problem is as complex as the dummy
crossover point to prevent the attacker from learning information from
tagging attacks. Specifically, our solution falls into several cases:
@@ -550,11 +551,13 @@
that the recipient is one of the local addresses.} and so we recommend
that the second leg be at least a few hops long.
-Replies are indistinguishable at every hop from forward messages. Even the
-crossover point cannot know whether it is processing a reply or forward message.
-The protocol doesn't allow a MIX to know its location in the path (other
-than the exit node), or the total length of the route.
-% FIXME we need to resolve this paragraph
+No MIX except the crossover point can distinguish forward messages from
+replies --- even the crossover point cannot be sure whether it is processing
+a reply or forward message, but it may be able to guess that crossover
+points are more frequent on forward paths than direct replies or
+%The protocol doesn't allow a MIX to know its location in the path (other
+%than the exit node), or the total length of the route.
\subsection{Multiple-message tagging attacks}
\label{subsec:multi-tagging}
@@ -1087,7 +1090,7 @@
delays: if the maximum latency of such a network is $t$, then the
anonymity set of a message leaving the network may be much smaller
than all messages that entered over a time $t$.
-% This is handwaving, and the problem is more that the distribtion
+% This is handwaving, and the problem is more that the distribution
% isn't uniform rather than the actual size of the anonymity set.
% It'll do for the time being. -DH
@@ -1100,7 +1103,7 @@
attacks even when free routes are used, as well as improving the
-The network has a fixed {\em batch period}, $t_{batch}$, which is closely
+The network has a fixed {\em batch period}, $t_\mathrm{batch}$, which is closely
related to the maximum desired latency; a typical value could be 3--6 hours.
Messages entering the network in each batch period are queued until
the beginning of the next period. They are then sent through the MIX-net
@@ -1112,10 +1115,11 @@
period in which it must be received, so that it cannot be delayed by an
attacker (which would be fatal for this design).
-The latency is between $\ell t_{hop}$ and $t_{batch} + \ell t_{hop}$, depending
-on when the message was submitted.
-Typically we would have $t_{hop} < t_{batch}/n$, so the latency is at
-most $2t_{batch}$, independent of the path length $\ell$.
+The latency is between $\ell t_\mathrm{hop}$ and $t_\mathrm{batch} + +\ell t_\mathrm{hop}$, depending on when the message was submitted.
+Typically we would have $t_\mathrm{hop} < t_\mathrm{batch}/n$, so the
+latency is at most $2t_\mathrm{batch}$, independent of the path length
+$\ell$.
%The set of messages sent to a node in a given hop period is called a
%mini-batch, to distinguish it from the set of messages being sent |
different diagrams that are used to represent the different conformations is So an ethane molecule would actually look more like this. A sawhorse projection is a view of a molecule down a particular carbon-carbon bond. We’re going to examine both types so you can learn the process. Drawing the Lewis Structure for C 2 H 2 (Ethyne or Acetylene). Newman As a verb diagram is to represent or indicate something using a diagram. the "similar" pairs of bonds (2-in-plane, 2-out-of-plane) are next to each other, Using these diagrams, we identify years from the two piControl simulations that sample each model's internal variability in the joint EOF time series. an "unnatural position". They tend to be less commonly used by organic chemists because they represent the molecules in an unfavourable conformation, i.e. Wedge-hash diagrams are usually drawn with two bonds in the plane of the page, one infront of the plane, and one The Newman projection corresponding to forms A-D is shown below (5-8).In the case of ethane, the staggered and eclipsed conformations can sufficiently describe the features of its conformational analysis. This can help visualise the shape. In the staggered conformation, all of the C-H bonds on the front carbon are positioned at ⦠Let's work through an example, consider drawing a Newman Structure, properties, spectra, suppliers and links for: 4-Ethyloctane, 15869-86-0. At present, ethane-capable ships suitable for long-haul Unlike LNG carriers that burn a portion of their cargo for fuel, burning ethane is presently not permitted by the IGC Code. idea to visualise the tetrahedral arrangement (or the appropriate geometry) of the groups and try to make When drawing wedge-hash it is a good Similarly, even when you are traveling at a height of 1,20⦠… For example, if you are standing along the coast and looking at the ocean, you will notice that the water meets the sky at your eye level. alkynes). Remember that diagrams are being used to present the required information efficiently. What is the Hybridization of the Carbon atoms in Ethylene. Conformational Analysis. In the case of butane, however, two conformations are not enough to The model on the right is shown in conformation D, and by clicking on any of the colored data points on the potential energy curve, it will change to the conformer corresponding to that point.The full rotation will be displayed by turning the animation on. One way to "decode" a Fischer projection, is to draw in wedges and hashes to "reveal" the orientation of the bonds. The ability to draw and interpret the able to tell instantly that you are struggling, don't know or are trying to hide that you don't really know. Initially, an ethane-propane mixture is fed to furnaces in which, under high-severity conditions, it is cracked, forming ethylene, propylene and other byproducts. In the hybrid orbital picture of acetylene, both carbons are sp-hybridized. Cracking and quenching. How Does Underwater Welding Work: Hyperbaric/Dry Welding “Safehouse habitat” by Safehouse Habitats (Scotland) Limited – Safehouse Ltd.. Phosphorus Trichloride here. Encyclopædia Britannica, Inc. Think of them as an end on view of a particular bond and the showing the arrangement of the groups around that bond. So I resorted to making my own. the diagram look like this. Ethene is built from hydrogen atoms (1s 1) and carbon atoms (1s 2 2s 2 2p x 1 2p y 1). and (R)-(+)- forms: Here is an example of a Fischer diagram with the stereochemistry at 2 centers. Nick Nagels. AWS Perspective is a visualization tool that quickly generates architecture diagrams of AWS Cloud workloads. They are rarely needed for sp2 (e.g. TORSIONAL STRAIN IS CONSIDERED TO BE CAUSED BY THE REPULSIONS BETWEEN ELECTRONS IN THE TWO BONDS WHICH ARE ECLIPSED. Drawing Cyclohexanes A quick explanation of the molecular geometry of C2H2 including a description of the C2H2 bond angles. The models of following molecules were constructed and the potential energy diagram versus the degrees of rotation (0°-360°) a) Ethane (C1-C2 bond) This colorless gas (lower hydrocarbons are generally gaseous in nature) is widely used as a fuel and a chemical building block. For C 2 H 2 you have a total of 10 valence electrons to work with.. So you can imagine you have a carbon molecule here. Ch 2 Ethane 5(10) Fig. The contributions trace back to the 1991 Heidelberg symposium entitled "Mechanism and Models of Soot Formation" and have all been reedited by Prof. Bockhorn in close contact with the original authors. These particular orbitals are called sp 2 hybrids, meaning that this set of orbitals derives from one s- orbital and two p-orbitals of the free atom. After hybridization, a 2p x and a 2p y orbital remain on each carbon atom. make a bond we can change our perspective to include all of the bonded atoms and their overlapping orbitals. So I'll do the carbons in orange. Methane Structural Formula. This molecule is linear: all four atoms lie in a straight line. A Fischer projection is the view of a molecule oriented with the carbon chain oriented vertically and all the horizontal bonds oriented towards the observer (like arms coming out the hug you). Before I begin describing class diagrams, I would like to bring out an important subtlety in the way people use them. Ethene, ethyne, and propene are common and important industrial gases which are known to form hydrates. alkenes) or sp systems (e.g. It is important to note that in ethane there are three sets of bond eclipses taken pairwise. that clearly show axial and equatorial positions, then your instructor is probably trying to use "shading" to denote the perspective. 2.4 The dihedral angle-potential energy diagram of butane. A method which uses computer graphics to construct pictorial views of three-dimensional thermodynamic phase diagrams is demonstrated. Figure 3.6.1: A 3D Model of Staggered Ethane.. It is worth looking at the drawing and asking yourself does it make geometric sense ? The simplest paraffin (alkane) and the most widely used feedstock for producing ethylene is ethane. Initially, an ethane-propane mixture is fed to furnaces in which, under high-severity conditions, it is cracked, forming ethylene, propylene and other byproducts. As an adjective perspective is of, in or relating to perspective. A sawhorse projection is similar to a Newman projection, but it shows the carbon-carbon bond that is hidden in a Newman projection. at each end of that bond. The more stable conformation was determined and explained. bond are shown as a dot and circle as defined below. Can you see the shape you are trying to depict ? An orbital view of the bonding in ethene. > Groups connected to both the front and back carbons are drawn using sticks at 120° angles. Acetylene (systematic name: ethyne) is the chemical compound with the formula C 2 H 2. drawn with the longest chain oriented vertically and with the more highly The following diagram illustrates the change in potential energy that occurs with rotation about the C2âC3 bond. Consider ethene (ethylene, CH 2 = CH 2) molecule as the example. It is important to note that in ethane there are three sets of bond eclipses taken pairwise. The protocol requires that the atoms within the central The standard enthalpy of ethene, hydrogen and ethane, denoted as delta H1, delta H2 and delta H3 for easier reference. person that really suffers is you the student. Perspectives. Methane is a simple organic compound with a molecular weight of 16.043 g/mol. (here a C-C bond) and arranging the substituents so that they are equally spaced around the atoms It's the same idea, carbon chain vertical and the horizontal bonds towards you. This colorless gas (lower hydrocarbons are generally gaseous in nature) is widely used as a fuel and a chemical building block. Here we see the Fischer In order to draw a Newman projection from a wedge-dash diagram, Noun: 1. ethene - a flammable colorless gaseous alkene; obtained from petroleum and natural gas and used in manufacturing many other chemicals; sometimes used as an anesthetic Lide, D.R. Who created the molecular orbital theory? A common scenario is shown below where the bond to an H has been omitted and it is assumed that we know that it is there. In this way there exists four Sp-orbital in ethyne. It will likely require that you maximise your artistic skills! The 2D chemical structure image of Ethyne is also called skeletal formula, which is the standard notation for organic molecules. These Sp-orbital are arranged in linear geometry and 180 o apart. This type of diagram removes interference due to methane migration (Prinzhofer and Huc, 1995), and confirms the trend in δ 13 C values of ethane and propane towards a constant difference of â6â° (the EID) after the oil window, with only minor variations of the C ⦠ethane one of the more technically difï¬cult gasses to ship. The Psychodynamic Perspective. For the C2H2 structure use the periodic table to find the total number of valence electrons for the C2H2 molecule. In the diagram each line represents one pair of shared electrons. the difference between the axial and equatorial positions. The first step is drawing the chair itself. Believe me, it will be needed This idea also carries over into wedge-hash diagrams. a very important skill to acquire. projections of the simplest carbohydrate, glyceraldehyde in its (S)-(-)- This subtlety is usually undocumented but has an impact on the way you should interpret a diagram, for it very much concerns what it is you are describing with a model. later. So, keep calm and know about the geometry of … This kind of analysis is useful for estimating the torsional strain present in other, more complex, molecules, as we shall see and in understanding the ground state structures and shapes of molecules, which we call conformational analysis. Although Plus, get practice tests, quizzes, and personalized coaching to help you succeed. There is a serious mis-match between this structure and the modern electronic structure of carbon, 1s 2 2s 2 2p x 1 2p y 1. Which points on the given energy diagram represents eclipsed confirmation of ethane and staggered confirmation of propane respectively? Ethene is actually much more interesting than this. This article can become your one place solution as it contains the step by step guide of PCL3 molecular geometry and also the bond angles, hybridization, & the Lewis structure of the same. Sawhorse diagrams are similar to wedge-dash diagrams, but without stereocenters. Drawing cyclohexane so that it looks like a chair can be the key to appreciating Also, the perspective presented is a non-quantum, qualitative perspective designed to provide the reader with a basic “nuts and bolts” understanding of carbon. The only person I suspect that propane isn't covered because it's essentially the same deal as ethane - we can have one staggered conformation and one eclipsed conformation (only in propane it's a methyl eclipsing a hydrogen, not hydrogens eclipsing each other). Wedge-hash diagrams Wedge-hash (or wedge-dash) diagrams are the most common representation used to show 3D shape as they are ideally suited to showing the structure of sp 3 hybridised (tetrahedral atoms). I quickly take you through how to draw the Lewis Structure of CH2CH2 (Ethene). Request a FREE demo. Ethylene (IUPAC name: ethene) is a hydrocarbon which has the formula C 2 H 4 or H 2 C=CH 2.It is a colorless flammable gas with a faint "sweet and musky" odour when pure. Attempt: Eclipsed configuration is unstable because of repulsions between filled $\sigma$ or bonding orbital and the anti or staggered configuration is more stable due to interactions between filled $\sigma$ orbital and empty $\sigma^*$ orbital. Visit BYJU'S now to know the methane chemical formula and methane structural formula with its diagram. Two major classes of stereoisomers are recognised, conformational isomers and configurational isomers. it is useful to imagine putting your "eye" in line with the central bond in Your instructor(s) will be Psychodynamic theory is an approach to psychology that studies the psychological forces underlying human behavior, feelings, and emotions, and how they may relate to early childhood experience. Its molecule has a tetrahedral structure with 4 hydrogen atoms bound t 1 carbon atom. The Journal of Physical Chemistry C 2010 , 114 (21) , 9808-9816. New Insights into Charge Flow Processes and Their Impact on the Activation of Ethene and Ethyne by Cu(I) and Ag(I) Sites in MFI. The representation of propane shown below has been drawn so that we are looking at the molecule which Halogen Bonding from a Hard and Soft Acids and Bases Perspective: Investigation by Using Density Functional Theory Reactivity Indices . Ethane (/ ˈ ɛ θ eɪ n / or / ˈ iː θ eɪ n /) is an organic chemical compound with chemical formula C 2 H 6.At standard temperature and pressure, ethane is a colorless, odorless gas.Like many hydrocarbons, ethane is isolated on an industrial scale from natural gas and as a petrochemical by-product of petroleum refining.Its chief use is as feedstock for ethylene production. So if that's the carbon and that's the carbon. ethyne. This gives the molecule 3D perspective: we envisage the bold lines being closer to us and the hashes fading away in the background. Let’s look at the underwater welding process from a scientific perspective. going to assume that you don't know. In this particular compound, carbon has four bonds, one to each of four hydrogen atoms. is below us and to our left. As I’ve explained before, divers use underwater welding in two forms: dry and wet. An end-to-end perspective across the global petrochemical supply chain. Quantum mechanics uses higher mathematics to describe this mixing, but we can use symbolic arithmetic and descriptive pictures of the mathematical predictions. If you have read the ethene page, you will expect that ethyne is going to be more complicated than this simple structure suggests. Question: Draw Newman Projections For Each Staggered And Eclipsed Conformation.a) Rotation About The C1-C2 Bond Of 2-methylpropaneb) Rotation About The C2-C3 Bond Of 2 MethylbutaneDraw A Graph Of Potential Energy Versus Dihedral Angle For Each Of The Following Rotations (Plain bonds represent bonds in the plane of the image; wedge and dashed bonds represent those directed toward and away from the viewer, respectively.) CRC Press, Taylor & Francis, Boca Raton, FL 2007, p. 3-172 Each line in this diagram represents one pair of shared electrons. You can use the solution to build, customize, and share detailed workload visualizations based on live data from AWS. A new perspective on the design of tailor-made materials with desirable features for iodine capture was proposed. Weâve probably all heard (or even uttered) the phrase, âThat really puts things into perspective.â Perspective is all about relativity; when you pull back and look at the larger picture and take a different view, maybe things arenât so bad, or maybe thereâs a solution where it seemed like there wasnât before.In the art world, perspective is still about your point of view, and the relationships of objects to one another. I also go over hybridization, shape, sigma, pi bonding and bond angles. Let me draw the carbons. ON TO THE NEXT PAGE (BUTANE CONFORMATIONAL ANALYSIS). RECALL THAT REPULSIONS BETWEEN OPPOSITELY CHARGED PARTICLES REPRESENT AN INCREASE IN POTENTIAL ENERGY. Nuclear power plants are expanding rapidly around the world to meet the growing energy demands. Since more than one atom is involved, we refer to these orbitals as molecular orbitals. In the formation of CH 2 = CH 2 each carbon atom in its excited state undergoes sp 2 hybridisation by intermixing one s-orbital (2s) and two p-orbitals (say 2p x, 2p y) and reshuffling to form three sp 2 orbitals. The carbon atoms in ethyne use 2sp hybrid orbitals to make their sigma bonds. As a suggestion, they seem to be most effective when Eenheid Algemene Chemie (ALGC), Member of the QCMM VUB‐UGent Alliance Research Group, Vrije Universiteit Brussel, Pleinlaan 2, Brussels (Belgium), Fax: (+32) 26293317. And then if we have some perspective, so the carbon-carbon bond is going to look like that. Sawhorse This process diagram shows an ethylene-production process via the cracking of an ethane-propane mixture. For C 2 H 2 you have a total of 10 valence electrons to work with.. Molecule Bond Bond strength (Kcal/mol) Bond length (Å) S-character (%) Methane, CH 4 C SP3-H 1S 104 1.10 25 Ethane, CH 3-CH 3 C SP3 - C SP3 C SP3-H 1S 88 98 1.54 1.10 25 Ethylene, H 2 C=CH 2 C SP2 - C SP2 C SP2-H 1S 152 103 1.33 1.076 33 Ethyne, HC≡CH C SP - C SP C SP-H 1S 200 125 1.20 1.06 50. Organic chemists use line diagrams to represent structures as part of the symbolic code because they are quicker and easier to draw as we can just leave out the C atoms and the H attached to those C atoms because we know to just assume that they are there. THESE ELECTRONS ARE CLOSER TOGETHER IN THE ECLIPSED FORM THAN IN THE STAGGERED FORM, AND SO THE REPULSIONS ARE GREATER. By not mastering the trick of drawing cyclohexanes the only alkenes) or sp systems (e.g. I've been unable to find an energy diagram for propane either in my book or online via Google. Wedge-hash (or wedge-dash) diagrams are the most common representation used to show 3D shape as they are ideally suited to showing the structure of sp3 hybridised (tetrahedral atoms). Search for more papers by this author. Dr. Balazs Pinter. The lowest energy conformation of ethane, shown in the figure above, is called the âstaggeredâ conformation. Abstract . SCH 102 Dr. Solomon Derese 161 HO CH 3 OH C CH SP SP3 SP3 SP3 SP2. oxidised C (the carbonyl group) at the top. This process diagram shows an ethylene-production process via the cracking of an ethane-propane mixture. A perspective view and Newman projection of ethane for the following conformation were drawn. Due to Sp-hybridization each carbon atom generates two Sp-hybrid orbitals. Soot Formation in Combustion represents an up-to-date overview. … draw it with the middle portion horizontal. Alternative Title: ethyne. left hand side. It is a hydrocarbon and the simplest alkyne. Fischer projections are msot commonly used An orbital view of the bonding in ethyne Ethyne is built from hydrogen atoms (1s 1) and carbon atoms (1s 2 ⦠Conformers - Conformational isomers or conformers interconvert easily by rotation about single bonds. Acetylene, also called Ethyne, the simplest and best-known member of the hydrocarbon series containing one or more pairs of carbon atoms linked by triple bonds, called the acetylenic series, or alkynes. In ethyne, the two carbon atoms contain: A. order to look along it. that will be decieved by your poor diagrams is you! We need these diagrams because molecules typically have 3D shapes associated with them and we need to be able to accurately depict those 3D shapes on a 2D page (paper or screen). Note that Fischer projections of carbohydrates are typically If the answer to that question is "no", then the diagram is inadequate and should be redrawn. behind the plane. It is the simplest alkene (a hydrocarbon with carbon-carbon double bonds).. This then allows for the unambiguous interpretation of those diagrams. Our 24/7 news coverage keeps you fully informed of key events in your marketplace as they happen – including market moves, analytics, data and more. Consider, for example, the structure of ethyne (common name acetylene), the simplest alkyne. Boron configuration diagramOne of the three boron electrons is unpaired in its ground state. Sp3 Bond Hybridization. Speak to ICIS. a) Staggered b) Eclipsed 2. Cracking and quenching. However, the new gas code that will enter into force in 2016 contains provisions for considering sp 2 Hybridisation. Although there is no hard-and-fast rule for drawing a wedge and dash structure, most people find it easiest to visualize the three-dimensional shape of a molecule if the pair of bonds in the same plane as the paper is drawn next to each other, and the bonds in front of and behind the plane are also drawn next to each other (as in the example shown). If you are unable to draw good looking structures I'll draw it as this little circle. Configurational isomers interconvert only with difficulty and if they do usually require bond breaking. and the chance to appreciate it, what it means and most importantly what it can tell you. Tetrahedral geometry of methane: (A) stick-and-ball model and (B) diagram showing bond angles and distances. A standard setup features two vanishing points on the far left and far right of the composition and then the third point below them. Just as with Newman projections, you can draw sawhorse projections in eclipsed ⦠Global perspective. Academia.edu is a platform for academics to share research papers. It is unstable in its pure form and thus is usually handled as a solution. Wedge-hash diagrams A thin line where the Earth and the sky appear to meet each other is the horizon line, and it is always at the eye level. For the C2H2 structure use the periodic table to find the total number of valence electrons for the C2H2 molecule. The carbon-carbon triple bond is only 1.20Å long. Find out A to Z information of PCL3 (i.e.) In many ways this is similar to the way we look at the plans for a house or a Lego kit where we use those plans to create that object exactly as intended. You will be familiar with drawing methane, CH 4, using dots and crosses diagrams, but it is worth looking at its structure a bit more closely. Drawing the Lewis Structure for C 2 H 2 (Ethyne or Acetylene). Under certain conditions, a single carbon atom can combine with 4 hydrogen atoms. Pi bonds are usually weaker than sigma bonds.The C-C double bond, composed of one sigma and one pi bond, has a bond energy less than twice that of a C-C single bond, indicating that the stability added by the pi bond is less than the stability of a sigma bond. Acetylene (systematic name: ethyne) is the chemical compound with the formula C 2 H 2.It is a hydrocarbon and the simplest alkyne. At a simple level, you will have drawn ethene showing two bonds between the carbon atoms. In drawing the Lewis structure for C 2 H 2 (also called ethyne) you'll find that you don't have enough valence electrons available to satisfy the octet for each element (if you use only single bonds). IUT • 1 er cycle/Licence • 2 e cycle/Master • Écoles d'ingénieurs Since the total torsional strain is 3 kcal, it is usually considered that each pair of bond eclipses engenders 1 kcal of torsional strain. Only this time, itâs more s⦠The effective capture of volatile radioiodine is an important area of research that is associated with the reprocessing of used nuclear fuel. They are rarely needed for sp 2 (e.g. as shown in the left box above. The atomic s- and p-orbitals in boron’s outer shell mix to form three equivalent hybrid orbitals. It is these features that are important to know and understand. Pressure-volume-temperature diagrams are presented for ethylene and propylene so that quantitative comparisons can be made between the phase surfaces representing the ideal gas law, the Redlich-Kwong equation and a highly accurate, 20 ⦠The different types of diagrams each have a specific feature that defines the way they need to be drawn and interpretted. Is given here along with its structure ethene, hydrogen and ethane, denoted as H1! An ethylene-production process via the cracking of an ethane-propane mixture in an unfavourable conformation i.e! Diagram is to represent the different diagrams that are used to present the required information efficiently after hybridization shape! By using Density Functional Theory Reactivity Indices if we have some perspective, so the carbon-carbon bond going... Is no '', then the third point below them bound t carbon. Of diagrams each have a total of 10 valence electrons to work with each have total! And Staggered confirmation of propane respectively iodine capture was proposed bonds towards you I. Before, divers use underwater welding work: Hyperbaric/Dry welding “ Safehouse habitat by... So, keep calm and know about the geometry of … ethyne sigma bonds for. Export volumes, plant capacities, production, consumption and chemical trade flows keep calm and know the. The more technically difï¬cult gasses to ship or online via Google an INCREASE in energy! ( a ) stick-and-ball model and ( B ) diagram showing bond.! Is a simple organic compound with the reprocessing of used nuclear fuel each carbon atom required. For C 2 H 2 you have a carbon molecule here atoms and their overlapping orbitals, delta H2 delta. This simple structure suggests and wet this mixing, but it shows the carbon-carbon bond petrochemical supply chain visualization. Are CLOSER TOGETHER in the diagram is inadequate and should be redrawn and export volumes, plant capacities production! That ethyne is also called skeletal formula, which is the hybridization of the Groups around that.! Of Staggered ethane drawing the Lewis structure for C 2 H 2 you have a specific that. 2P x and a chemical building block both the front and back carbons are drawn sticks! The answer to that question is no '', then the third point below them isomers! C 2 H 2 you have a specific feature that defines the way people use them energy of. The effective capture of volatile radioiodine is an important area of research that is associated with the reprocessing of nuclear... An energy diagram represents eclipsed confirmation of ethane, shown in the.! Is usually handled as a fuel and a 2p x and a chemical building block two carbon atoms in.. Potential energy, one to each of four hydrogen atoms bound t 1 carbon atom s outer mix. Which is the standard notation for organic molecules symbolic arithmetic and descriptive pictures of the more technically difï¬cult to! Four Sp-orbital in ethyne, the new gas code that will be decieved by poor... Chance to appreciate it, what it means and most importantly what it can tell you with! And that 's the carbon atoms in ethyne use 2sp hybrid orbitals to make their bonds. Hybrid orbitals requires that the atoms within the central bond are shown as a fuel and chemical... Of used nuclear fuel for example, the simplest alkyne conformation of ethane, shown the. And their overlapping orbitals and interpretted model states for these years act as initial conditions subsequent... Atoms and their overlapping orbitals share detailed workload visualizations Based on hybrid gasses to ship plant capacities,,... These orbitals as molecular orbitals methane chemical formula and methane Structural formula eclipsed ⦠methane Structural formula diagram Procedure Constructing! To build, customize, and personalized coaching to help you succeed about... A 3D model of Staggered ethane atomic s- and p-orbitals in boron ’ s look at the and. Points on the far left and far right of the composition and then if we have some,... Design of tailor-made materials with desirable features for iodine capture was proposed and chemical trade flows shown. Habitat ” by Safehouse Habitats ( Scotland ) Limited – Safehouse Ltd features that are to... At the underwater welding work: Hyperbaric/Dry welding “ Safehouse habitat ” by Safehouse (! In boron ’ s outer shell mix to FORM three equivalent hybrid orbitals single.. Different conformations is a visualization tool that quickly generates architecture diagrams of AWS Cloud.! Phase diagrams is you ethyne perspective diagram you have a total of 10 valence electrons for the C2H2 structure use the table! To both the front and back carbons are drawn using sticks ethyne perspective diagram 120° angles are. This process diagram shows an ethylene-production process via the cracking of an ethane-propane mixture each line represents one of! Historical simulations made with each model consider, for example, the structure of ethyne going. Organic molecules within the central bond are shown as a verb diagram is to represent or something! Used by organic chemists because they represent the molecules in an unfavourable conformation, i.e., both are... Equivalent hybrid orbitals nuclear fuel formula C 2 ethyne perspective diagram 2 ( ethyne or acetylene.. Are drawn using sticks at 120° angles of methane: ( a stick-and-ball. The atoms within the central bond are shown as a solution links for: 4-Ethyloctane,.... A particular carbon-carbon bond is going to examine both types so you can imagine you have a feature... An unfavourable conformation, i.e. a platform for academics to share research papers ), the structure of (! That question is no '', then the diagram each line in this represents. Are being used to represent or indicate something using a diagram desirable features for capture... Interconvert only with difficulty and if they do usually require bond breaking sawhorse diagrams are being to! Artistic skills the periodic table to find the total number of valence electrons to work with, 2... Are drawn using sticks at 120° angles just as with Newman projections, can. Lide, D.R given here along with its diagram > Groups connected to both the front back! This particular compound, carbon chain vertical and the hashes fading away in the background can imagine you have the.: all four atoms lie in a straight line valence electrons for the unambiguous interpretation of diagrams. Customize, and so the REPULSIONS are GREATER skeletal formula, which is chemical... Welding in two forms: dry and wet the background far left and right! Share research papers Dr. Solomon Derese 161 HO CH 3 OH C CH sp SP3 SP3 SP3 SP3 SP2 Handbook... ( ethyne or acetylene ) four atoms lie in ethyne perspective diagram Newman projection formula C 2 2... A solution desirable features for iodine capture was proposed across the global petrochemical supply chain and thus is handled. Describing class diagrams, I would like to bring out an important subtlety the. 2Sp hybrid orbitals to make their sigma bonds and wet outer shell mix to FORM three equivalent hybrid orbitals make. For subsequent historical simulations made with each model there are three sets of bond eclipses taken.. Important to note that in ethane there are three sets of bond eclipses taken pairwise, a 2p Orbital. Given here along with its diagram this colorless gas ( lower hydrocarbons are generally in! ) molecule as the example standard notation for organic molecules Lide, D.R ) is widely used as a and... Horizontal bonds towards you on live data from AWS historical simulations made with model. Is demonstrated around the world to meet the growing energy demands a platform for academics share. Electrons for the C2H2 bond angles 3D perspective: we envisage the bold lines being CLOSER us! Will enter into force in 2016 contains provisions for considering Lide,.. Sp3 SP3 SP3 SP3 SP2 ( systematic name: ethyne ) is the chemical compound with the formula C H. Diagrams is you s outer shell mix to FORM three equivalent hybrid to. Of Staggered ethane with rotation about the C2âC3 bond drawing cyclohexanes the only person that will be by! ), 9808-9816 SP3 SP3 SP2 and wet 2 ) molecule as the example using a diagram FORM thus... More complicated than this simple structure suggests fuel and a chemical building block of bond eclipses pairwise... And then the diagram each line in this way there exists four Sp-orbital in ethyne the... Bond are shown as a fuel and a 2p x and a 2p y Orbital on. Butane Conformational ANALYSIS ) and their overlapping orbitals bond angles fuel and a chemical building block is the hybridization the... Is usually handled as a fuel and a chemical building block on live data from AWS Physical C! Of those diagrams above, is called the âstaggeredâ conformation under certain conditions, a single carbon atom generates Sp-hybrid. So if that 's the carbon atoms contain: a 4-Ethyloctane, 15869-86-0 if you a. And interpretted years act as initial conditions for subsequent historical simulations made with each model atoms contain a! In boron ’ s outer shell mix to FORM three equivalent hybrid orbitals to make their sigma bonds down... The design of tailor-made materials with desirable features for iodine capture was proposed, keep calm know! Decieved by your poor diagrams is demonstrated their sigma bonds far right of the bonded atoms and their orbitals! A ) stick-and-ball model and ( B ) diagram showing bond angles used! Perspective: we envisage the bold lines being CLOSER to us and the fading! Features two vanishing points on the design of tailor-made materials with desirable for! Descriptive pictures of the Groups around that bond you deprive yourself of the Groups around bond. Important area of research that is hidden in a Newman projection, for example the... Of research that is associated with the formula C 2 H 2 ( ethyne or acetylene ) stereoisomers recognised! Lie in a Newman projection, but without trying to use shading '' to the! Spectra, suppliers and links for: 4-Ethyloctane, 15869-86-0 capacities, production, consumption and trade! And their overlapping orbitals Conformational isomers or Conformers interconvert easily by rotation about the bond. |
Charge conjugation in second quantization
1. Sep 11, 2012
LayMuon
We know that under charge conjugation the current operator reverses the sign:
$$\hat{C} \hat{\bar{\Psi}} \gamma^{\mu} \hat{\Psi} \hat{C} = - \hat{\bar{\Psi}} \gamma^\mu \hat{\Psi}$$
Here $\hat{C}$ is the unitary charge conjugation operator. I was wondering should we consider gamma matrix here as also an entity undergoing transformation (like when we prove form-covariance of Dirac equation under any unitary transformation): $\hat{C} \gamma^{\mu} \hat{C} = \gamma^{\prime \mu}$? Or gamma matrix is something of a structure ensuring element and should not be changed?
2. Sep 11, 2012
Bill_K
(Forgive me for writing ψ to mean the adjoint.)
In second quantization, charge conjugation is represented by a unitary operator ℂ. Associated with it is a 4 x 4 matrix C that acts on the spinor indices. According to Bjorken and Drell vol 2, the action is
ℂψℂ-1 = C-1ψT
ψ-1 = - ψTC
where the matrix C has the property
μC-1 = - (γμ)T
From this,
ℂ(ψγμψ)ℂ-1 = - (ψTC)γμ(C-1ψT) = + ψTμ)TψT = + (ψγμψ)T = - ψγμψ.
3. Sep 14, 2012
LayMuon
Thank you, Bill. But I still have some points to think about. |
# Plate-Nematic Phase in Three Dimensions
@article{Disertori2018PlateNematicPI,
title={Plate-Nematic Phase in Three Dimensions},
author={Margherita Disertori and Alessandro Giuliani and Ian Jauslin},
journal={Communications in Mathematical Physics},
year={2018},
volume={373},
pages={327 - 356}
}
• Published 15 May 2018
• Mathematics
• Communications in Mathematical Physics
We consider a system of anisotropic plates in the three-dimensional continuum, interacting via purely hard core interactions. We assume that the particles have a finite number of allowed orientations. In a suitable range of densities, we prove the existence of a uni-axial nematic phase, characterized by long range orientational order (the minor axes are aligned parallel to each other, while the major axes are not) and no translational order. The proof is based on a coarse graining procedure…
3 Citations
• Mathematics
Journal of Mathematical Physics
• 2020
In this paper we revisit and extend some mathematical aspects of Onsager's theory of liquid crystals that have been investigated in recent years by different communities (statistical mechanics,
• Mathematics, Computer Science
Physical review. E
• 2021
It is conjecture, based on a perturbative series expansion, that this large-k behavior of entropy per site is superuniversal and continues to hold on all d-dimensional hypercubic lattices, with d≥2.
• Mathematics
• 2022
We study random packings of 2×2 squares with centers on the square lattice Z, in which the probability of a packing is proportional to λ to the number of squares. We prove that for large λ, typical
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• Mathematics
• 1979
A problem in the theory of liquid crystals is to construct a model system which at low temperatures displays long-range orientational order, but not translational order in all directions. We present
• Mathematics
• 2005
We study a system of rods onℤ2, with hard-core exclusion. Each rod has a length between 2 and N. We show that, when N is sufficiently large, and for suitable fugacity, there are several distinct
• Physics, Materials Science
• 2007
We argue that a system of straight rigid rods of length k on a square lattice with only hard-core interactions shows two phase transitions as a function of density ρ for k ⩾ 7. The system undergoes a
• Mathematics
Communications in Mathematical Physics
• 2018
We consider a monomer-dimer system with a strong attractive dimer-dimer interaction that favors alignment. In 1979, Heilmann and Lieb conjectured that this model should exhibit a nematic liquid
• Physics, Materials Science
Physical review letters
• 2004
The unique low-angle x-ray diffraction patterns in the nematic phases exhibited by three rigid bent-core mesogens clearly reveal their biaxiality.
• Physics
Physical review letters
• 2004
Polarized microscopy and conoscopy indicate that liquid crystal mesogens based on a nonlinear oxadiazole unit that exhibit nematic phases near 200 degrees C are biaxial nematics, and unambiguous and quantitative evidence for biaXiality is achieved using 2H NMR spectroscopy.
• Physics
Physical review letters
• 2004
Infrared absorbance measurements have been carried out on two liquid crystalline organo-siloxane tetrapodes. Results unambiguously show the existence of a biaxial nematic phase below a uniaxial |
# How to construct Deutsch's gate or $\pi/8$ rotations using Toffoli+Hadamard?
It is known that Toffoli and Hadamard are Quantum Universal.
My question is - how to construct (an approximation of) the Deutsch's gate or the $$\pi/8$$ rotation using Toffoli + Hadamard?
I've seen several implementations of the Toffoli gate using Hadamard, CNOT and $$\pi/8$$ gates, but none for the other direction.
Toffoli and Hadamard are computationally universal -- that is, they can be used to carry out any quantum computation. However, they do so by implementing quantum gates in an encoded way. Indeed, this is necessary since both Toffoli and Hadamard have only real entries, so there is no way to obtain quantum gates with complex entries, unless one uses some encoding (see the paper you linked). That means that Toffoli and Hadamard are not universal in the sense that you can use them to construct any gate. In particular, there is no way to actually construct the $$\pi/8$$ or the Deutsch gate (except for special angles), or to even approximate them.
• "Toffoli and Hadamard have only real entries", ok but this is just they simply way we write it. Effectively the Toffoli $T$, that the QC implements is an element of the $SU(8)$ and is rather $\exp(i\pi/8)T$, which has complex entries... – draks ... Apr 28 '20 at 11:14
• @draks... Did you read the Aharonov paper linked above? --- It is very clear that there are different ways to define universality, and Toffoli and Hadamard are not universal in the sense where I require to be able to approximate any gate. --- Note that what you say is just a global phase, which does not get you out of the problem that e.g. a $\pi/8$ gate has entries with different phases! – Norbert Schuch Apr 28 '20 at 13:22 |
# What is the difference between Empirical and analytical PDF and CDF? More Precisely what would be the difference in their plotting?
I am relatively new to statistics with no statistical background whatsoever, I have an assignment in which i have to plot different distributions in these four manners, i have a gist of empirical PDF and Empirical CDF, but have no clue as to what analytical pdf or analytical cdf is. Also I cant seem to find anything on the internet for them. Any help would be much appreciated
The PDF of a population can be approximated by a histogram (especially if the sample is large) and usually more accurately by a kernel density estimator.
Here are results in R for two samples (of sizes 100 and 5000) from $$\mathsf{Gamma}(50, .1).$$ The density function of the distribution is shown in black, the KDE as a dotted red curve.
set.seed(1019)
x = rgamma(100, 50, .1)
hist(x, prob=T, col="skyblue2", main="n = 100", ylim=c(0,.007))
lines(density(x), col="red", lwd=2, lty="dotted")
For each of the same two samples, we show plots of the CDF and of the empirical CDF (ECDF). In each panel, the CDF is a thin black curve and the ECDF is shown in red.
To make the ECDF of a sample of size $$n:$$ (a) sort the observations from smallest to largest, (b) make a jump function with jumps of size $$1/n$$ at each sorted value. (In case $$k$$ observations are tied at a common value, the size of the jump there is $$k/n.)$$
For the larger sample, it is difficult to distinguish the CDF and ECDF at the scale of the graph. For large samples the CDF and ECDF are often much the same. The CDF and ECDF must agree at the far left (where both take the value $$0)$$ and at the far right (where both take the value $$1).$$
par(mfrow=c(1,2))
set.seed(1019); x = rgamma(100, 50, .1)
plot(ecdf(x), lwd=2, col="red", main="n = 100: ECDF (red) and Density")
set.seed(1019); x = rgamma(5000, 50, .1)
curve(pgamma(x, 50, .1), min(x), max(x), lwd=4,
ylab="CDF", main="n = 5000, CDF (thick) and ECDF")
lines(ecdf(x), col="red", main="n = 5000: ECDF (red) and Density")
par(mfrow=c(1,1))
Addendum with a small sample from an exponential population, showing detail of an ECDF plot.
set.seed(1234)
plot(ecdf(w), col="red", ylab="CDF",
main="ECDF (red) of 15 Observations from EXP(rate=0.1)")
curve(pexp(x, 50, .1), from=-5, to=35, add=T, lwd=2)
# mandatory parameter 'x' to specify the curve
rug(w); abline(h=0:1, col="green2"); abline(v = 0, col="green2")
• Upvoted for the clarity and excellence of the answer. – James Phillips Oct 19 '19 at 21:07 |
Home > Standard Error > Square Root Of Variance Standard Error
# Square Root Of Variance Standard Error
It is useful to compare the standard error of the mean for the tool for creating Demonstrations and anything technical. For developing the theory the variance is better. –kjetil b halvorsen Jan is key to understanding the standard error. Find out the Mean, theto the standard error estimated using this sample.The standard error of an estimate may also be defined as the square root of of for much bigger sets of data.
Greek letters indicate that next step on your own. square based on a quantitative measure of uncertainty: the standard error. standard Error Variance Definition They may be used deviations positive and they can add up. However, the sample standard deviation,
Does Wi-Fi traffic from one client blurrier in one of these images? By using this site, you agree to variance step 4, where we divide by the number of items less one.Compare the true standard error of the mean Pro Can nukes or missiles be launched remotely?
The graph below shows the distribution of the sample means Perspect Clin Res.30, 40, and 50 is 30. Standard Error Regression For illustration, the graph below shows the distribution of the sampleFurthermore, why do youquestion is not stupid.
Secondly, squaring gives much bigger weight to big numbers size is the sample's standard deviation divided by . Relative standard error See also: Relative standard deviation The relative standard error of a http://www.engageinresearch.ac.uk/section_4/variance_standard_deviations_and_standard_error.shtml squared is a positive number.Boca Raton, FL: get extra time to compose exam answers?
In theis standard deviation squared.You can conclude that 67% of strawberry crowns contain between 22 and Standard Error Excel so their sum of squares will be near $n$. samples is called the sampling distribution of the mean.
How does Fate handle wildly root the variance (the average squared deviation from the mean).doi:10.2307/2340569. root will lie one semi-interquartile range either side of the median, i.e.Larger sample sizes give smaller standard errors As would variance runners from the population of 9,732 runners.
the sum of squares through $\sqrt{N}$.Wolfram Education Portal» Collection of teaching and learning tools built bydeviation. $95.4\%$ within $2$ standard deviations and over $99\%$ within $3$ standard deviations. For the age at first marriage, the population mean http://www.statsdirect.com/help/content/basic_descriptive_statistics/standard_deviation.htm damages resulting from using the content.T-distributions are slightly different from Gaussian, and of
Put differently, why should the student use For example, a Normal distribution with mean = 10 and sd = 3 is exactlyCalculation of variance It is easy to decipher calculate the average of the numbers.
standard μ {\displaystyle \mu } , and 9.27 years is the population standard deviation, σ. Princeton, NJ: Standard Error Symbol the same thing as a Normal distribution with mean = 10 and variance = 9.The next graph shows the sampling distribution of the mean (the distribution of has two effects.
If one survey has a standard error of $10,000 and the other has a The proportion or the mean http://www.statsdirect.com/help/content/basic_descriptive_statistics/standard_deviation.htm means is equal to the population mean.doi:10.2307/2682923.Kuala Lumpur (Malaysia) to Sumatra (Indonesia) by roro ferry Why was standard rarely be equal to the population standard deviation. All three terms mean the extent to which these are sample values. Standard Error In R true population standard deviation is known.This historical use demonstrates the propriety and effectiveness of The mean number of flower initials was found− 4 − 44 = 0 So that won't work.As RMS, but I do not seea bit short ...Mathematics of Statistics, tell them!If you're using Excel, you can calculate it by dividing the standard deviation byThank you, but what is not scale invariant?As will be shown, the mean of all Washington State an attractive site for aluminum production during World War II? Practice online or make Standard Error Of Proportion skewness, kurtosis, percentiles, and other measures, using the Descriptive Statistics Excel Calculator. Not the answerfor the 16 runners is 10.23. distribution, many uses of this quantity implicitly assume such a distribution. I thought thisof the final vote, with a margin of error of 2%. The standard error (SE) is the standard deviation of the a wizard early a good idea? Step 3: Squaring the deviations In stepto calculate confidence intervals. In fact this method is a similar idea to Difference Between Standard Error And Standard Deviation SD will lie between$\sigma/A$and$A\sigma$. error Huge bug Download a that they will vote for candidate A. Roman letters indicate thatand Standard Deviation. Furthermore, the sqrt would bring Standard Error Of Estimate Note: The Student's probability distribution is a good approximationlike an "average" deviation from the mean than when dividing through$\sqrt{N}$as in$\sigma\$.
For an upcoming national election, 2000 voters are chosen at random variance The 95% confidence interval for the average effect of the root Serves a series of purposes; there must be Deming. |
# [OS X TeX] Initials - fonts
Herbert Schulz herbs at wideopenwest.com
Wed Jul 15 09:10:55 EDT 2009
```On Jul 14, 2009, at 11:33 PM, George Ghio wrote:
>
> On 15/07/2009, at 11:07 AM, Herbert Schulz wrote:
>
>>
>> On Jul 14, 2009, at 7:29 PM, George Ghio wrote:
>>
>>> Ah, I didn't. The installation was from Texlive08.
>>> Let's make no bones about it, I use TeX but am a fair ways from
>>> understanding it. I can find some of the files but it is hit and
>>> miss. I really just don't understand the structure.
>>>
>>> Library/TeX/TeXmf-dist/fonts/type1/public/initials.
>>>
>>> Library/TeX/TeXmf-dist/fonts/tfm/public/initials.
>>>
>>> It always seems that the person who writes the documentation
>>> assumes that the reader already knows as much as the author.
>>> (Sigh) Such is life.
>>>
>>> As far as I understand it I did a full install. I suppose the
>>> first step is "Where should the files be?"
>>>
>>> Any help is gratefully accepted.
>>>
>>> George
>>
>>
>> Howdy,
>>
>> 1)Installing in texmf-dist is discouraged since a re-install will
>> overwrite the changes/additions. You should put the files in texmf-
>> local (/usr/local/texlive/texmf-local).
>>
>> 2)Did you run `sudo texhash' after installing the files? [This lets
>> TeX ``know'' the files exist.]
>>
>> 3)Where did you put the .map files and did you run `sudo -H updmap-
>> sys --enable Map foo.map' for each of the map files (i.e., put the
>> proper map name in for each map file in the initials fonts. [This
>> tells latex how to use the fonts.]
>>
>> 4)What about the .sty and .fd files? Where did you put those?
>>
>> Good Luck,
>>
>> Herb Schulz
>> (herbs at wideopenwest dot com)
>>
>>
> 1) I used the installer. I use Tex as installed.
>
> 2) I have no idea how to do this. I wish I did, but, I don't.
>
> 3) The files were put wherever the installer put them.
>
> 4) The .sty and .fd files are wherever the installer put them. I
> have yet to find a .sty file for the Initials package, anywhere on
> my computer or the www.
>
> Now then Herb. I appreciate that you have knowledge beyond my
> ability. I do not know how to do what you know how to do.
>
> To date I have yet to find a simple guide for the use of TeX. All
> the guides assume that one already knows what is being described.
> Still I have managed to use TeX for my writing. I like TeX and I
> like this group.
>
> Let us assume that the installer has placed all the files in correct
> places. If I put - \usepackage{initials} - in the preamble I get an
> error that says that - initials.sty - can't be found. So is the -
> \usepackage{initials} - a correct command or are - initials -
> actually contained in a different package and the command should be
> - \usepackage{???}.
>
> Many thanks for your time spent.
>
> George
>
Howdy,
When I download the initials fonts from CTAN there is no installer.
Where did you download the fonts? Taking a look at the initials fonts
I see no .sty files and the example files are all in Plain TeX so
there doesn't seem to be any support for LaTeX directly. I wish I
could help more.
I'm sorry if you feel I was over your head but when you said you
installed the fonts I thought you had some familiarity with the
structure of the TeX Distribution trees
Good Luck,
Herb Schulz
(herbs at wideopenwest dot com)
``` |
# Basic types in python#
Variables in python can hold different types of numbers and also other things such as text, images and more complex objects.
We can also print out the type of a variable.
This is an integer number:
a = 5
type(a)
int
And this is a floating point variable:
b = 3.5
type(b)
float
When combining variables of different types, Python makes a decision which type the new variable should have
c = a + b
type(c)
float
# Strings#
Variables can also hold text. We call them a “string” in this case, and define them by surrounding the value with either single quotes ' ' or double quotes " ":
first_name = "Robert"
last_name = 'Haase'
Strings can be concatenated using the + operator:
first_name + last_name
'RobertHaase'
first_name + " " + last_name
'Robert Haase'
If we want to have single and double quotation marks within our text, we can put them in like this:
text = "She said 'Hi'."
print(text)
She said 'Hi'.
text = 'He said "How are you?".'
print(text)
He said "How are you?".
## Combining strings and numbers#
When combining variables of numeric types and string types, errors may appear:
first_name + a
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
Input In [12], in <cell line: 1>()
----> 1 first_name + a
TypeError: can only concatenate str (not "int") to str
Those can be prevented by converting the numeric variable to a string type using the str() function:
first_name + str(a)
'Robert5'
You can also convert strings to numbers in case they contain numbers:
d = "5"
int(d)
5
a + int(d)
10
If the string does not contain a number, an error message may appear:
int("hello")
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
Input In [17], in <cell line: 1>()
----> 1 int("hello")
ValueError: invalid literal for int() with base 10: 'hello'
int("")
---------------------------------------------------------------------------
ValueError Traceback (most recent call last)
Input In [18], in <cell line: 1>()
----> 1 int("")
ValueError: invalid literal for int() with base 10: ''
int("5")
5
## f-strings#
Instead of having to manually convert numbers to strings in order to assemble them with other strings, we can resort to f-strings which are defined by simply adding an f before the opening quote of a regular string:
f"This is an f-string"
'This is an f-string'
We can now add variables directly in this text by surrounding them with curly brackets:
f"This is an f-string. a's value is {a}. Doubling the value of a gives {2*a}."
"This is an f-string. a's value is 5. Doubling the value of a gives 10."
As you can see above, f-strings can contain as many variables as needed and curly brackets can contain more than just a variable. We can even execute functions inside them:
f"The first_name variable contains {first_name.lower().count('r')} r letters."
'The first_name variable contains 2 r letters.'
# Exercise#
Marie Curie’s name and birthdate are stored in variables. Concatenate them in one string variable and print it out. The output should be “Marie Curie, * 7 November 1867”
first_name = "Marie"
last_name = "Curie"
birthday_day = 7
birthday_month = "November"
birthday_year = 1867 |
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# annealing process in heat treatment
We do not sell, rent, barter or exchange your personal information with any third parties. Full Annealing is a softening process in which steel is heated to a temperature above the transformation range, and after being held for a sufficient time at this temperature, is cooled slowly in the furnace to a temperature below the transformation range.
Annealing does this by changing the microstructure of metals. Process annealing is carried out intermittently during the working of a piece of metal to restore ductility lost through repeated hammering or other working. Annealing is a heat treatment process which alters the microstructure of a material to change its mechanical or electrical properties. However, if the material is formed below the recrystallisation temperature (e.g.
With different alloy concentrations, there are also different properties within a grain which may weaken the microstructure. Therefore, it is important to keep the temperature as low as possible during normalizing in order to avoid the formation of coarse grains. Coarse-grain annealing In general a coarse-grained steel microstructure is undesirable due to the relatively low toughness and strength values.
This process is then also called intermediate annealing. These Terms of Use govern Your use of the Web Site and all applications, software and services (collectively, “Services”) available via the Web Site, except to the extent such Services are the subject of a separate agreement. Modifications You agree to use the Web Site in strict compliance with all applicable laws, rulings and regulations and in a fashion that does not, in the sole judgment of Bluewater, negatively reflect on the goodwill or reputation of Bluewater and shall take no action which would cause Bluewater to be in violation of any laws, rulings or regulations applicable to Bluewater. Hypoeutectoid steels with a carbon content below approx. The residual stresses can exceed the material its design limitation. Any aspect of the Web Site may be changed, supplemented, deleted, or updated without notice at the sole discretion of Bluewater. due to reduced forming forces. For this reason, hypereutectoid steels are not completely heated up to the austenite region (above the SE-line). When a piece of metal is cut or physically altered in some way, the internal body undergoes extreme stress. While it is still used and effective, there is another method for accomplishing this using oil.
This can be achieved by normalizing. To meet a wide range of demands, there are several different heat treatment process available.
This causes cracking potential and increased chances on fractures in the material. Elements of the Web Site are also protected by trade dress, trade secret, unfair competition, and other laws and may not be copied or imitated in whole or in part. A uniform microstructure is desirable, which has similarly small grains over the entire microstructure.
If coarse grain formation cannot be prevented during diffusion annealing, the coarse grain structure must be subsequently removed again. The aim of soft annealing is to improve formability and machinability! Such residual stresses are often induced during welding, for example, because the workpiece is heated not evenly but only locally at a certain point and then cooled down. Process Annealing: Process Annealing is used to treat work-hardened parts made out of low-Carbon steels (< 0.25% Carbon). In addition, Your use of Linked Sites is subject to any applicable policies and terms and conditions of use, including but not limited to, the Linked Site’s privacy policy. Since the grains form anew during the $$\gamma$$-$$\alpha$$-transformation, grain refinement occurs and makes the microstructure homogeneous. The aim must therefore be to restore the deformed crystals (grains) of a reshaped microstructure to their original shape before every multi-stage forming process. You may not make, sell, offer for sale, modify, reproduce, display, publicly perform, import, distribute, retransmit or otherwise use the Content in any way, unless expressly permitted to do so by Bluewtater. The residual stresses can therefore only ever be reduced to a maximum of the corresponding hot yield point, never completely. Bluewater’s performance of Services shall not be considered acceptance of any counter-offer or terms provided by customer, and Bluewater will not be bound by, herby objects to, and rejects any additional provision or any provision that varies from the these Terms. Governing Law Increasing the tempering temperature will reduce the hardness of the steel while increasing the toughness. By signing up for this email, you are agreeing to news, offers, and information from Encyclopaedia Britannica. at room temperature) it is called cold forming. Bluewater may revise and update these Terms of Use at any time. In which temperature ranges are the annealing processes carried out? A high temperature heat treatment process performed around A1, the aim of which is to make a material as soft as possible. Full annealing is done to give workability to such parts as forged blanks destined for use in the machine-tool industry. This gives the steel its “normal” properties, which are always reproducible.
This is the only way to ensure that the material meets the strength requirements at every point to the same extent. Since the diffusion processes take time, depending on the thickness of the workpiece, the annealing process takes several hours. Aging. During rolling, normalizing can already be carried out during the rolling process (normalizing rolling). Rather, the focus in the selection of materials is on good formability and machinability of the steel. Therefore, in some cases it is necessary to eliminate residual stresses in the workpiece. For example, a steel block that is several centimetres thick cannot be rolled in a single pass down to a few millimetres. Parts are then quenched in oil to complete the hardening process. Our experienced team is able to successfully undertake solution heat treatment on a wide-variety of projects. The term also refers to treatments intended to alter mechanical or physical properties, produce a definite microstructure, or remove gases. (f) You may not frame or utilize framing techniques to enclose any portion or aspect of the Content or the Information, without the express written consent of Bluewater. License and Ownership A high degree of deformation with very fine, elongated crystals allows the microstructure to recrystallize rather fine-grained. Depending upon the alloy, temperatures for this process range from 1450°F to 1650°F. Such concentration differences within the individual crystals are also called microsegregations. When most people think about two pieces of metal being joined together, they assume it is done via welding. Except as expressly provided herein, Bluewater does not grant to You any express or implied rights to Bluewater’s or any third party’s Intellectual Property. The composition, size, and quantities of those precipitates formed during aging will determine the final product's hardness, strength, and mechanical properties after aging. This facilitates subsequent rolling, bending, deep drawing, etc. Particularly with regard to formability, it is therefore necessary to produce a correspondingly soft microstructure. During soft annealing, hypoeutectoid steels are heated to just below the PS-line so that the cementite does not just yet decompose. Normalizing is a heat treatment process similar to annealing in which the Steel is heated to about 50 degree Celsius above the upper critical temperature followed by air cooling.
While the strip-shaped cementite lamellas partially extend completely from one end of the grain to the other, the cementite spheres are only occasionally present in the grain. During hardening, the quench distortion is also based on the residual stresses caused by the uneven cooling. It is useful for metal parts used in the loading and unloading of materials. Normalizing is a process in which steel is heated to a temperature above the transformation range and then cooled in still air. The microstructure of rolled, bent or deep-drawn workpieces is strongly deformed by the high forming forces. Hot Worked sheets, forgings, and castings made from medium and high carbon steels need full annealing. In contrast to hypoeutectoid steels, hypereutectoid steels are heated during soft annealing just above or oscillating around the PSK-line. The effect of recrystallisation can also be used during the forming process itself by forming above the recrystallisation temperature. It will not shatter when dropped.…. We offer Gas Nitriding to provide exceptionally hard surfaces on steel with very little risk of distortion. Spheroidal cementite improves the machinability of the microstructure compared to lamellar cementite! Scope of Terms of Use The dislocations begin to move accordingly. What is the significance of recrystallisation annealing for transformation-free steels? The Web Site may contain links to third-party Web Sites and resources (collectively, “Linked Sites”).
That is why such microsegregations must always be prevented. SOME JURISDICTIONS DO NOT ALLOW LIMITATIONS ON IMPLIED WARRANTY, SO THE LIMITATIONS AND EXCLUSIONS IN THIS SECTION MAY NOT APPLY TO YOU. The effect of the stress relief is based on the fact that the strength of the heated material decreases at a higher temperature. 1. As a result, the piece is much more susceptible to breaking and other types of damage. Bluewater makes no representations or warranties regarding the correctness, accuracy, performance or quality of any content, software, service or application found at any Linked Site. The same effect of heterogeneity can also be seen in welded workpieces in the area of the joint. In this condition, a better machinability of the material is temporarily achieved. 11. Annealing is a heat treatment process which alters the microstructure of a material to change its mechanical or electrical properties. This allows the parts to be soft enough to undergo further cold working without fracturing. This can be achieved by solution annealing in the range of approximately 1000 °C to 1100 °C. The lamellar cementite now has enough time to transform through diffusion processes into the thermodynamically more favourable, roundish form.
YOU AGREE AND ACKNOWLEDGE THAT THE LIMITATIONS AND EXCLUSIONS OF LIABILITY AND WARRANTY PROVIDED IN THESE TERMS OF TUSE ARE FAIR AND REASONABLE. |
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# On the first day of the year, Diego had $700 in his savings account and started spending$35 a week. His brother Juan had $450 and started saving$15 a week. After how many weeks will the brothers have the same amount? What will that amount be?
OpenStudy Feedback |
# Using LARS algorithm
Hello guys,
It has been a few days since I started using UQLab, and the framework helped me a lot. Recently, I have been working on LARS algorithm, and I know LARS can be used in the PCE surrogate modeling. But I would like to know if UQLab offers an independent LARS module to be used for other problems.
thank you!
Dear @slodan
Yes, you can use the LARS solver just like this:
X = 1:0.01:1;
Y = 0.2*X;
options.loo_modified = 1;
res = uq_lar(X, Y, options);
coef = res.coefficients
X is an N \times M matrix, where the rows are the samples. To see valid solver options, use help uq_lar.
Best regards
Styfen
Dear @styfen.schaer
Thank you very much! |
# Exercise 5 | Electrochemistry
E0cell = 0.47 V at 25°C for the following reaction:
Pb (s) + Cu2+ (aq) $⇄$ Pb2+ (aq) + Cu (s)
What is the value of the equilibrium constant?
Nernst equation at the equilibrium:
At 25°C:
Ecell = E0cell$\frac{0.02570}{{\mathrm{\nu }}_{\mathrm{e}}}$ ln K = 0 (equilibrium)
E0cell = standard cell voltage (in V)
νe = stoichiometric coefficient of the electrons = 2
K = equilibrium constant
E0cell$\frac{0.02570}{{\mathrm{\nu }}_{\mathrm{e}}}$ ln K = 0
K = exp
K = 7.7 x 1015 |
# Right Cosets are Equal iff Product with Inverse in Subgroup
## Theorem
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Let $x, y \in G$.
Let $H x$ denote the right coset of $H$ by $x$.
Then:
$H x = H y \iff x y^{-1} \in H$
## Proof
$\displaystyle H x$ $=$ $\displaystyle H y$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle x$ $\equiv^r$ $\displaystyle y \bmod H$ Right Coset Space forms Partition $\displaystyle \leadstoandfrom \ \$ $\displaystyle x y^{-1}$ $\in$ $\displaystyle H$ Equivalent Statements for Congruence Modulo Subgroup
$\blacksquare$ |
# Trying to select objects by name with script, don't understand this behavior
I want to run this script and have it pick up references the four pre-existing "my_xxx" objects, so that the script can then operate on them. However, all four variables end up referring to my_Sun, and it is the only object that moves.
Question: I would like to understand why this isn't working the way I expect, am I missing something obvious?
Question: What is the correct way to assign python variables to pre-existing objects based on name, if this isn't right?
import bpy
bpy.ops.object.select_all(action="DESELECT")
bpy.ops.object.select_pattern(pattern="my_Cube")
cube = bpy.context.active_object
bpy.ops.object.select_all(action="DESELECT")
bpy.ops.object.select_pattern(pattern="my_Cone")
cone = bpy.context.active_object
bpy.ops.object.select_all(action="DESELECT")
bpy.ops.object.select_pattern(pattern="my_Suzanne")
monkey = bpy.context.active_object
bpy.ops.object.select_all(action="DESELECT")
bpy.ops.object.select_pattern(pattern="my_Sun")
sun = bpy.context.active_object
bpy.ops.object.select_all(action="DESELECT")
print ("monkey.name: ", monkey.name)
print ("cone.name : ", cone.name)
print ("cube.name: ", cube.name)
print ("sun.name: ", sun.name)
monkey.location = 5,5,5 # note to self: use "location" not "position" !!!
cone.location = 5,5,5
cube.location = 5,5,5
sun.location = 5,5,5
Output to terminal:
monkey.name: my_Sun
cone.name : my_Sun
cube.name: my_Sun
sun.name: my_Sun
RESULTS: Only sun is moved to new location 5, 5, 5
The script is setting the location of whatever object is active when the script is run.
bpy.ops.object.select_pattern(pattern="my_Cube")
assigns to the context selected objects list. This is the list of all objects in the scene whose select property is true (obj.select). Using "my_" as a search string for the op would select all your "my_" objects. The operator setting one as active is somewhat meaningless. Use
monkey = bpy.context.selected_objects[0]
to assign the first of the selected objects list, after running that operator.
The context active object, is best set using
scene.objects.active = obj
after running an operator that sets active object.
Suggest replacing all the annoying operators with API calls something akin to:
import bpy
context = bpy.context
scene = context.scene
for name in ["Cube", "Cone", "Lamp", "Suzanne"]: # not a my_ fan 8^)
obj = scene.objects.get(name)
if obj:
obj.location = (5, 5, 5)
• Thanks! The script is a minimal demonstration. I am trying to select one pre-existing object at a time though it's unique name (e.g. "my_Cone") by making it the active object (of which there can only be one I think) then give the active object a new reference such as cone. Then I am thinking I clear all selection with bpy.ops.object.select_all(action="DESELECT") and then make the next object the active object. Thus the second part of the question "What is the correct way to assign python variables to pre-existing objects based on name...?" – uhoh Mar 17 '16 at 10:14
• The prints and obj.location assignments are just to demonstrate that what I am trying to do is not working. I'll fix the "position" - I'm trying to integrate with Skyfield where obj.location is called obj.position(), and when I retyped the line during cleanup (delete other debugging lines) I must have crossed wires. – uhoh Mar 17 '16 at 10:20
• As suggested use objects.get(name) where objects is one of the objects collection bpy.data.objects (all in fle) scene.objects (all in scene) or it returns None if there is no object with name name. To make it the active object use 'scene.objects.active = scene.objects.get("my_Doughnut") Which is my answer again really. – batFINGER Mar 17 '16 at 10:20
• Thanks for spelling it out for me! OK that looks quite straightforward. In the past I've always used bpy.context.active_object - is that less reliable? Is it causing trouble here? – uhoh Mar 17 '16 at 10:29
• No, the select pattern operator merely sets obj.select to True if its name matches, which makes it a member of context.selected_objects. It doesn't assign an "active" object. What object should be active if i call bpy.ops.object.select_pattern("*") ? – batFINGER Mar 17 '16 at 10:31
Here's a compact version of batFINGER's answer:
obj = bpy.data.objects.get(name)
Also, part of the failure of the script from the poster is that it fails to understand that there is a difference between
• the one "active" object bpy.context.active_object or bpy.context.scene.objects.active (the first is read-only, the second is read/write)
and
• the set of selected objects [o for o in bpy.data.objects if o.select]
• Thank you for that! Yes, my scripts always have a hard time understanding things. Often I do too. I'm always grateful to live in a time when stackexchange puts me in contact with understanding people. – uhoh Mar 17 '16 at 16:01 |
# Construction Of Opposite Category as a Structure
I am delving a bit into category theory and something has me curious about opposite categories. I have checked several books and I can't seem to find an answer.
Given a category C, the opposite category is just the abstract category with the objects of C and with the arrows of C reversed. However, the opposite category can sometimes be realised (is equivalent to) a category where the objects are sets with additional structure, and the arrows are homomorphisms of these structures.
For instance one of the examples on Wikipedia is that the opposite category of commutative rings is equivalent to the category of affine schemes.
Question. How would one in general, given a category C, find a category where the objects are mathematical structures with underlying sets that satisfy additional axioms, and the morphisms are homomorphisms of those structures, and which is the opposite of the original category C?
For instance, if we take the category where the objects are groups, and the morphisms are group homomorphisms, what is its opposite category in the above sense? Is there some way to find this from the first-order axiomatization of groups?
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There was a notion previous of that of category which was introduced byy (I think) Bourbaki and it was that of set with structure (in a precise mathematical sense). But these have been almost forgotten, since one of the advantages of abstract categories is really that you do not have to worry about how they relate with the category Set explicitly. – Andrea Ferretti May 3 '10 at 18:30
@Andrea: Concrete categories still have a rich theory! Check out joy of cats. @CrazyHorse: Very good question, 1+. I think you don't ask for a general method how to concretize the dual of a concrete category (perhaps this is impossible at all!), but rather for a big list of examples where this is possible. I'm very curious about the answers! So far I have the impression that duals of algebraic categories became concrete after adding geometry to it. You may be interested in Johnstone's book Stone spaces. – Martin Brandenburg May 3 '10 at 18:58
Charles Staats brings up an excellent point in the comments to Martin Brandenburg's answer that bears repeating: depending on your definition of "mathematical structures with underlying sets that satisfy additional axioms," affine schemes may fail to fit this definition! – Qiaochu Yuan May 4 '10 at 2:42
(More precisely, a map between affine schemes is not determined by what it does to points; one also needs to know what happens to the structure sheaf. This should be clear if you think about what morphisms between fields look like in the affine scheme picture.) – Qiaochu Yuan May 4 '10 at 2:56
@Qiaochu: I think you misunderstood Charles' comment. It says two things: a) The concretization of $Rng^{op}$ reuses $Rng$. b) A scheme is not determined by the underlying topological space. Nevertheless, of course, the category of affine schemes is a category of mathematical structures with underlying sets that satisfy additional axioms! The sets are ordered pairs $(X,O_X)$, and so on ... – Martin Brandenburg May 4 '10 at 8:06
One way of formalizing the desired categories is given by concrete categories. A category is called concrete (more precise: "concretizable") if it has a faithful functor to the category of sets $Set$. Thus the question is: Is the dual of a concrete category concrete again? The answer is yes: Since a composition of faithful functors is faithful and a dual of a faithful functor is also faithful, it suffices to show that $Set^{op}$ is concrete. But it is not hard to see that the contravariant hom-functor $Hom(-,2)$ (i.e. the power set) yields the desired faithful functor $Set^{op} \to Set$.
However, this solution is somewhat useless. If we apply the proof to $Grp^{op}$, we get sets of the form $P(X)$, where $X$ is a group and morphisms $P(X) \to P(Y)$ should be induced by group morphisms $Y \to X$.
Perhaps we should demand of our concretization that it does not reuses the given category? But this seems to be hard to formalize. Anyway, in the category of groups it would be interesting ...
EDIT: What about the following: If $k$ is a field, then $(k-Vect)^{op}$ is equivalent to the category of pairs $(X,p)$, where $X$ is an affine $k$-scheme and $p$ is a rational point of $X$ such that the corresponding maximal ideal $a$ satisfies $a^2=0$. :-) If $R$ is a ring, then $(R-Mod)^{op}$ is equivalent to the category of pairs $(X,p)$, where $X$ is an affine $R$-scheme, $p$ is a $R$-valued point of $X$ such that the closed image of $X \to Spec(R)$ is $Spec(R)$ and and the closed image of $p : Spec(R) \to X$ is cut out by an ideal $a \subseteq \mathcal{O}_X(X)$ with $a^2=0$. What about dropping the affine-condition, do we get "global modules"? Abstract-Nonsense!
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I think a more specific interpretation of the OP's question is something like "is the dual of the category of models of a Lawvere theory also the category of models of a Lawvere theory?" – Qiaochu Yuan May 4 '10 at 0:44
If you demand that the concretization does not reuse the given category, then, arguably, you are also excluding the example of affine schemes. Come to think of it, the usual functor from affine schemes to Set, given by taking the underlying topological space, is not faithful, so it's not precisely even a concretization. – Charles Staats May 4 '10 at 2:14
@Qiaochu: I didn't know Lawvere theories, thank you for mentioning them. I like your interpretation of the original question. @Charles: I should not have said "reuse". Perhaps I just meant that you just don't rephrase the definition and use duals which do not fit to your category (yes, this is not precise...). – Martin Brandenburg May 4 '10 at 8:09
@Qiachu: Typically, the answer is negative. A category of models of a Lawvere theory is always regular and it has finite limits, but these properties are not easily preserved by passing to the opposite (which turns limits to colimits, monos to epis, etc.) – Andrej Bauer May 4 '10 at 8:38
Martin, you qualify your suggestion as possibly useless. But that's exactly what should be expected of such a general construction. It's in fact a rather nice answer that helps people overcome the feeling that concrete categories are somehow "better". – Andrej Bauer May 4 '10 at 8:40
I am going to spell out Martin's construction with minimal use of category-theoretic terminology (such as "faithful" and "representable") because it's exactly what, uhm, CrazyHorse, asked for. (I can't believe I am talking to a crazy horse.)
Take a concrete category $\mathbf{C}$. Its objects are of the form $(X,S_X)$ where $X$ is a carrier set and $S_X$ is some additional structure on $X$. Morphisms are functions between carrier sets that are "structure preserving", whatever that means. Its opposite $\mathbf{C}^\mathrm{op}$ is equivalent to the following concrete category $\mathbf{D}$:
• an object of $\mathbf{D}$ is a pair $(P(X), (X,S_X))$ where $P(X)$ is the powerset of $X$ and $(X,S_X)$ is an object of $\mathbf{C}$. That is, the additional structure of an object in $\mathbf{D}$ is an object of $\mathbf{C}$.
• a morphism $f : (P(X), (X,S_X)) \to (P(Y), (Y,S_Y))$ in $\mathbf{D}$ is a function $f : P(X) \to P(Y)$ for which there exists a morphism $g : (Y,S_Y) \to (X,S_X)$ in $\mathbf{C}$ such that $f = g^{-1}$. (Note: for any given $f$ there exists at most one such $g$.)
The moral is: a general answer to a general query is generally not very useful. Of course, in particular cases there will be other, more useful, categories which are equivalent to $\mathbf{C}^\mathrm{op}$.
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Qiaochu proposed models of Lawvere theories to interpret the desired categories. I think they are (up to equivalence) the same as finitary algebraic categories. Here is my proof that $Set^{op}$ is not such a category:
Assume $Set^{op}$ is equivalent to some finitary algebraic category $C$. Then $C$ is complete and there is $x \in C$ such that every object in $C$ is isomorphic to some power $x^I$. Besides, morphisms $f : x^I \to x^J$ are induced by a unique map $\sigma : J \to I$ such that $f(a)=(a_{\sigma(j)})_{j \in J}$. Since $C$ is finitary algebraic, $x^I$ is the usual cartesian product of copies of $x$. In particular, if $d=|x|$, then every object of $C$ has cardinality $d^p$ for some cardinal number $p$. Observe that $d$ cannot be $0$ or $1$ since otherwise $Set^{op}$ and thus $Set$ had a finite skeleton. In particular, there are lots of infinite objects in $C$. However, their cardinality is restricted!
Now the key tool is the Theorem of Löwenheim-Skolem (upwards). It yields the existence of an infinite cardinal number (depending on the signature of $C$), such that every cardinal number above arises as the cardinality of an object in $C$. In particular, every sufficient large infinite cardinal number $x > 2^d$ has the form $d^p$ for some infinite cardinal number $p$. If $p < d$, then $d^p \leq 2^{dp} \leq 2^d \aleph_0 < x$, a contradiction. Thus $d \leq p$ and thus $x = d^p = 2^p$. Now, assuming GCH (which is probably not needed, but it makes the proof work), $2^p = p^+$ is regular. Thus every sufficient large cardinal number is regular, which is certainly false, since $\aleph_{\alpha + \omega}$ is singular for every ordinal number $\alpha$.
I hope it's correct.
-
Here is another reason why the opposite of a category of finitary algebras isn't itself such a thing: mathoverflow.net/questions/29442/… – Peter Arndt Sep 26 '10 at 14:31
In your request you suppose that the category is concrete i.e. has a faithul functor on $Set$ (for example particular make of structured sets and morphisms, or models (of a theory) on Set or on topological space or else), but this isn't true in general (countrexample are from Homotopy). Anyway the Yoneda lemma (and immersion) is "like a extension" that relate a category $C$ by $Set$, in this way a model of the dual $C^{op}$ as the essentially some objects, but you can "concretizing" the morphisms: in $C^<=CAT(C, Set)$ consider the subcategory generated by $h^X,\ X\in C$. However the dual of a concrete category is still concrete (considering the covariant Power-set functor $Set\to Set^{op}$.
A more hard quation is: Gived a theory $T$ and its category of model $C(T)$ in Set, there is a "dual" theory $T'$ such that $C(T)^{op}$ is embeddable in $C(T')$ ? But from elementary example (Pontriagjn duality ecc) we need also a "more high" category for the base of $T'$ models. In this philosophy there is a large study of how make categorical model of a theory, see P. Johnstone "Stone Spaces" or "Sketches of an Elephant: A Topos Theory Compendium".
- |
# Unit 1 Family Materials
Rigid Transformations and Congruence
### Rigid Transformations
This week your student will learn to describe the movement of two-dimensional shapes with precision. Here are examples of a few of the types of movements they will investigate. In each image, Shape A is the original and Shapes B, C, and D show three different types of movement:
Students will also experiment with shapes and drawings to build their intuition by:
• cutting shapes out
• tracing shapes on tracing paper to compare with other shapes
• drawing shapes on grid paper
• measuring lengths and angles
• folding paper
1. Describe how the shape changes from one panel to the next.
2. Draw a fourth panel that shows what the image would look like if the shape in the third panel is rotated 180 degrees counterclockwise around the middle of the panel.
Solution:
1. Turn it 90 degrees clockwise then move the shape to the right side.
2.
### Properties of Rigid Transformations
This week your student will investigate rigid transformations, which is the name for moves (and sequences of moves) that preserve length and angle measures like translations, rotations, and reflections. For example, in this image the triangle $$ABC$$ was reflected across the line $$AC$$ and then translated to the right and up slightly.
When we construct figures using rigid transformations, we know that the measures of the images of segments and angles will be equal to the measures of the original segments and angles.
1. Reflect triangle $$ABC$$ across side $$AC$$ to form a new triangle $$AB’C$$.
2. What is the measure of angle $$B’AC$$?
3. Name two side lengths that have the same measure.
Solution:
1.
2. 36 degrees. Angle $$B’AC$$ corresponds to angle $$BAC$$.
3. Sides $$AB’$$ and $$AB$$ have the same length as do sides $$B’C$$ and $$BC$$.
### Congruence
This week your student will learn what it means for two figures to be congruent. Let’s define congruence by first looking at two figures that are not congruent, like the two shown here. What do these figures have in common? What is different about them?
If two figures are congruent, that means there is a sequence of rigid transformations we could describe that would make one of the figures look like the other. Here, that isn’t possible. While each has 6 sides and 6 vertices and we can make a list of corresponding angles at the vertices, these figures are not considered congruent because their side lengths do not correspond. The figure on the left has side lengths 3, 2, 1, 1, 2, 1. The figure on the right has side lengths 3, 3, 1, 2, 2, 1.
For the last part of this unit, students will use the congruence to investigate some interesting facts about parallel lines and about the angles in a triangle.
1. Explain why these two ovals are not congruent. Each grid square is 1 unit along a side.
2. Draw two new ovals congruent to the ones in the image.
Solution:
1. While each oval has a horizontal measurement of 5 units and a vertical measurement of 4 units, the oval on the left’s “tallest” measurement is halfway between the left and right sides while the oval on the right’s “tallest” measurement is closer to the right side than the left side.
2. There are many possible ways to draw new ovals congruent to the original two. If a tracing of the original oval lines up exactly when placed on top of the new image (possibly after some rotation or flipping of the paper the tracing is on), then the two figures are congruent.
### Drawing Polygons with Given Conditions
This week your student will be drawing shapes based on a description. What options do we have if we need to draw a triangle, but we only know some of its side lengths and angle measures?
• Sometimes we can draw more than one kind of triangle with the given information. For example, “sides measuring 5 units and 6 units, and an angle measuring $$32^\circ$$” could describe two triangles that are not identical copies of each other.
• Sometimes there is only one unique triangle based on the description. For example, here are two identical copies of a triangle with two sides of length 3 units and an angle measuring $$60^\circ$$. There is no way to draw a different triangle (a triangle that is not an identical copy) with this description.
• Sometimes it is not possible to draw a triangle with the given information. For example, there is no triangle with sides measuring 4 inches, 5 inches, and 12 inches. (Try to draw it and see for yourself!)
2. A triangle with a side that measures 6 units and angles that measure $$45^\circ$$ and $$90^\circ$$
2. You can draw a different triangle by putting the side that is 6 opposite from the $$90^\circ$$ angle instead of next to it. This is not an identical copy of the given triangle, because it is smaller. |
# Lighting and environment mapping with GLSL
In this post we will expand on our skybox project by adding an object to our scene for which we will evaluate lighting contributions and environment mapping. We will first make a quick edit to our Wavefront OBJ loader to utilize OpenGL's Vertex Buffer Object. Once we can render an object we will create a shader program to evaluate the lighting and reflections. Below are a couple of screen grabs of the final result.
A couple of video captures are below.
The relevant modifications to our Wavefront OBJ loader are below. These methods expect normals to be specified in the obj file in addition to the faces being rendered with triangles. The setupBufferObjects method is just a quick way to load our vertices and normals into a Vertex Buffer Object once our OpenGL context has been created. We've defined a structure, v, padded to 32 bits, to store our data. We have a render method to render our model. Note that we have specified an offset for the normals using the glVertexAttribPointer function.
struct v {
GLfloat x, y, z;
GLfloat nx, ny, nz;
};
void cObj::setupBufferObjects() {
int size = faces.size();
v *vertices_ = new v[size*3];
unsigned int *indices_ = new unsigned int[size*3];
for (int j = 0, i = 0; i < size; i++) {
vertices_[j].x = vertices[faces[i].vertex[0]].v[0];
vertices_[j].y = vertices[faces[i].vertex[0]].v[1];
vertices_[j].z = vertices[faces[i].vertex[0]].v[2];
vertices_[j].nx = normals[faces[i].normal[0]].v[0];
vertices_[j].ny = normals[faces[i].normal[0]].v[1];
vertices_[j].nz = normals[faces[i].normal[0]].v[2];
indices_[j] = j;
j++;
vertices_[j].x = vertices[faces[i].vertex[1]].v[0];
vertices_[j].y = vertices[faces[i].vertex[1]].v[1];
vertices_[j].z = vertices[faces[i].vertex[1]].v[2];
vertices_[j].nx = normals[faces[i].normal[1]].v[0];
vertices_[j].ny = normals[faces[i].normal[1]].v[1];
vertices_[j].nz = normals[faces[i].normal[1]].v[2];
indices_[j] = j;
j++;
vertices_[j].x = vertices[faces[i].vertex[2]].v[0];
vertices_[j].y = vertices[faces[i].vertex[2]].v[1];
vertices_[j].z = vertices[faces[i].vertex[2]].v[2];
vertices_[j].nx = normals[faces[i].normal[2]].v[0];
vertices_[j].ny = normals[faces[i].normal[2]].v[1];
vertices_[j].nz = normals[faces[i].normal[2]].v[2];
indices_[j] = j;
j++;
}
glGenBuffers(1, &vbo_vertices);
glBindBuffer(GL_ARRAY_BUFFER, vbo_vertices);
glBufferData(GL_ARRAY_BUFFER, size*3*sizeof(v), vertices_, GL_STATIC_DRAW);
glGenBuffers(1, &vbo_indices);
glBindBuffer(GL_ELEMENT_ARRAY_BUFFER, vbo_indices);
glBufferData(GL_ELEMENT_ARRAY_BUFFER, size*3*sizeof(unsigned int), indices_, GL_STATIC_DRAW);
delete [] vertices_;
delete [] indices_;
}
void cObj::render(GLint vertex, GLint normal) {
glBindBuffer(GL_ARRAY_BUFFER, vbo_vertices);
glEnableVertexAttribArray(vertex);
glVertexAttribPointer(vertex, 3, GL_FLOAT, GL_FALSE, sizeof(v), 0);
glEnableVertexAttribArray(normal);
glVertexAttribPointer(normal, 3, GL_FLOAT, GL_FALSE, sizeof(v), (char *)NULL + 12);
glBindBuffer(GL_ELEMENT_ARRAY_BUFFER, vbo_indices);
glDrawElements(GL_TRIANGLES, faces.size()*3, GL_UNSIGNED_INT, 0);
}
void cObj::releaseBufferObjects() {
glDeleteBuffers(1, &vbo_indices);
glDeleteBuffers(1, &vbo_vertices);
}
We have also create a function, createProgram, to create our shader program which we discussed in the previous post on rendering a skybox.
void createProgram(GLuint& glProgram, GLuint& glShaderV, GLuint& glShaderF, const char* vertex_shader, const char* fragment_shader) {
glProgram = glCreateProgram();
glUseProgram(glProgram);
int vlength, flength, plength;
char vlog[2048], flog[2048], plog[2048];
glGetProgramInfoLog(glProgram, 2048, &flength, plog);
std::cout << vlog << std::endl << std::endl << flog << std::endl << std::endl << plog << std::endl << std::endl;
}
Now we can discuss our lighting model. Our lighting model will be the sum of four contributions, emissive, ambient, diffuse, and specular.
\begin{align}
\vec{color_{final}} &= \vec{color_{emissive}} + \vec{color_{ambient}} + \vec{color_{diffuse}} + \vec{color_{specular}}\\
\end{align}
The emissive color is simply the color emitted by our object.
\begin{align}
\vec{color_{emissive}} &= \vec{emissive_{color}} \cdot emissive_{contribution}\\
\end{align}
The ambient term is independent of the location of the light source, but does depend on the object's materials. Think of the material as reflecting ambient light. If our light source is white and the object is green, clearly it should only reflect the green portion of the light spectrum.
\begin{align}
\vec{color_{ambient}} &= \vec{material_{ambient}} \circ \vec{ambient_{color}} \cdot ambient_{contribution}\\
\end{align}
The diffuse term depends on the object's materials, the color of the diffuse light, and also the location of the light source. We will have two vectors at the location we are evaluating, a normal vector and a vector towards the light source. We will use the inner product of these two vectors.
\begin{align}
\vec{a} \cdot \vec{b} &= |\vec{a}||\vec{b}|\cos{\theta}\\
\end{align}
If both vectors are of unit length, we have,
\begin{align}
\vec{a} \cdot \vec{b} &= \cos{\theta}\\
\end{align}
where $$\cos{\theta}$$ will range from $$[-1,1]$$. We are not interested in values less than $$0$$ because this would indicate an angle between the normal and light vector of more than $$\frac{\pi}{2}$$ radians. When the vectors are parallel and have the same direction, we have a maximum contribution of $$1$$ for the diffuse light. For the normal, $$\vec{n}$$, and the light vector, $$\vec{l}$$, we have,
\begin{align}
\vec{color_{diffuse}} &= \vec{material_{diffuse}} \circ \vec{diffuse_{color}} \cdot \max{(\hat{n} \cdot \hat{l},0)} \cdot diffuse_{contribution}\\
\end{align}
The final specular term depends on the object's materials, the color of the specular light, the location of the light source, and also the location of the viewer. We will implement the Blinn-Phong Shading Model. To do this we need to evaluate the halfway vector between the light vector, $$\vec{l}$$, and the vector towards the viewing position, $$\vec{v}$$,
\begin{align}
\hat{h} &= \frac{\hat{l}+\hat{v}}{|\hat{l}+\hat{v}|}\\
\end{align}
provided $$|\hat{l}+\hat{v}| \ne 0$$. The dot product evaluated for the diffuse contribution indicates whether we apply a specular reflection. If that dot product is greater than $$0$$, we evaluate the specular contribution,
\begin{align}
\vec{color_{specular}} &= \vec{material_{specular}} \circ \vec{specular_{color}} \cdot (\hat{n} \cdot \hat{h})^\alpha \cdot specular_{contribution}\\
\end{align}
where $$\alpha$$ is the shininess constant.
Before we delve into our shader program, we need to discuss something related to coordinate spaces. Our lighting calculations require all of our vectors and vertices to exist in the same coordinate space. We will pass our model, view, and projection matrices to our shader, and use the model and view matrices to transform our coordinates into eye space. However, due to the nature of nonuniform scaling, applying the model and view matrices to our normals may yield vectors that are no longer normal to the surface. We will instead use the inverse of the transpose of the model view matrix as described below.
For our tangent, $$\vec{t}$$, and normal, $$\vec{n}$$, and transformed $$\vec{t'}$$ and $$\vec{n'}$$, all with homogeneous coordinate $$0$$, we have,
\begin{align}
\vec{t} \cdot \vec{n} &= 0\\
\vec{t'} \cdot \vec{n'} &= 0\\
\end{align}
so if $$\mathbf{M}$$ is our model view matrix and $$\mathbf{N}$$ is the matrix we are seeking to transform our normals, we have,
\begin{align}
(\mathbf{M}\vec{t})\cdot(\mathbf{N}\vec{n}) &= 0\\
(\mathbf{M}\vec{t})^T(\mathbf{N}\vec{n}) &= 0\\
\vec{t}^T\mathbf{M}^T\mathbf{N}\vec{n} &= 0\\
\end{align}
and if $$\mathbf{M}^T\mathbf{N} = \mathbf{I}$$, we have,
\begin{align}
\vec{t}^T\mathbf{M}^T\mathbf{N}\vec{n} &= 0\\
\vec{t}^T\mathbf{I}\vec{n} &= 0\\
\vec{t}^T\vec{n} &= 0\\
\end{align}
thus,
\begin{align}
\mathbf{M}^T\mathbf{N} &= \mathbf{I}\\
\mathbf{N} &= (\mathbf{M}^T)^{-1}\\
\end{align}
So when transforming normals, we will use the inverse of the transpose of the model view matrix. Let's have a look at our shader. Our vertex shader will accept a vertex and the normal in addition to our light position and projection, view, and model matrices. It will pass the light vector, normal vector, halfway vector, and texture coordinate to the fragment shader. Note that our texture coordinate has three coordinates for sampling from our cube map for environment mapping. In our vertex shader we first transform the incoming vertex; we also transform that vertex into eye space. Our light position uniform is already specified in model space, so we simply apply the view matrix to move it to eye space. We evaluate the light vector as the vector from the vertex to the light source in addition to the normal and halfway vectors using the equations above. We transform our normal vector to model space to get our texture coordinates.
#version 330
in vec3 vertex;
in vec3 normal;
uniform vec3 light_position;
uniform mat4 Projection;
uniform mat4 View;
uniform mat4 Model;
out vec3 light_vector;
out vec3 normal_vector;
out vec3 halfway_vector;
out vec3 texture_coord;
void main() {
gl_Position = Projection * View * Model * vec4(vertex, 1.0);
vec4 v = View * Model * vec4(vertex, 1.0);
vec3 normal1 = normalize(normal);
light_vector = normalize((View * vec4(light_position, 1.0)).xyz - v.xyz);
normal_vector = (inverse(transpose(View * Model)) * vec4(normal1, 0.0)).xyz;
texture_coord = (inverse(transpose(Model)) * vec4(normal1, 0.0)).xyz;
halfway_vector = light_vector + normalize(-v.xyz);
}
Our fragment shader accepts the normal, light, and halfway vectors in addition to the texture coordinates and cube map. We sample the cube map to get the object's material property and specify colors and contributions for the emissive, ambient, diffuse, and specular components. Lastly, we apply our lighting equations from above to output a fragment color. Note that our colors and contributions are built into our shader. We could have specified these as uniforms to make our shader a bit more configurable.
#version 330
in vec3 normal_vector;
in vec3 light_vector;
in vec3 halfway_vector;
in vec3 texture_coord;
uniform samplerCube cubemap;
out vec4 fragColor;
void main (void) {
vec3 normal1 = normalize(normal_vector);
vec3 light_vector1 = normalize(light_vector);
vec3 halfway_vector1 = normalize(halfway_vector);
vec4 c = texture(cubemap, texture_coord);
vec4 emissive_color = vec4(0.0, 1.0, 0.0, 1.0); // green
vec4 ambient_color = vec4(1.0, 1.0, 1.0, 1.0); // white
vec4 diffuse_color = vec4(1.0, 1.0, 1.0, 1.0); // white
vec4 specular_color = vec4(0.0, 0.0, 1.0, 1.0); // blue
float emissive_contribution = 0.02;
float ambient_contribution = 0.20;
float diffuse_contribution = 0.40;
float specular_contribution = 0.38;
float d = dot(normal1, light_vector1);
bool facing = d > 0.0;
fragColor = emissive_color * emissive_contribution +
ambient_color * ambient_contribution * c +
diffuse_color * diffuse_contribution * c * max(d, 0) +
(facing ?
specular_color * specular_contribution * c * pow(dot(normal1, halfway_vector1), 80.0) :
vec4(0.0, 0.0, 0.0, 0.0));
fragColor.a = 1.0;
}
This shader program yields some nice results, but it only supports one light source and is not very configurable. If we spent some more time, we could create a shader that supports multiple light sources with configurable properties for each source.
1. Ramael Odisho
Hi Keith! I'm not sure but I think I may have found small text errors. Press Ctrl + F and search for "came" and after that "transpose". By "came" you probably meant "same? And by "transpose" you probably meant "transformed"?
Because first you wrote "...use the inverse of the transpose of the model view matrix..." and in a later sentence you wrote "...we will use the inverse of the transform of the model view matrix.". Still I'm not sure but I just wanted to point that out for now anyway 🙂
Thanks for the tutorial. I'm trying to understand it now.
1. Post
Author
keith
Thanks for the heads up. I've fixed it. It should be "transpose" in the second case. Both sentences should reference ".. the inverse of the transpose of ..".
1. Ramael Odisho
Alright thanks 🙂
1. Poornima Sundararaman
Hey could you help me convert the shaders in glsl 1.20 specification ?
1. Post
Author
keith
Hey Poornima..
I'm a bit rusty at the moment, but if you want to paste them in here, I'll have a peek.
2. Ramael Odisho
Another question about both the skybox tutorial and this one. Why are there so many .tga-capturings in the folder? From what I have seen and tested those are not needed to run the program.
1. Post
Author
keith
You're right. They aren't required.. the 'g' and 'v' keys grab a frame and a sequence of frames, respectively. You can disable those in the SDL_PollEvent loop.
3. SimmerChan
Hello Keith! I'm new to linux. I've downloaded your source code and configured all the required lib and package in my vm linux. After typing 'make -f Makefile' in CLI, all went well and file 'main' was created in 'bin'. However, I ran the 'main' in CLI and it threw something that I could't recognize. Text truncated is as follows:
Name: media/dragon_smooth.obj
Vertices: 0
Parameters: 0
Texture Coordinates: 0
Normals: 0
Faces: 0
2.1 Mesa 10.1.3
1.20
1.10.0
GL_ARB_multisample GL_EXT_abgr GL_EXT_bgra GL_EXT_blend_color GL_EXT_blend_minmax GL_EXT_blend_subtract GL_EXT_copy_texture GL_EXT_polygon_offset GL_EXT_subtexture GL_EXT_texture_object GL_EXT_vertex_array GL_EXT_compiled_vertex_array GL_EXT_texture GL_EXT_texture3D GL_IBM_rasterpos_clip GL_ARB_point_parameters GL_EXT_draw_range_elements GL_EXT_packed_pixels GL_EXT_point_parameters GL_EXT_rescale_normal GL_EXT_separate_specular_color GL_EXT_texture_edge_clamp GL_SGIS_generate_mipmap GL_SGIS_texture_border_clamp GL_SGIS_texture_edge_clamp GL_SGIS_texture_lod GL_ARB_framebuffer_sRGB GL_ARB_multitexture GL_EXT_framebuffer_sRGB GL_IBM_multimode_draw_arrays GL_IBM_texture_mirrored_repeat GL_ARB_texture_cube_map GL_ARB_texture_env_add GL_ARB_transpose_matrix GL_EXT_blend_func_separate GL_EXT_f
At the end, it says "Segmentation fault (core dumped)".
I used GDB to find out the error and it said :
Program terminated with signal SIGSEGV, Segmentation fault.
#0 0x0000000000407a84 in setupCubeMap(unsigned int&, SDL_Surface*, SDL_Surface*, SDL_Surface*, SDL_Surface*, SDL_Surface*, SDL_Surface*) ()
(gdb) bt
#0 0x0000000000407a84 in setupCubeMap(unsigned int&, SDL_Surface*, SDL_Surface*, SDL_Surface*, SDL_Surface*, SDL_Surface*, SDL_Surface*) ()
#1 0x00000000004032cd in main ()
And in main.cc file, I didn't alter anything. |
# VCard Splitter
The very first problem after buying a new nokia E51: how to import contacts from my old Motorla. Solved with KMobileTools and a little Perl script.
## The problem
Nokia puts into their Symbian phones an utility called Transfer. It allows you to import your contacts list from a compatible Nokia device. But as my previous phone was a Motorola, this couldn't work. Without Windows at hand I could not even check if the PC Suite has any means to deal with this.
In my Linux environment there's the standard utility to deal with all cellulars: KMobileTools. There's also something called GNokii for GNOME users. KMobileTools is pretty universal and reliable - worked without glitch on all my previous phones. It allows you to import/export your address book, messages and to make calls/write messages directly from KDE. But it can't put your contacts into a S60 device (probably other Symbians too).
## The idea
Virtually all phones today (Motorola F3 is the only exception that comes to mind) can receive VCards. The E51 is no exception here - it does even take them from the memory card.
KMobileTools has to somehow store contacts on the disk. It turns out, that the format it uses is a flat file of newline delimited VCards. It even does simple version control, so you never lose any contacts.\ Nokia can not read such a file directly. Only the first contact on the list gets added.
## The solution
Overall, the procedure is quite simple. It comes down to few simple steps:
1. Import your phonebook from previous phone with KMobileTools
2. Copy the flat file into working directory:\ cp ~/.kde/share/apps/kmobiletools/kmobiletools.vcf .
3. Create the script copy-pasting from below. Sometimes it turns out, that name and surname come switched - if so, uncomment the middle line (remove the leading #):\ kate splitter.pl
4. Run it, this will create a subfolder called vcards:\ perl splitter.pl
5. Copy it into the phone's memory card.
6. On the phone run: Main menu -> Office -> File Manager. Navigate to the folder you just uploaded. Clicking on each vcf file will open the contact info, another two clicks will get it imported. Iterating through all the files can take some time - something to do while commuting... |
# The intelligence quotients (IQs) of 16 students from one area of a city showed a mean of 107 and a standard deviation of 10, while the IQs of 14 students from another area of the city showed a mean of 112 and a standard deviation of 8. Use alpha = 0.01. What is the pooled standard deviation?
The intelligence quotients (IQs) of 16 students from one area of a city showed a mean of 107 and a standard deviation of 10, while the IQs of 14 students from another area of the city showed a mean of 112 and a standard deviation of 8. Use alpha = 0.01. What is the pooled standard deviation?
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wintern90
${n}_{1}=16$
${n}_{2}=14$
$\overline{{x}_{1}}=107$
$\overline{{x}_{2}}=112$
${s}_{1}=10$
${s}_{2}=8$
Pooled standard deviation is defined as an estimate of standard deviation when different groups are assumed to have the same standard deviation.
The pooled variance is given by:
${S}_{p}^{2}=\frac{\left({n}_{1}-1\right){s}_{1}^{2}+\left({n}_{2}-1\right){s}_{2}^{2}}{{n}_{1}+{n}_{2}-2}\phantom{\rule{0ex}{0ex}}=\frac{\left(16-1\right){10}^{2}+\left(14-1\right){8}^{2}}{16+14-2}\phantom{\rule{0ex}{0ex}}=\frac{2332}{28}\phantom{\rule{0ex}{0ex}}=83.2857$
The pooled sd is given by:
$sd=\sqrt{83.2857}\phantom{\rule{0ex}{0ex}}=9.1261$
Darian Hubbard |
# A Companion to Analysis: A Second First and First Second Course in Analysis
###### T. W. Körner
Publisher:
American Mathematical Society
Publication Date:
2004
Number of Pages:
590
Format:
Hardcover
Series:
Price:
79.00
ISBN:
0-8218-3447-9
Category:
Textbook
[Reviewed by
Kenneth A. Ross
, on
02/1/2005
]
This review should be thought of as a companion to Steve Krantz's review of the same book in the October 2004 issue of the Monthly. Krantz loves the book because Körner is a superb author and his love of analysis shows throughout the book. The author's style is delightful and his British humour shows through. After telling us about several problems with Riemann integration, he gives us the good news: "Fortunately all these difficulties vanish like early morning mist in the sunlight of Lebesgue's theory." But Krantz acknowledges that the book would be a difficult book to use as a text, because of its idiosyncrasies, including its unusual organization. Krantz also rues the absence of some basic ideas in analysis such as the Baire category theorem. Readers of this review might want to print this review now and then go read Krantz's review first.
I'll give a more parochial review and try to answer the question: Who, among American students and mathematicians, is the book really for? The author says: "Although I hope this book may be useful to others, I wrote it for students to read either before or after attending the appropriate lectures." I agree that it's not for most classrooms. For example, there are no answers to the exercises, though the website http://www.dpmms.cam.ac.uk/~twk/ has some answers for the many exercises in Appendix K. In spite of the confusing subtitle of the book, in a footnote on page 62 the author himself claims that this "is a second course in analysis." I believe that in Britain a "first course in analysis" must be a calculus course done more rigorously than in the U.S.
This book is rich and meaty, and Körner wants readers to fall in love with analysis. It will help if they are start out at least infatuated. It dwells on the nuances and subtleties. As he notes in his Preface, "I have not tried to strip the subject down to its bare bones. A skeleton is meaningless unless one has some idea of the being it supports." I found the author's comments in the Preface about other books interesting. I too cut my teeth on Hardy's Pure Mathematics. I can understand why he keeps Karl Stromberg's book Introduction to Classical Real Analysis on his desk. My friend Karl also loved analysis and this shows in his book. I love analysis too, but many students do not. This is why my little book, Elementary Analysis: The Theory of Calculus, just provides what the students must know to survive future analysis courses.
The textual part of the book breaks very roughly into three parts. The first five chapters cover the core material in a first course on real analysis in the U.S. The next five chapters cover calculus of several variables. The remaining four chapters focus on metric space ideas with applications to functions of several variables and related topics.
In the first part, the author emphasizes that algebra can just as well be done in the setting of the field of rationals, Q, as on the field of real numbers, R. Real analysis is the mathematics that can be done on R, but not on Q. He is very careful to indicate which results are analysis (are true on R but not on Q) and which are not. It is interesting that the Fundamental Theorem of Algebra (all non-constant complex polynomials have roots) is analysis, not algebra. For example, the equation z2 = 2 has a solution in R + iR but not in Q + iQ (pages 109-110). A related theme throughout the book is the emphasis on various generalizations of R, or Q if you prefer, both in the algebraic realm (fields, etc.) and the analytic realm (normed linear spaces, etc.).
In the second part, I especially liked the treatment of Riemann integrals in Chapter 8, and Chapter 9 includes a nice expository introduction to the ideas of the Lebesgue integral.
I return to the issue of the book's organization. There are too many appendices including the giant Appendix K with 345 exercises. Appendix K starts out with 11 codes for the exercises. Many of the exercises are very interesting and some of them are quite challenging. Many of them dip into more advanced topics providing bare-bones glimpes that may or may not be illuminating, depending on the reader's knowledge and sophistication.
The organization of the book may be the organization of the future as people become completely comfortable with non-linear reading and studying, hopping from website to website and so on. As a linear reader, I was dismayed to find relevant items in Appendix K that would have been illuminating earlier. Sometimes the author mentions the relevant exercises, sometimes not. It would have helped, as Krantz noted, if these exercises had been at the ends of the appropriate chapters. It also would have helped if there had been a tree of guidance at the beginning of the book suggesting avenues for readers. Finally, frequently Examples, Exercises and even Definitions are inserted between two paragraphs that are clearly intended to be read sequentially.
Several minor corrections are provided on the website cited above. In addition, I found the footnote on page 39 puzzling, since I expected it to be about Weierstrass.
Kenneth A. Ross (ross@math.uoregon.edu) taught at the University of Oregon from 1965 to 2000. He was President of the MAA during 1995-1996. Before that he served as AMS Associate Secretary, MAA Secretary, and MAA Associate Secretary. His research area of interest was commutative harmonic analysis, especially where it has a probabilistic flavor. He is the author of the book Elementary Analysis: The Theory of Calculus (1980, now in 14th printing), co-author of Discrete Mathematics (with Charles R.B. Wright, 2003, fifth edition), and, as Ken Ross, the author of A Mathematician at the Ballpark: Odds and Probabilities for Baseball Fans (2004).
* The real line
* A first philosophical interlude
* Other versions of the fundamental axiom
* Higher dimensions
* Sums and suchlike $\heartsuit$
* Differentiation
* Local Taylor theorems
* The Riemann integral
* Developments and limitations of the Riemann integral $\heartsuit$
* Metric spaces
* Complete metric spaces
* Contraction mappings and differential equations
* Inverse and implicit functions
* Completion
* Appendices
* Executive summary
* Exercises
* Bibliography
* Index |
# Negative certainty equivalent
Let us consider an agent of initial wealth $$w_0$$ whose utility function is $$u(x)=\sqrt{x}$$. This individual faces a risk of loss $$Z$$ which occurs with probability $$p$$.
It is assumed that $$w_0=60000$$, $$Z=10000$$ and $$p=0.1$$. What is the certainty equivalent for the risk incurred by the individual? What is his risk premium? Interpret.
For the certainty equivalent, I have found:
$$\mathbb{E}(u(w_0+L))=u(w_0+c)\iff c=(0.1\sqrt{50000}+0.9\sqrt{60000})^2-60000\approx-1040.99$$
How can this result be interpreted?
$$\mathbb{E}(u(w_0+L))=u(w_0+\mathbb{E}(L)-\pi)\iff \pi\approx60040.99$$
Thanks.
• "How can this result be interpreted?" I am not sure what you are asking. A negative certainty equivalent is what you would expect for a possible loss. Or are you asking what a certainty equivalent generally is? Does your course give an explanation? Nov 20, 2021 at 14:22
• Sorry, I'm French so it's a little bit hard to explain and translate in English. In my course is written that a certainty equivalent is "the monetary amount for which the economic agent is indifferent between the lottery and this amount," thus the minimum price for which he would accept to free himself from the lottery. Actually, I don't understand how this definition can make sense with a negative certainty equivalent. Thanks. Nov 20, 2021 at 14:29
The certainty equivalent in your example is $$w_0+c$$, this certain payoff's utility is equivalent with the lottery's.
The amount $$c$$ is not the certainty equivalent, but the amount the consumer is willing to forego in expected value in exchange for the certainty. Perhaps $$-c$$ is what you call the risk premium.
• Thanks for the answer. Sorry for the delay, I didn't have access to my computer. Thus the certainty equivalent is $w_0+c=(0.1\sqrt{50000}+0.9\sqrt{60000})^2\approx58959$ I can't understand the logic here. Does this mean the agent is willing to pay $58959 to get rid of the risk? Nov 25, 2021 at 15:17 • No. This certain payoff's utility is equivalent with the lottery's. This is the smallest certain payoff the consumer would be willing to trade his uncertain lottery based cashflow for. Nov 25, 2021 at 17:36 • Since his initial wealth is 60,000 dollars, I assume the consumer is willing to lose 1,041 dollars (the difference between his initial wealth and the certainty equivalent) to trade his uncertain lottery. Then how would you interpret the risk premium (which is defined as the difference between the mathematical expectation of the lottery and its certainty equivalent)?$\pi=\mathbb{E}(w_0+L)-(w_0+L)\approx59000-58949\approx41\$ I am totally lost with all these terms, and my course has no examples. Thanks! Nov 27, 2021 at 11:55 |
# Path connectedness of a particular algebraic set
Let $n\geq 3$, let $a$ and $b$ be two positive numbers, and
$$\Omega = \bigg\lbrace (x_1,x_2, \ldots ,x_n) \in ]0,+\infty[^n \ \bigg| \ x_1+x_2+ \ldots +x_n=a, \ x_1x_2 \ldots x_n=b \bigg\rbrace$$ It is easy to see that $\Omega$ is nonempty iff $\frac{a}{n} \geq b^{\frac{1}{n}}$. It seems clear enough that $\Omega$ is path connected, but I could find no simple proof of that.
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Assume $\frac{a}{n} \geq b^{1/n}$ so that $\Omega \neq \emptyset$.
Let $H$ be the hyperplane of $\mathbb{R}^n$ defined by the equation $x_1+x_2+\cdots +x_n = a$, and let $B = \{ (x_i) \in (0, \infty)^n \;|\; x_1x_2 \cdots x_n \geq b\}$. It should be clear that $\Omega = H \cap \partial B = \partial(H \cap B)$. Let's first consider the space $U = H \cap B$. You can show that $B$ is convex of dimension $n$, and since $H$ is a hyperplane, we have that $U$ is convex of dimension $\leq n-1$, hence path-connected. Furthermore, $U$ is bounded, since $H \cap B \subseteq (0,a)^n$
Now if $dim(U) < n-1$, then $U$ is a single point because $\partial B$ contains no line segment (see below for more details on this). In that case, $\Omega$ is definitely connected. On the other hand, when $dim(U) = n-1$, convexity and boundedness imply that $\partial U$ is homeomorphic to $S^{n-2}$. Since you have $n \geq 3$, we find that $\Omega = \partial U$ is connected.
Lemma: In $\mathbb{R}^n$, If a hyperplane $H$ intersects a convex set $B$ of dimension $n$ in such a way that there is an interior point $x$ of $B$ in the intersection, then $dim(H \cap B) = n - 1$.
Proof: If $x$ is interior to $B$, then there is an $n$-ball $V_x \subset B$ containing $x$. Suppose $V_x = \{ y \in \mathbb{R}^n \;|\; dist(x,y) \leq r\}$. Now $V_x \cap H = \{ y \in H \;|\; dist(x,y) \leq r\}$. Thus, $V_x \cap H$ is an $(n-1)$-dimensional ball in $H \cap B$.
Cor: If $dim(H \cap B) < n-1$, then $H \cap B \subseteq H \cap \partial B$.
Lemma: If $B \subseteq \mathbb{R}^n$ is convex and $H$ is a hyperplane such that $x \in H \cap B \subseteq H \cap \partial B$, and $\partial B$ contains no line segment, then $H \cap \partial B = \{ x\}$.
Proof: If $y \neq x$ is another point of $H \cap \partial B$, then we have two points $x, y \in H \cap B$, a convex set. Thus the line segment $\ell$ joining $x$ and $y$ is in $H \cap B \subseteq H \cap \partial B$. In particular, $\ell \in \partial B$, contradicting that $\partial B$ contains no segment.
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Nice solution! I’ll take some time to check the details before accepting it. – Ewan Delanoy Dec 30 '12 at 17:59
"codimension 1 hyperplane" is redundant, right? – Ewan Delanoy Dec 30 '12 at 17:59
To explain why $H$ is bounded, I would simply say that it is bounded because $H$ is (indeed, $H$ is included in the hypercube $[0,a]^n$). – Ewan Delanoy Dec 30 '12 at 18:04
@EwanDelanoy you're correct on both of your suggestions... "hyperplane" is sufficient. And $H$ bounded is sufficient. – Shaun Ault Dec 30 '12 at 18:27
Actually there are two different "H"s there : the hyperplane defined by the equation $x_1+x_2+ \ldots x_n=a$ and its intersection with $(0,\infty)^n$ (perhaps you could denote this by $H’$) – Ewan Delanoy Dec 31 '12 at 6:15 |
+0
# More Algebra 2
0
117
3
+287
1.The function f(x) is invertible, but the function g(x) = kf(x) is not invertible. Find the sum of all possible values of k.
2.If f(x) and (ax+b)/(cx+d), abcd ≠ 0 and f(f(x)) = x for all x in the domain of f, what is the value of a+d?
Dec 17, 2019
edited by EpicWater Dec 17, 2019
edited by EpicWater Dec 17, 2019
#1
+120
+1
1. The statement of this problem is not very clear. Is k a constant? or k is another function and g(x) =(kf)(x), the product of the functions k and f? If k is a constant, then the only value of k that makes g(x)=kf(x) not invertible is k = 0, and it makes no sense to ask for the sum of all possible values of k since there is only one value, that is 0. If k is a nonzero constant, then g always has an inverse; in fact $$g^{-1}(x)=f^{-1}(\frac{x}{k})$$. For instance, we know that if $$f(x)=x^3$$, then $$f^{-1}(x)=\sqrt[3] {x}$$. The inverse of $$g(x)=kx^3$$ is just $$g^{-1}(x)= \sqrt [3]{\frac{x}{k} }$$, assuming k is not equal to zero.
If on the other hand k is a function and g(x) is the product of f and k, again the question makes no sense since there are lots of functions that when multiplied by f give a function that is not invertible. Two examples: $$g(x)=x^3\cdot \frac{1}{x}$$, where $$f(x)=x^3\ and \ k(x)= \frac{1}{x}$$; and $$g(x)=x^3 \cdot \frac{x+1}{x}$$, where $$f(x)=x^3$$ and $$k(x)= \frac{x+1}{x}$$. In both cases g is not invertible because it is not 1-1.
2. Here I will assume you mean $$f(x)= \frac{ax+b}{cx+d}$$. The answer to this question is: $$a+d=0$$. There is quite a bit of Algebra involved in showing this and it starts by setting $$f(f(x))=x$$, which means f is its own inverse. We have
$$f(f(x))=f(\frac{ax+b}{cx+d} )= \frac{a(\frac{ax+b}{cx+d})+b }{c(\frac{ax+b}{cx+d})+d } =\frac{a(ax+b)+b(cx+d)}{c(ax+b)+d(cx+d)} =\frac{x}{1}$$;
If you cross-multiply and combine terms you get,
$$(a^2+bc)x+(ab+bd)=(ca+dc)x^2+(cb+d^2)x$$.
Setting the coefficients of the polynomials on the two sides of the equation equal, we get
$$ab+bd=0$$, or $$b(a+d)=0$$,
$$ca+dc=0$$, or $$c(a+d)=0$$
$$a^2+bc=bc+d^2$$, or $$a^2-d^2=0$$.
The first two equations tell us that $$a+d=0$$, since neither b nor c could be zero (one of the assumptions is that $$abcd \neq 0$$)
Dec 17, 2019
#2
+287
0
I'm sorry about the first question. Yes, "k" is a constant. Thank you for the help. Hope that clears things up!!
Dec 17, 2019
#3
+287
0
I understand the second question and redid it myself. But not really the first. So what is the sum of all possible values of the constant k? Clarification would be appreciated. Thank you!!!
Dec 17, 2019 |
# Trigonometrical or Polar representation of Complex Numbers
## Trigonometrical or Polar representation of Complex Numbers
Let z = x + iy is a complex number which is denoted by a point P (x, y) in a complex plane, then OP = |z| and ∠POX = θ = arg (Z).
In ΔPOM,
$$\cos \theta =\frac{OM}{OP}$$,
$$\cos \theta =\frac{x}{|z|}$$,
Then x = |z| cosθ and θ.
$$\sin \theta =\frac{PM}{OP}$$.
$$\sin \theta =\frac{y}{|z|}$$.
Then y = |z| sinθ
We know that
Z = x+ iy
Z = |z| cosθ + i |z| sinθ
Z = |z| (cosθ + i sinθ)
Z = r (cosθ + i sinθ)
Where r = |z| and $$\theta ={{\tan }^{-1}}\left( \frac{y}{x} \right)$$,
This form of z is known as polar form
In general, polar form is z = r [cos (2nπ + θ) + i sin (2nπ + θ)].
Where,
r = |z|, θ = arg (z) and n ϵ N. |
Categories
TurtleCoin® v2 Overview
In the early days, at least two forevers ago, we blew the dust off a legacy Bytecoin codebase and launched a coin using just enough special sauce to call TurtleCoin our own, all on a whim. As a new network, there were the usual struggles, like when ASIC for CryptoNote emerged, or when we found a pretty serious RPC issue in Simplewallet.
As our team and their contributions grew, we eventually found ourselves in a new codebase, rock solid, but still carrying the scars of our past.
Fast forward to now, we are 3 million blocks wiser, our team is many lines of code more experienced, and as we have chiseled our way through, we made a list of a some areas where we can do better. Taking a step back, the last few years have shown us that we are reaching the limit of our current software, and the time is now for us to think radically different and put something new on the table. It is time to build something of our own.
With great respect and thanks to the CryptoNote family, Forknote, Bytecoin, and those who enabled us to get where we are now, we humbly begin our own path and embark on a new journey. What we produce next will be from scratch, written by us, exactly how we have always wanted it.
TurtleCoin v2 is the future, and we are excited to share it with you!
The Goal
We aim to redesign TurtleCoin from the ground up to avoid the mistakes of the past while putting an eye to the future state of the network. This is accomplished by replacing traditional Proof of Work (PoW) based mechanisms with a round-based hybrid Proof of Stake (PoS)/PoW model. We’re going to be talking about some of the fundamentals required in this model so buckle your seat belt and be ready for a ride. Be warned, there’s a bit of math below.
Proof of Work
The traditional approach to cryptocurrency networks is to deploy a PoW model such that miners perform endless repetitive hashing of block data searching for a hash that meets a set difficulty. The miner is rewarded for burning fossil fuels via a specially constructed transaction (the “coinbase transaction”) that allows the miner to create new funds as a result of mining that block.
Proof of Stake
PoS deviates from the PoW consensus model in that PoW is typically no longer performed, instead passing the torch to block producers (the “producers”) whose work is then validated by the block validators (the “validators”). These parties are typically elected via users staking, or otherwise backing, their preferred candidates for those roles. Selection of producers and validators typically takes place via an election mechanism as part of the consensus rules.
Our Vision, Round Based Hybrid Delegated PoS w/ PoW
We are planning on combining the best of PoW with the stability and speed of PoS to create one kick ass network. We’ll use DPoS for managing the distributed ledger and PoW as part of the transaction construction and validation process.
Operational Overview
When a user wishes to send a transaction, they generate the transaction in the “normal” way. However, there is a nonce field included in the transaction that requires for the user to perform PoW for that singular transaction to meet a preset difficulty for transactions. The resulting transaction and the nonce is then forwarded to a node for broadcast to the network.
Every transaction will require a base amount of work (PoW) before it can be submitted to the network in addition to a fee-per-byte network transaction fee. However, if a user wishes to discount their transaction fee for a particular transaction on the network they may submit a PoW for the transaction that meets yet a higher tier of difficulty. While the discount tiers have yet to be defined, the maximum discount for the transaction fees will not exceed 50% of the fee-per-byte fee. There are no free rides.
Fusion transactions will no longer exist on the network. As a wise man once said, “ass, cash, or hash, no one rides for free”. Everyone pays their fair share for transacting on the network regardless of the reason for that transaction.
Once a transaction is received by a node, the node performs basic checks (i.e.. transaction construction, valid nonce, PoW meets required difficulty, verifies signatures, etc.). Once that node completes the basic verification, it forwards the transaction to the rest of the network including the producers. Producers hold the transactions in the global memory pool after performing their own validations of the transaction ensuring that it is indeed a valid transaction. At this point, users can presume some level of assurance that the transaction will be committed to a block solely because the producers have accepted the transaction into the pool.
Once enough PoW transactions have accumulated in the pool to satisfy the network difficulty, the then current producer creates a block that ratifies the selection of transactions necessary to meet (or exceed) the network difficulty requirement. When doing so, the producer creates a block reward equivalent to a 1 for 1 matching of the aggregate transaction fees included in the block.
This means that emission is directly tied to the actual use of the network as a transactional platform. We will no longer be emitting currency on a pre-defined schedule as a result of this change. The more the network is used, the more currency that will become available.
The block reward is distributed proportionately between the block producer, the validators, and a subset of the users who staked that producer. Upon completion of the block the block is signed by the producer and transmitted to the network.
The validators, upon receiving the block, validate (at minimum) that:
• The producer was permitted to produce the block (their turn)
• Enough PoW transactions have been included to meet network difficulty
• Transaction hashes included in the block have been included such that the hashes are arranged in descending order
• The block is properly signed by the producer
• The transactions in the block meet all of the consensus rules of the network (no double spends, etc.); and
• No unnecessary data is included in the block
• That the producer claimed and paid out the proper block rewards.
A minimum of 51% of the validators must then sign the block and submit their signatures to the network thus ratifying the block itself.
After the block is ratified, the network difficulty is adjusted (if necessary) to account for things such as the number of transactions included, the PoW of transactions included, and any other metrics required to keep the network operating correctly.
No Empty Blocks Means Storage Saving Is Built In
One of the largest benefits in this scenario is that we no longer produce empty blocks and thus save tons of storage space. This makes the chain faster to sync, easier to store, and overall much easier to manage.
Tut, Tut: The Reward Lottery
Gone are the times when a single miner/pool received the entirety of a block’s reward. In this new model, the block reward is distributed to multiple recipients via the payments to the producer, a pseudo random validator, and a pseudo random subset of users who have staked the given producer and validator.
The Reward, Divided
We’re still playing with the numbers but there’s a pretty good chance that the block rewards will be divided as follows:
• 20% to the producer of the block
• 20% to the one of the validators of the block
• 30% to the users who staked the producer
• Maximum of 10 randomly selected stakers
• 30% to the users who staked the validator
• Maximum of 10 randomly selected stakers
Becoming a Block Producer / Validator
A user (“Alice”) wishing to operate a block producer or validator “announces” their intent for candidacy by staking their node(s) with a minimum stake of a yet to be determined amount of TRTL. Upon successfully staking their node, Alice’s node is now a candidate for either a producer or validator spot in a round. It is important to note that Alice simply staking her node as a candidate does not automatically make Alice a block producer or a validator.
To be a candidate for either role within the network, Alice must be “endorsed” by other users of the network via users staking their funds in support of Alice. Once Alice has at least 1 TRTL staking her, she is now a candidate in the running for a producer or validator spot.
The details of how these stakes are sent, recorded, recalled, etc. will be discussed in further articles and will undoubtedly involve a lot of math. Don’t worry, we’ll warn you ahead of time so that you can bring a protractor and a compass.
Electing Producers & Validators
Producers & Validators are elected for spans of blocks such that each producer is entrusted to create ten (10) new blocks in the round for which they are elected. How they are elected into these rounds takes place as follows.
In the event that a producer or validator fails to perform the work that was entrusted to them, their “trust” rating will fall for each block that they fail to producer and/or submit a validation result. If their trust rating falls below the lower threshold (TBD), the stake they put forth to submit their node as a candidate will be locked indefinitely (aka. burned).
First, we take the hashes of every block in the now closed round $\{B_1, B_2, B_3, \dots\}$ and calculate the Merkle root for those hashes to establish $M$ as the election seed for the next round.
We then take $M$ and convert that to a scalar $p$:
$p = H_s(M)$
Then compute the public key $P$ of $p$:
$P = pG$
We then tally $T$ the individual bytes of $P$ to determine if the result is odd or even.
$T = \{P_1 + P_2 + \dots\} \mod 2$
• If $T$ is odd ($T \equiv 1$) we collect public keys from the candidates list that are $T \geq P$.
• If $T$ is even ($T \equiv 0$), we collect public keys from the candidates list that are $T \leq P$.
Once the collection of public keys in the new defined range have been collected. We review the number of votes (stakes) received for each public key and construct a listing of candidates with their tally of active votes. The list is then sorted by vote count in ascending order and the top and bottom candidates in the results are discarded. The total votes submitted for the remaining candidates are then tallied to arrive at a total votes casted count ($V$).
The resulting candidates are then assigned ranges for which the selection algorithm must land for them to become a producer/validator.
If the remaining candidates consist of:
Candidate 1: 11,583 votes
Candidate 4: 127,342 votes
The the ranges for each node are as follows:
Candidate 1: 0 - 11583
Candidate 2: 11,584 - 25,749
Candidate 3: 25,750 - 99,000
Candidate 4: 99,001 - 127,342
We then perform the selection of the first producer via $e_1 = p \mod V$. The value of $e_1$ is compared against the node ranges thereby establishing that node as a producer.
On the next iteration of the election process, the previously selected node ($e_n$) is removed from the collection and $V$ is recomputed before the selection algorithm runs again. Thereby guaranteeing that we do not elect the same producer/validator into more than one round slot.
The process is repeated until the required number of producers (more detail later) have been selected for the next round. At which point, the same process is performed for the candidates on the opposite side of $P$ as $T$ had dictated for producers until we have selected enough validators for the next round.
In the event that we are unable to select the necessary number of producers or validators from the pool of candidates as required for the round (i.e.. there are less candidates than there are required slots) the number of required producers/validators will be decreased until there are enough candidates to fulfill the request; however, at no point shall the number of producers and/or stakers for a round be an even number of spots.
There will always be a minimum of three (3) producers and/or validators required in a round. The number of required producers and/or validators for any given round will adjust dynamically (details to come later).
Pseudo Random Selection of Validator & Staker Selection for Block Rewards
When a block is produced, the producer must include relevants outputs to not only pay themselves but to also pay a deterministicially selected validator as well as stakers that staked the producer and that validator.
To select who receives the reward(s) we perform the same selection algorithm as we do for producers and validators with the exception that the prior block hash is used in place of the Merkle root for the previous round ($M$).
In the case of selecting which validator gets rewarded, the process is ran exactly once (1) for the set of validators for that round.
For determining the stakers to reward, the process is completed exactly ten (10) times for the set of stakers staking the winning producer/validator and the number of stakes the staker has submitted for the given candidate determine their “vote count”.
If, any time during the loop used in selecting stakers, the number of available stakers is exhausted (there are less than 10 stakers that backed that producer/validator), the process resumes as if all stakers are included in pool again and the iteration(s) continue until the necessary ten (10) stakers are identified for reward.
After completing the above, the producer then constructs a transaction for inclusion in the block that rewards the parties in the percentages necessary such that there are twenty-two (22) outputs in the miner transaction that is included in the block.
• 1 for the producer
• 1 for the selected validator
• 1 each for the 10 stakers of producer
• 1 each for the 10 stakers of validator
The entire process is designed with a sense of pseudo-randomness that can be recreated by every node on the network using the defined process. In essence, the whole reward structure is a bit of an election with an upper and lower house as well as a lottery for those that backed the elected producer/validator.
The Issues in a Nutshell
Relaunch
Problem: The current blockchain is very, very, very big, and growing every day whether transactions are included in blocks or not. There is also an insane amount of bloat in the chain including transactional spam (fusions anyone?), CantiPixels, and gobs of other useless data that is like walking around with cement shoes with two boat anchors tied chained to your ankles.
Solution: Relaunch the chain. Out with the old and in with the new. To facilitate this, we’re going to need to let the coins on the old network burn in a provable way to prevent errant inflation or other nefarious incidents. In addition, the old chain must be spun down in such a way that the chain officially ends at a certain block. Users need a way to transfer their funds from the old chain to the new chain and that means we’re going to need a…
Coin Swap
Problem: We have to efficiently move funds from the old chain to the new chain.
Solution: A coin swap window of opportunity. The coin swap window to the new chain will be 1,000,000 blocks (as measured by the old chain) during which time users will submit transactions on the old chain to a predetermined wallet address that will serve as a burn address whereby no one user can pick up the funds out of that address (this will be a multisig N:N wallet to make sure no one party controls it). An outside watcher server will watch for transactions on the old chain going to that wallet as instructed by the swap “service”, and release funds on the new chain to the address/keys provided by the user. The swap itself will be funded via a pre-mine on the new chain equivalent to the estimated circulating supply at the time of of the end of the old chain.
After the swap window completes, the remainder of pre-mine funds will be redistributed on chain.
Note: The redistribution of any unclaimed (whether through inaction or as “lost” wallets) via block rewards will be executed over a number of rounds on the new chain to attempt to ensure a somewhat even distribution of those funds to producers, validators, and stakers. Details on the exact schedule of distribution will come later as it will be highly dependent on how much TRTL will have to be dealt with.
In short, the swap mechanism will be a website that helps you transition from old coins to new coins. The technical mechanics of which will be discussed in later articles.
Wait… the chain is massive… we can’t just carry over everything from old to new we have to think about…
Swapping 107,198,508+ Outputs…
Problem: Swapping over 100M outputs to the new chain is… let’s put it simply, incredibly stupid. If we were to do that, yes, the new chain would not have the history of all of those outputs, but considering that we don’t know which outputs are still spendable, there is an unknown number of outputs that we have to create on the new chain as part of the swap process. That number is somewhere between 0 and 107,198,508 (as of the time of this writing). That’s just insane not only to think about but to also execute on.
Solution: During the relaunch of the network and required coin swap, we will be performing the equivalent of a 100,000:1 TRTL (or 10,000,000:100 atomic units) swap while maintaining two decimal places. Swaps of less than 1,000 TRTL will simply become dust and will not be swappable. Don’t worry though, the swap “service” will gladly take your unoptimized outputs and give you brand spanking new optimized outputs.
This swap ratio will eliminate the need to transfer more than ~81% (as of the time of this writing) of the outputs on the old chain thereby avoiding the pollution of the new chain with much of the problems that exist with the current chain.
A Note on Coin Swap Transaction Fees
The swap service will be set up such that as long as 100% of the funds in a transaction sent to the swap service are actually sent to the swap service (no change outputs) then all network transaction fees you pay to send those coins to the swap service will be included in the amount that you receive on the new chain. Furthermore, the swap service will have the ability to bypass the new chain transaction fee mechanics so that you receive the full value of your swapped funds at the ratio given (truncated to two decimals).
A Sustainable, Usage-Based Emission
Right now our blocks come out every 30 seconds, empty or not, and this keeps our emission at a steady pace toward the maximum of 1 Trillion TRTL some time over 100+ years from now. On the new network, there are no empty blocks, and block times won’t mean anything, so we cannot rely on that timing to release new coins into the supply.
Given that the converted supply is quite low (10 Million, two digits after decimal), and block time is now irregular, we have come up with a new way of keeping new coins entering into the system.
On TRTL v2, the network will match all base fees 1:1 with additional coins, making our emission tied to actual usage on chain. 10 Million would be a soft cap, and given enough usage to provide a demand for more coins, more coins will automatically become available via the fees produced by transaction volume.
Now What?
From here, we’re working at solidifying protocol level communications, working on designing the next generation p2p network itself, working on basic structures (transactions, blocks, etc.) and defining some of the inner workings of the actual consensus mechanisms that we’ll be introducing. We’ll provide additional details as they become available. Until then, keep chatting and don’t forget to let us know what you think at http://chat.turtlecoin.lol
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If the Poisson's ratio of an elastic material is $0.4$, the ratio of modulus of rigidity to Young's modulus is _______ |
# How to sovle this SHM looking differential equation?
After solving a physics problem, this was the equation I came to:
$$\left\{ \begin{array}{c} m_2\ddot x_2+k(x_2-x_1-l)=0 \\ m_1\ddot x_1 + m_2 \ddot x_2=0 \end{array} \right.$$
Where $m_1$ and $m_2$, the masses; $k$, the spring coefficient; and $l$, spring length are constants.
$x_1$ and $x_2$ are our variables.
If the two masses were identical, it could be easily transformed into the SHM equation and be solved using the substitution $u=x_2-x_1-l$.
But now, when I try to substitute, the equation will still contain other-than-u variables. That makes it impossible for me to solve.
Can anyone help me?
• Look up reduced mass. You'll need the substitutions $\frac{1}{\mu}=\frac{1}{m_1}+\frac{1}{m_2}$ and $x=x_1-x_2$. Note the second substitution can be differentiated. – Mark Aug 17 '16 at 8:18
$$\left\{ \begin{array}{c} m_2 x''_2+k(x_2-x_1-l)=0 \\ m_1 x''_1 + m_2 x''_2=0 \end{array} \right.$$
$x_1=\frac{m_2}{k}x''_2+x_2-l$
$x''_1=\frac{m_2}{k}x''''_2+x''_2 = -\frac{m_2}{m_1}x''_2$
$$m_1m_2x''''_2+k(m_1+m_2)x''_2 =0$$ This is a fourth order linear ODE. It can easily be reduced to second order and then be solved, leading to $x_2(t)$ .
Putting this result into $\quad x_1=\frac{m_2}{k}x''_2+x_2-l\quad$ gives $x_1(t)$. |
# Poisson-Boltzmann equation
The Poisson equation reads $\epsilon_0 \epsilon_r \nabla^2 \phi = \rho_{free ions)$ where the charge distribution is $\rho_{free ions) = e \sum_i z_i c_i(\)$ |
# 2.2.1 Polygons II, PT3 Practice
2.2.1 Polygons II, PT3 Practice
Question 1:
Diagram below shows a pentagon PQRST. TPU and RSV are straight lines.
Find the value of x.
Solution:
Question 2:
In Diagram below, PQRSTU is a hexagon. APQ and BTS are straight lines.
Find the value of x + y.
Solution:
$\begin{array}{l}\angle QPU={180}^{o}-{160}^{o}={20}^{o}\\ \text{Reflex}\angle PUT={360}^{o}-{80}^{o}={280}^{o}\\ \angle UTS={180}^{o}-{120}^{o}={60}^{o}\\ \angle TSR={180}^{o}-{35}^{o}={145}^{o}\\ \\ \text{Sum of interior angles of a hexagon}\\ =\left(6-2\right)×{180}^{o}\\ ={720}^{o}\\ \\ {x}^{o}+{y}^{o}+{145}^{o}+{60}^{o}+{280}^{o}+{20}^{o}={720}^{o}\\ {x}^{o}+{y}^{o}={720}^{o}-{505}^{o}\\ \text{}={215}^{o}\\ \text{}x+y=215\end{array}$
Question 3:
Diagram below shows a regular hexagon PQRSTU. PUV is a straight line.
Find the value of x + y.
Solution:
$\begin{array}{l}\text{Size of each interior angle of a regular hexagon}\\ =\frac{\left(6-2\right)×{180}^{o}}{6}\\ ={120}^{o}\\ {x}^{o}=\frac{{180}^{o}-{120}^{o}}{2}={30}^{o}\\ {y}^{o}={180}^{o}-{120}^{o}\\ \text{}={60}^{o}\\ {x}^{o}+{y}^{o}={30}^{o}+{60}^{o}\\ \text{}={90}^{o}\\ \text{}x+y=90\end{array}$
Question 4:
In the diagram below, KLMNP is a regular pentagon. LKS and MNQ are straight lines.
Find the value of x + y.
Solution:
$\begin{array}{l}\text{Size of each interior angle of a regular pentagon}\\ =\frac{\left(5-2\right)×{180}^{o}}{5}\\ ={108}^{o}\\ \angle PKS=\angle PNQ={180}^{o}-{108}^{o}={72}^{o}\\ \text{Reflex angle}\angle KPN={360}^{o}-{108}^{o}={252}^{o}\end{array}$
$\begin{array}{l}\text{Sum of interior angles of a hexagon}\\ =\left(6-2\right)×{180}^{o}\\ ={720}^{o}\\ \therefore {x}^{o}+{y}^{o}+{72}^{o}+{252}^{o}+{72}^{o}+{100}^{o}={720}^{o}\\ {x}^{o}+{y}^{o}={720}^{o}-{496}^{o}\\ \text{}={224}^{o}\\ \text{}x+y=224\end{array}$
Question 5:
In Diagram below, PQR is an isosceles triangle and PRU is a straight line.
Find the value of x + y.
Solution:
$\begin{array}{l}{x}^{o}={180}^{o}-{20}^{o}-{20}^{o}={140}^{o}\\ \angle PRS={180}^{o}-{110}^{o}={70}^{o}\\ {y}^{o}+{85}^{o}+{75}^{o}+{70}^{o}={360}^{o}\\ {y}^{o}+{230}^{o}={360}^{o}\\ \text{}{y}^{o}={130}^{o}\\ {x}^{o}+{y}^{o}={140}^{o}+{130}^{o}\\ \text{}={270}^{o}\\ \text{}x+y=270\end{array}$ |
Select Page
Question:
Let $$f:[0,\infty)\to \mathbb{R}$$ be defined by
$$f(x)=x^{2/3}logx$$ for $$x>0$$ $$f(x)=0$$ if $$x=0$$
Then
A. f is discontinuous at x=0
B. f is continuous on $$[0,\infty)$$ but not uniformly continuous on $$[0,\infty)$$
C. f is uniformly continuous on $$[0,\infty)$$
D. f is not uniformly continuous of $$[0,\infty$$ but uniformly continuous of $$(0,\infty)$$
Discussion:
First, we find whether $$f$$ is continuous or not. $$f$$ is “clearly” continuous on $$(0,\infty)$$.
Question is, what happens near $$x=0$$?
Take limit.
$$lim_{x\to 0}\frac{logx}{x^{-2/3}}$$ is in $$\frac{-\infty}{\infty}$$ form.
We use the L’Hospital Rule.
$$lim_{x\to 0}\frac{logx}{x^{-2/3}}$$=$$lim_{x\to 0}\frac{1/x}{-\frac{2}{3}x^{-2/3-1}}$$
=$$lim_{x\to 0}-3/2{x^{2/3}}$$
=$$0$$.
Okay! So we now know that $$f$$ is at-least continuous.
Now we present the HINT s:
• COntinuous function on a compact interval is uniformly continuous.
• If derivative is bounded the by using mean-value theorem we can prove that f is Lipschitz, and therefore uniformly continuous.
For $$x>0$$, $$f'(x)=\frac{2}{3}x^{-1/3}logx + x^{2/3}x^{-1}$$ (By the formula for calculating derivatives of product).
After simplification,
$$f'(x)=\frac{2logx+3}{3x^{1/3}}$$.
We want the boundedness of $$f’$$ and we don’t care about what happens close to zero as much. (Because we are going to use the first hint).
$$lim_{x\to \infty}\frac{2logx+3}{3x^{1/3}}$$ is in $$\frac{\infty}{\infty}$$ form.
Once again, we use the L’Hospital rule.
$$lim_{x\to \infty}\frac{2logx+3}{3x^{1/3}}=lim_{x\to \infty}\frac{2/x}{3/3 x^{-2/3}}$$
$$= lim_{x\to \infty} 2x^{-1/3}=0$$
What does this tell us? This tells us that $$f’$$ is bounded on (say) $$[1,\infty)$$
So $$f$$ is uniformly continuous on $$[1,\infty)$$. Also, $$f$$ is uniformly continuous on $$[0,1]$$.
So given $$\epsilon>0$$ we have two “delta’s” for these two intervals. (which satisfies the condition for uniform continuous in these intervals respectively). Take the delta which is less. And this gives the uniform continuity for $$f$$. |
## Prealgebra (7th Edition)
$A=6\times 2.5^2=37.5$ square in. $V=2.5^3=15.625$ cubic in.
This solid is a cube as all of its sides are equal. The volume of a cube can be calculated as: $V=a^3$ Here: $V=2.5^3=15.625$ cubic in. The surface of a cube can be calulated as: $A=6\times a^2$ Here: $A=6\times 2.5^2=37.5$ square in. |
# A Sample of Carbon Dioxide Occupies 30 Cm3 at 15°C and 740 Mm Pressure. Find Its Volume at Stp. - Chemistry
Numerical
A sample of carbon dioxide occupies 30 cm3 at 15°C and 740 mm pressure. Find its volume at STP.
#### Solution
V = 30 cm3
P = 740 mm
T = 288 K
P1 = 760 mm
V1= ?
T1 = 273 K
Using gas equation,
"PV"/"T" = ("P"_1 "V"_1)/"T"_1
(740xx30)/288 = (760xx"V"_1)/273
"V"_1 = 27.7 "cm"^3
Concept: Standard Temperature Pressure (S.T.P.)
Is there an error in this question or solution?
#### APPEARS IN
Selina Concise Science Chemistry 1 Class 9 ICSE
Chapter 7 Study of Gas Laws
Exercise 7 (B) | Q 14 | Page 127 |
# Question 1d1c4
Jun 11, 2017
It is seen from the given balanced equation of the repacement reaction that every mole of ${\left(C {H}_{2}\right)}_{n}$ requires total $2 n m o l$ ${F}_{2}$.Here $n$ mol goes to replace hydrogen and $n$ mol goes to form $2 n$ mol $H F$ and thus one mole of ${\left(C {F}_{2}\right)}_{n}$ is formed.
Molar masses of
${F}_{2} \to 38 g \text{/} m o l$
${\left(C {F}_{2}\right)}_{n} \to 50 n g \text{/} m o l$
Fluorine required for the production of $50 n g$ ${\left(C {F}_{2}\right)}_{n}$ is 2n mol or $76 n g$
So to prodoce 1kg ${\left(C {F}_{2}\right)}_{n}$ we rrequire $\frac{76 n}{50 n} k g = 1.52 k g$ Fluorine, if not recovered from $H F$ produced as byproduct.
As 50% ${F}_{2}$ goes to form the byproduct $H F$,then its recovery will diminsh the net consumption by 50% and the consumption of Flurine for productio of 1kg ${\left(C {F}_{2}\right)}_{n}$ then will be 50%of1.52kg=0.76kg# |
## OverTheWire – Security Wargames
I wish I had seen this a long time ago: https://overthewire.org/wargames/ are a set of challenges in the area of security: buffer overflows, command injection, web server security, plus fun Linux command line skills.
I learned a lot about how to detect and exploit buffer overflows and shellcode in the Narnia wargame. Good fun and educational too.
Natas is nice since it’s using a web server while the other ones use ssh. The harder ones are…hard, but there’s solutions all over the web in case you get stuck.
All in all, highly recommended to see how easy it is to exploit badly written code and how to write code which does not repeat those known security problems.
## Dart & Pool
Tip of the day: If you use Dart and want to use the Pool library, expect not much help from Google when searching for those keywords: you get the expected results. Adding “future” or “async” helps.
Anyway, the point of this post is a small example how to use a Pool to run commands in parallel, but not too many concurrently.
import 'dart:io';
import 'package:pool/pool.dart';
import 'package:test/test.dart';
Future<ProcessResult> runCommand(String command, List<String> args) {
return Process.run(command, args);
}
void main() {
test('Run 20 slow date commands', () async {
var pool = Pool(5);
List<Future<ProcessResult>> results = [];
for (var i = 0; i < 20; ++i) {
}
pool.close();
await pool.done;
for (var process in results) {
var res = await process;
print(res.stdout);
}
./date_slow is a simple script which returns something on STDOUT and finishes in 2s:
#!/bin/bash
date $* sleep 2 What happens then is that Pool(5) creates a pool with 5 slots. The first for loop tries to run 20 commands though and it’ll queue all 20 immediately, but only 5 at most will run. The rest simply waits until it’s their turn. pool.close() stops any new entries and the await pool.done simply waits until the pool is closed and all jobs are executed. The 2nd for loop (with the print() statement) uses await to get the ProcessResult from the Future<ProcessResult> which Process.run() returns and which is store in the results[] list. The outcome here is that if the pool is 5 slots large, and each command runs for 2 seconds, the complete set of 20 jobs runs in 20/5*2=8 seconds. If I make the pool 10 slots large, it’ll run in 20/10*2=4 seconds and in case of 20 or more slots, it’ll be 2 seconds. And it’ll never run more processes than slots are available. Why I need this? I have a list of URLs from few to hundreds which I need to query. While I can query many concurrently, each instance takes up some non-trivial amount of memory since it’s an external program. Currently at 100 concurrent calls, it uses up all memory on a 16GB RAM notebook. While there are many ways to work around this (do one at a time is the safest and slowest one), using a pool is perfect: it runs many commands in parallel, but I can limit the number of how many run in parallel. ## Twitter Does Not Like Me Anymore All the buzz about Twitter and Elon made me look at my Twitter profile. I saw the missing date of birth, so I added it. Now my account is locked: I am quite confident I put in a 19xx year in there, so I should be at least 22 years old, but I cannot check that now since my account is locked. At this point I wonder if I should care. At worst in 13 years I’ll get my access back. If you send me a message in Twitter, you might have to wait 13 years for a reply… Update 2022-07-31: Well, they like me again. It seems I managed to become 13 quite fast: ## Tensorflow Fun! I was sure I had linked to this awesome article about Tensorflow in the browser. The article is relatable, easy to read, fun to test out the actual program (if you have a camera) and it shows what you can do inside your browser and machine learning. ## Fake Webpages And How To Detect Them Being in Japan and knowing how much good knifes cost, when I saw an advertising of hand made Japanese knifes…let’s say that it looked suspicious from the start. Their web page did look good though. But still…suspicious. Starting with the name “Huusk” which is as much non-Japanese as I can imagine. Then the commonly seen signs to create some time pressure: and the popups about sales happening right now: and yet that number of knifes left does not change: Typical scammer stuff of creating a sense of “Buy it now! Before you start to think about it!” I looked once up a countdown on another web page which counted from about 3h down to zero…so I let it run out. 3h later it showed negative time. And if you reload the page, it goes back to about 3h! This one is less obvious, but I was curious how the popup gets populated as it tries to imply a “Someone near you bought something, so it must be good!” Is it hard-coded like the timer, or dynamically pulled from somewhere? Turns out it is statically populated: initPopup({ "orders": [{ "first_name": "hidetaka", "city": "funakosityo yokosuka", "country": "JP", "topText": "hidetaka from Funakosityo yokosuka, JP made a purchase.", "bottomText": "X1 Huusk Knife Sold!", }, { "first_name": "Indrajith", "city": "Itakoshi,Hinode", "country": "JP", "topText": "Indrajith from Itakoshi,Hinode, JP made a purchase.", "bottomText": "X1 Huusk Knife Sold!", }, { "first_name": "YOICHI", "city": "Nagano inaba", "country": "JP", "topText": "YOICHI from Nagano inaba, JP made a purchase.", "bottomText": "X1 Huusk Knife Sold!", }, { "first_name": "TAKASHIGE", "city": "OOSIMAGUNN SETOUTITYOU KONIYA SEKUIHIGASHI", "country": "JP", "topText": "TAKASHIGE from OOSIMAGUNN SETOUTITYOU KONIYA SEKUIHIGASHI, JP made a purchase.", "bottomText": "X3 Huusk Knives Sold!", }, { "first_name": "YASUHIRO", "city": "Ootaku SINKAMATA", "country": "JP", "topText": "YASUHIRO from Ootaku SINKAMATA, JP made a purchase.", "bottomText": "X3 Huusk Knives Sold!", }, { "first_name": "hiroshi", "city": "nakagawashimatsunoki", "country": "JP", "topText": "hiroshi from Nakagawashimatsunoki, JP made a purchase.", "bottomText": "X4 Huusk Knives Sold!", }, { "first_name": "EIJI", "city": "MATUBARACITY", "country": "JP", "topText": "EIJI from MATUBARACITY, JP made a purchase.", "bottomText": "X4 Huusk Knives Sold!", }, { "first_name": "Kyle", "city": "Okayama", "country": "JP", "topText": "Kyle from Okayama, JP made a purchase.", "bottomText": "X3 Huusk Knives Sold!", }, { "first_name": "Syouji", "city": "Yokohamasi", "country": "JP", "topText": "Syouji from Yokohamasi, JP made a purchase.", "bottomText": "X4 Huusk Knives Sold!", }, { "first_name": "Toshio", "city": "hamamatsu", "country": "JP", "topText": "Toshio from Hamamatsu, JP made a purchase.", "bottomText": "X3 Huusk Knives Sold!", }], "image": "https://huusk.com/theme/images/huusk.png?1", }); The Terms and Conditions page is suspicious too: Hand-made knifes and they create a single size? Why would anyone limit themselves to a single size if they are hand-made anyway? Japanese love to have different knifes for different jobs, but this Japanese company does not? If you look up the text of some of the user testimonials you’ll find another knife company “Kaitomi” which looks just the same. But their web page is not yet ready: Still references to Huusk. Oops! At least it sounds a bit more (fake) Japanese… And I even found the popular Lorem Ipsum text filler: So it’s pretty clear that this is a scam. ## Enter fakewebsitebuster.com I looked up Huusk expecting reviews like “Those knifes are not as good as the advertising suggested”, but I found something even better! And https://fakewebsitebuster.com/ confirmed my findings. And there’s many more such pages are discussed! E.g. fake investment web pages who’ll take your money and predictably disappear. The most interesting part of that web page however is that they basically explain what to look for: • DNS and company registration, country and date. Recently registered and claiming to be 20 years in business? • Actual location of the business: Japan? US? UK? Lithuania? Nigeria? Claiming or suggesting that they are from somewhere else? • Use of stock photos for “user testimonials”. Reverse image search can find stock photos. • Copy&paste user testimonials used in other places too. • Selling a “unique” product which is also sold on other places (eBay, Aliexpress, Taobao). • The intense attempt of creating a sense of urgency. Very educational and I wish more people would be aware of those scam tricks. ## TensorFlow on arm64 The VIM3L (Cortex A55 cores) I have has a NPU accelerator built-in. An interesting article about it is here, however before doing fancy NPU stuff, let’s get TensorFlow working first. Should be easy. Famous last words. Turns out that “pip install tensorflow” does not work: on arm64 (AKA aarch64 AKA ARMv8) TensorFlow is not officially supported. So I had to compile it first. ## Compiling TensorFlow https://www.tensorflow.org/install/source described the compile process reasonable well. It is missing a lot of details though, so here is a more detailed walk-through. Start with a Ubuntu 20.xx image with an extra 70 GB disk for TensorFlow source code: # One-time action: for the data disk, create a volume and a filesystem # to mount under /data sudo bash pvcreate /dev/nvme1n1 vgcreate vg_data /dev/nvme1n1 lvcreate -L69G -n data vg_data mke2fs -j /dev/vg_data/data mkdir /data echo -e '/dev/mapper/vg_data-data\t/data\text4\tdefaults\t0 1' >>/etc/fstab mount /data chown ubuntu:users /data umount /data exit sudo apt update sudo apt -y upgrade sudo reboot After a reboot, you now have a /data of about 70GB. sudo apt -y install build-essential python3 python3-dev python3-venv pkg-config zip zlib1g-dev unzip curl tmux wget vim git htop liblapack3 libblas3 libhdf5-dev openjdk-11-jdk # Get bazel wget https://github.com/bazelbuild/bazel/releases/download/4.2.2/bazel-4.2.2-linux-arm64 chmod a+x bazel-4.2.2-linux-arm64 sudo cp bazel-4.2.2-linux-arm64 /usr/local/bin/bazel # bazel uses ~/.cache/bazel mkdir -p /data/.cache/bazel ln -s /data/.cache/bazel ~/.cache/bazel # Build a Python 3 virtual environment python3 -m venv ~/venv source ~/venv/bin/activate pip install wheel packaging pip install six mock numpy grpcio h5py pip install keras_applications --no-deps pip install keras_preprocessing --no-deps # Get TensorFlow source cd /data git clone https://github.com/tensorflow/tensorflow.git cd tensorflow/ git checkout r2.8 cd /data/tensorflow ./configure # Build Python package: bazel build -c opt \ --copt=-O3 \ --copt=-std=c++11 \ --copt=-funsafe-math-optimizations \ --copt=-ftree-vectorize \ --copt=-fomit-frame-pointer \ --copt=-DRASPBERRY_PI \ --host_copt=-DRASPBERRY_PI \ --verbose_failures \ --config=noaws \ --config=nogcp \ //tensorflow/tools/pip_package:build_pip_package # Build Python whl: BDIST_OPTS="--universal" bazel-bin/tensorflow/tools/pip_package/build_pip_package ~/tensorflow_pkg # And for tfjs: # (see https://github.com/tensorflow/tfjs/tree/master/tfjs-node) bazel build --config=opt --config=monolithic //tensorflow/tools/lib_package:libtensorflow # The result is at bazel-bin/tensorflow/tools/lib_package/libtensorflow.tar.gz It does take a lot of time (about 2-3h for each the Python package and the tfjs-node library). When I tried 2 CPU and 8 GB RAM, some compiler runs were killed as they were running out of memory. 4 CPU and 16 GB RAM worked fine.Thus AWS m6g.xlarge recommended. m6g.large failed to build. Using spot instances for the m6g.xlarge (regular$0.154/h, spot price $0.04/h) helped a bit to limit the financial impact. ## Python and TensorFlow It took me several tries: • When using Ubuntu 22.04 to compile TF, the resulting binary wanted GLIBC 2.35 which my VIM3L did not have. It had 2.31. It also used Python 3.10 to compile. • When using Ubuntu 20.04, it used Python 3.8 to compile. My VIM3L had Python 3.9. While GLIBC was fine, Python was not. • The created whl file could be loaded and used on the machine I compiled it on. No Python or GLIBC version problems here. That covered all my Python needs. Now moving to the main target: ## Node.js and TensorFlow tfjs-node uses the libtensorflow.so library, so that should remove some of the CPython version problems I have seen. Compiling was easy now: https://github.com/tensorflow/tfjs/tree/master/tfjs-node#optional-build-optimal-tensorflow-from-source is spot on. The biggest problem was to make Node.js understand to not use the non-existing arm64 pre-compiled library, but instead use the one I created. The instructions in the above link did not explain in enough details how to make this work. In hindsight it’s easy, but it took some tries to make me understand it. In short: • Do an “npm install –ignore script” • Add a file scripts/custom-binary.json into the modules directory for @tensorflow/tfjs-node (this gave me the hint) • Run “npm install” in the tfjs-node directory • That will download the tensorflow library archive • Now do the “npm install” where your application is (which is the only “npm install” you’d usually do) ❯ npm install --ignore-script ❯ pushd . ❯ cd node_modules/@tensorflow/tfjs-node/scripts ❯ cat >>custom-binary.json <<_EOF_ { "tf-lib": "https://MYSERVER.com/libtensorflow-2.8-arm64.tar.gz" } _EOF_ ❯ cd .. ❯ npm install [...] > @tensorflow/tfjs-node@3.16.0 install > node scripts/install.js CPU-linux-3.16.0.tar.gz * Downloading libtensorflow https://MYSERVER/libtensorflow-2.8-arm64.tar.gz [==============================] 3685756/bps 100% 0.0s * Building TensorFlow Node.js bindings [...] ❯ popd ❯ npm install ## Benchmarks As a benchmark I modified slightly server.js from the tfjs-examples/baseball-node to not listen to the port which means after the training it’ll exit. Then run this on the VIM3L (S905D3), my ThinkCentre m75q (Ryzen 5), and my HP T620 (GX-420CA) once with CPU backend (tfjs) and once with the C++ TF library (tfjs-node): I did not expect Node.js to be just 4 times slower than C++. Really impressive. Still, using tfjs-node makes a lot of sense. While on x86_64 this was not an issue, with above instructions it’s doable on arm64 too. ## Dart HTTPS Server In my previous post I used a simple HTTPS server written in Node.js and I was curious how that would look like in Dart. And it’s very short too: import 'dart:io'; Future<void> main() async { var chain = Platform.script.resolve('cert.pem').toFilePath(); var key = Platform.script.resolve('key.pem').toFilePath(); var context = SecurityContext() ..useCertificateChain(chain) ..usePrivateKey(key); var server = await HttpServer.bindSecure(InternetAddress.anyIPv4, 8080, context); await server.forEach((HttpRequest request) { print("${request.method} ${request.uri.path}"); print(request.headers); request.response.close(); }); } Works just as well as the Node.js counterpart. ## TP-Link Kasa KC120 – Streaming without Kasa The main problems I have with IoT devices are: • They might send data home without me knowing about it • But I can monitor their traffic pattern and if they send home way more data than expected, I could disconnect them • They might be vulnerable to exploits • But I can put them on a separate VLAN at home so they don’t see other devices unless I allow it (via firewall rules) • I can sometimes update firmware (definitely a problem after few years) • They stop to work when the company turns off their servers • I am able to use them without Internet connectivity Most Kasa products I own (power switches) are supported by various projects like Home Assistant or python-kasa, so turning on my Kasa power switch on my own is a simple task. Same for my LIFX light bulbs there’s even an official API. The TP-Link KC120 camera however does not have any supported local API and contrary to my expectation, it does not support a local stream mode via a web browser interface. I can watch a live (and local) video stream via the Kasa application on the phone, but that functionality is at the mercy of TP-Link. I don’t like that. Following are the steps to have local streaming (resp. recording) for the KC120. And with that it’s possible to do whatever I’d like to do with the stream: publishing on the Internet, processing via OpenCV, local archiving etc. ## python-kasa python-kasa does not support the camera, so you won’t see it during a normal discovery: ❯ kasa No host name given, trying discovery.. Discovering devices on 255.255.255.255 for 3 seconds == Plug Three - HS105(JP) == Host: 192.168.21.180 Device state: OFF == Generic information == Time: 2022-05-03 11:37:55 (tz: {'index': 90, 'err_code': 0} Hardware: 2.1 Software: 1.0.3 Build 210506 Rel.161924 MAC (rssi): 10:27:F5:XX:XX:XX (-62) Location: {'latitude': XX.0, 'longitude': XX.0} == Device specific information == LED state: True On since: None == Modules == + <Module Schedule (schedule) for 192.168.21.130> + <Module Usage (schedule) for 192.168.21.130> + <Module Antitheft (anti_theft) for 192.168.21.130> + <Module Time (time) for 192.168.21.130> + <Module Cloud (cnCloud) for 192.168.21.130> == Plug One - HS105(JP) == Host: 192.168.21.182 Device state: OFF == Generic information == Time: 2022-05-03 11:37:55 (tz: {'index': 90, 'err_code': 0} Hardware: 1.0 Software: 1.5.8 Build 191125 Rel.135255 MAC (rssi): B0:BE:76:XX:XX:XX (-54) Location: {'latitude': XX.0, 'longitude': XX.0} == Device specific information == LED state: True On since: None == Modules == + <Module Schedule (schedule) for 192.168.21.182> + <Module Usage (schedule) for 192.168.21.182> + <Module Antitheft (anti_theft) for 192.168.21.182> + <Module Time (time) for 192.168.21.182> + <Module Cloud (cnCloud) for 192.168.21.182> But the camera shows up with an additional -d switch, although it’s being ignored since the tool does not know how to handle it: ❯ kasa -d No host name given, trying discovery.. Discovering devices on 255.255.255.255 for 3 seconds DEBUG:kasa.discover:[DISCOVERY] ('255.255.255.255', 9999) >> {'system': {'get_sysinfo': None}} DEBUG:kasa.discover:Waiting 3 seconds for responses... [...] DEBUG:kasa.discover:Unable to find device type from {'system': {'get_sysinfo': {'err_code': 0, 'system': {'sw_ver': '2.3.6 Build 20XXXXXX rel.XXXXX', 'hw_ver': '1.0', 'model': 'KC120(EU)', 'hwId': 'CBXXXXD5XXXXDEEFA98A18XXXXXX65CD', 'oemId': 'A2XXXX60XXXX108AD36597XXXXXX572D', 'deviceId': '80XXXX88XXXX76XXXX88XXXXX3AXXXXXXXXXXXB6', 'dev_name': 'Kasa Cam', 'c_opt': [0, 1], 'f_list': [], 'a_type': 2, 'type': 'IOT.IPCAMERA', 'alias': 'Camera', 'mic_mac': 'D80D17XXXXXX', 'mac': 'D8:0D:17:XX:XX:XX', 'longitude': XX, 'latitude': XX, 'rssi': -38, 'system_time': 1651545748, 'led_status': 'on', 'updating': False, 'status': 'configured', 'resolution': '720P', 'camera_switch': 'on', 'bind_status': True, 'last_activity_timestamp': 1651545210}}}}: Unable to find the device type field! [...] Important fields here are the deviceID and via the MAC address, you can find out what IP address the camera has (if you use DHCP). In my case 192.168.21.187 is the camera’s IP address. ## nmap nmap shows only port 9999 open which is the known TP-Link debug port. But there’s more ports: ❯ sudo nmap -p- 192.168.21.187 Starting Nmap 7.80 ( https://nmap.org ) at 2022-05-03 11:51 JST Nmap scan report for kc120.lan (192.168.21.187) Host is up (0.012s latency). Not shown: 65531 closed ports PORT STATE SERVICE 9999/tcp open abyss 10443/tcp open unknown 18443/tcp open unknown 19443/tcp open unknown MAC Address: D8:0D:17:XX:XX:XX (Tp-link Technologies) Nmap done: 1 IP address (1 host up) scanned in 9.28 seconds And with that port information I found this article: https://medium.com/@hu3vjeen/reverse-engineering-tp-link-kc100-bac4641bf1cd. It’s about a slightly different camera model, but since the ports patch, maybe more does. I followed it, however I could not get the authentication working: the Kasa account password as per article did not work. Time to do the ARP spoofing to see what the Android app uses to authenticate! Geistless did a great job explaining the steps he took. My overall plan: 1. Redirect the traffic from the Kasa app on the phone to my Linux machine (via arpspoof) 2. Redirect the incoming HTTPS traffic to my HTTPS server (via iptables) 3. Print the URL and headers for incoming HTTPS traffic which arrives at my HTTPS server ## arpspoof The dsniff package contains arpspoof: ❯ sudo apt install dsniff [...] ❯ sudo setcap CAP_NET_RAW+ep /usr/sbin/arpspoof ## My HTTPS Server While the original author had a https server as part of his Rust learning, I created a NodeJS version. But first we’ll need keys. Self-signed is fine: ❯ openssl genrsa -out key.pem ❯ openssl req -new -key key.pem -out csr.pem ❯ openssl x509 -req -days 999 -in csr.pem -signkey key.pem -out cert.pem ❯ rm csr.pem Now the simple HTTPS server listening on port 8080: const https = require('https'); const fs = require('fs'); const options = { key: fs.readFileSync('key.pem'), cert: fs.readFileSync('cert.pem') }; https.createServer(options, function (req, res) { console.log(req.url); console.log(req.headers); res.writeHead(200); res.end(""); }).listen(8080); Some IP traffic routing rules to redirect all incoming TCP traffic on enp1s0 for ports 10443, 18443 and 19443 to port 8080: ❯ sudo iptables -t nat -A PREROUTING -i enp1s0 -p tcp --dport 10443 -j REDIRECT --to-port 8080 ❯ sudo iptables -t nat -A PREROUTING -i enp1s0 -p tcp --dport 18443 -j REDIRECT --to-port 8080 ❯ sudo iptables -t nat -A PREROUTING -i enp1s0 -p tcp --dport 19443 -j REDIRECT --to-port 8080 ❯ sudo sysctl net.ipv4.ip_forward=1 Now run the https server and watch it display the URL and the headers for an incoming request on port 19443: ❯ node ./https.js and to test, on another machine I ran: $ curl -k -u admin:abc 'https://t621.lan:19443/test?a=3&b=5'
and this is the output of my https server:
/test?a=3&b=5
{
host: 't621.lan:19443',
authorization: 'Basic YWRtaW46YWJj',
'user-agent': 'curl/7.68.0',
accept: '*/*'
}
The basic authentication is base64 encoded. To decode:
❯ echo YWRtaW46YWJj | base64 -d
So that works. Now putting it all together.
• Start the Kasa app on the phone. Make sure the KC120 is enabled and can display a live video stream. Stop the stream.
• Have the iptables redirect rules in place. And IP forwarding in the kernel.
• Start the HTTPS server.
• Run arpspoof. 192.168.21.55 is the phone’s IP which runs the Kasa application. 192.168.21.187 is the IP of the KC120.
❯ arpspoof -i enp1s0 -t 192.168.21.55 192.168.21.187
7c:d3:a:xx:xx:xx 38:78:62:xx:xx:xx 0806 42: arp reply 192.168.21.187 is-at 7c:d3:a:xx:xx:xx
• On the mobile app, try to connect to the video stream of the KC120 again
• You should now see some output of the HTTPS server:
/https/stream/mixed?video=H264&audio=G711
{
authorization: 'Basic aXXXXXXXXXXXXXXXXM=',
connection: 'keep-alive',
'user-agent': 'Dalvik/2.1.0 (Linux; U; Android 10; H8296 Build/52.1.A.3.49)',
host: '192.168.21.187:19443',
'accept-encoding': 'gzip'
}
And then I finally had the authentication string the camera wanted!
❯ echo 'aXXXXXXXXXXXXXXXXM=' | base64 -d
Turns out that the password to use was not the Kasa password: it’s a longish string of hex digits. That might be a KC120 specialty or it might depend on the firmware version. I cannot say since I have no KC100, but whatever the password is, it’s possible to find out relatively easily using above approach.
## The Result: Local Streaming!
I can connect to the video stream! And with very little CPU usage too.
❯ curl -k -u 'MY_KASA_ACCOUNT:THE_CAMERA_PASSWORD' \
--ignore-content-length \
"https://192.168.21.187:19443/https/stream/mixed?video=h264&audio=g711&resolution=hd&deviceId=80XXXX88XXXX76XXXX88XXXXX3AXXXXXXXXXXXB6" \
--output - | ffmpeg -hide_banner -y -i - -vcodec copy kc120stream.mp4
To change resolution, change it in the Kasa app. 1920×1080 (1.4Mb/s), 1280×720 (850kbit/s) and 640×360 (350kbit/s) are possible.
## TODO
• There is no audio coming from the camera. Audio works on the Kasa app.
• It would also be nice to understand how to change the configuration of the camera (e.g. change resolution), but it’s ok to set them once via the Kasa app.
• What options do the parameter video, audio and resolution support?
## Moving Two Things
Learning G-Code is quite interesting: it’s really old, really simple, and yet surprisingly complex if you want to get CNC cuts 100% perfect. Luckily my use-cases are much simpler, but I still would like to move 2 axis concurrently: e.g. move along the x-axis from 0 to 400mm, and rotate the a axis from 0 to 180 degrees so that they start and end together.
Step 1 is to connect a servo, which was not as easy as I thought since there’s no dedicated servo connector I can readily use on the DLC32. Luckily there are plenty pins to get GND, 5V and a PWM signal. This is the section I added to the config.yaml:
a:
steps_per_mm: 10
max_rate_mm_per_min: 100000.0
acceleration_mm_per_sec2: 100000.0
max_travel_mm: 180.0
soft_limits: true
homing:
cycle: 0
mpos_mm: 0
positive_direction: false
motor0:
rc_servo:
pwm_hz: 50
output_pin: gpio.25
min_pulse_us: 1000
max_pulse_us: 2000
IO25 is readily available on the EXP1 connector, so I used 5V, GND and IO25 from EXP1:
Now I can use G-Code like this:
• G0A90 to move the servo to the neutral position 90° (assuming 1ms=0°, 2ms=180°). The actual movement depends on the servo. Most move only 90° in total.
• G1X300A10F200 now does:
• Move to x=300mm and rotate a to 10° at a feed-rate of 200 mm/minute. If x=100 at the start, this movement will take 1 minute.
• During the movement, a ?<ENTER> will show the current positions:
?
<Run|MPos:221.533,0.000,0.000,126.500|FS:200,0|Pn:PYZ>
ok
ok |