induction
search
number theory unsolved
number theory

[quote="Jumbler"]Prove that for any integer $ n\geq 2$ there is a set $ A_n$ of $ n$ distinct positive integers such that for any two distinct elements $ i, j \in A_n$, $ |i \minus{} j|$ divides $ i^2 \plus{} j^2$.[/quote]

It's easy to show thru induction on $ n$ :

1) it's true for $ n\equal{}2$ : choose for example $ \{1,2\}$

2) Suppose it's true for $ n$ and that we have a set $ A_n\equal{}\{a_k\}_{k\equal{}1}^n$ with $ a_i<a_{i\plus{}1}$

Let $ a\equal{}\prod_{k\equal{}1}^n(a_n\plus{}1\minus{}a_k)$

Let $ A_{n\plus{}1}\equal{}\{b_k\}_{k\equal{}1}^{n\plus{}1}$ defined as :
$ \forall k\in[1,n]$ : $ b_k\equal{}a\cdot a_k$
$ b_{n\plus{}1}\equal{}a(a_n\plus{}1)$

Then,$ \forall 1\le i<j\le n$ : $ b_j^2\plus{}b_i^2\equal{}a^2(a_j^2\plus{}a_i^2)$
But $ a_j^2\plus{}a_i^2\equal{}\alpha_{ij}(a_j\minus{}a_i)$ thru induction hypothesis
So $ b_j^2\plus{}b_i^2\equal{}a^2\alpha_{ij}(a_j\minus{}a_i)$ $ \equal{}a\alpha_{ij}(b_j\minus{}b_i)$ and so $ b_j\minus{}b_i|b_j^2\plus{}b_i^2$

And, $ \forall 1\le i\le n$ : $ b_{n\plus{}1}^2\plus{}b_i^2\equal{}a^2((a_n\plus{}1)^2\plus{}a_i^2)$
But $ b_{n\plus{}1}\minus{}b_i\equal{}a(a_n\plus{}1\minus{}a_i)$. We know that $ a_n\plus{}1\minus{}a_i|a$ and so $ b_{n\plus{}1}\minus{}b_i|a^2$ and so $ b_{n\plus{}1}\minus{}b_i|b_{n\plus{}1}^2\plus{}b_i^2$
Hence the result.

3) example of such a construction :
$ A_2\equal{}\{1,2\}$
$ a\equal{}(3\minus{}1)(3\minus{}2)\equal{}2$ and so $ A_3\equal{}\{2,4,6\}$
$ a\equal{}(7\minus{}2)(7\minus{}4)(7\minus{}6)\equal{}15$ and so $ A_4\equal{}\{30,60,90,105\}$
...

Take any set $ S$ of $ n$ distinct positive integers $ s_a, a\equal{}1,...,n$.

Let $ K \equal{} \prod_{a,b, a<b} |s_a \minus{} s_b| \implies K > 0$ and $ \forall a,b,  |s_a \minus{} s_b|$ divides $ K$

Now take $ A_n$ to be $ \{r_a \equal{} K s_a, a \equal{} 1,...,n\}$

Then for $ i,j \in A_n, |i \minus{} j|$ divides $ K^2$ divides $ i^2 \plus{} j^2$

[quote="PhilG"]or simply take $ A_n \equal{} \{ n! a, a \equal{} 1,...,n \}$[/quote]

Cool and nice, congrats !   :)

added problem:(a little stronger)
prove that a set $ A$ exists such that for each $ a>b;a,b \in A$ and $ |A|\equal{}n$ we have $ a\minus{}b | a$.
( http://www.mathlinks.ro/viewtopic.php?search_id=1541109136&t=294638 )


$(7-w)(7-x)(7-y)(7-z) = 4$ such that $w \neq x \neq y \neq z$.

Find $x+y+w+z$.

there can be more than one solution I think:

[hide]w=6, x=8, y=4, z=25/3[/hide]

[hide]I think you're forced to use 1, -1, 2, and -2.[/hide]

[quote="Hikaru79"][quote="dk2"]I guess you have to assume that x, y, z, and w, are integers.
Even then I think there might be more than one solution.[/quote]

Assuming that we're talking integers here, how could there be more than one possible solution? We know that there are four factors in there equal to four, and that [b]none of them are equal[/b]. The factors of 4 are $\displaystyle (4),(2),(1),(-1),(-2),(-4)$. However, if you use $\displaystyle (4)$ or $\displaystyle (-4)$, you must use its complementary $\displaystyle 1$, and that will only be two factors, not four. Therefore, you MUST use $\displaystyle 4=(-2)(2)(-1)(1)$. No other way will get you four factors, since I have already eliminated four (as a factor). So these are the only possible ways to write 4 as a product of 4 factors. And since $\displaystyle 7-x=y$ obviously only has one solution, there can only be one solution for this whole thing.

Or am I missing something? How could there possibly be more than one solution?[/quote]

Sorry about that. Yes, you're right there's only 1 solution to that question if x, y, z, and w are integers. Note to self: the next time I post something, I should actually DO the question.

lol


$ \frac{\sqrt{3}\minus{}1}{sinx}\plus{}\frac{\sqrt{3}\plus{}1}{cosx} \equal{} 4\sqrt{2}$

[hide="hint"]
Note that $ \sqrt {3} \minus{} 1 \equal{} \frac {4sin\frac {\pi}{12}}{\sqrt {2}}$ and $ \sqrt{3}\plus{}1 \equal{} \frac {4cos\frac {\pi}{12}}{\sqrt {2}}$
[/hide]


calculus
integration
function
calculus computations

Hmm.. Did I see you last Tuesday? I recognize the problem, and you claim to be from Washington.

[hide="hint"]
replace u=x+1 in the second part, you will get the inverse fuction of the first[/hide]


calculus
derivative
integration
algebra
binomial theorem

Let S := the sum you gave.

Take a look at S/2 and S-S/2.

[quote="frznfire219"]Hi there, I'm kind of stumped on this so I thought this might be a good place to ask, thanks for any help that you can provide!

what is 1/2 + 2/(2^2) + 3/(2^3) + 4/(2^4)...[/quote]

just fiddling around with some derivatives:

you want f(1/2) where f(x)=:Sigma:n*x^n (n from 1 to :inf:) = d/dx :Sigma:x^n+1 = d/dx (x:^2:/(1-x)) = ((1-x)(2x)-(x:^2:)(-1))/(1-x):^2:=(2x-x:^2:)/(1-x):^2:. When we plug in x=1/2, we get (1-1/4)/(1/4)=3.

erg where'd I go wrong?

[quote="MysticTerminator"]
just fiddling around with some derivatives:

you want f(1/2) where f(x)=:Sigma:n*x^n (n from 1 to :inf:) = d/dx :Sigma:x^n+1 = d/dx (x:^2:/(1-x)) = ((1-x)(2x)-(x:^2:)(-1))/(1-x):^2:=(2x-x:^2:)/(1-x):^2:. When we plug in x=1/2, we get (1-1/4)/(1/4)=3.

erg where'd I go wrong?[/quote]

d/dx (xn+1) = (n+1)xn
= nxn + xn

You should be able to get it from there.

oh whoops ok so I have to subtract off :Sigma:x^n=1, so 3-1=2 and it all works out.

Yeah, if you want 1+2x^2+3x^3+..+nx^n simpler method (simpler than differentiation) would be multiplying it by 1-x and seeing what happens...

that's essentially what the first method is, no? just cancelling out terms until we get a geometric series. I just wanted to see if I could come up with another method (as in, I agree that the first method is probably better).

For yet another method, let S = 1/2 + 1/22 + 1/23 + 1/24...
And then look at S + S/2 + S/22 + S/23...

But Mystic's method of manipulating the power series with calculus is particularly powerful. You can find the sum a wide variety of different series that way.

For example, try finding the sum of the series 1 + 1/(2*22) + 1/(3*23) + 1/(4*24) + ...

should that first term be 1/2 in order to fit the pattern?

Yes. Make that first term 1/2. So find the sum of the series 1/2 + 1/(2*22) + 1/(3*23) + 1/(4*24) + ...

okay...so that's :sigma:x^n/n (n=1 to infinity) with x=1/2 = integral :sigma:(x^(n-1))(n=1 to inf)=integral 1/(1-x)=-ln(1-x). plugging in x=1/2 gives -ln(1/2)=ln 2. wow what a totally random number to come from that sequence.

Pretty cool huh? Try these...

:Sigma: [n*(100Cn)], from n = 0 to 100
:Sigma: [1/(n+1)*(100Cn)], from n = 0 to 100

generalization of the first one:

binomial theorem (for integers) states that (x+1)^n=:Sigma:C(n,i)x^i (i=0 to n). let's take derivatives=> n(x+1)^(n-1)=:Sigma:i*C(n,i)x^(i-1)(i=0 to n). let's now plug in x=1 => n*2^(n-1)=:Sigma:i*C(n,i). so...answer for your specific problem would be ... 100*2^99?

another way to get the identity n*2^(n-1)=:Sigma:i*C(n,i) that i thought up would be to interpret the left side as choosing a committee of i people from n people and then choosing a leader of that committee. So, we can also choose a leader first (n ways) and then choose the rest of the committee (2^(n-1) ways).

generalization of the second one: we start again with the binomial theorem, and now integrate, giving 2^(n+1)/(n+1)=:Sigma:C(n,i)/(i+1). So the specific answer would be 2^101/101. And another way to get that identity combinatorially would be...hm. i dunno what represents dividing by (i+1)?


In general, this forum has drifted far too difficult once again.  I encourage the mods to be more aggressive about moving posts to Intermediate, and I will be doing so frequently myself as well.  Those of you posting in this Forum should be careful not post items that are too difficult.  Read the guidelines, and if you're still not sure, post in Intermediate instead.  I will continue to move threads I think are too difficult for Getting Started.  If you posted a question in this forum, but don't see it here anymore, look in Intermediate or Pre-Olympiad for it.

what kinds of subjects or courses are for this high school basics one?

Read this topic: http://www.mathlinks.ro/Forum/viewtopic.php?t=32837

Does Mod Arithmetic go here? Because i had a thread that had modular arithmetic in it and now its nowhere to be seen!

Looking at your old posts up to Nov. 4th, the only one I've found that has the word CRT in it is [url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=237900[/url], unless I screwed up.

No, this post is like old. Like old old. Like when i first started posting in high school basics. I just wanted to know for future reference

Is the trigonometry allowed here restricted to right triangles?

unless you consider 45-45-90 and 30-60-90 trig, which you shouldn't.

It specifically says "Precalculus" in the High school Intermediate Topics description.

[quote="awesomeone"]Does the AIME fall under basic or intermediate?[/quote]

The first few questions of AIME might be fine. I'm not sure though.

I agree, although I think AIME is more on the intermediate side.

AIME is definitely an intermediate topic.


Let $ a,b,c$ be rational numbers such that $ a \plus{} b \plus{} c \equal{} a^2 \plus{} b^2 \equal{} c^2 \equal{} 1$. Prove that there exists $ m,n\in\mathbb{Z}$ such that $ abc \equal{} \frac {m^2}{n^3}$.

We have $ c^2 \equal{} 1$, then $ c \equal{} \pm 1$.

If $ c\equal{}1$, we have $ a\plus{}b\equal{}0$ and $ a^2\plus{}b^2\equal{}1$. Hence, $ abc \equal{} ab \equal{} \frac{(a\plus{}b)^2 \minus{} (a^2\plus{}b^2)}{2} \equal{} \frac{0\minus{}1}{2} \equal{} \minus{}\frac{1}{2}$. Taking $ m\equal{}2$ and $ n\equal{}\minus{}2$ we get $ \frac{m^2}{n^3} \equal{} \frac{4}{\minus{}8} \equal{} \minus{}\frac{1}{2}$.

If $ c\equal{}\minus{}1$, then $ a\plus{}b\equal{}2$ and $ a^2\plus{}b^2\equal{}1$. Again, $ abc \equal{} \minus{}ab \equal{} \frac{(a\plus{}b)^2 \minus{} (a^2\plus{}b^2)}{2} \equal{} \frac{4\minus{}1}{2} \equal{} \frac{3}{2}$. Taking $ n\equal{}6$ and $ m\equal{}18$ we have $ \frac{m^2}{n^3} \equal{} \frac{3^4.2^2}{2^3.3^3} \equal{} \frac{3}{2}$.

We didn't use that $ a,b,c$ were rational numbers. Are you sure the problem statement is correct? Maybe it was $ a\plus{}b\plus{}c \equal{} a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$.

I am extremely sorry. There was a typo.  :blush:
The condition of the problem should have been:

$ a \plus{} b \plus{} c \equal{} a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$

[quote="Mashimaru"]Let $ a,b,c$ be rational numbers such that $ a \plus{} b \plus{} c \equal{} a^2 \plus{} b^2 \plus{} c^2 \equal{} 1$. Prove that there exists $ m,n\in\mathbb{Z}$ such that $ abc \equal{} \frac {m^2}{n^3}$.[/quote]

$ abc\in\mathbb Q$ and so :

If $ abc\equal{}0$ : $ abc\equal{}\frac{0^2}{1^3}$

If $ abc\ne 0$, $ abc\equal{}\frac pq$ with $ p\ne 0$ and then $ abc\equal{}\frac pq\equal{}\frac{(p^2q)^2}{(pq)^3}$


[quote="236factorial"]1. The square root of what number is two times the number 8?

2. What is the largest perfect square less than 225 that is a multiple of 9?

3. What number is closest to the square root of 584?[/quote]

[hide="Question 1"]$\sqrt{x}=2*8$

$x=16^2=256$[/hide]

[hide="Question 2"]The perfect squares underneath 225 are:

196, not a multiple of 9

169, not a multiple of 9

$\boxed{144}$, is a multiple of 9[/hide]

[hide="Question 3"]
We try and find the closest perfect square to 584, which are 625 ($25^2$) and 576 ($24^2$) 584 is closer to 576 so the number closest to $\sqrt{584}$ is 24.[/hide]

[hide="1"]
 $2 \cdot 8=16\\\\16^2=\boxed{256}$
[/hide][hide="2"]
196 doesn't work,  169 doesn't work, but \boxed{144}$ works.[/hide][/hide]

[hide="#1"]Write an equation to represent this, $\sqrt{n}=2\times8$  Solve it to get:  $n=\boxed{256}$[/hide]

[hide="#2"]In order for a perfect square to be divisible by 9, the square root must be a multiple of 3.  $\sqrt{225}=15$  The lowest multiple of 3 less than 15 is 12.  $12^2=144$  That makes the answer $\boxed{144}$[/hide]

[hide="#3"]The 2 squares on either side of 584 are  $24^2=576$ and $25^2=625$.  576 is obviously closer to 584 than 576, so the answer is $\boxed{24}$[/hide]


calculus
derivative
LaTeX
calculus computations

how do i diffrentiate in terms of y in the following equation:
y = 1/2x^2

is this classroom math level?  I'll let you think about that

Errrrrr... this isn't really middle-school level please post this on the High-school forums.

But in any case, the derivative is x because you do (1/2)x^(2-1)* 2

well, the reason for LaTeX is: is that $ \frac{1}{2x^2}$ or $ \frac{1}{2}x^2$?


http://www.stripcreator.com/comics/inducted_into_math/469815

(I spelled "Involving" wrong in the comic's
title, but it's permanent now.  I am sorry.)

Indeed, it does explain the basic premise of induction.  That does seem useful for someone who has never seen induction before.  Good work.


Let $\epsilon>0$.Prove that for almost all $x \in [0,1]$ the following holds: there doesn't exist finite number of rationals $\frac{p}{q}$ such that $|x-\frac{p}{q}|<\frac{1}{q^{2+\epsilon}}$.

take $E_n$ the set of real $x \in [0;1]$ such that there exists $k,n$ satisfying: $|x-\frac{k}{n}|<\frac{1}{n^{2+e}}$.
Then it is clear that $m(E_n) \leq n\frac{2}{n^{2+e}}=\frac{2}{n^{1+e}}$.
Suppose that for $x \in [0;1]$ there exist infinitely many $p,q$ satisfying $|x-\frac{p}{q}|<\frac{1}{q^{2+e}}$. Hence $x$ is in infinitely many $E_n$. But the set of real that are in infinitely many $E_n$ is of Lebesgue measure $0$ since $\sum m(E_n) < +\infty$.
Hence, for almost every real there is only a finite number of fraction $\frac{p}{q}$ satisfying $|x-\frac{p}{q}|<\frac{1}{q^{2+e}}$


Solve system of equation :

$ \left\{ \begin{array}{l}\sqrt[3]{{x^2 }} \plus{} \sqrt[3]{{y^2 }} \equal{} b \\ \sqrt {x^2  \plus{} \sqrt[3]{{x^4 y^2 }}}  \plus{} \sqrt {y^2  \plus{} \sqrt[3]{{x^4 y^2 }}}  \equal{} a\\ \end{array} \right.$

Put $ x \equal{} w^3,\; y \equal{} v^3$.

Get: $ \left\{\begin{array}{l}w^2 \plus{} v^2 \equal{} b \\
\sqrt {w^6 \plus{} w^4v^2} \plus{} \sqrt {v^6 \plus{} w^4v^2} \equal{} a \\
\end{array} \right.$

$ \left\{\begin{array}{l}w^2 \plus{} v^2 \equal{} b \\
w^2\sqrt {w^2 \plus{} v^2} \plus{} |v|\sqrt {v^4 \plus{} w^4} \equal{} a \\
\end{array} \right.$

$ \left\{\begin{array}{l}w^2 \plus{} v^2 \equal{} b \\
w^2\sqrt {b} \plus{} |v|\sqrt {v^4 \plus{} w^4} \equal{} a \\
\end{array} \right.$

$ \left\{\begin{array}{l}w^2\sqrt {b} \plus{} v^2\sqrt {b} \equal{} b\sqrt {b} \\
w^2\sqrt {b} \plus{} |v|\sqrt {v^4 \plus{} w^4} \equal{} a \\
\end{array} \right.$

$ v^2\sqrt {b} \minus{} |v|\sqrt {v^4 \plus{} w^4} \equal{} b\sqrt {b} \minus{} a$

$ v^2\sqrt {b} \minus{} b\sqrt {b} \plus{} a \equal{} |v|\sqrt {v^4 \plus{} w^4}$

$ (v^2\sqrt {b} \minus{} b\sqrt {b} \plus{} a)^2 \equal{} v^2(v^4 \plus{} w^4)$

$ w^4 \equal{} \frac {1}{v^2}(v^2\sqrt {b} \minus{} b\sqrt {b} \plus{} a)^2 \minus{} v^4$ From the other hand $ w^4 \equal{} (b \minus{} v^2)^2$

$ \frac {1}{v^2}(v^2\sqrt {b} \minus{} b\sqrt {b} \plus{} a)^2 \minus{} v^4 \equal{} (b \minus{} v^2)^2$

$ (v^2\sqrt {b} \minus{} b\sqrt {b} \plus{} a)^2 \minus{} v^6 \equal{} v^2(b \minus{} v^2)^2$ - equation of the power = $ 6$  :wow:


I've always wondered and no one has answered this.  What exactly does a mathematician do?  Like we all know what someone means when they say "I'm a doctor." But what does it mean to be a mathematician?  Any ideas?

I think it would be a term for someone in a field so specialized that he would rather you just call him a mathematician. Or it might mean a math researcher.

There is one sentence which I read in some book: "What does a mathematician do? he solves problems. One may wonder what's the difference between a student and a mathematician then? The first solve problems already solved while the latter to which solution is not known." (it's not exact but it expresses well what was meant). Personally I like this opinion (though I guess it's not exact - it describes research mathematician rather than "theoretical" [as I found them called this way] mathematicians).

Another similar opinion was given in a book covering number theory (though I can't remind myself what the title or author was - only that it was published by Springer in Graduate Texts series) where were giving some motivating examples (e.g.: Ramanujan) in foreword.

Anyway I realize that above (ie: the very first) statement is not just - there surely is a lot of people who prefer teaching than actual research and undoubtedly they can call themselves mathematician as well.

Though still I am too young to know anything about exact tasks of mathematician - just the vision appearing from the books built my opinion.

My question is: how does a mathematician get money? For example, how did Euler find the time and money to think of all that interesting stuff? Did he have to go to work, or what?

[quote="zeb"]For example, how did Euler find the time and money to think of all that interesting stuff? Did he have to go to work, or what?[/quote]
Read this: [url=http://www-history.mcs.st-andrews.ac.uk/history/Mathematicians/Euler.html]Euler.[/url]

The stories of 17th or 18th century mathematicians aren't particularly pertinent to today's situation. In the specific case of Euler, he was supported throughout his career in academic institutions, and he did do some teaching. His primary institution, the St. Petersburg Academy of Sciences, was largely supported by royal patronage - the Czar of Russia wanted to have a collection of internationally distinguished scholars. Euler obtained this academic position at an extremely young age, as a result of what you might call cronyism: Daniel Bernoulli, who had known Euler in Basel, recommended him for the post. (Originally it was to be a position in physiology, but by the time Euler reached St. Petersburg, he wound up in the mathematical-physical division of the Academy.)

[quote="Megus"]There is one sentence which I read in some book: "What does a mathematician do? he solves problems. One may wonder what's the difference between a student and a mathematician then? The first solve problems already solved while the latter to which solution is not known." (it's not exact but it expresses well what was meant). Personally I like this opinion (though I guess it's not exact - it describes research mathematician rather than "theoretical" [as I found them called this way] mathematicians).

<paragraph snipped>

Anyway I realize that above (ie: the very first) statement is not just - there surely is a lot of people who prefer teaching than actual research and undoubtedly they can call themselves mathematician as well.[/quote]

A mathematician is an expert in solving problems with mathematics. The mathematician might teach these problem-solving techniques or invent new problem-solving techniques rather than solve problems, but the definition rests on being an expert in the mathematical techniques.

Most mathematicians are college professors, spending their time teaching and researching mathematics. Some engineering firms and scientific research departments have mathematicians on staff to perform complex calculations beyond the usual mathematical techinques known to their engineers and scientists. Other firms use mathematics more directly and hire mathematicians for that. For example, Google hires mathematicians to research search algorithms. Some financial insitutions use mathematicians to model the stock market. Insurance companies use statisticians, called actuaries, to model the risk in their insurance policies. I myself work as a cryptologist, using mathematics to analyze ciphers.

Some mathematicians are more like philosophers, seeking truth through mathematics. Others are more like artists, creating beauty in elegant abstract constructions. But being an applied mathematician myself, I see a mathematician as someone who carries around a big mental toolkit and knows how to use those tools.

Erin Schram

"Euler obtained this academic position at an extremely young age, as a result of what you might call cronyism"

This is the first time I see cronyism have a positive result!

Cool, I wanted to be a cryptologist too! But I don't see where their money comes from either...

I'm sure Erin could comment on this better than me, but I bet there are lots of companies and agencies that have a need for qualified cryptologists. But you need to learn how to program also.

[quote="Erin J. Schram"]
A mathematician is an expert in solving problems with mathematics. The mathematician might teach these problem-solving techniques or invent new problem-solving techniques rather than solve problems, but the definition rests on being an expert in the mathematical techniques.

Most mathematicians are college professors, spending their time teaching and researching mathematics...
[/quote]

I like this definition---it's much clearer than my usual lame attempt at it---but I don't agree that most mathematicians are college professors.  Of course, this disagreement might stem from different levels of qualification for "expert."  College professors almost all have Ph.D.s, but there are relatively few mathematical experts who have gone that far with their formal educations.  Many bachelors-degree-level mathematicians work in industry, often in actuarial sciences.  I'd consider them to be experts.

As for the person who asked how cryptologists make a living... the single largest employer of mathematicians in the U.S. is the National Security Agency (or at least that was true a few years ago).  Cryptology is one of the many different mathematical things done at the NSA, so that's one option to investigate.  (I personally wouldn't work for the NSA as it's part of the Department of Defense, but many people I know have been happy working there.)

so mathematics would be a good way to get a government job?

I'd like a government job, good pay, and no risk of getting laid off

unless the government is overthrown...then... :|

CheesyP18, you can still be laid off from a government job.  Congress has to decide every year how much money goes to each part of the government, and whatever lab you're working for could get a lot less money that it was expecting.  I think the pay is also not extraordinary.

However, you're right that the job security is probably better that it would be in a company, and that the pay is probably better than it would be in a university.

[quote="smbelcas"]As for the person who asked how cryptologists make a living... [/quote]

Cryptographers can now make a living in industry or academia, not only in military or government work.  It is a small field, but then again so is mathematics.

[quote] the single largest employer of mathematicians in the U.S. is the National Security Agency [/quote]

I think the majority of employees so designated are not necessarily what would be considered mathematicians (i.e. researchers) in academia, but include people with less than a doctorate, as well as PhD-holders who for various reasons also would not fit the description.

[quote="CheesyP18"]so mathematics would be a good way to get a government job?
[/quote]

One of my friends from college, who was an economics major, went directly to a job with the Bureau of Labor Statistics.  He told me at the time that he got the job partly because his math background was so strong, and that really there was a desire for mathematics/statistics majors in some of the government Departments (such as Labor). Of course, that was 15 years ago, so things might have changed.  But I'm guessing that there will always be some need for mathematics experts with the government.

We had a student, a math/economics double major, also go directly from a bachelors degree to a job in the Bureau of Labor Statistics. That was about five years ago.

Well, my dad is a mathematician who has worked at 2 (or maybe 3, I was too young to remember) universities, so I hope this does his job justice:

Mathematicians who are into research, as others have already stated, solve problems in existing fields of math and create new problems to be solved. They publish papers (sometimes in math journals, etc) with their results. A mathematician tends to have work groups made up of collegues who like to work in the same fields, and therefore may travel a lot, especially because recently, because of how easy it is to travel and comunicate these days, their work groups tend to be international. (My dad, for example, travels frequently to Canada, Spain and Japan)
As for the "college profesors" thing: in most universities, researchers must teach at least one course to students of the university they work at, at any level and/or help student with their thesis, etc.

On the other hand, my dad had fellow math students who ended up doing very different things:
a friend of his ended up doing statistics for the UN.

:)


Source: Czech and Slovak Match 1999
For arbitrary positive numbers $a,b,c$, provet the inequality
$\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b}\geq 1$.

[quote="kunny"]Since $(a+b+c)^{2}\geq 3(ab+bc+ca)\Longleftrightarrow (a-b)^{2}+(b-c)^{2}+(c-a)^{2}\geq 0$ and $a>0,\ b>0,\ c>0,$ we have
 $\frac{a^{2}}{ab+2ca}+\frac{b^{2}}{bc+2ca}+\frac{c^{2}}{ca+2bc}\geq \frac{(a+b+c)^{2}}{(ab+2ca)+(bc+2ab)+(ca+2bc)}$
$=\frac{(a+b+c)^{2}}{3(ab+bc+ca)}\geq 1.\ Q.E.D.$[/quote]
yes, that's how i did..
I learned that
$\frac{x^{2}}{a}+\frac{y^{2}}{b}\geq \frac{(x+y)^{2}}{a+b}$ from Math Olympiad Treasure...
and i found this very useful.. :)

[quote="kimby_102"]
I learned that
$\frac{x^{2}}{a}+\frac{y^{2}}{b}\geq \frac{(x+y)^{2}}{a+b}$ from Math Olympiad Treasure...
and i found this very useful.. :)[/quote]

In fact, this is a special case of Cauchy in Engel form.

$\sum \frac{a_{i}^{2}}{b_{i}}\ge \frac{ (\sum a_{i})^{2}}{ \sum b_{i}}$

:)

More information about CS in Engel form, see:
http://www.mathlinks.ro/Forum/viewtopic.php?highlight=engel&t=15125

Thanks, [b]Lovasz[/b]! This is some really good stuff on inequalities.

For arbitrary positive numbers $a,b,c$, provet the inequality\[\frac{a}{b+\lambda c}+\frac{b}{c+\lambda a}+\frac{c}{a+\lambda b}\geq \frac{3}{1+\lambda}.(\lambda >0)\]

[hide="Solution to sqing's"]
By Cauchy, $\left(\sum_{cyc} a(b + \lambda c)\right)\left(sum_{cyc}\dfrac{a}{b + \lambda c}\right) \ge (a + b + c)^2$. But then $(\lambda + 1)(ab + bc + ca)\left(sum_{cyc}\dfrac{a}{b + \lambda c}\right) \ge (a + b + c)^2$. It follows from $(a + b + c)^2 \ge 3(ab + bc + ca)$.
[/hide]

[quote=sqing]For arbitrary positive numbers $a,b,c$, prove the inequality\[\sqrt{\frac{a}{b+\lambda c}}+\sqrt{\frac{b}{c+\lambda a}}+\sqrt{\frac{c}{a+\lambda b}}\geq \frac{3}{\sqrt{1+\lambda}}.(\lambda >0)\][/quote]

We have
$\sqrt{\frac{a}{b+\lambda c}}+\sqrt{\frac{b}{c+\lambda a}}+\sqrt{\frac{c}{a+\lambda b}}=\frac{a}{\sqrt{ab+\lambda ca}}+\frac{b}{\sqrt{bc+\lambda ab}}+\frac{c}{\sqrt{ca+\lambda bc}}$
By chebchev
$\geq \frac{a+b+c}{3} ( \frac{1}{\sqrt{ab+\lambda ca}}+\frac{1}{\sqrt{bc+\lambda ab}}+\frac{1}{\sqrt{ca+\lambda bc}} )$
By Jensen
$\geq \frac{a+b+c}{3} \frac{3\sqrt{3}}{\sqrt{ab+bc+ca}\sqrt{1+\lambda}}$
$= \frac{\sqrt{3}(a+b+c)}{\sqrt{ab+bc+ca}\sqrt{1+\lambda}}\geq \frac{3}{\sqrt{1+\lambda}}.$
Then we need to prove that
$(a+b+c)^2\geq 3(ab+bc+ca)$
wich is obviousely true.

[quote=sqing]For arbitrary positive numbers $a,b,c$, prove the inequality\[\sqrt{\frac{a}{b+\lambda c}}+\sqrt{\frac{b}{c+\lambda a}}+\sqrt{\frac{c}{a+\lambda b}}\geq \frac{3}{\sqrt{1+\lambda}}.(\lambda >0)\][/quote]
Your inequality is not true for $a=\frac{8}{13}, b=\frac{1}{230848}, c=1,\lambda=\frac{3}{8}$.

[hide = Solution]By the Cauchy-Schwarz Inequality, we know that:
$$\left( \frac{a}{b + 2c} + \frac{b}{c + 2a} + \frac{c}{a + 2b} \right) \left(a(b+2c)+b(c+2a)+c(a+2b) \right) \ge (a+b+c)^2 \Longleftrightarrow \frac{a}{b + 2c} + \frac{b}{c + 2a} + \frac{c}{a + 2b} \ge \frac{(a+b+c)^2}{3ab+3bc+3ca}$$Since
$$a^2+b^2+c^2 \ge ab + bc + ca \Longleftrightarrow (a+b+c)^2 \ge 3(a+b+c) \Longleftrightarrow \frac{(a+b+c)^2}{a+b+c} \ge 3$$it follows that
$$\frac{a}{b + 2c} + \frac{b}{c + 2a} + \frac{c}{a + 2b} \ge \frac{(a+b+c)^2}{3ab+3bc+3ca} \ge 1$$as desired.[/hide]


probability
function
parameterization
geometric series

Suppose you have 20 lock-picking devices. Every time you use one it successfully picks the lock, but it can either break or stay intact; if it stays intact it can be used again. Suppose the probability of a device breaking is $p$ for $0<p<1$ and each device stays that way until it breaks. What is the probability in terms of $p$ that the 20 picks last through 100 locks?

If you're up for it, generalize for $k$ lock-picking devices over $n$ locks.

I do not know the answer, by the way; I'm asking. My first thought is to use generating functions but I haven't yet made any headway.

[hide="With generating functions"]For a given pick, the generating function for the probability that the pick lasts at least for $n$ locks is $f(x) = x+(1-p)x^{2}+(1-p)^{2}x^{3}+\ldots = x\sum_{n \geq 0}(x(1-p))^{n}= \frac{x}{1-x(1-p)}$.  Thus, the generating function for the probability that $k$ picks last for at least $n$ locks is $f(x)^{k}= \frac{x^{k}}{(1-x(1-p))^{k}}$, and we want to find the coefficient of $n$ in this expression.  But this is just $x^{k}\sum_{n \geq 0}\binom{n+k}{k}x^{n}(1-p)^{n}$, and the coefficient of $x^{n}$ in this expression is $\binom{n}{k}(1-p)^{n-k}$.[/hide]

One could have done this perfectly well without generating functions, as the niceness of the answer suggests.  I'll leave a purely combinatorial solution to others.

JBL, are you sure? Two reasons I ask:

1) for n=21, k=20, and p=.6, the given formula yields a probability greater than 1. I constructed several similar examples.

2) I do not understand the deduction:

[quote="JBL"]But this is just $x^{k}\sum_{n \geq 0}\binom{n+k}{k}x^{n}(1-p)^{n}$, and the coefficient of $x^{n}$ in this expression is $\binom{n}{k}(1-p)^{n-k}$.[/quote]

Perhaps that is where the mistake may be found? I am not sure.

Okay, I must've done something stupid with my probabilities -- probably in the use of the "at least this many" generating function, if not many other places.  (My expression also gives 0 whenever $k > n$ when of course it should give 1.)

For that deduction: there's an $x^{k}$ out in front, so we need the $n-k$-th term from the sum.  The mistakes I've made are much more fundamental to my solution than that, I think.

So, I made several low-level mistakes, actually.  Let's do it from scratch by a slightly different method.  Let $q = 1-p$ to save some typing.  Then
\[f(x) = px+pqx^{2}+pq^{2}x^{3}+\ldots \]
is the generating function whose $n$-th coefficient is the probability that a given pick lasts for [i]exactly[/i] $n$ locks.  This is just a geometric series, so $f(x) = \frac{px}{1-qx}$.  Then the generating function whose $n$-th coefficient is the probability that $k$ picks last for exactly $n$ locks is
\[f(x)^{k}= \frac{x^{k}p^{k}}{(1-qx)^{k}}= x^{k}p^{k}\sum_{i = 0}^\infty \binom{i+k-1}{k-1}x^{i}q^{i}. \]
Bringing the terms out front inside and re-indexing, this gives us
\[G(x) = \sum_{i = k}^\infty p^{k}q^{i-k}\binom{i-1}{k-1}x^{i}. \]
The coefficient of $x^{i}$ is the probability $k$ picks last for $i$ locks, so the value we want is
\[\sum_{i = n}^\infty p^{k}q^{i-k}\binom{i-1}{k-1}= 1-\sum_{i=k}^{n-1}p^{k}q^{i-k}\binom{i-1}{k-1}. \]
In principle, I think the original approach I took (with the "at least this many" generating function) should work, but I haven't been able to make it work out properly.  In any event, one correct answer should be sufficient :)

We can apply a relatively standard probability problem, called a "negative binomial distribution," (according to my discrete math book).

It is equivalent to: What is the sum of the probabilities of getting exactly $j$ heads in $n$ flips, $j=0,1,2,...,k-1$ where p(head)=p.

The negative binomial distribution is the problem for a single $j$.

The answer is trivially: $\sum_{j=0}^{k-1}\binom{n-1}{j}p^{j}(1-p)^{n-j}$

I doubt there is a closed form because that would imply that there is a simple closed form for $\sum_{j=0}^{k}\binom{n}{j}$ for any $k$.

Altheman, I think you have some parameter problems -- compare your answer to mine.  In particular, it's not clear at all what you think the equivalence is between the two problems.

Once you've seen the answer, explaining it combinatorially isn't very hard: just imagine that you use each pick in sequence until it runs out.  Then the total number of locks opened (the dummy variable $i$ in my answer) must be at least $n$; of these, we must select $k-1$ to be the locks on which a given pick broke (the last pick must break on the last lock, so that's why $\binom{i-1}{k-1}$) and the probability of any particular such outcome is $p^{k}q^{i-k}$.

You have to get exactly k locks breaking [which is the same as getting a head], and you can have 0 to 19 lock picks break [or get 0 to 19 heads].

Couldn't you just use something like this?

$\sum_{k=0}\binom{100}{k}p^{k}(1-p)^{100-k}$

That's a rather weak explanation -- in particular, it doesn't account for the $n-1$.  (Your formula also has an extra factor of $1-p$.)

Problem: show algebraicly (i.e., without reference to the fact that they count the same quantity) that Altheman's expression and my expression are equal.


Here's a somewhat interesting problem:

A sphere of radius R is placed near a long, straight wire that carries a steady current I.  The magnetic field generated by the field is B.  What is the total magnetic flux passing through the sphere?

since $ \nabla .\vec B \equal{} 0$ we get $ \int_{V} \nabla.\vec B d\tau \equal{} \int_{S} \vec B.d\vec a \equal{} 0$


Sequence $ \{a_{n}\}$ is defined by $ a_{1}= 2007,\, a_{n+1}=\frac{a_{n}^{2}}{a_{n}+1}$ for $ n \ge 1.$ Prove that $ [a_{n}] =2007-n$ for $ 0 \le n \le 1004,$ where $ [x]$ denotes the largest integer no larger than $ x.$


This is a general heads-up for contest managers for homeschooler groups to let them know that AMC 10/12 registration is annoying. The AMC 10/12 registration forms, both online and on paper, request a CEEB high school I.D. number as the first step in filling out the form, but the College Board's annoying Web site does not make it easy to look up the appropriate school codes for homeschoolers. (Moreover, AMC's Web site displays the College Board look-up page in a SMALL, nonresizable pop-up window, which is very bad usability.) When I used my son's registration for the SAT I last year to look up the number, I then found out that AMC's online form doesn't accept that number anyway. GRRRRR.

After phoning the helpful staff at the AMC office with AMC's 800 number, I was told to simply use the paper form to register and not fill in any CEEB number at first. Apparently AMC will assign a mock CEEB number to our homeschool group after receiving our registration. I calmed down after talking to the AMC staff member who answered my phone call  :)  and am now filling out the paper form. But I did have to post this to let other homeschooling parents know how to avoid the wasted time I encountered in tracking down a number that turned out to be useless anyway.

Best wishes for a fun year of participation in the AMC 10 and 12 contests.

Well, I'm certainly glad you got good service and helpful information over the phone. And of course, I'm glad to hear you are registered for the AMC 10 and/or AMC 12.

We'll probably stay with the existing on-line registration forms and procedure this year, it's simply too chancy to change the software in the middle of registration,  with only 6 weeks or so left.  But if you ccontact us, we would definitely be interested in hearing your suggestions for improvements to the on-line registration process.   Over the summer, we make updates and changes, and we can try to include your suggestions.

The College Board's CEEB number has been useful to us over the years, it is a well-known unique identifier for schools.  But there are a number of entities, including homeschools, for which it doesn't exist and so as we get more of these entities into the AMC database we end up splicing our substitute into an existing system.
It makes for a bit of a rough spot.  Thanks for working with us over that rough spot.

Steve Dunbar
AMC Director


[quote="Peter"]An integer sequence $\{a_{n}\}_{n \ge 1}$ is defined by \[a_{1}=2, \; a_{n+1}=\left\lfloor \frac{3}{2}a_{n}\right\rfloor.\] Show that it has infinitely many even and infinitely many odd integers.[/quote]

[b]Case 1:[/b] Let us suppose that the set of odd integers in $a_{n}$ is finite. It means that there is an $m$ so that for all $n\geq{m}$  $a_{n}$ is even. Then for all $n\geq{m}$
\[a_{n+1}=\frac{3}{2}a_{n}\]

so $ a_{m+k}=(\frac{3}{2})^{k}a_{m}$. But if $r$ is the greatest number so that $2^{r}|a_{m}$ then $a_{m+r+1}$ is not an integer,contradiction!

[b]Case 2:[/b] Now let us suppose that the set of even numbers in $a_{n}$ is finite. That means we can find an $m$ so that for all $n\geq{m}$ $a_{n}$ is odd, which means that all the numbers $b_{n}=a_{n}-1$ are even integers. Then for all $n\geq{m}$
\[a_{n+1}=\frac{3a_{n}-1}{2}=\frac{3}{2}(a_{n}-1)+1\]

and so $ b_{n+1}=\frac{3}{2}b_{n}$,but this case was discussed above. :wink:


I have $L.H.S \geq \frac{3}{4} > \frac{2}{3}$.Have I done?
I'm not sure because I showed that $L.H.S  > \frac{2}{3}$ not $L.H.S \geq \frac{2}{3}$

[quote="zhaobin"]see what Mamat said. :)[/quote]
Do you want to say that I had done?  :D

no,I only want to there must be something wrong in your solution :D  :D

[quote="hungkhtn"]Prove for all $a,b,c$ nonnegative we have:

$\frac{a^2}{3b^2+3c^2-2bc}+\frac{b^2}{3a^2+3c^2-2ac}+\frac{c^2}{3a^2+3b^2-2ab} \ge \frac{2}{3}$

 :)

I hope It's nice to relax![/quote]
This ineq. is a particular case of the following inequality:
If $p\ge -1$ and $\frac{2p+1}{2} \le q \le 4(p+2)^2$, then for all $a,b,c$ nonnegative numbers we have:

$\frac{a^2+qbc}{b^2+c^2+pbc}+\frac{b^2+qca}{c^2+a^2+qca}+\frac{c^2+qab}{a^2+b^2+qab} \ge \frac{q}{p+2}+2$.

 :)

[quote="spdf"]I think it is easy
By Cauchy-swars,we have [tex]\sum \frac{a^2}{3b^2+3c^2-2bc}.\sum(a^2(3b^2+3c^2-2bc)\geq\ (a^2+b^2+c^2)^2[/tex]
We need to prove[tex]3(a^4+b^4+c^4)+4abc(a+b+c)\geq\ 6(a^2b^2+b^2c^2+c^2a^2)[/tex]
By schur[tex]3(a^4+b^4+c^4)+3abc(a+b+c)\geq\ 3\sum a^3b\geq\ 3\sum a^2b^2[/tex](symetric sum).equality take when (a;b;c)=(0;t;t)[/quote]

It's what I want to said.
The better inequality is:

$\frac{a^2}{2b^2-bc+2c^2}+\frac{b^2}{2a^2-ac+2c^2}+\frac{c^2}{2a^2-ab+2b^2} \ge 1$

It's use similar way!

[quote="siuhochung"][quote="hungkhtn"]For the problem:

$\frac{a^2}{b^2-bc+c^2}+\frac{b^2}{a^2-ac+c^2}+\frac{c^2}{a^2-ab+b^2} \ge 2$

My friend tell me a supper nice and short solution.It's also funny![/quote]
WLOG assume $a \le b \le c$.
Then $a^2-ac \le 0$, $a^2-ab \le 0$, so
$\frac{a^2}{b^2-bc+c^2}+\frac{b^2}{a^2-ac+c^2}+\frac{c^2}{a^2-ab+b^2} \ge \frac{a^2}{b^2-bc+c^2} + \frac{b^2}{c^2}+\frac{c^2}{b^2} \ge 2$.
Equality holds when a=0 and b=c
Is this short enough? :P[/quote]

OK.It's nice and short.It's all I want to say. :lol:

[quote="siuhochung"]can you tell us your friend's solution?[/quote]
You solution is same as my friend's.


Prove : For every natural number n larger than 2  equation a^x+b^x=c^x has no natural solutions.

I have nice and short proof but i don't know tex or how it's called.

1. there is no $n$ in the equation
2. show your prove, I claim the prove is wrong or fake
3. and iff you have a prove: wrong section
 :rotfl:

@ZetaX: I believe (hope ?) it's a joke.
If not then show your proof, zaohing369. :)

I saw it as joke, too (cause of the last line), thus the smiley ;)

Hey guys I suppose(think) we must wait a little!You can expect anyting from a chinese! ;)


algebra
function
domain
real analysis
real analysis unsolved

Denote $ S\equal{}[0,1] \times [0,1]$. For each real number $ t$ with $ 0<t<1$, let $ C_t$ be the set of all points $ (x,y) \in S$ that are on or above the line joining $ (t,0)$ to $ (0,1\minus{}t).$ Prove that the points common to $ C_t$ are those points in $ S$ which are on or above the curve: $ \sqrt{x}\plus{}\sqrt{y}\equal{}1$.

For $ 0 < t < 1$, the equation of the line $ PQ: \frac {x}{t} \plus{} \frac {y}{1 \minus{} t} \equal{} 1$.
$ \therefore y \equal{} 1 \minus{} t \minus{} \left(\frac {1}{t} \minus{} 1\right)x \equal{} 1 \plus{} x \minus{} \left(t \plus{} \frac {1}{t}x\right)\leq 1 \plus{} x \minus{} 2\sqrt {x}$, yielding $ y\leq (1 \minus{} \sqrt {x})^2$.
The equality holds when $ t \equal{} \sqrt {x}$. Therefore we have $ 0\leq y\leq (1 \minus{} \sqrt {x})^2,\ 0\leq x\leq 1\Longleftrightarrow \sqrt {x} \plus{} \sqrt {y}\leq 1$,
that is the domain in which the line $ PQ$ can pass through.


What is EV Theorem,can anyone explain for me? :)

and how can you use this in olympiad inequalities?? do you know where to find aplications??


Can you help me with these absolute value questions and explain each step clearly? Thank you!

Find the sum of the solutions of the equation $|2x+1|+|x-4|=12$

How many elements does the solution set of $|x-1|-2|x|=3|x+1|$

The inequality $|x-2|>x+1$ is true for wha values of x?

[hide="third problem"]Either $x-2>x+1$ or $-(x-2)>x+1$ (absolute values).
Obviously the first equation won't work out since the x's cancel, so let's look at the second equation.

$-x+2>x+1$
$-2x>-1$

$x<\frac{1}{2}$[/hide]

[quote="SnowStorm"]#1
[hide]
There are four cases:
$2x + 1 + x - 4 = 12$
$2x + 1 - (x - 4) = 12$
$-(2x+1) + x + 4 = 12$
$-(2x+1) - (x-4) = 12$

Then it is just a matter of solving them individually:
1)
$2x + 1 + x - 4 = 12$
$3x - 3 = 12$
$3x = 15$
$x = 5$

2)
$2x + 1 - (x - 4) = 12$
$2x - x + 1 + 4 = 12$
$x + 5 = 12$
$x = 7$

3)
$-(2x+1) + x + 4 = 12$
$-2x - 1 + x - 4 = 12$
$-x - 5 = 12$
$x = -17$

4)
$-2x - 1 - x +4 = 12$
$-3x + 3 = 12$
$x = -3$

And summing them:
$-17 + -3 + 7 + 5 = -8$
[/hide][/quote]
SnowStorm, your approach is corect, but you need to go back to the original equation and check if the answers work. For example, if you plug in $x=7$ to the original equation... ;)


Your family traveled 260 miles to the beach at an average speed of 65 mph. Traffic on the return trip slowed your speed to an average of 40 mph. To the nearest whole number, what was the mean speed for the entire trip? Express your answer in miles per hour.

[hide]Okay, he spent 260/65=4 hours on the way there.  He spent 260/40=6 1/2 hours on the way back.  He traveled a total of 520 miles in 10.5 hours, so he traveled about 49.5, or 50 miles in an hour[/hide]

[hide]
Simple.
Use harmonic mean.
$\frac{1}{65}+\frac{1}{40}=\frac{8}{520}+\frac{13}{520}=\frac{21}{520}$
Divide by 2:
$\frac{21}{1040}$
Take inverse:
$\frac{1040}{21}$
which is approx 50
[/hide]

High schoolers are discouraged from posting in the middle school forum, especially if their solution is the same as other people's.


How is the best way to study for MATHCOUNTS?  :)  :) ???
If you have a opinion , please post!

best way for anything math related is to do lots and lots of problems
i also suggest doing it in the time frame given, to practice moving along quickly and pacing yourself

AoPS books are wonderful for mathcounts, so i'd suggest those too


trigonometry
rotation
geometry
angle bisector
similar triangles
geometry unsolved

[quote="Mathias_DK"]Let $ \triangle ABC$ be a equilateral triangle. Suppose there exists $ P$ such that$ |AP| \equal{} 1$, $ |BP| \equal{} 2$ and $ |CP| \equal{} 3$. Find the length of the sides of $ \triangle ABC$. (No trigonometry allowed - i.e. sinus, cosinus, etc.)[/quote]

A big hint is rotate the triangle 60 degrees about a certain vertex.  Then, you should be able to do some nice calculations to figure out the side.

I don't remember right now how to prove that such a triangle exists but if it exists I can let the side length of $ \triangle ABC$ be $ s$. Then $ 3(1^4\plus{}2^4\plus{}3^4\plus{}s^4)\equal{}(1^2\plus{}2^2\plus{}3^2\plus{}s^2)^2$. Therefore $ s\equal{}\sqrt{7}$.

[quote="puuhikki"]I don't remember right now how to prove that such a triangle exists but if it exists I can let the side length of $ \triangle ABC$ be $ s$. Then $ 3(1^4 \plus{} 2^4 \plus{} 3^4 \plus{} s^4) \equal{} (1^2 \plus{} 2^2 \plus{} 3^2 \plus{} s^2)^2$. Therefore $ s \equal{} \sqrt {7}$.[/quote]
I don't understand nothing of your solution. How do you get $ 3(1^4 \plus{} 2^4 \plus{} 3^4 \plus{} s^4) \equal{} (1^2 \plus{} 2^2 \plus{} 3^2 \plus{} s^2)^2$ ?

[quote="puuhikki"]A proof of this fact can be found in http://www.artofproblemsolving.com/Forum/viewtopic.php?t=67254 but unfortunately it uses some trigonometry.[/quote]
Ahh... Ok then ^^

Here is my solution [hide="Without trigonometry"]
Let $ \triangle ABC$ be the triangle and denote $ AB \equal{} BC \equal{} CA \equal{} s$.
Then according to Ptolemys theorem we have $ |CP| |AB| \ge |AC| |BP| \plus{} |AP| |BC|$ $ \iff$ $ 3s \ge 2s \plus{} s \equal{} 3s$. With equality iff $ \square CAPB$ is cyclic. So $ \square CAPB$ is cyclic. Denote $ O \equal{} AB \cap CP$.
Because $ \square CAPB$ is cyclic we get that $ \angle BPA \equal{} 120$, and $ \angle CPA \equal{} 60$, so $ AO$ is the angle bisector of $ \angle BPA$. Then $ \frac {BO}{AO} \equal{} \frac {BP}{AP} \equal{} 2$. So since $ AB \equal{} s$ we get $ BO \equal{} \frac {2s}{3}$ and $ AO \equal{} \frac {s}{3}$.
Since $ \triangle COA$ has same angles as $ \triangle BOP$ we get: $ \frac {OA}{OP} \equal{} \frac {AC}{BP}$ $ \iff$ $ OP \equal{} \frac {2}{3}$. So $ OC \equal{} \frac {7}{9}$. And thus $ \frac {OC}{OP} \equal{} \frac {7}{2}$ yielding $ \frac {T_{\triangle CBA}}{T_{\triangle PBA}} \equal{} \frac {7}{2}$.

From here we easily obtain $ \frac {T_{\triangle CBA}}{T_{\triangle PBA}} \equal{} \frac {s^2}{2} \equal{} \frac {7}{2}$ by looking at similar triangles, and thus $ s \equal{} \sqrt {7}$. QED :P
[/hide]


for x,y,z>0
xy+yz+zx=1

prove:

$(x+y+z)^5\geq\frac{27}{4}(x^3+y^3+z^3+x+y+z)$
really attractive :rotfl:  :ninja:  :huh:  :blush:  :D

Put $ x+y+z =a ; xyz =b $
 The ineq changes into : $ 4a^5-27a^3+54a \geq 81b $
  We split problem to 2 cases :
 Case 1 : Prove that $ 4a^5 -27a^3+54a \geq 9\sqrt3 $
 Case 2 : Prove that $ 4a^5 -30a^3+54a \geq 0 $ ;)

nttu can you it completely?? :P  :P  :rotfl:  :ninja:

if you solved ,it would be interesting

wait for my next post :P  :idea:

Don't pay attention to my 2 cases !  :lol:
 We just need to prove :
 $ 4a^5 -27a^3+54a \geq \frac{27}{a} $ (1) . Put $ t=a^2 \geq 3 $
 Indeed , $ (1) \iff 4t^3-27t^2+54t-27 \geq 0 \iff (t-3)^2(4t-3) \geq 0 $ , which is true .
 The rest is to prove : $ 1 \geq 3ab \iff (xy+yz+xz)^2 \geq 3xyz(x+y+z) $ , which is also true  :D


function
integration
calculus
derivative
inequalities
real analysis
real analysis unsolved

Let $I$ an interval of real numbers and $f:I\to R$ a primitivable function.
 If   $ f(\frac{x+y}{2})\leq\frac{f(x)+f(y)}{2}$ for any $x,y$ in $I$, then $f$ is continuous. ;)

Well, since primitivable is not a standard mathematical term on this side of the world, why don't you NOT leave me guessing and give a definition.

We say that a function $f$ has a primitive $F$ if $F(x)=\int f(x)dx$ ;)

I have not managed to prove it. But the following statements may be helpful.

Recall that a derivative has the following two properties:

(1) It doesn't have a jump discontinuity;
(2) It has the intermediate value property (Darboux property).

And the two propositions:

(1) If $g(x^+), g(x^-)$ exist for all $x$, and $g$ satisfies the Jensen inequality ($n=2$) then it is continuous.
(2) If $g$ is bounded and satisfies the Jensen inequality ($n=2$), then it is continuous.

I have another solution. Let $F$ be a primitive. Since $F$ is Jensen convex, as in the proof of the Jensen inequality we can prove that $ F'(\frac{x_1+...+x_n}{n})\leq\frac{F'(x_1)+...+F'(x_n)}{n}$. Thus $F'(ax+(1-a)y)\leq aF'(x)+(1-a)F'(y)$ for all $a\in [0,1]$ rational and any $x,y$.  Now, fix $a$ rational in $[0,1]$ and $x$  (but observe they are arbitrary) and consider the function $ g(y)=aF'(x)y+(1-a)F(y)-\frac{1}{1-a} F(ax+(1-a)y)$. This function is increasing thus for all $y_1<y_2$ we have $ g(y_1)\leq g(y_2)$. This means that for any $x$, any rational $a$ and any $y_1<y_2$ we have $ aF'(x)y_1+(1-a)F(y_1)-\frac{1}{1-a}F(ax+(1-a)y_1\leq aF'(x)y_2+(1-a)F(y_2)-\frac{1}{1-a}F(ax+(1-a)y_2)$.
    Since $F$ is continuous, for $y_1,y_2$ fixed we make $a$ to tend to some arbitrary real $b\in [0,1]$. It follows that the last inequality holds in fact for any real $a\in [0,1]$. Next, make $y_1$ to tend to $y_2$ in the last inequality and this will give you that $f$ is in fact convex. From here I think you know how to proceed. ;)

Nice solution, harazi.

[quote="harazi"]Next, make $y_1$ to tend to $y_2$ in the last inequality and this will give you that $f$ is in fact convex.[/quote]
In fact, I don't see how it works.

But instead, I see that $g(y)$ is increasing for all reals $a\in (0,1)$. And $g'(y)\geq 0$ gives that $f$ is convex.

Ok, it seems that each one of us doesn't see what the other one sees. :)  Why $g$ is increasing for any $a$? And why it follows that $f$ is convex from my solution? It suffices to write the inequality in the form $a F'(x)+(1-a)\frac{ F(y_2)-F(y_1)}{y_2-y_1}\geq (1-a)\frac{F(ax+(1-a)y_2)-F(ax+(1-a)y_1)}{y_2-y_1}$. Now make $y_2$ to tend toward $y_1=y$.

[quote="harazi"]Why $g$ is increasing for any $a$? And why it follows that $f$ is convex from my solution? [/quote]

It holds \[aF'(x)y_1+(1-a)F(y_1)-\frac{1}{1-a}F(ax+(1-a)y_1)\leq aF'(x)y_2+(1-a)F(y_2)-\frac{1}{1-a}F(ax+(1-a)y_2)\] all $y_1<y_2$ and rational $a\in (0,1)$.

Suppose $b\in (0,1)$ ($b$ may be irrational here). Pick up a sequence of rational numbers converging to $b$. Taking limit of both sides of the equality above (where we replace $a$ by the terms of the rational sequence), we see that it also holds if we replace $a$ by $b$.

Since $y_1<y_2$ and they are arbitrarily chosen, we conclude that $g(y)$ is increasing for any $b\in (0,1)$. And $g'(y)\geq 0$ tells us that $f$ is convex.

Furthurmore, a convex function is continuous inside the interval. In order to complete the proof, let us consider the endpoints of the interval. Suppose that $I=[a,b)$ and we want to prove that $f(x)$ is right continuous at
$x=a$. If not, then there exists $\varepsilon > 0$, and for every $\delta>0$, there exist $x_n\in (x,x+\delta)$ such that $|f(x_n)-f(x)| > \varepsilon$. Hence there exists a subsequence $\{x_{n_k}\}$ such that $f(x_{n_k})> f(x)+\varepsilon$ or $f(x_{n_k})< f(x)+\varepsilon$. Without loss of generality, suppose $f(x_{n_k})> f(x)+\varepsilon$. Since $f(x)$ is a derivative, it has intermediate value property, so there exist a sequence $y_n\searrow a$ such that $f(y_n)=f(a)+\varepsilon$. From the convexity, $f(x)\leq f(a)+\varepsilon$ on $(a,a+y_1)$, which contradicts with $f(x_{n_k}) > f(a)+\varepsilon$. Now we conclude that $f(x)$ is right continuous at $x=a$.


real analysis
calculus
integration
real analysis solved

Suppose $f(x)$ is increasing on $[0,1]$. Prove that it holds
$\int_{[0,m(E)]} f(x) dx \leq \int_E f(x) dx$
for any measurable set $E\subset [0,1]$?

Assuming $f\geq0$, looks like it boils down to proving $m(E\cap[m(E),1]) = m([0,m(E)]\cap E^c)$ which follows from $m(E) = m([0,m(E)]\cap E) + m([0,m(E)]\cap E^c)$.  Is this wrong?

[b]Edit[/b]: Then we get

 $\displaystyle \int_{[0,m(E)]} f(x) dx = \int_{E \cap [0,m(E)]} f(x) dx + \int_{E^c \cap [0,m(E)]} f(x) dx \leq \int_{E \cap [0,m(E)]}f(x) dx + \int_{E \cap [m(E),1] } f(x) dx = \int_E f(x) dx$ because f is increasing.

yes..that is right...
could you please go on?

Again, you can consider $E$ to be the limit of a decreasing sequence of open sets, and since the conclusion holds whenever $E$ is an interval, we also get it in the general case, I think.

[quote="liyi"]Can the moderator move this topic to 'unsolved problems' section?
I didn't solve it...[/quote]
Ok!

Now it can be moved to 'solved problems' section.
:D

And what solution is correct, in your opinion? ;)

Limpet, you surely need more explanation on the integral part of your proof.:wacko:  The part about measure is absolutely correct!!

[quote="dickchimney"]Limpet, you surely need more explanation on the integral part of your proof.[/quote]
It is my opinion too ;)

I was looking through some old problems on this forum. Here's what I meant in post number $3$ (although I think limpet's approach can be used to get a nicer solution :)):

Let $G_n$ be a sequence of open sets which tends to a $G_{\delta}$ set which is whithin a set of measure zero of $E$. Each $G_n$ is the union $\bigcup_{m\ge 1}I_m^n$ of some disjoint open intervals. Now translate each interval $I_{m_0}^n$ to the left so that its left endpoint becomes $a$, where $a$ is the measure of the set formed by the union of all the intervals $I_m^n$ which are to the left of $I_m^n$ in $G_n$. By using the monotonicity, we get $\int_{[0,m(G_n)]}f dx\le \int_{G_n}f dx$. By taking this to the limit when $n\to\infty$, we get the desired result.

Hope it's Ok.


limit
trigonometry
logarithms
calculus
calculus computations

find this limit ! :|

$ \lim_{k\to\infty}\;\prod_{j=k}^{2k}\;\left(\frac{\pi}{2}\cdot\frac{1}{\tan^{-1}j}\right)$

[ answer [i]hidden [/i]below ]

[hide]
$ answer: \;\;\;\quad\boxed{4^{\frac{1}{\pi}}}$
[/hide]

Let $ s_{n}= \prod_{k=0}^{n}\left(\frac{\pi}{2}\cdot\frac{1}{\tan^{-1}(n+k)}\right)$. Taking log to both side,

\begin{eqnarray*}\ln s_{n}& = &-\sum_{k=0}^{n}\ln \left( \frac{2}{\pi}\tan^{-1}(n+k) \right)\\ & = &-\sum_{k=0}^{n}\ln \left( 1-\frac{2}{\pi}\tan^{-1}\frac{1}{n+k}\right)\\ \end{eqnarray*}

But since $ -\ln \left( 1-\frac{2}{\pi}\tan^{-1}x \right) = \frac{2}{\pi}x+o(x)$, we have

$ \ln s_{n}= \left(\frac{2}{\pi}\sum_{k=0}^{n}\frac{1}{n+k}\right)+no\left(\frac{1}{n}\right)$

and $ \ln s_{n}\; \to \; \frac{2}{\pi}\ln2$.


The total number of digits it takes to number the pages of a book from 1 to 250 inclusive is:

a. 637 b. 638 c. 640 d. 644 e. NG

[hide=" click to reveal hidden content"]

I make this $ 642$.

$ 1$ to $ 9$: $ 9$ digits;

$ 10$ to $ 99$: $ 2\times 90 \equal{} 180$ digits;

$ 100$ to $ 250$: $ 3\times 151\equal{} 453$ digits.

Therefore $ 9\plus{}180\plus{}453 \equal{} 642$.

[/hide]

First, let's cover the unit digits, 1 - 9. These are 9 numbers, each with 1 digit, so that's $ 9 \times 1  \equal{} 9$ digits.

Now the tens, 10 - 99. These are 90 numbers, each with 2 digits, so that's $ 90 \times 2 \equal{} 180$.

Now for the hundreds, 100 - 250. That's 151 numbers, each with 3 digits, so we have $ 151 \times 3 \equal{} 453$.

Thus, we have $ 453 \plus{} 9 \plus{} 180 \equal{} 644$, or $ \boxed{\textbf{D}}$.

isabella2296, you have the same solution as mine, except that you make $ 453 \plus{} 9 \plus{} 180 \equal{} 644$, but $ 3 \plus{} 9 \equal{} 12$, so the units digit must be $ 2$.

Oops, you must have posted while I was still typing, and yeah, I made an arithmetic error. Thanks for pointing it out! :)


We consider the series (a_n)n>=1 with elements from R+\{0} so that:

lim n-->oo (1/a_(n+1)^2-1/a_n^2) =a.

i) show that a>=0.
ii) if a>0 then lim n-->oo a_n=0

bye bye!

i) This isn't hard at all: assume a<0. Then there exist N natural and b<0 such that (1/a_(n+1)^2-1/a_n^2) < b for any n>=N. This means that for any n>=N we have 1/(a_(n+1))^2< 1/(a_n)^2+b. Because b<0 this means that there exists k such that 1/(a_k)^2<0, which is false. We got a contradiction.

ii)Denote 1/(a_n)^2=b_n. Then b_(n+1)-b_n->a>0, so b_n->infinity, so a_n->0.


inequalities
linear algebra
matrix
linear algebra unsolved

[b]Let [/b]$ A,B \in M_{n}$ $ \mathbb{(R)}$,   $ Rank(B) = 1$ ,
Find :
         $ 2.det^{2}(A) \leq (det^{2}(A + B) + det^{2}(A - B))$
"..I will try to become a good student of jean ..."

[quote="studen of jean"][b]Let [/b]$ A,H \in M_{n}$ $ \mathbb{(R)}$,   $ Rank(H) = 1$ ,
Find :
         $ 2.det^{2}(A) \leq (det^{2}(A + H) + det^{2}(A - H))$
"..I will try to become a good student of jean ..."[/quote]

there exist P,Q invertible $ H=PJ_{1}Q$ with

$ J_{1}=(k_{ij})$ with $ k_{11}=1$ and $ k_{ij}=0$ for others

let $ B=P^{-1}AQ^{-1}$

$ det(A\pm H)^2=det(PQ)^2(det(B \pm J_{1})^2$

then

$ det(A+H)^2+det(A-H)^2 = det(PQ)^2(2detB^2 + 2x^2)\geq 2det(A)^2$

Use $ \det(I\plus{}M)\equal{}1\plus{}tr M$ for $ rk M\equal{}1$.

I think ! you can prove that with :[b] rank(B)=1[/b]then :
             $ det(A\plus{}B)\plus{}det(A\minus{}B)\equal{}2.detA$ [b](A Very Nice !!!)[/b]
[b]I Love Ho Chi Minh but don't understand they call ...?[/b]
[b]They don't understand nice of Math !!!![/b]

[quote="studen of jean"][b]I Love Ho Chi Minh but don't understand they call ...?[/b]
[b]They don't understand nice of Math !!!![/b][/quote]What do you mean? Who is Ho Chi Minh?

[quote="studen of jean"]I think ! you can prove that with :[b] rank(B)=1[/b]then :
             $ det(A \plus{} B) \plus{} det(A \minus{} B) \equal{} 2.detA$[/quote]

Yes, you can. (despite the mistaken appearance that I gave of having disproved this in another thread...)

Now ,if [b]Rank(B)=2[/b]
    $ det(A \minus{} B) \plus{} det(A \plus{} B) \equal{} 2(det(A) \plus{} det(A_{3})$ ([b]Very Nice !![/b])
$ A_{3}$[b] is matrix don't have colum 1,2 .don't row 1,2 [/b] [b]of matrix A[/b]

[quote="studen of jean"]Now ,if [b]Rank(B)=2[/b]
    $ det(A - B) + det(A + B) = 2(det(A) + det(A_{3})$ ([b]Very Nice !![/b])
$ A_{3}$[b] is matrix don't have colum 1,2 .don't row 1,2 [/b] [b]of matrix A[/b][/quote]

Why are all of your matrices 3x3? :)

Also, what you're saying isn't true, as your $ A_3$ always has determinant $ 0$.  Do you mean ${ \frac {\det(A - B) + \det(A + B)}{2} = \det(A) + \det(A_3 + B)}$?  At least that's true when $ B$ is in a certain Jordan normal form. :)

Anyway, start a new thread if you want to talk about these other topics.


quadratics
superior algebra
superior algebra unsolved

This is a very simple problem that I can't figure out for some reason.

a) Find the order of $[3]$ and $[4]$ in $Z_{37}$.
b) Find a primitive element in $Z_{37}$.

If you don't want to do the calculations in a) both orders are 18. However, I can't figure out what to do for b. It's embarrassing. Please help.

I think 5 might work,as it can't have order 2,3,6,9,18 since its not a quadratic residue and the only cases are order 12,4,36 and when i checked i felt that order ca'nt be 12 and 4.
perhaps i am wrong.

hmmm me again proved my stupisity  :blush:

[quote="ZetaX"]When $4$ has order $18$, what can you say about $2$¿[/quote]
Thanks!


calculus
integration
number theory unsolved
number theory

Give a sequence ${a_n}$ bounded such that:
$0\leq 10a_n+7a_{n-1}+a_{n-2}\leq 9$
Show that:There exists $M$ so that $\forall n\geq M$ then :$a_n=0$

that's impossible!
counterexample:
$a_n=0.1$


modular arithmetic
number theory proposed
number theory

Suppose $n \geq 2$ is an integer. Prove:
\[n \mid \big(1^{n-1}+2^{n-1}+\dots +(n-1)^{n-1} \big)+1\]
if and only if for every prime divisor $p$ of $n$,
\[p \mid {n \over p}-1 \ \ \ \ \ \ \ \ ,\ \ \ \ \ \ \ \ p-1 \mid {n \over p}-1 \]

The second condition is clearly equivalent to $p-1|n-1$, so let's write it like this from now on :).

First, given that $p|\frac np-1,p-1|n-1$ for all $p|n$, it means that $n$ is square-free, so that divisibility is equivalent to $p|1^{n-1}+\ldots+(n-1)^{n-1}+1$ for all primes $p|n$, which is clear, since $1^{n-1}+\ldots+(n-1)^{n-1}\equiv (n-p+1)^{n-1}+(n-p+2)^{n-1}+\ldots+(n-1)^{n-1}$, which is $\equiv -1\pmod p$ (we deduce this from $p-1|n-1$).

Conversely, given that $n|1^{n-1}+\ldots+(n-1)^{n-1}+1$, by using an argument similar to the one above, and the fact that $1^a+2^a+\ldots+(p-1)^a$ is $\equiv 0\pmod p$ for a prime $p$, when $p-1\not|a$, and $\equiv -1\pmod p$ when $p-1|a$, we conclude that $p-1|n-1,\ \forall p|n$.

We must now show that we also have $p|\frac np-1$. However, I don't think this is true :?. Take, for example, $n=561=3\cdot 11\cdot 17$, the smallest Carmichael number. We have $17\not|\frac n{17}-1=32$. In fact, the first relation you wrote seems to be equivalent to the fact that $n$ is a Carmichael number (the proof in the first paragraph only uses the fact that $n$ is square-free and $p-1|n-1$ for all $p|n$, it doesn't use the stronger $p|\frac np-1$), while the next two relations form a stronger condition.
What am I missing here? :?

I think you're wrong. The second condition is correct. I solved the problem at exam and I'm sure the problem is correct  :D  :D  :D

Ok, let's take it from the top :).

Assume first that $p-1|n-1\ (*),p|\frac np-1\ (**)$ for all primes $p|n$. Since $(*)$ implies that $n$ is square-free, it suffices to show that $p|S$ for all primes $p|n$, where $S=\sum_{i=1}^{n-1}i^{n-1}+1$. For $i\in\overline{1,n-1}$, we have $i^{n-1}\equiv 1\pmod p$ if $p\not|i$, and $0\pmod p$ otherwise. This means that $S\equiv -\frac np+1\pmod p$, and we get the conclusion.

Conversely, if $n|S$, then from $p|S$ we get $p-1|n-1$ (I believe I explained why in post number $2$), and, using $S\equiv-\frac np+1\pmod p$ again, we get $p|\frac np-1$.

[quote="grobber"]For $i\in\overline{1,n-1}$, we have $i^{n-1}\equiv 1\pmod p$ if $p\not|i$, and $0\pmod p$ otherwise. This means that $S\equiv -\frac np+1\pmod p$, and we get the conclusion.[/quote]

Isn't it $S\equiv +\frac np+1\pmod p$? Am I missing something?

Each sum of $p$ consecutive powers in that sum is $-1\pmod p$. Since there are $\frac np$ such sums (I considered that the sum goes from $1$ to $n$ instead of $n-1$, since it makes no difference anyway), we get $-\frac np$, and we add that $1$ at the end.


Find all $z$ such that $\sqrt{5z+5}-\sqrt{3-3z}-2\sqrt{z}=0$


Haha I have been wanting to make this thread in a while.

What adjustments were made to get used to the freshman year, or high school? Was it a lot different from junior high or middle school? Were the classes much more difficult for you? Did homework take a lot of time? Did you still have free time to do math, etc.?

I guess you get the point.  :lol:

Err...I went from like 10 mins of hw a night to like 3-4 hours... Also my english teacher was impossible...

Like there was nothing scary about the campus or anything, it was just that there was a lot of work.

It became significantly easier to get A's because my high school did not have A-'s... classes were moderately difficult, but not shockingly. I always have free time to do math.

I became about a bazillion times more motivated to get A's.

I started spending less time on math and more time actually having a social life. Because spending one hour on understanding concepts of math and three hours screwing around with friends is approximately as effective as slaving for four hours on a couple math questions.

I went to Red MOP.

and I started getting bad grades and getting generally pwned by myself and my parents.

Darn.

Well, all of a sudden:

i. I had some control over which classes I would take
ii. I had the opportunity to meet many more people, since all the classes were different
iii. High school clubs are a few levels above anything you'll see in middle school
iv. The school orchestra could play in time (on good days)
v. Time management becomes an issue for those with multidimensional interests (and for pilers)
vi. Fewer cut-and-paste assignments, and such. Thank God.

In terms of workload, it tends to increase because in high school, you have the opportunity to take above-average classes, such as AP's. However, the workload becomes, in most cases, slightly more useful (most of it is still busywork).

However, on the whole, the school day becomes a bit more productive, so it's really much, much better.

mostly really easy, I realized that there's no point in taking a class that has a lot of work, so I basically scouted my choices for each year, and I had to actually do homework at home maybe a total of ten times the entire school year? The only thing that was hard to do at school was english essays, but I only had 3-4 a year, and I usually did them during chesmitry or aide period, so that worked out really well

Schoolwise, I found it much less difficult than middle school. Maybe it's because there wasn't as much of an "Get a B and you DIE" mentality, but I found that I had to put significantly less effort into schoolwork. Less effort = more good.

Should I talk about my college freshman year?

I found that people were a lot richer, the number of minorities went from about 60% to about 2% yet and hence, while racial jokes were perfectly acceptable like any other joke before, they were not anymore. The number of Jews went from 2% to like 30%.

And suddenly lacrosse, crew, and Abercrombie became the norm.


trigonometry
function
real analysis
real analysis solved

Let $\alpha$ be the root of the equation of $\cos x=x$.

Prove that $\frac{2\pi}{9}<\alpha<\frac{\pi}{4}$.

$cos(x) - x $ is a decreasing function , so $cos(\frac {\pi}{4}) = \frac {1}{\sqrt{2}} < \frac {\pi}{4}\Longrightarrow \alpha < \frac {\pi}{4} $
$ cos(x) > 1-x^2/2 \geq  x $  if  $ 0 < x < \sqrt{3}-1 $,  so  $ \frac{2\pi}{9}<\sqrt{3}-1<\alpha$
 :cool: Do you know " Hiroshima , mon amour " ? :cool:

Is this a song which sung by Alcatrazz or a movie? :?

It's a movie,  by Alain Resnais ..
 :cool:


calculus
derivative
function
complex numbers
real analysis
real analysis solved

Please help me with this problem, I've tried everything!

How do I solve the derivative of functions f(z)=Arg(z) and g(z)=x+y (as z = x + iy) by using the definition. Meaning [f(z)-f(w)]/(z-w) -> f'(w) as z->w... And where is this derivative defined?

1) This is a question of complex analysis.  Very few high school students know complex analysis.

2) These look very much like homework problems -- do you have to do these for a class?

3) What do you mean, "Where is this derivative defined?"  You gave the definition of the derivative within your post.

Here are some hints.

If you know the Cauchy-Riemann equations, you have necessary and sufficient conditions for these derivatives to exist.  Both of these functions are (continuous) functions from the complex numbers to the (strictly) real numbers.

In particular, the Cauchy-Riemann equations assert certain things about where functions from $\mathbb{C}$ to $\mathbb{R}$ can be differentiable.  Besides that, I probably shouldn't say any more.

>1) This is a question of complex analysis. Very few high school students know >complex analysis.

Here in Sweden this is high school-stuff.

>2) These look very much like homework problems -- do you have to do these for a >class?

These are not my homework. I'm studying for exam and these are basic exercises in my studybook - still I don´t know how to solve these. I think I should figure out how to simplify f(z)-f(w)/z-w so that z-w wouldn't approach to 0... just can´t do that!

>3) What do you mean, "Where is this derivative defined?" You gave the definition of >the derivative within your post.

I meant that for example the derivative of |x| is "defined" in R\{0}. Wrong word I guess.

And for what I've heard Arg(z) isn't continuous in negative x-axis, if it's defined (-pi,pi) as it normally is.[/b]

Using the Cauchy-Riemann equations on $f(z)=\Re z+\Im z$, we get

$u_x(x,y)=1$
$v_y(x,y)=0$

So right away the function is nowhere differentiable.

[quote="JBL"]1) This is a question of complex analysis.  Very few high school students know complex analysis.

[/quote]

Just to go off topic for a second, I guess teaching derivatives of complex numbers isn't so ridiculous for a high school curriculum that doesn't completely ignore complex numbers, huh?  I mean, the derivatives are defined exactly the same; all you need to worry about is going to zero in different 'directions', which is exactly analogous to how it's handled in multivariable calculus.

I can infer that this is not HS level in Sweden.

Calculus of complex functions is not an "Intermediate Topic" for high-school students under any reasonable definition of "Intermediate".

Thus, this topic has been moved.

If $f(z)=u(z)+iv(z)$ is smooth (i.e., has whatever properties of the existence and continuity of real derivatives you would want to ask for), then it is complex analytic if and only if is statisfies the Cauchy-Riemann equations:

$u_x=v_y$
$u_y=-v_x.$

If the function does satisfy these equations, then $f'=u_x+iv_x=v_y-iu_y.$

Now suppose $f$ takes on only real values. Note that this applies to both of your examples. Then $v$=0 and thus $v_x=v_y=0$ everywhere. By the Cauchy-Riemann equations, we then have that $u_x=u_y=0,$ which makes $u$ and hence $f$ constant.

There are no non-constant real-valued complex analytic functions.

Ok, now I got it!

So using the definition I can first approach by keeping Re(z) constant and then Im(z):

1) As Re(z)=Re(w): f(z)-f(w)/z-w = (Im(z)-Im(w))/i(Im(z)-Im(w)) = 1/i

But 2) As Im(z)=Im(w): f(z)-f(w)/z-w = (Re(z)-Re(w))/(Re(z)-Re(w)) = 1

=> f(z) isn't differentiable anywhere. Or is it in (0,0)?

Can this kind of approach be done somehow to the other function Arg(z)?

[quote="MarieKaasi"]Can this kind of approach be done somehow to the other function Arg(z)?[/quote]
Yes, certainly. All you really need to know about Arg(z) is that it is real valued; the  limit 1 in your argument would become some other number, but you wouldn't even need to know what that number is specifically.

An yes, these functions fail to have a complex derive everywhere. There aren't any points that are different.


spontaneous prose was done by the beatniks where there would be a cirlce of people each with a pen and paper.   they would write and after about a sentence worth of time and then pass their prose around.  the beatniks wouldnt read what was written on the sheet and would just start writing, but we dont have to do it that way.  i'll start with a sentence or two, then someone else read what the previous person wrote and (copy and paste what the other person wrote, preferably and) write down the first thing that comes to mind that ought to  follow.  (it doesnt have to make logical sense, it just has to make sense to you when you write it! the settings ushually change erratically.)  so, to no further ado:

The coffee has lost it's hot vigor and has now cooled lukewarm and disatifying.

The coffee has lost it's hot vigor and has now cooled lukewarm and disatifying.
I went to drink the coffee, but it was disgusting. I spit it back into the cup.

i like tora's line better so...

The coffee has lost it's hot vigor and has now cooled lukewarm and disatifying.
I probe it, briefly, with my swotty pen, and am greatly amused at the flying pigs of yore.
The dungenese mushrooms that grew every spring ceased to reveal itself.

The coffee has lost it's hot vigor and has now cooled lukewarm and disatifying.
I probe it, briefly, with my swotty pen, and am greatly amused at the flying pigs of yore.
The dungenese mushrooms that grew every spring ceased to reveal itself.
I lay on my back and dream of the lady of Shalott who longed to go to Camelot.
Just then suddenly, my pants seem to have been stucken with rigor mortis.
I jump into the shower, still thinking about the lady of Shalott.
And whom should I find taking a cool shower but the lady of Shalott herself!
I say "Aren't you supposed to be looking down to Camelot?"
To which she replied, "I will, as soon as you are good enough to remove yourself from my presence so that I can step out of the bath and put something on."
I quickly get out of the bathroom, and the lady put on a dress that most envy, she steps down the stairway, and vanishes.
So I get back to my shower, now thinking of dungeons and dragons.
I get out thinking "I should play that"
But I find to my distress that I don't have it so I start playing Brothers in Arms instead.
It turns out that I am bad at that game, so I start playing Armagetron instead.
I come to the sickening realization that I am bad at any game whatsoever, and sit brooding.
I then realize that USAMO is today. I hurry and get on the road.
i find that i've left all my things back at home and rush back only to find i've left my key inside.

The coffee has lost it's hot vigor and has now cooled lukewarm and disatifying.
I probe it, briefly, with my swotty pen, and am greatly amused at the flying pigs of yore.
The dungenese mushrooms that grew every spring ceased to reveal itself.
I lay on my back and dream of the lady of Shalott who longed to go to Camelot.
Just then suddenly, my pants seem to have been stucken with rigor mortis.
I jump into the shower, still thinking about the lady of Shalott.
And whom should I find taking a cool shower but the lady of Shalott herself!
I say "Aren't you supposed to be looking down to Camelot?"
To which she replied, "I will, as soon as you are good enough to remove yourself from my presence so that I can step out of the bath and put something on."
I quickly get out of the bathroom, and the lady put on a dress that most envy, she steps down the stairway, and vanishes.
So I get back to my shower, now thinking of dungeons and dragons.
I get out thinking "I should play that"
But I find to my distress that I don't have it so I start playing Brothers in Arms instead.
It turns out that I am bad at that game, so I start playing Armagetron instead.
I come to the sickening realization that I am bad at any game whatsoever, and sit brooding.
I then realize that USAMO is today. I hurry and get on the road.
I find that i've left all my things back at home and rush back only to find i've left my key inside.
I wondered how such a feat could be accomplished.
Then I remembered that I threw the key in through an window to shoo the cat.
I went to the car where the cat got shooed, but the key wasn't there.
I went to the cat that I shooed, thinking that he ate it.
My suspicions turned out well founded.
I took the cat to the emergency room.

bubka Posted: Fri May 12, 2006 6:46 am Post subject:

--------------------------------------------------------------------------------

The coffee has lost it's hot vigor and has now cooled lukewarm and disatifying.
I probe it, briefly, with my swotty pen, and am greatly amused at the flying pigs of yore.
The dungenese mushrooms that grew every spring ceased to reveal itself.
I lay on my back and dream of the lady of Shalott who longed to go to Camelot.
Just then suddenly, my pants seem to have been stucken with rigor mortis.
I jump into the shower, still thinking about the lady of Shalott.
And whom should I find taking a cool shower but the lady of Shalott herself!
I say "Aren't you supposed to be looking down to Camelot?"
To which she replied, "I will, as soon as you are good enough to remove yourself from my presence so that I can step out of the bath and put something on."
I quickly get out of the bathroom, and the lady put on a dress that most envy, she steps down the stairway, and vanishes.
So I get back to my shower, now thinking of dungeons and dragons.
I get out thinking "I should play that"
But I find to my distress that I don't have it so I start playing Brothers in Arms instead.
It turns out that I am bad at that game, so I start playing Armagetron instead.
I come to the sickening realization that I am bad at any game whatsoever, and sit brooding.
I then realize that USAMO is today. I hurry and get on the road.
I find that i've left all my things back at home and rush back only to find i've left my key inside.
I wondered how such a feat could be accomplished.
Then I remembered that I threw the key in through an window to shoo the cat.
I went to the car where the cat got shooed, but the key wasn't there.
I went to the cat that I shooed, thinking that he ate it.
My suspicions turned out well founded.
I took the cat for surgery to try to find that obnoxious piece of metal
I use that obnoxious piece of metal to cure cancer while spelling supercalifragilisticexpialadocious.

The coffee has lost it's hot vigor and has now cooled lukewarm and disatifying.
I probe it, briefly, with my swotty pen, and am greatly amused at the flying pigs of yore.
The dungenese mushrooms that grew every spring ceased to reveal itself.
I lay on my back and dream of the lady of Shalott who longed to go to Camelot.
Just then suddenly, my pants seem to have been stucken with rigor mortis.
I jump into the shower, still thinking about the lady of Shalott.
And whom should I find taking a cool shower but the lady of Shalott herself!
I say "Aren't you supposed to be looking down to Camelot?"
To which she replied, "I will, as soon as you are good enough to remove yourself from my presence so that I can step out of the bath and put something on."
I quickly get out of the bathroom, and the lady put on a dress that most envy, she steps down the stairway, and vanishes.
So I get back to my shower, now thinking of dungeons and dragons.
I get out thinking "I should play that"
But I find to my distress that I don't have it so I start playing Brothers in Arms instead.
It turns out that I am bad at that game, so I start playing Armagetron instead.
I come to the sickening realization that I am bad at any game whatsoever, and sit brooding.
I then realize that USAMO is today. I hurry and get on the road.
I find that i've left all my things back at home and rush back only to find i've left my key inside.
I wondered how such a feat could be accomplished.
Then I remembered that I threw the key in through an window to shoo the cat.
I went to the car where the cat got shooed, but the key wasn't there.
I went to the cat that I shooed, thinking that he ate it.
My suspicions turned out well founded.
I took the cat for surgery to try to find that obnoxious piece of metal.
I take the key, and finally enter my room, perspiring heavily.
I probe myself, with my sweaty pin, an I am no longer greatly amused at the walking pigs of lore.
To be precise, I am of the opinion that the walking pigs of lore bore.
I look out, and see the flames of the sky have been drenched.


geometry
trapezoid
rectangle
ratio
trigonometry
similar triangles
AMC

As I said the problem is 20 my friends think that the answer is 98 can someone give a step by step solution?
20:
Trapezoid ABCD has bases AB and CD and diagonals intersecting at K. Suppose that AB=9 and DC=12, and that the area of Triangle AKD is 24. What is the area of trapezoid ABCD?

[hide]Draw isoceles trapezoid ABCD. Draw diagonals intersecting at K. Obviously, AKD is congruent to BKC. Likelywise, AKB is similar to CKD. Call the altitude to base AB of triangle AKB 3x and the altitude to base CD of triangle CKD 4x. Now the area of the trapezoid is the sum of areas AKB, CKD, AKD, and BKC which equals $ 13.5x\plus{}24x\plus{}48$. Now take isosceles trapezoid ABCD again. Drop altitudes to the base CD from point A and point B to get a rectangle and two triangles. The area of the rectangle is $ 63x$ and the area of the two triangles is $ 10.5x$ as the lengths of the legs are 1.5 and 7x. So now you have $ 63x\plus{}10.5x\equal{}13.5x\plus{}24x\plus{}48$ and simple arithmetic gives you $ x\equal{}\frac{4}{3}$. $ \frac{4}{3}*73.5\equal{}98$.[/hide]

why dividedby 7?  is it because the ratio is 4/3 and 4+3=7?

Yes, because the two bases add to 21 and the formula for a trapezoid is $ (b_1\plus{}b_2)*h/2$.  None of the other answer choices make much sense if you look at it this way.

You can make it more shorter by applying the following formula:
S(ABC)=1/2.AB.AC.sinBAC
Now just using the similar triangle to get the ratio and then use the area of S(AKD), we will get the result as above!

btw, ignite, BKC will have area of 24 (or area(BKC)=area(AKD)) no matter what the trapezoid is in this problem.

of course to assume its isoscles, you can make it easier, but with two area ratios, you will find that the area of BKC is always 24.

infact, i think for any trapezoid thats the case.

Similar triangles for the top and bottom, and use the ratio of the bases for the left and top one... $ 24\plus{}\frac{3}{4}*24\plus{}24\plus{}\frac{16}{9}*\frac{3}{4}*24\equal{}98$.

Pretty simply.. right one is by laws of sines or the ratio of the bases of the right and top triangles.  Bottom one is simply $ \frac{4^2}{3^2}$

[quote="infinity4ever"]Also, when in doubt, go with the answer that makes the most sense.  (NOT 96 though).  I remember that none of the answer choices divided 7 except for 98  :D

However, guessing < doing the problem and getting it right.  Ignite's solution is perfectly fine.[/quote]

Now AMC is going to put all choices multiple of 7 next year.


We have a number of towns, with bus routes between some of them (each bus route goes from a town to another town without any stops). It is known that you can get from any town to any other by bus (possibly changing buses several times). Mr. Ivanov bought one ticket for each of the bus routes (a ticket allows single travel in either direction, but not returning on the same route). Mr. Petrov bought n tickets for each of the bus routes. Both Ivanov and Petrov started at town A. Ivanov used up all his tickets without buying any new ones and finished his travel at town B. Petrov, after using some of his tickets, got stuck at town X: he can not leave it without buying a new ticket. Prove that X is either A or B.

In the graph with towns as the vertices and routes as the edges, Ivanov's journey implies $deg(A)$ and $deg(B)$ are odd and the rest even. If Petrov ends on $X \neq A,B$ then $n deg(X)$ must be odd so $deg(X)$ must be odd. $\bot$


$P$ is a point inside the triangle $ABC$. Lines are drawn through point $P$ such that they are parallel to the sides of the triangle. If the areas of the three resulting triangles with a vertex at $P$ have areas $4$, $9$, and $49$, find the area of $ABC$.

[hide="Solution"]First of all, we see that the three triangles of which we are given areas are similar to triangle $ABC$.

If we let the three triangles of which have areas $4$, $9$, and $49$, be $x$,$y$, and $z$, respectively, we see that the ratios of the sides are $2$, $3$, and $7$.

Now, we can apply those parallel lines!! Notice that if we add together all three of the corresponding sides of the three smaller trangles, the length of the new segment is equal to correspoinding side on triangle $ABC$.

Thus, the ratios of the side lengths of $ABC$ to our three smaller triangles is $2: 3: 7: 12$. Thus, the desired area is $12^2=144$. [/hide]

[hide="Solution"]
Notice that it is possible to "slide" the second and third triangles to the left side of $\triangle ABC$, so that all triangles have one side adjacent to $\triangle ABC$, and the sum of their sides is equal to the left side of $\triangle ABC$.

Notice that all the triangles are similar. Since the ratio of their areas is $4: 9: 49$, the ratio of their sides must be $2: 3: 7$, the left side of $\triangle ABC$ must be $2+3+7=12$. Thus the area of $\triangle ABC$ must be $12^2=144$.
[/hide]


calculus
integration
algebra unsolved
algebra

Find all positive reals $a$ such that every positive integer can be expressed as a finite sum of distinct integral powers of $a$.

for all natural m \[ m=\sum_{j=1}^k a^{i_j} \], were ${i_j}$ distinct non negative integers, or distinct any integers?
If non-negative then $a=2^{1/m}$, or more solutions, for example $a=\frac{\sqrt 5 +1}{2}$.

distinct any integer.
Easy to see if $a$ is a solution,then for any positive integer $m$,$a^{\frac{1}{m}},\frac{1}{a}$ also a solution.
So we only need to show all basic solution bigger than 1.
If $a$ is rational,$a=2$.
But otherwise,I can't solve

well i know that $a=3$ works but i have no idea on how to find other solutions. :oops:

No,$a=3$ doesn't work.For example,we can't express 2.

woops i thought it meant any combination of addition and subtraction. sorry.


search
inequalities
inequalities unsolved

I know it!  :P
(if 2 pages of results are too much to you, why not restrict search to "Inequalities Theorems & Formulas"?)
[hide=":D"]you are strange, you even know that smilies can contain hidden text but you don't know how to search for "cauchy in engel form"??  :wink: [/hide]

[quote="conejita"]What is the Cauchy in the Engel form???
I only know Cauchy-Schwarz:  $(a+b)(c+d)\geq (ac+bd)^{2}$[/quote]
See [url=http://www.mathlinks.ro/Forum/viewtopic.php?&t=99058]here[/url].

I'll be nice onlike some other people (lol jk  :o )
Cauchy in Engel form is just another tool to help with fractions:
$\frac{a^{2}}{b}+\frac{c^{2}}{d}\geq \frac{(a+c)^{2}}{b+d}$
of course, this is just a particular case of Cauchy :D but its helpful..


algebra
polynomial
algorithm
inequalities
absolute value
real analysis
real analysis unsolved

Is there an infinite sequence $a_{0},a_{1},\dots$ of nonzero real numbers such that for $n=1,2,\dots$ the polynomial
$p_{n}(x)=a_{0}+a_{1}x+a_{2}x^{2}+\dots+a_{n}x^{n}$ has exactly $n$ distinct real roots?

There are obviously two possibilities here- either there is a sequence or there isn't. If there is, we want to construct such a sequence; an explicit sequence or at least an algorithm is the most convincing way to prove existence.
If there isn't, we'll have to use inequalities to get there; the coefficients of a polynomial with only real roots satisfy a family of inequalities by Viete's relations and various classical inequalities.

I won't spoil this with a full solution, but here's the [hide="answer"]Yes, there is such a sequence.[/hide]

After peeking at the spoiler, the following idea came to my mind: We construct the sequence inductively. A polynomial of degree $1$, $a_{1}X+a_{0}$ always has real roots (take $a_{0}\neq 0$). Next, given a "good" polynomial $P_{n}= a_{0}+a_{1}X+\ldots+a_{n}X^{n}$, I thought of letting $a_{n+1}$ tend to $0^{+}$. $n$ of the roots of the new polynomial should be more or less the same as the roots of $P_{n}$ and the last root is also real, but distinct from the others. That's because:
(a) we can make $a_{n+1}X^{n+1}$ grow locally a lot slower than $P_{n}$;
(b) there are only a finite number of roots to worry about;
(c) the last root should be a big one in absolute value (that's why I took $a_{0}\neq 0$).

It's more of an intuitive proof, but I hope that the idea is correct.

what if i wrote this: fourier series to the nth term.

let the sequence be the fourier coefficients

is this right??????

[quote="perfect_radio"](a) we can make $ a_{n \plus{} 1}X^{n \plus{} 1}$ grow locally a lot slower than $ P_{n}$;[/quote]

The problem has been posted before and this is the key concept.  In fact, we can find bounds on $ a_{n\plus{}1}$ to make this proof rigorous.


floor function
limit
calculus
calculus computations

Discuss the continuity of $f$
$f(x)=\left\{ \begin{array}{ll}(-1)^{\lfloor x^{2}\rfloor}&{ \text{if}\ \ x < 0}\\ \lim_{n \to \infty}\frac{1}{1+x^{n}}&{ \text{if}\ \ x \geq 0}\end{array}\right.$

$f$ is discontinuous at $x=-\sqrt{k}$ for all positive integers $k$, because it jumps from $-1$ to $1$ or back at such points. Also discontinuous at $x=1$, where the left limit is $1$, right limit is $0$, and $f(1)=1/2$. But $f$ is continuous at $0$, where both definitions yield the value $1$.


algorithm
number theory
modular arithmetic

I know this question has to do with mods, but I don't get it

If n/5 leaves a remainder of 1, what would 3n/5 leave a remainder of.

If $ \frac{n}{5}$ leaves a remainder of $ 1$ and a quotient of $ q$, then $ n \equal{} 5q \plus{} 1$.  What is $ 3n$?

15q+3, but how would you do it with mods?

Mods are just a shorthand for talking about the remainders alone.  The point is that the "part that is divisible by $ 5$" is still divisible by $ 5$ after we multiply by $ 3$, so to find the remainder after multiplication by $ 3$ it suffices to multiply the old remainder by $ 3$.  Modular addition works the same way.  The main thing I'm trying to get across here is that the above reasoning is not essentially different from modular arithmetic.

In terms of modular equivalence, $ n \equiv 1 \bmod 5 \implies 3n \equiv 3 \bmod 5$.

[quote="Urc"]Oh, thanks!

I actually understand modular arithmetic and the equivalent now![/quote]
Do I detect a bit of a sarcastic undertone? Don't do that. When someone answers your questions here, they're trying to help you.

[quote="Urc"]If n/5 leaves a remainder of 1, what would 3n/5 leave a remainder of.[/quote]

First, in modulo (as opposed to "modular") arithmetic the symbol  $ \,\displaystyle{\equiv}\,$  means [i]congruent[/i], not equivalent.

Second, the statement $ \displaystyle{n\equiv 1\bmod{5}}$ translates as "$ n$ is congruent to $ 1$ modulo $ 5$" in English.

Two numbers are congruent to each other modulo some number, $ m$, if $ m$ divides their difference. For example, in your problem $ \displaystyle{n\equiv 1\bmod{5}}$ means $ \displaystyle{\frac{(n - 1)}{5}}$ leaves a zero remainder.

Another way to look at this is that if you divide $ n$ by $ 5$ it leaves a remainder of $ 1$.

Now, if $ \displaystyle{\frac{(n-1)}{5}}$ leaves a remainder of zero, then $ \displaystyle{3\cdot\frac{(n-1)}{5}}$ also leaves a remainder of zero, does it not?

Carrying out this multiplication gives: $ \displaystyle{\frac{(3n - 3)}{5}}$.

Remember though, that we said that $ 5$ divides $ (n-1)$ is the same as saying $ \frac{n}{5}$ leaves remainder $ 1$, so that $ 5$ divides $ (3n - 3)$ means $ \frac{3n}{5}$ leaves remainder $ 3$. The shorthand way of writing that in modulo arithmetic is: $ \displaystyle{3n\equiv 3\bmod{5}}$.

The benefit of modulo arithmetic is that multiplication and addition work the same way they do for an ordinary equation. In other words,
\[ n\equiv 1\bmod{5} \Rightarrow 3\cdot n\equiv 3\cdot 1\bmod{5} \Rightarrow 3n\equiv 3\bmod{5}\]

Therefore, $ n/5$ leaves remainder $ 1\implies 3n/5$ leaves remainder $ 3$.

[quote="Dr. No"][quote="Urc"]Oh, thanks!

I actually understand modular arithmetic and the equivalent now![/quote]
Do I detect a bit of a sarcastic undertone? Don't do that. When someone answers your questions here, they're trying to help you.

[quote="Urc"]If n/5 leaves a remainder of 1, what would 3n/5 leave a remainder of.[/quote]

First, in modulo (as opposed to "modular") arithmetic the symbol  $ \,\displaystyle{\equiv}\,$  means [i]congruent[/i], not equivalent.

Second, the statement $ \displaystyle{n\equiv 1\bmod{5}}$ translates as "$ n$ is congruent to $ 1$ modulo $ 5$" in English.

Two numbers are congruent to each other modulo some number, $ m$, if $ m$ divides their difference. For example, in your problem $ \displaystyle{n\equiv 1\bmod{5}}$ means $ \displaystyle{\frac {(n - 1)}{5}}$ leaves a zero remainder.

Another way to look at this is that if you divide $ n$ by $ 5$ it leaves a remainder of $ 1$.

Now, if $ \displaystyle{\frac {(n - 1)}{5}}$ leaves a remainder of zero, then $ \displaystyle{3\cdot\frac {(n - 1)}{5}}$ also leaves a remainder of zero, does it not?

Carrying out this multiplication gives: $ \displaystyle{\frac {(3n - 3)}{5}}$.

Remember though, that we said that $ 5$ divides $ (n - 1)$ is the same as saying $ \frac {n}{5}$ leaves remainder $ 1$, so that $ 5$ divides $ (3n - 3)$ means $ \frac {3n}{5}$ leaves remainder $ 3$. The shorthand way of writing that in modulo arithmetic is: $ \displaystyle{3n\equiv 3\bmod{5}}$.

The benefit of modulo arithmetic is that multiplication and addition work the same way they do for an ordinary equation. In other words,
\[ n\equiv 1\bmod{5} \Rightarrow 3\cdot n\equiv 3\cdot 1\bmod{5} \Rightarrow 3n\equiv 3\bmod{5}
\]
Therefore, $ n/5$ leaves remainder $ 1\implies 3n/5$ leaves remainder $ 3$.[/quote]

How would you detect sarcastic tones online...? Anyway, I actually do understand it now.

Sarcastic undertones are found usually by the person who reads it. Sometimes you don't type something that is sarcastic, but then other people interpret that it is (This case included).

Straying away from english. Here is a good link for more info on Modular Arithmetic

[url=http://www.artofproblemsolving.com/Wiki/index.php/Modular_arithmetic/Introduction]Introduction to Modular Arithmetic[/url]

[quote="Urc"]How would you detect sarcastic tones online...? Anyway, I actually do understand it now.[/quote]

Context, I suppose. But in as much as I was wrong, I apologize. Glad to hear that your understanding is genuine.

Or, for this problem, you could be cheap and just assume n=6 or something. Of course, that wouldn't help you learn modular arithmetic, but you don't really need it for this problem.

Modular arithmetic is a system of arithmetic for integers (note the importance - integers only) where numbers "wrap around" to 0 after they reach a certain value. For example, in mod 6, the numbers go from 0, 1, 2, 3, 4, 5, and by the time it reaches 6, it goes back to 0. A famous example is the clock. :D The minutes go modulo 60 and the hour goes modulo 12 or 24 depending on your clock configuration.

That's a good example. Another one I can think of right now is poker. Sometimes people play going "around the horn", meaning that in a straight, the order can be K, A, 2, 3, 4.

(This might not be the best example... :( )

So, using Yongyi's clock example, modulo 12 would be: 0,1,2,3,4,5,6,7,8,9,10,11,0,1,2,3....?

How can you use modular arithmetic in math?

By the way, thanks!  :)

[quote="isabella2296"]What exactly is modular arithmetic?[/quote]

Modular arithmetic (I actually prefer saying modulo arithmetic) is arithmetic done on remainders, the left over bits when doing division on integer numbers.

When one says $ a$ is congruent to $ b$ [b]modulo[/b] $ m$, written in symbols as $ a\equiv b\bmod{m}$, one means that $ a$ and $ b$ have the same remainder when divided by $ m$.

The arithmetic part of modulo arithmetic comes from the fact that multiplication and addition work the same for modulo congruences as they do for ordinary equations.

For example, $ 10\equiv 3 \bmod{7}$.

Since that is true, $ 3\cdot 10\equiv 3\cdot 3\bmod{7}\Rightarrow 30\equiv 9\bmod{7}$. Of course $ 9\equiv 2\bmod{7}$ as is $ 30\equiv 2\bmod{7}$.

Modulo arithmetic is useful for clocks, calendars, and computer encryption algorithms among other things.

As a fun thing, see if you can figure out what day Christmas 2009 falls on without looking at a calendar.

[quote="isabella2296"][quote="Yongyi781"]Uh... December 25th?[/quote]

I think he(or she) means day of the week. I'm not exactly sure how to figure that out though...[/quote]
Yes, that is what he means.

Let's see if we can figure it out.

December 25, 2008 is day 360 (this is a leap year -- ordinarily it would be 359).

$ 360\equiv 10\equiv 3\bmod{7}$.

Today, June 2, 2008, is day 154.

$ 154\equiv 0\bmod{7}$

$ (360 \minus{}154)\equiv (3 \minus{} 0)\bmod{7}\Rightarrow 206\equiv 3\bmod{7}$.

$ (206 \plus{} 365)\equiv (3 \plus{} 365)\equiv 4\bmod{7}$;  Christmas 2009 is 4 days after whatever today is.

Now let Sunday $ \equiv 0\bmod{7}$, Monday $ \equiv 1\bmod{7}$, Tuesday $ \equiv 2\bmod{7}$, Wednesday $ \equiv 3\bmod{7}$, Thursday $ \equiv 4\bmod{7}$, Friday $ \equiv 5\bmod{7}$, and Saturday $ \equiv 6\bmod{7}$.

$ \text{Monday } \plus{} 4\equiv (1 \plus{} 4)\equiv 5\bmod{7}\equiv \text{ Friday}$.

For clock problems, because they are in mod 12, you can subtract 12 until you reach a value less than 12. They will be congruent in mod 12. (In mod n, adding or subtracting kn, where k is an integer, leaves the value unchanged.)

This would also work here because the numbers are relatively small. For more complex calculations, you'll have to go into a whole other branch of math.

[quote="Urc"]I also have a questin about how to do most clock problems...I just can't understand them...[/quote]

Let's work an example of a clock problem.

First, suppose tomorrow is your last day of school before summer break, and your school gets out at 3:15 pm.  Suppose tomorrow morning at 8:23 am, while sitting in math class, you decide you want to know how long it is until you are dismissed for the summer. The trick is that clocks use two different moduli: 12 for the hours, and 60 for the minutes. So you have to do the calculations for hours and minutes separately.

Starting with the minutes: $ (15 \minus{} 23)\equiv \minus{}8\bmod{60}$.

Now for the hours: $ (3 \minus{} 8)\equiv \minus{}5\bmod{12}$.

The problem now is that both these numbers are negative.

If we subtract an hour from the hours, and add 60 minutes to the minutes, we don't change the nature of the result since we subtract an hour and add an hour (60 minutes), we effectively add nothing.

\[ (\minus{}5 \minus{} 1)\equiv \minus{}6\bmod{12}\quad (\minus{}8 \plus{} 60)\equiv 52\bmod{60}\]

The hours are still negative, but $ 12\equiv 0\bmod{12}$, so we can add $ 12$ or any multiple of $ 12$ to the hours, and it doesn't change the remainder modulo $ 12$ at all.
\[ (\minus{}6 \plus{} 0)\equiv (\minus{}6 \plus{}12)\equiv 6\bmod{12}\]

Now, both the hours and the minutes are positive; hence, you have $ 6\text{ hours, }52\text{ minutes }$ until you get out of school for the summer.

I'd like to send a big thanks to Hunter 34, for providing the link to the Introduction of Modular Artimetic and Dr. No, for your explainations regarding Modular Arithmetic ^^. It really cleared up quite a few clouds for me in respects to Modular Arithmetic. After reading the Excerpt on one of the AoPS Introductory texts regarding Modular Arithmatic and getting a ">.<" reaction to it, you both helped me out a great deal. Glad I stumbled on this thread while looking around. You have my gratitude ^^.

~lZerol