topology
calculus
integration
invariant
advanced fields
advanced fields theorems

How do we prove it? I'm not looking for something compeltely rigorous. something between intuitive and rigorous.

http://math.etsu.edu/MultiCalc/Chap5/Chap5-7/index.htm

with a simple google search....

For Stokes' theorem in full generality:
It's a trivial calculation on a box; break into components and apply the 1-dimensional Fundamental Theorem in each direction.
Everything looks like a box; any compact smooth manifold with boundary is diffeomorphic to a nonoverlapping finite union of boxes, and the integrals we're looking at are diffeomorphism invariant (up to orientation).

All of the work is in the definitions.


Question 1:

Let $ N ={a_{m}}10^{m}+\cdots+a_{2}{10^{2}}+a_{1}{10}+a_{0}$, where $ 0 \leq a_{k}\leq 9$, be the decimal expansion of a positive integer $ N$. Prove that $ 7$, $ 11$, and $ 13$ all divide $ N$ if and only if $ 7$, $ 11$ and $ 13$ divide the integer

$ M = (100a_{2}+10a_{1}+a_{0})-(100a_{5}+10a_{4}+a_{3})+(100a_{8}+10a_{7}+a_{6})-\cdots$

Question 2:

Find all the solutions of the linear congruence $ 3x-7y \equiv 11 (mod\,13)$

[hide="Further Hint, #1"]
Try out a couple examples where $ 1001|N$ and where $ 1001 \not | N$, and think about what happens to $ M$ every time you add $ 1001$ to $ N$. ;)[/hide]

[hide="Solution, #2"]
Multiply both sides by the inverse of $ 7$ in $ \bmod 13$ (i.e. $ 7^{-1}\equiv 2$, since $ 7 \cdot 2 \equiv 1 \bmod 13$) to obtain $ \boxed{y \equiv 6x+4 }\ \bmod 13$, from which we obtain solutions for each residue $ x$.[/hide]

That's an awful lot more work than needs to be done.  Try just dividing $ N$ into chunks of three digits as per the problem statement.

t0rajir0u...I originally wrote the solution without bothering with the sigma notation steps, and breaking into groups of 3 as you suggested. However, the sign of the congruences that I obtained contained the expression $ (-1)^{k}$, so I then had to look at when k even and odd. For this reason I just looked at groups of 6 instead of every second group of 3. Many steps can be omitted. For instance in the converse proof, the introduction of k is completely unneccessary, but I thought I would include it to show that we are not restricted to the choice of p.

[quote="sludgethrower"]However, the sign of the congruences that I obtained contained the expression $ (-1)^{k}$[/quote]

This is precisely the pattern obtained in the required expression.  What is the problem?

There was no problem with doing it like that, it was just that I wanted an opportunity to use sigma notation. I love sigma notation :)


Define set A to be a set in which every element is distinct and fixed in its position, and define subset F* to be a subset of set A (the elements of F* are also fixed in position.)

F* is RISING if and only if each element of F* is greater than the one to its left, and FALLING if each element of F* is less than the one to its left.

Prove that $\forall K \in \mathbb{Z}$, a set A of cardinality $K^{2}+1$ has either one RISING subset F* or one FALLING subset F* of length $K+1$ elements

<bump>

interesting problem Matt :)
i wanna see how this works

I tried attacking this one with ordered pairs, and inducting, and I think I made some progress. If anyone has a solution, I'd like to see it.

You can find a proof [url=http://mathforum.org/library/drmath/view/55942.html]here[/url]. That's where I stole it from.

I'm changing the notation so that $n = K$ and $k$ is a totally different variable. Also, let $a_{i}$ be the element in the $i$th position.

The basic idea is to show that if there is no rising subset of length $k+1$, then there must be a falling subset of length $k+1$.

Let $m(k)$ be the length of the longest rising subset starting at $a_{k}$. Assume that $m(k) \le n$ for all $k$, i.e. there is no rising subset of length $n+1$. Also, we trivially have $m(k) \ge 1$.

Since there are $n^{2}+1$ values of $m(k)$ (at each of the positions in the set), all of which are elements of the set $\{1, 2, \ldots, n\}$, by pigeonhole we know $n+1$ of the $m(k)$'s are equal. So $m(k_{1}) = m(k_{2}) = \cdots = m(k_{n+1})$ with $k_{1}< k_{2}< \cdots < k_{n+1}$.

If $a_{k_{i}}< a_{k_{j}}$ with $i < j$, we can add $a_{k_{i}}$ to the longest rising subset starting with $k_{j}$, so that $m(k_{i}) > m(k_{j})$, contradicting their equality. Hence we must have $a_{k_{1}}> a_{k_{2}}> \cdots > a_{k_{n+1}}$, forming a falling subset with $n+1$ elements.

Excellent! I didn't think of using pidgeonhole, but that makes sense now. Good call.

Generalization:  In a sequence of length $mn+1$ either there is an increasing subsequence of length $m+1$ or a decreasing subsequence of length $n+1$.

Is the proof given in the link presented?


On the AMC 10, how many points do you get for leaving a question blank?
Each place I look says a different number. :(

1.5, it used to be 2.5, but they changed it last year.

For the definitive, authoritative, complete answer, see the AMC 10/12 Teachers' Manual,
http://www.unl.edu/amc/d-publication/d1-pubarchive/2007-8pub/1012TM/2008AMC1012A-TM.pdf
Section II, Paragraph 5, on page 5 and instruction 4 on the facsimile front page of the AMC 10 and AMC 12 on pages 26 and 27.

Steve Dunbar
MAA Director, American Mathematics Competitions


search
number theory proposed
number theory

Hi,

find all integers $ p$ and $ q$ such that $ p\vert q^{2}\plus{}1$ and $ q\vert p^{2}\plus{}1$.

this is a standart problem with infinit descent
if $ \frac{x^{2}\plus{}y^{2}\plus{}1}{xy}$ is an integer then it is equal to $ 3$
the solutions are $ F_{2m\minus{}1},F_{2m\plus{}1}$ (fibbonaci sequence)
you should search for this problem because it has been posted also before...


[quote="ckck"]There are purple and green marbles in a bag in an unknown ratio, but when 15 green marbles are removed, the ratio becomes 2 purple marblesfor each green marble. Then, when 45 purple marbles are additionally removed, the ratio becomes 5 green marbles for each purple marble. How many green marbles were in the bag before any were reomved?[/quote]
[hide]$\frac{g-15}{p}=\frac{1}{2}$
$p=2g-30$

$\frac{g-15}{p-45}=\frac{5}{1}$
$5p-225=g-15$
$10g-150-225=g-15$
$9g=360$
$g=\boxed{40}$
$p=50$[/hide]


group theory
superior algebra
superior algebra unsolved

Hey guys,

I found this amongst my brothers work, and being brand new to group theory, I was hoping someone could show me how to solve it. I'm trying to get better with my maths so that I can do well after I finish first year of uni.

"Let G be a group. Definite a map o:G x G --> G by (x,y) = x. Is (G,o) a group?

Now, I know that there are three axioms that need to hold:-
1. Associativity
2. Existance of an identity
3. Existance of inverse(s)

I'm just terrible at trying to prove it. Can anyone post some lines or a solution that I can run through to try and understand?

$ (G,\circ)$ is not a group provided $ G$ has more than one element since, e.g., the second axiom fails. To see this assume that $ (G,\circ)$ were a group with identity element $ e$. Then consider $ e\circ x$ to get a contradiction.

I'm having some trouble. I understand what you are saying, but the difficulty is getting it from head to paper.

Assume e is in G, but ive got no idea how to weed out a contradicftion from this :(

Well, you need to pay attention that $ e$ is considered as identity element with respect to the operation $ \circ$!. We should not confuse it with the identity element $ e'$ with respect to the original group structure which we denote by, say, $ (G,\cdot)$.

So if there is an identity element $ e$ with respect to $ \circ$, then, for all $ x\in G$, we have $ x\equal{}e\circ x$. Now evaluate the last expression to see that the set $ G$ has just one single element. And this is a contradiction to the additional assumption I have given above.

Does this help you?


calculus
integration
logarithms
trigonometry
quadratics
calculus computations

[hide="Hint 2"]To speed this up, in the above hint $a=\frac18$,$b=\frac14$,$c=-\frac18$ and $d=\frac14$[/hide]

[quote="7h3.D3m0n.117"]...what did I do wrong?[/quote]
You've got two mistakes in there either one of which would get you yelled at in a first semester calculus class.

In the first place, when you make a substitution, you must carry through the implications of the chain rule.

That is, if you make the substitution $u=x^{4}+4$ then you must also say that $du=4x^{3}\,dx.$ Hence

$\int_{0}^{1}\frac{x^{3}}{x^{4}+4}\,dx$ becomes $\frac14\int_{4}^{5}\frac{du}{u}$

But if that $x^{3}$ isn't there, then you can't do it the same way; what you did with that substitution is simply false.

The second issue: The integral of $u^{-1}$ is the logarithm of $u.$ That's the one exception to the "power rule" that you must know.

I'm going to suggest that you stay away from problems of this difficulty and work on more basic issues.

[quote="Kent Merryfield"][quote="7h3.D3m0n.117"]...what did I do wrong?[/quote]
You've got two mistakes in there either one of which would get you yelled at in a first semester calculus class.
[/quote]

thanks...i'm only in 8th grade  :lol:

yeah i saw my mistake after my post when my dad threw me off the computer but i couldn't get back on to re-do it  :(

[quote="Path_finder"]Simple inverse trigonometric integral, that is a inverse tan if memory serves me right.[/quote]

I don't see any trig  :maybe:

[quote="Path_finder"]Simple inverse trigonometric integral, that is a inverse tan if memory serves me right.[/quote]

$\int \frac{1}{x^{2}+1}dx = \arctan{x}$

It is $x^{2}$, not $x^{4}$... I can't see how that would help with

$\int \frac{1}{x^{4}+4}dx$

[quote="Kent Merryfield"]You need to use coffeym's hints to get quadratic denominators, then complete the squares in the denominators to determine useful linear changes of variables. At that point, arctangents will start to appear.[/quote]

Oh, I see how to do it now...

[hide="Edit"]
I found that:

$\int \frac{dx}{x^{4}+4}=-\frac{1}{8}\arctan(1-x)+\frac{1}{8}\arctan(x+1)-\frac{1}{16}\ln(x^{2}-2x+2)+\frac{1}{16}\ln(x^{2}+2x+2)$

So I determined that:

$\int_{0}^{1}\frac{dx}{x^{4}+4}= \frac{1}{8}\arctan{2}+\frac{1}{16}\ln{5}$

Is that correct?[/hide]


Fie $\ f: \mathbb{N}\rightarrow\mathbb{N}$ astfel incat $\ f(n+1)>f(f(n))$.Daca $\ n$ este natural, aflati $\ f(n)$.
OIM 1977

solutia problemei o gasesti la http://www.mathlinks.ro/Forum/viewtopic.php?t=75980


integration
calculus
calculus computations

$\sum^n_{i=1}(-1)^{n-i}\frac{n+1}{i+1}\binom{n}{i-1}$?

The answer is: $1$ if $n$ odd, $\frac{n}{n+2}$ if $n$ even. How do I get there?

Thx!

When you see a binomial coefficient, it is often a good idea to add the factor $x^i$. And every time you see $i+1$ in the denominator, it is, probably, coming from $\int_0^1 x^i\,dx$ ;)


Euler
geometry
incenter
trigonometry
geometry unsolved

[quote="Mathx"]Let $ABC$ be a triangle with all its angles less than $120^{\circ}$, and let $P$ be its Fermat point. Prove that the Euler lines of triangles $PBC$, $PCA$, $PAB$ are concurrent.[/quote]

See http://groups.google.de/group/geometry.puzzles/browse_thread/thread/e4d2502291c8be2c/58f5aec0798166cc or http://www.mathlinks.ro/Forum/viewtopic.php?t=6188 .

  darij

Dear Darij,how to prove the case $I$(the incenter) instead of $P$ ?Thanks

How to prove that the Euler lines of  ABC,ABI,BCI,ACI meet at a point called Schiffler point,I think it is very hard.

[quote="darij grinberg"]See http://groups.google.de/group/geometry.puzzles/browse_thread/thread/e4d2502291c8be2c/58f5aec0798166cc[/quote]
The theorem shown at the bottom is truly nice!

([i]Due to Barry Wolk[/i])
Theorem. Let A, B, C, D be four points in the plane, no three
collinear. Consider the Euler lines of each of the four triangles
BCD, ACD, ABD, ABC. If three of these four lines are concurrent,
then all four lines are concurrent.

We may assume D lies inside $\triangle ABC$(actually unnecessary). Let $G_1,G_2,G_3$ be the centroids of DBC,DCA,DAB, $D_1,D_2,D_3$ the midpoints of AD,BD,CD, $O_1,O_2,O_3$ the circumcenters. Now apply Ceva to the triangle $D_1D_2O_3$ wrt $G_3$. We find that $\frac{\sin\angle D_1O_3G_3}{\sin\angle D_2O_3G_3}=|\frac{AB^2-BD^2}{AB^2-AD^2}|$. Combining the other 2 like expressions, the remaining is trivial.

To prove I also satisifies this property, a direct brute-force solution is to subsitute I=D...


As the figure shown, an equilateral triangle is formed by nine circles, and each side contains 4 circles. Now $ 1,2,3,4,5,6,7,8,9$ are put into each circle and making the sum of the four circle on each side of the triangle are equal.In the current picture, the the numbers in three veritices are $ 1,5,9$.
Other than this combination in the three vertices, we believe that there will be some possible combiantions also have such property as the above mentioned.
Please find all possible solutions to these combinations.

well, label the circles (from top, going clockwise) a, b, c, d, e, f, g, h, i.  let S=the sum of a row.

Then:
a+b+c+d=S
d+e+f+g=S
g+h+i+a=S
3S=a+b+c+a+b+c+d+e+f+g+h+i=a+b+c+45, since a+b+c+d+e+f+g+h+i=1+2+3+4+5+6+7+8+9 in some order.

From this we see a+b+c is a multiple of 3.  I think there is a solution is a+b+c=6,9,12,15,18,21

I could be wrong though  :oops:

[quote="abcak"]
3S=a+b+c+a+b+c+d+e+f+g+h+i=a+b+c+45[/quote]
i think you mean a + d + g + 45 = 3S  :P


Euler function integration calculus logarithms limit inequalities

1) Is there an infinite series for $ \Gamma \left(\frac{1}{3}\right)$?

2) Could you provide an hint on how to show that $ \Gamma '(1) \equal{} \minus{} \gamma$?

3) How can we compute $ \Gamma ' \left(\frac{1}{2}\right)$?

4) Let $ F(a) \equal{} \frac{\Gamma (a)}{\Gamma(a \plus{} 1/2)}$. It's easy to compute F'(1/2) (having at hand the numerical answer to the previous question). But how can we calculate F''(1/2), F'''(1/2), and so on? Could you provide an hint?

Thanks in advance

1) Yes, there is such a series representation, for $ \Gamma(z\plus{}1)$, with the Taylor coefficients given by a recurrence formula.

2) and 3).  Use the formula:

$ \psi(z)\equal{}\minus{}\frac{1}{z}\minus{}\gamma\minus{}\sum\limits_{n\equal{}1}^{\infty}\left(\frac{1}{z\plus{}n}\minus{}\frac{1}{n}\right)$,   $ z\neq 0,\minus{}1,\minus{}2,\cdots,$

Differentiate, set $ z\equal{}1/2$, and play with it.

hello carcul , howzit goin'  ?  :D

(1) unfortunately, there isn't.  :(  it is not possible to get such a series. it has even frustrated both euler and weierstrass for years, since they could get to an infinite product but never to an infinite series. hell,  forget taylor,  no one has even gotten an "asymptotic" series for $ \Gamma(z)$ ! [ refer to whittaker and watson for details ] also,  Lanczos approximation is probably the best we have today as far as a numerical recipe goes.... i use the one in C to lecture in class. btw/ an "asymptotic" series is possible for $ \frac{1}{\Gamma(z)}$

the only thing one can  do is break the ordinary gamma function into what is called upper incomplete and lower incomplete gamma functions, and then you can actually write series representations for each one of them.

$ \Gamma(m)\;\equal{}\;\gamma(m,x)\plus{}\Gamma(m,x)$ where

$ \gamma(m,x)\equal{}\int_0^x\;u^{m\minus{}1}\cdot e^{\minus{}u}\;\textbf du$ and $ \Gamma(m,x)\equal{}\int_x^\infty\;u^{m\minus{}1}\cdot e^{\minus{}u}\;\textbf du$

given $ m$ is an integer, you can then do integration by parts to eventually arrive at

$ \gamma(m,x)\equal{}(m\minus{}1)! \cdot\left(1\;\minus{}\; e^{\minus{}x}\cdot \sum_{n\equal{}0}^{m\minus{}1}\frac{x^n}{n!}\right)$

$ \Gamma(m,x)\equal{}(m\minus{}1)! \cdot e^{\minus{}x}\cdot \sum_{n\equal{}0}^{m\minus{}1}\frac{x^n}{n!}$

(for a given $ m$, the $ \sum$ on the RHS is actually finite, and the taylor expansion for $ e^{\minus{}x}$ is what will eventually yield the [b]infinite[/b] series for each one)

(2) $ \Gamma(x)\equal{}\int_0^\infty\;u^{x\minus{}1}\cdot e^{\minus{}u}\;\textbf du$

differentiate both sides w.r.t. $ x$

$ \Rightarrow \quad \Gamma'(x)\equal{}\int_0^\infty\;u^{x\minus{}1}\cdot e^{\minus{}u}\cdot \ln\;u\;\textbf du$

put $ x\equal{}1$ to get

$ \Gamma'(1)\equal{}\int_0^\infty\;e^{\minus{}u}\cdot \ln\;u\;\textbf du$

$ \equal{}\lim_{p\to\infty}\;\int_0^p\left(1\minus{}\frac{u}{p}\right)^p \cdot\ln u \;\textbf du\;\equal{}\; \lim_{p\to\infty}\;\sum_{q\equal{}0}^p \left\{ \frac{(\minus{}1)^q}{p^q} \cdot  \binom{p}{q}\cdot \int_0^p u^q\cdot   \ln\;u\;\textbf du\right\}$

$ \equal{}\lim_{p\to\infty}\; p\left[ \ln\; p\cdot \sum_{q\equal{}0}^p  \frac{(\minus{}1)^q}{q\plus{}1} \cdot  \binom{p}{q} \;\minus{}\;  \sum_{q\equal{}0}^p \frac{(\minus{}1)^q}{ (q\plus{}1)^2} \cdot  \binom{p}{q}  \right]$

$ \equal{}\lim_{p\to\infty}\; \frac{p}{p\plus{}1}\left[ \ln\; p \cdot \sum_{q\equal{}0}^p (\minus{}1)^q \cdot  \binom{p\plus{}1}{q\plus{}1} \;\minus{}\;  \sum_{q\equal{}0}^p \frac{(\minus{}1)^q}{q\plus{}1} \cdot  \binom{p\plus{}1}{q\plus{}1}   \right]$

now since,

$ \sum_{q\equal{}0}^p(\minus{}1)^q \cdot  \binom{p\plus{}1}{q\plus{}1} \;\equal{}\;1$

and

$ \sum_{q\equal{}0}^p \frac{(\minus{}1)^q}{q\plus{}1} \cdot  \binom{p\plus{}1}{q\plus{}1}\;\equal{}\; \sum_{q\equal{}0}^p \binom{p\plus{}1}{q\plus{}1} \cdot \int_{\minus{}1}^0\;x^q\;\textbf dx\;\equal{}\;\int_{\minus{}1}^0\;\frac{(1\plus{}x)^{p\plus{}1}\minus{}1}{x} \;\textbf dx$

simplify everything to get

 $ \Gamma'(1)\equal{}\lim_{p\to\infty}\;  \frac{p}{p\plus{}1}\left[ \ln\; p  \;\minus{}\;  \sum_{q\equal{}1}^{p\plus{}1}\frac{1}{q}  \right]\;\equal{}\; \boxed{\minus{}\gamma}$

(3) i will give you a hint (like i gave to students)

first show this relationship (it is   easy)

$ \frac{\Gamma'(1)}{\Gamma(1)}\;\minus{}\; \frac{\Gamma'(\frac{1}{2})}{\Gamma(\frac{1}{2})}\;\equal{}\;\ln \;4$

all you   need to do then is   put  $ \Gamma'(1)\equal{}\minus{}\gamma,\qquad \Gamma(1)\equal{}1,\qquad \Gamma\left(\frac{1}{2}\right)\equal{}\sqrt \pi$

oh ! look  :o  my lunch break is over, i will try and come back tonight or tomorrow and post the answer for (4).  got to go now....  :oops:

btw/ i am not sure if you know, but there is a contest within the math community as to who can come up with a closed form for $ \boxed{\Gamma\left(\frac{\pi}{2}\right)}$

i am trying but no luck so far.  :rotfl:

The series I was referring to is the following one

$ \Gamma(z\plus{}1)\equal{}\sum\limits_{n\equal{}0}^{\infty}a_{n}z^{n}$, $ |z|<1$,

where $ a_{0}\equal{}1$ and $ na_{n}\equal{}\minus{}\gamma a_{n\minus{}1}\plus{}\sum\limits_{k\equal{}2}^{n}(\minus{}1)^{k}a_{n\minus{}k}\zeta(k)$.

Note that $ a_{n}\equal{}\frac{\Gamma^{(n)}(1)}{n!}$ and we get that

$ \Gamma^{(n\plus{}1)}(1)\equal{}\minus{}\gamma\Gamma^{(n)}(1)\plus{}n!\sum\limits_{k\equal{}1}^{n}(\minus{}1)^{k\plus{}1}\frac{\zeta(k\plus{}1)}{(n\minus{}k)!}\Gamma^{(n\minus{}k)}(1)$.

Now, if Carcul is interested in calculating $ \Gamma(1/3)$ exactly by using some power series expansions, exactly as $ \pi$ is via the BBP formula,  then, yes, you are right, this formula is not of too much use.

Thanks Didilica and Misan for your comments.

[quote="Misan"](3) i will give you a hint (like i gave to students)[/quote]

Yes, that's easy, thanks.

[quote="Misan"]btw/ i am not sure if you know, but there is a contest within the math community as to who can come up with a closed form for $ \boxed{\Gamma\left(\frac{\pi}{2}\right)}$[/quote]

Why $ \Gamma\left(\frac{\pi}{2}\right)$ and not, for example, $ \Gamma (\pi)$? Is there anything special about that value?

[quote="Carcul"]
 Why $ \Gamma\left(\frac {\pi}{2}\right)$ and not, for example, $ \Gamma (\pi)$? Is there anything special about that value?[/quote]

you know what, that is a darn good question!  :maybe:  i absolutely have no idea why they picked  $ \Gamma\left(\frac {\pi}{2}\right)$ .... i think i will try to ask around for some background on this....

you see, problem with me is when i encounter a problem, i try solving it and when i am done, i move on.
what i think i should do more often, at least for the interesting problems, is question the intent and motivation for it as well.
a little history never hurts, right ?  :oops:

Yes, that's right. But, for a start, is that value even expected to have a closed form?

5) Can anyone explain to me from where come the inequality

$ \frac{n^x n!}{x(x\plus{}1) ... (x\plus{}n)} \leq \Gamma(x) \leq \frac{n^x n!}{x(x\plus{}1) ... (x\plus{}n)} \frac{x\plus{}n}{n}$?

Thank you very much.

[quote="Carcul"]5) Can anyone explain to me from where come the inequality

$ \frac {n^x n!}{x(x + 1) ... (x + n)} \leq \Gamma(x) \leq \frac {n^x n!}{x(x + 1) ... (x + n)} \frac {x + n}{n}$?

Thank you very much.[/quote]
the gamma function $ \Gamma(x)$ is log-convex in $ x$ hence it follows that $ (\log \Gamma(x))'$ is increasing in $ x$  define

$ \Psi(x) = \frac {\log \Gamma(x + n) - \log \Gamma(n)}{(x + n) - n},n\geq2 \in \mathbb{N}$
by the mean value theorem and using the $ \log$ convex property we have
$ \Psi( - 1) < \Psi(x) \leq \Psi(1)$ for $ x \in (0,1)$
this gives $ x\log(n - 1) \leq \log (\Gamma(x + n)) - \log (n - 1)! \leq \log n$
since $ \log$ is monotonic increasing we have
$ \frac {n^x n!}{x(x + 1) ... (x + n)} \leq \Gamma(x) \leq \frac {n^x n!}{x(x + 1) ... (x + n)} \frac {x + n}{n}$
for $ x \in (0,1),n \geq 2 \in \mathbb{N}$

actually Emil Artin has a brilliant account about Gamma functions :)

Thanks for the explanation.

[quote="Mythili"]actually Emil Artin has a brilliant account about Gamma functions[/quote]

Yes, I have noted that, but obviously I don't have access to that book.

6) What is the value of $ \Gamma ''\left(\frac{1}{2}\right)$? (I am just interested in the value for a check)

[quote="Carcul"]Thanks for the explanation.

[quote="Mythili"]actually Emil Artin has a brilliant account about Gamma functions[/quote]

Yes, I have noted that, but obviously I don't have access to that book.

6) What is the value of $ \Gamma ''\left(\frac {1}{2}\right)$? (I am just interested in the value for a check)[/quote]

Dont worry Carcul, mythili alias pardesi  :P   :rotfl:   has an ebook i believe n will give you that   :)

sorry Carcul neither do i have it...read that 2-3 years back ...very thin a day is enuf for that book but very nice :) ...lemme see if i can find the book  :)

[quote="Carcul"]
6) What is the value of $ \Gamma ''\left(\frac {1}{2}\right)$? (I am just interested in the value for a check)[/quote]

start with the basics, first show the following (takes only a few lines using the fundamental definition given by weierstrass)

$ \dfrac{\Gamma'(x)}{\Gamma(x)}\; \equal{} \; \minus{} \gamma \plus{} \sum_{m \equal{} 1}^\infty\;\left( \dfrac{1}{m} \minus{} \dfrac{1}{m \plus{} x \minus{} 1}\right)$
[ where $ x\;\ne 0, \minus{} 1, \minus{} 2, \minus{} 3,...$]

now differentiate both sides w.r.t. $ x$

$ \dfrac{\Gamma(x)\Gamma''(x) \minus{} \Gamma'^{2}(x)}{\Gamma^2(x)}\; \equal{} \; \sum_{m \equal{} 1}^\infty\;\dfrac{1}{(m \plus{} x \minus{} 1)^2}$

put $ x \equal{} \dfrac{1}{2}$ to get

$ \dfrac{\Gamma\left(\dfrac{1}{2}\right)\Gamma''\left(\dfrac{1}{2}\right) \minus{} (\Gamma'\left(\dfrac{1}{2}\right))^2}{(\Gamma\left(\dfrac{1}{2}\right))^2}\; \equal{} \; 4\sum_{m \equal{} 1}^\infty\;\dfrac{1}{(2m \minus{} 1)^2}$

RHS is $ \dfrac{\pi^2}{2}$ and you already know the values of $ \Gamma\left(\dfrac{1}{2}\right)$ and $ \Gamma'\left(\dfrac{1}{2}\right)$

btw/ you also know how to show $ \Gamma''(1) \equal{} \zeta(2) \plus{} \gamma^2$, right ? :oops:

Misan, thanks for your explanation, but please read carefully what I write:

[quote="Carcul"](I am just interested in the value for a check)[/quote]

[quote="Misan"]btw/ you also know how to show $ \Gamma''(1) \equal{} \zeta(2) \plus{} \gamma^2$, right ?[/quote]

Right, and I have devised a general formula for $ \Gamma''(n)$, with n natural.


analytic geometry
conics
parabola
quadratics
algebra
quadratic formula

[quote="pololoco"]The daily production cost, in pesos, for a desk manufacturer is y=0.3x squared -5.3x+546.8, where y is the daily production cost and x is the number of desks produced daily.

a.what is the daily production cost if the number of desks produced daily is 4,6,8,20 and 12?

b. What happens to the daily production cost as the company increases its daily production of desks?

c. How many desks should be the manufacturer produce to minimize the daily production cost?

d. What is the minimum production cost.

PLEASE HELP ME. :(

Kindly show the solution and answers.

Over and out.[/quote]

Well the first step to solving every single math problem is latex.  :lol:

$ y \equal{} 0.3x^2 \minus{} 5.3x \plus{} 546.8$

Basically for part "a" you plug in the values 4, 6, 8, 12, and 20 for $ x$ and then evaulate the expression for $ y$.

You should get the following production cost for each of the following numbers;

4 --> 530.4
6 --> 525.8
8 --> 523.6
12 --> 526.4
20 --> 560.8

b.) Well it depends. If you sell one desk you will get a price that is higher than the price of two desks. So it appears that the price is decreasing, but according to the information above, somewhere between 8 and 12 is the point at which the price is the lowest. Once it reaches this point, it begins getting higher.

c.) Well if you subtract $ y$ from both sides of the equation you get;

$ 0 \equal{} 0.3x^2 \minus{} 5.3x \plus{} (546.8 \minus{} y)$

Now let's imagine a coordinate grid in which the $ x$ values are the number of desks and $ y$ values are the prices. We want $ y$ to be as low as possible, so we are looking for the lowest price when we graph this parabola. Or in other words we are looking for the smallest $ y$ value. The lowest point on the parabola is the one such that $ b^2 \equal{} 4ac$. Where $ a$, $ b$, and $ c$ are the $ x^2$, $ x$, and $ x^0$ coefficients. We want $ b^2 \equal{} 4ac$ because that will make the value under the square root (using the quadratic formula) equal to zero. We want it to be equal to zero because when it is equal to zero because this only gives us one $ x$ value for that $ y$ value. The lowest point on the parabola is going to be the only $ x$ value for that $ y$ value.

So now we look at the equation $ b^2 \equal{} 4ac$ we know that $ b \equal{} \minus{}5.3$, $ a \equal{} 0.3$, and $ c \equal{} 546.8 \minus{} y$. So now we can solve for $ y$.

$ (\minus{}5.3)^2 \equal{} 4(0.3)(546.8 \minus{} y)$
For now lets just keep $ 546.8 \minus{} y$ as $ c$, so that it is easier to work with.
$ 28.09 \equal{} 1.2c$
$ 23.41 \equal{} c$
Plugging in $ 546.8 \minus{} y$ in for $ c$, we get;
$ 23.41 \equal{} 546.8 \minus{} y$
So $ y \equal{} 523.39$
This is just the minimum production cost. And the answer to D.).

To find $ x$ or the number of desks, we just simply use our brain. We know that $ b^2 \minus{} 4ac \equal{} 0$, so the value of $ x$ is just $ \minus{}b/2a$.

[b]NOTE: In any quadratic, the tip  of the parabola can be found by evaluating[/b] $ b/2a$.

$ b/2a$ happens to be about 8.83.


function
complex analysis
complex analysis unsolved

sin(1/z) = [e^(i/z) - e^(-1/z)]/2i = [e^[y/(x^2 + y^2)]*[(cos(x/(x^2 + y^2)) + i(sin(x/(x^2 + y^2))] - e^[-y/(x^2 + y^2)]*[(cos(-x/(x^2 + y^2)) + i(sin(-x/(x^2 + y^2))]] / 2i for z = x + iy for x,y, in R I think.

Where is this function analytic?
Thanks for your help.

[hide="Hint"]For all $ z\neq 0$ - because of two different arguments working separately:
1.) the function is a composition of analytic functions;
2.) checking the Cauchy-Riemann differential equations.

... I must confess that I did not check the rather lengthy expression. ;)[/hide]


linear algebra
matrix
geometry
geometric transformation
dilation
superior algebra

what are the generators and relations of the group $ GL(n, Z_2)$?

What do you mean, "the" generators and relations? There are an extremely large number of choices, and nothing canonical about any of them.

For $ GL_n(F)$, one possible set of generators is the set of elementary matrices: row swaps such as $ \begin{bmatrix}0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1\end{bmatrix}$, shears such as $ \begin{bmatrix}1 & a & 0 \\
0 & 1 & 0 \\
0 & 0 & 1\end{bmatrix}$, and one-axis dilations such as $ \begin{bmatrix}a & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1\end{bmatrix}$. That generates everything, and a representation of each element can be constructed with Gaussian elimination.

Restricting to the field of two elements, the third type of elementary matrix disappears. The first two can be trimmed, as $ \begin{bmatrix}1 & 1 & 0 & \cdots \\
0 & 1 & 0 & \cdots \\
0 & 0 & 1 & \cdots \\
\vdots & \vdots & \vdots & \ddots\end{bmatrix}$ is similar to all other shears by a permutation matrix, which is itself a product of row swaps. These permutations can be reduced down to a generating set for $ S_n$, such as the transposition $ \begin{bmatrix}0 & 1 & 0 & \cdots \\
1 & 0 & 0 & \cdots \\
0 & 0 & 1 & \cdots \\
\vdots & \vdots & \vdots & \ddots\end{bmatrix}$ and the cycle $ \begin{bmatrix}0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\
0 & 0 & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
1 & 0 & 0 & \cdots & 0\end{bmatrix}$.

That's a three-element generating set, which is probably as small as it gets. I don't want to go through and find the long list of relations; matrix groups are a lot easier to understand anyway.

[quote="rulin"]whether does there exist the two generators?[/quote]

http://www.informaworld.com/smpp/content~content=a778400498~db=all

hi,randomgraph,many thanks for you!

Unfortunately, I cannot get the fulltext because my university donot buy the electronic papers of this Publishing House. can you post this article if you can get it? Thanks again.

Nope, I can't access it either, and there doesn't seem to be a preprint available online. There are many papers on the problem of generating the finite simple groups with two elements (I believe this has been settled by now, in the affirmative), so you should be able to find articles on the linear case, and perhaps modify the argument to suit $ GL$ as well.


calculus
integration
logarithms
trigonometry
calculus computations

Calculate the following indefinite integrals.

[1] $\int \ln (x^2-1)dx$

[2] $\int \frac{1}{e^x+1}dx$

[3] $\int (ax^2+bx+c)e^{mx}dx\ (abcm\neq 0)$

[4] $\int \left(\tan x+\frac{1}{\tan x}\right)^2 dx$

[5] $\int \sqrt{1-\sin x}dx$

[5]

$(x+1)\text{ln}(x+1)+(x-1)\text{ln}(x-1)-2x+C$

I deleted #4, i think my answer for 5 should be right, kunny.

[quote="adidasty"][5]

$(x+1)\text{ln}(x+1)+(x-1)\text{ln}(x-1)-2x+C$ :(

I deleted #4, i think my answer for 5 should be right, kunny.[/quote]

[quote="erdos"][b][4][/b]

$\int \left( \tan x+\frac{1}{\tan x}\right) ^{2}dx=\int \left[ \left( \tan ^{2}x+1\right) +\left( \frac{1}{\tan ^{2}x}+1\right) \right] dx$

$=\tan x+\int \frac{1+\tan ^{2}x}{\tan ^{2}x}dx=\tan x+\int \left( -\frac{1}{\tan x}\right) ^{\prime }dx$

$=\tan x-\frac{1}{\tan x}$[/quote]

 :)

im glad erdos got #4, that one was killing me

[quote="erdos"]and what about [b][1][/b] ?? is it all right ?  :D[/quote]

Looks right, nice job  :D .

Sorry, I don't have any great way of formating math on a forum, if anyone has any suggestions  please tell me.

In any case, #5,

Integral[Sqrt[1-Sin(x)] dx]     with x=arcsin(z) ; dx=dz/Sqrt[1-z^2]
Integral[Sqrt[1-z] dz/Sqrt[1-z^2]]
Integral[dz/Sqrt[1+z]]
2Sqrt[1+z] + C
2Sqrt[1+Sin(x)]+C

[quote="kunny"][quote="erdos"]and what about [b][1][/b] ?? is it all right ?  :D[/quote]

Sorry for my delay reply, erdos.

What's the sign of $x+1$? :([/quote]
Oh Sorry I forgot about that  :blush:  :blush:
I have corrected it

For fifth prob

$ \int \sqrt {1 \minus{} \sin x}: dx$

$ \equal{} \int \sqrt {\sin^2 \frac {x}{2} \plus{} \cos^2 \frac {x}{2} \minus{} 2 \sin \frac {x}{2}. \cos \frac {x}{2}} dx$

$ \equal{} \int ( \sin \frac {x}{2} \minus{} \cos \frac {x}{2} ) dx$

$ \equal{} \minus{}2 \cos \frac {x}{2} \minus{} 2 \sin \frac {x}{2}\plus{} c$

[quote="anandghegde"]For fifth prob

$ \int \sqrt {1 \minus{} \sin x}: dx$

$ \equal{} \int \sqrt {\sin^2 \frac {x}{2} \plus{} \cos^2 \frac {x}{2} \minus{} 2 \sin \frac {x}{2}. \cos \frac {x}{2}} dx$

$ \equal{} \int ( \sin \frac {x}{2} \minus{} \cos \frac {x}{2} ) dx$

$ \equal{} \frac { \minus{} \cos \frac {x}{2}}{2} \minus{} \frac {\sin \frac {x}{2}}{2} \plus{} c$[/quote]

$ ...\ \equal{} \int \left|\sin \frac {x}{2} \minus{} \cos \frac {x}{2}\right| dx\ \plus{}\ ...\ ?!$

[quote="misan"][quote="kunny"]:)[/quote]

[quote="kunny"]:([/quote]

extreme frustrated compensation mode.   :rotfl:[/quote]

What mean upper these ?! I didn't understand, sorry Misan !

Sorry Kunny sir,
irevived your two year old thread  :blush:

[b][1][/b] can also be done by factoring $ \ln{(x^2\minus{}1)}$ into $ \ln{(x\plus{}1)}\plus{}\ln{(x\minus{}1)}$ and integrating by parts.

[quote="kunny"]Calculate the following indefinite integrals.
[4] $\int \left(\tan x+\frac{1}{\tan x}\right)^2 dx$[/quote]
Generalization
Calculate
[1] $\int \left(\tan x+\frac{1}{\tan x}\right)^n dx$
[2] $\int \left(a\tan x+\frac{b}{\tan x}\right)^2 dx$

[quote="kunny"]Calculate the following indefinite integrals.
[2] $\int \frac{1}{e^x+1}dx$
[/quote]
Generalization
Calculate
$\int \frac{1}{(ae^x+b)^n}dx,a,b>0$


If $ P(x)$ is a polynomial of degree less than or equal to 2 and S is the set of all such polynomials such that $ P(1) \equal{} 1$
$ P(0) \equal{} 0$ and $ P'(x) > 0$ , x belongs to the interval $ [0,1]$
then

a)$ S \equal{} \phi$

b) S={$ (1 \minus{} a)x^2$+$ ax, 0 < a < 2$

c) S={$ (1 \minus{} a)x^2$+$ ax$ , $ a$ belongs to $ [0,\infty]$

d)S={$ (1 \minus{} a)x^2$+$ ax , 0 < a < 1$

[hide="Solution"]
First notice that $ P(x)\equal{}x$ is a solution, so answers a and d are out.

Also (Assuming $ P'(x)$ means the derivative of the polinomial), if $ a\equal{}0$ and $ P(x)\equal{}(1\minus{}a)x^2\plus{}ax$, then the derivative of $ P'(x)\equal{}2x\ge 0$ for $ x\in [0,1]$, which means that option c is out, and the answer is $ \boxed{\mathbf{(B)}}$.
[/hide]


:huh:
Hi everyone. Right now I am in desperate need for a number theory e-book which is for [b]fre[/b]e and corresponds to AMC12/AIME/USAMO. Or would roughly correlate to the curriculum of the AOPS class: Intermediate Number Theory Seminar. If you have any or know of any please put a link or attachment. Thanks
:(

I don't think it would be a good idea to post a whole book here, but I can send you one through pm.


calculus
derivative
function
complex analysis
complex numbers
complex analysis unsolved

Sorry for not using TeX... I'll hopefully update this posts soon to have it (I'm slow at it).

Anywhere, here's the problem:

Find the derivative f'(z), where it exists, and state where f is analytic.
f=(x+y) + i(sin x + cos y)

I really don't understand this well. I use the Cauchy-Riemann relations and get

ux=1, uy=-sin(y)
uv=1 -vx=-cos(x)

These are equal at y=-pi/2+2kpi and x=-pi+2kpi
They also seem to be continuous...

So from here I'm not sure how go get f' and I'm not sure where it's analytic.

You can find $ f'$ from $ f'(z)\equal{}\frac12(f_x\minus{}if_y)$. It exists only at those isolated points that you found.
Consequently, $ f$ is nowhere analytic: to be analytic at a point, it must satisfy the C-R equations in some neighborhood of that point.

[quote="mlok"]You can find $ f'$ from $ f'(z) \equal{} \frac12(f_x \minus{} if_y)$. It exists only at those isolated points that you found.
Consequently, $ f$ is nowhere analytic: to be analytic at a point, it must satisfy the C-R equations in some neighborhood of that point.[/quote]

So then you'd get f'(z) = 1-i?

Right, at the point where it exists, the derivative $ f'$ is equal to $ 1\minus{}i$.

This also follows from $ \mathrm{Re}(f(z))\equal{}\mathrm{Re}((1\minus{}i)z)$, because the complex derivative, if it exists, is determined by the real part of the function.

Okay, now I'm having another problem with this material. I need to find the branch points of $ f \equal{} sin(\sqrt{z})$. I know this is simpler than I'm making, but I'm not getting anywhere with it.

It's just $ 0$. The function is holomorphic in a neighborhood of any other point.


Find all triples of integers $ (x, y, z)$ such that $ x^2 \plus{} y^2 \equal{} 2z^2$.

Method 1:  the geometric approach, shown [url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=197236]here[/url].

Method 2:  note that $ x, y$ must have the same parity, hence

$ \left( \frac{x \plus{} y}{2} \right)^2 \plus{} \left( \frac{x \minus{} y}{2} \right)^2 \equal{} z^2$

and now we're just looking for Pythagorean triples.

Method 3:  the third technique uses factorization in $ \mathbb{Z}[ \sqrt{2} ]$.  We begin by rewriting $ y^2 \equal{} x^2 \minus{} 2z^2 \equal{} (x \minus{} \sqrt{2} z)(x \plus{} \sqrt{2} z)$.


number theory
greatest common divisor
superior algebra
superior algebra unsolved

The following question was confusing me so any help would be appreciated!

Find the solution in integers to the equation
155x+341y+385z=1 (where 1 is the gcd of 155,341, and 385).

There are many solutions.

One reliable approach (solving $ ax\plus{}by\plus{}cz\equal{}d$, where $ d$ is the gcd of $ a,b,c$):
Let $ s$ be the gcd of $ a$ and $ b$. Find a solution $ (x_{0},y_{0})$ to $ ax_{0}\plus{}by_{0}\equal{}s$. Next, find a solution $ (w,z)$ to $ sw\plus{}cz\equal{}d$. Combine the two; $ ax_{0}w\plus{}by_{0}w\plus{}cz\equal{}d$.


[quote="hanschinaman"]$a_{i}, i=1,2,...,n$ are positive real numbers such that $\sum_{i=1}^{n}{a_{i}}=1$, prove that
$(\sum_{i=1}^{n}{a_{i}a_{i+1}})(\sum_{i=1}^{n}{\frac{a_{i}}{a_{i+1}^{2}+a_{i+1}}})\geq \frac{n}{n+1}$[/quote]
$\sum_{i=1}^{n}{a_{i}a_{i+1}}\cdot\sum_{i=1}^{n}{\frac{a_{i}}{a_{i+1}^{2}+a_{i+1}}}\cdot\sum_{i=1}^{n}a_{i}(a_{i+1}+1)\geq\left(\sum_{i=1}^{n}a_{i}\right)^{3}$ by Holder's inequality.
Hence, it remains to prove that  $1\geq\frac{n}{n+1}\cdot\left(\sum_{i=1}^{n}a_{i}a_{i+1}+1\right)\Leftrightarrow\left(\sum_{i=1}^{n}a_{i}\right)^{2}\geq n\cdot\sum_{i=1}a_{i}a_{i+1},$
which true for all $2\leq n\leq 4.$  :(

Very good idea though not for all integers n. :)

I came up with the solution finally.Beautiful problem.We have from Cauchy
$ \left(\sum_{i=1}^{n}\frac{a_{i}}{a_{i+1}^{2}+a_{i+1}}\right)\left( \sum_{i=1}^{n}\frac{a_{i}(a_{i+1}+1)}{a_{i+1}}\right)\geq \left( \sum_{i=1}^{n}\frac{a_{i}}{a_{i+1}}\right)^{2}$.Now from hypotesis we have $ \sum_{i=1}^{n}\frac{a_{i}(a_{i+1}+1)}{a_{i+1}}=1+\sum_{i=1}^{n}\frac{a_{i}}{a_{i+1}}$
 Denoting $ S=\sum_{i=1}^{n}\frac{a_{i}}{a_{i+1}}$  we have
$ \frac{S^{2}}{S+1}\geq \frac{nS}{n+1}$,which is easy to prove.
So we just have to prove $ \left(\sum_{i=1}^{n}a_{i}a_{i+1}\right) \left(\sum_{i=1}^{n}\frac{a_{i}}{a_{i+1}}\right)\geq 1$ which is Cauchy inequality.

[quote="hanschinaman"]$ a_{i}, i \equal{} 1,2,...,n$ are positive real numbers such that $ \sum_{i \equal{} 1}^{n}{a_{i}} \equal{} 1$, prove that
$ (\sum_{i \equal{} 1}^{n}{a_{i}a_{i \plus{} 1}})(\sum_{i \equal{} 1}^{n}{\frac {a_{i}}{a_{i \plus{} 1}^{2} \plus{} a_{i \plus{} 1}}})\geq \frac {n}{n \plus{} 1}$[/quote]
My solution is not nice as mateivld, but let me post it
Applying Cauchy Schwarz Inequality, we have
\[ \sum^{n}_{i\equal{}1} \frac{a_i}{a^2_{i\plus{}1}\plus{}a_{i\plus{}1}} \equal{}\sum^{n}_{i\equal{}1} \frac{a_i}{a_{i\plus{}1}} \minus{}\sum^{n}_{i\equal{}1} \frac{a_i}{a_{i\plus{}1}\plus{}1} \ge \sum^{n}_{i\equal{}1} \frac{a_i}{a_{i\plus{}1}} \minus{}\frac{1}{(n\plus{}1)^2} \sum^{n}_{i\equal{}1} a_{i}\left( \frac{1}{a_{i\plus{}1} }\plus{}n^2\right) \equal{}\frac{n}{(n\plus{}1)^2} \left( (n\plus{}2) \sum^{n}_{i\equal{}1} \frac{a_{i}}{a_{i\plus{}1}} \minus{}n\right)\]
So, we have to prove
\[ \left( \sum^{n}_{i\equal{}1} a_ia_{i\plus{}1}\right) \left( (n\plus{}2) \sum^{n}_{i\equal{}1} \frac{a_{i}}{a_{i\plus{}1}} \minus{}n\right) \ge n\plus{}1\]
There are 2 cases to consider
Case 1. If $ \sum^{n}_{i\equal{}1} a_ia_{i\plus{}1} \ge \frac{1}{n}$, then it is trivial because
\[ (n\plus{}2) \sum^{n}_{i\equal{}1} \frac{a_{i}}{a_{i\plus{}1}} \minus{}n \ge n(n\plus{}1)\]
(by AM-GM)
Case 2. If $ \sum^{n}_{i\equal{}1} a_ia_{i\plus{}1} \le \frac{1}{n}$, applying Cauchy Schwarz Inequality again, we obtain
\[ \sum^{n}_{i\equal{}1} \frac{a_i}{a_{i\plus{}1}} \ge \frac{\left( \sum^{n}_{i\equal{}1} a_i\right)^2}{\sum^{n}_{i\equal{}1} a_ia_{i\plus{}1}} \equal{}\frac{1}{\sum^{n}_{i\equal{}1} a_ia_{i\plus{}1}}\]
It follows that
\[ \left( \sum^{n}_{i\equal{}1} a_ia_{i\plus{}1}\right) \left( (n\plus{}2) \sum^{n}_{i\equal{}1} \frac{a_{i}}{a_{i\plus{}1}} \minus{}n\right) \ge \left( \sum^{n}_{i\equal{}1} a_ia_{i\plus{}1}\right) \left( (n\plus{}2)\frac{1}{\sum^{n}_{i\equal{}1} a_ia_{i\plus{}1}} \minus{}n\right)  \equal{}(n\plus{}2) \minus{}n\sum^{n}_{i\equal{}1} a_ia _{i\plus{}1} \ge n\plus{}2\equal{}1\equal{}n\plus{}1.\]
Our proof is completed. :)


The circles $O_1\left(r_1\right)$ and $O_2\left(r_2\right)$ touch each other externally and the line l at the points A and B respectively. The circle $O_3\left(r_3\right)$ touches the line l and the circles $O_1\left(r_1\right)$ and $O_2\left(r_2\right)$.   Show that, changing $r_1$ and $r_2$, the circle $O_3\left(r_3\right)$ touches a fixed circle passing through A and B.

Invert with pole $B$. We get the following problem: we have a fixed line $d$ and $A\in d$. Consider a variable circle $\mathcal C_1$ tangent to $d$ in $A$ and the line $d'$ parallel to $d$ and tangent to the circle. Consider now the circle $\mathcal C_2$ tangent to $d,d'$ and the initial circle. Show that the tangents from $A$ to $\mathcal C_2$ are fixed.

Of course, it's one of the tangents is the line $d$ itself, which is, of course, constant, so we're only concerned with the other tangent. Let $O_2$ be the center of $\mathcal C_2$. Since $AO_2$ is the bisector of the angle formed by the tangents from $A$ to $\mathcal C_2$, we have to show that $AO_2$ is a fixed line. This is clear, and I'll let you finish it :).

Thanks Grobber   :D
How we can proof that the radius of the fixed circle is $\frac{5}{8}AB$ ?

Kind regards
___________
Leon

Hmm.. Maybe looking at the particular case when $r_1=r_2$ helps. One more thing: in what I wrote above, there are two choices for $\mathcal C_2$. When you move $d'$, choose $\mathcal C_2$ s.t. it's always on the same side of $\mathcal C_1$.

Your hint is good !

[b]Lemma 1.[/b] $AB^2  = 4r_1 r_2 $
[b]Proof.[/b]  From Pythagoras theorem follows:

$AB^2  + \left( {r_1  - r_2 } \right)^2  = \left( {r_1  + r_2 } \right)^2 {\rm{  }} \Rightarrow {\rm{  }}AB^2  = 4r_1 r_2 $ .

[b]Lemma 2.[/b] $\frac{1}{{\sqrt {r_3 } }} = \frac{1}{{\sqrt {r_1 } }} + \frac{1}{{\sqrt {r_2 } }}$
[b]Proof.[/b] Let $T = O_3 \left( {r_3 } \right) \cap \ell $. From Lemma 1 we have

$\begin{array}{c}
 4r_1 r_2  = AB^2  = \left( {AT + TB} \right)^2  = AT^2  + TB^2  + 2AT \cdot TB =  \\
  = 4r_1 r_3  + 4r_2 r_3  + 2 \cdot 2\sqrt {r_1 r_3 }  \cdot 2\sqrt {r_2 r_3 }  = 4r_3 \left( {\sqrt {r_1 }  + \sqrt {r_2 } } \right)^2  \\
 \end{array}$

Hence

$r_3  = \frac{{r_1 r_2 }}{{\left( {\sqrt {r_1 }  + \sqrt {r_2 } } \right)^2 }}{\rm{  }} \Rightarrow {\rm{  }}\frac{1}{{\sqrt {r_3 } }} = \frac{1}{{\sqrt {r_1 } }} + \frac{1}{{\sqrt {r_2 } }}$ .

[b]Theorem.[/b] The radius of fixed circle is $\frac{5}{8}AB$
[b]Proof.[/b]
We can suppose that $r_1  = r_2 =r$.
Let  $\gamma $  the fixed circle, let $M = O_3 (r_3 ) \cap d$ , let d' the perpendicular to d through M and $d' \cap \gamma  = \left\{ {C,D} \right\}$.
From Lemma 2 we have:
$\frac{1}{{\sqrt {r_3 } }} = \frac{2}{{\sqrt r }}{\rm{ }} \Rightarrow {\rm{ }}r_3  = \frac{r}{4}$
hence:
$\begin{array}{l}
 CM = 2r_3  = \frac{r}{2} \\
 AB = 2r \\
 AC^2  = r^2  + \frac{{r^2 }}{4} = \frac{5}{4}r^2  \\
 CO = \frac{1}{2}CD = \frac{1}{2}\frac{{AC^2 }}{{CM}} = \frac{5}{4}r = \frac{5}{8}AB \\
 \end{array}$


geometry
circumcircle
incenter
trigonometry
Euler
vector
geometric transformation

The bisector of angle A of triangle ABC meets the circumcircle of triangle ABC at point D. I is the incenter of triangle ABC , M the midpoint of segment BC , P the symmetric point of I in M . Extend DP to meet the circumcircle at point N. Prove among segments AN, BN , CN ,one equal s the sum of the other two.

I haven't managed to solve it, but here's an equivalent statement, which might be helpful (well, they're equivalent in the setting in which I'm looking at it, $BC\le AB\le CA$): Take points $X,Y$ on $(AB,(AC$ s.t. $AX=AY=a=BC$. Then $A,N,X,Y$ are concyclic.

Here's something extra: it seems that the Euler lines of $BCI_a, CAI_b, ABI_c$ concur at $N$ ($I_a$ and the others are the excenters).

I don't think $S[BNC]=S[INC]$ :? Maybe I'm wrong, though.

Here's what's going on (I think I have proofs of these things): Let $\omega_a$ be the circle passing through $A$ and $X\in(AB,Y\in(AC$ s.t. $AX=AY=a$. Define $\omega_b,\omega_c$ similarly. Then the circles $\omega_a,\omega_b,\omega_c$ and the Euler lines of $BCI_a, CI_b, ABI_c$ all pass through a point on the circumcircle of $ABC$, and this is the point $N$.

I'm taking advantage a few minutes at school.
It's very easy. I 'll post my solution right after school.

Here's a question: $N=X(?)$ in the ETC database. I bet Darij could be of some help here :).

Here is my solution for the concurence
    Let ABC be an acute triangle and orthic triangle (the pedal triangle of the orthocenter H). We 'll prove that the 3 Euler line of AB'C', BC'A', CA'B' are concurent.
    Let $O_a, O_b, O_c$ and $H_a, H_b, H_c$ be the circumcenters and orthocenters of AB'C', BC'A', CA'B'.

Noticing the 3 triangles are similar: AB'C' ~ A'BC'~ A'CB', we obtain
       The Euler line of of AB'C' intersects AB', AC', B'C' and forming respectively the acute angles $a, b, c$.
       The Euler line of of A'BC' intersects A'B, A'C', BC' and forming respectively the acute angles $a, b, c$.
       The Euler line of of A'CB' intersects A'C, A'B', B'C and forming respectively the acute angles $a, b, c$.
   We also have $O_aO_b // AB, O_cO_b // CB, O_aO_c // AC$. Hence applying the trig Ceva to triangle $O_aO_bO_c$, we 'll obtain our claim.
     Grobber 's problem is equivalent to mine by showing that A'B'C' have A, B, C its' excenters.
    Therefore, if you call $K_a, K_b, K_c$ such that there exists m, n, p such that
         $ mK_aA + nK_aB' + pK_aC' = 0 , mK_bA' + nK_bB + pK_bC' = 0, mK_cA' + nK_cC + pK_cB' = 0$ (they are all vectors) then
     $O_aK_a, O_bK_b, O_cK_c$ are concurent.

I did something very similar (angle-chase). Now try the rest of it, that thing with the $\omega$s. In fact, if I'm not wrong, I showed that the initial problem (the one posted by lqdschool) was equivalent to the fact that $N$ lies on $\omega_a$, so the problem is a consequence of this problem.

[quote="grobber"]Here's a question: $N=X(?)$ in the ETC database.[/quote]

N is the triangle center X(100), defined as the anticomplement of the Feuerbach point of triangle ABC. It is actually well-known that this point X(100) lies on the circumcircle of triangle ABC and on the Euler lines of triangles $BCI_a$, $CAI_b$, $ABI_c$.

What concerns the circles $\omega_a$, $\omega_b$, $\omega_c$, they were known to me under a different definition. In fact, applying Theorem 2.8 of [url=http://de.geocities.com/darij_grinberg]my paper "Generalization of the Feuerbach point"[/url] to triangle $I_aI_bI_c$, you see that if E and F are the orthocenters of the triangles $CAI_b$ and $ABI_c$, then the points E, F, A and N lie on one circle touching the line $I_bI_c$. (Hereby, the point N has to be [i]defined[/i] as the point of intersection of the Euler lines of triangles $BCI_a$, $CAI_b$, $ABI_c$ and not as the second intersection of the line DP with the circumcircle!) Now it remains to show that this circle through the points E, F, A and N is the circle $\omega_a$. In other words, you have to prove that this circle intercepts on the sides AB and AC of the triangle ABC two chords of length a = BC. Well, this is not so obvious, but I leave it as an exercise.

  Darij

I think this thread went off an a tangent, no pun intended.  Has anyone actually solved the original problem?

I think treegoner has, but there's something wrong with his notations (I don't know what exactly :)), and I have by proving those things in my previous messages: I first showed by Ptolemy and by using the condition we must prove that the problem is equivalent to the fact that the Euler line of $BCI_a$, $\omega_a$ (see one of my previous messages) and the circumcircle of $ABC$ are concurrent (at $N$, of course). Then I showed this last fact, but it's sort of long, so maybe I'll post it later (can't make any promises :)).

Actually, Grobber, I wonder how you got in the circles $\omega_a$, $\omega_b$, $\omega_c$ ! They are very nice but you really don't need them...

The line DP [i]is[/i] the Euler line of triangle $BCI_a$. This is because the point D is the circumcenter and the point P is the orthocenter of triangle $BCI_a$. [Why this? Well, the fact that D is the circumcenter of triangle $BCI_a$ is well-known - actually, the point D is the center of a circle through the points B, C, I and $I_a$. The fact that P is the orthocenter of triangle $BCI_a$ can be shown as follows: Since the point P is the reflection of the point I in the midpoint M of the segment BC, the point M is the midpoint of the segment IP. In other words, the segments BC and IP have the same midpoint M. But these two segments BC and IP are the diagonals of the quadrilateral BICP; hence, the diagonals of the quadrilateral BICP bisect each other, and this quadrilateral is a parallelogram. Thus, BP || CI. Now, $CI \perp CI_a$ (since the internal and the external angle bisector of an angle are perpendicular). Hence, $BP \perp CI_a$, and consequently, our point P lies on the B-altitude of triangle $BCI_a$. Similarly, the same point P lies on the C-altitude of triangle $BCI_a$. Hence, P is the orthocenter of triangle $BCI_a$, qed..]

Now, from Theorems 2.1, 2.5, 2.7 of [url=http://de.geocities.com/darij_grinberg]my paper "Generalization of the Feuerbach point"[/url], applied to the triangle $I_a I_b I_c$, we see that the Euler lines of triangles $BCI_a$, $CAI_b$, $ABI_c$ concur at one point Q on the circumcircle of triangle ABC, and that one of the segments AQ, BQ, CQ equals the sum of the two others. Now, the Euler line of triangle $BCI_a$ is the line DP, and thus we see that the point Q is the point of intersection of the line DP with the circumcircle of triangle ABC different from the point D. But this is exactly the way we defined the point N. Hence, Q = N, and we see that one of the segments AN, BN, CN equals the sum of the two others.

And the problem is dead!

  Darij

[quote=treegoner]Using the law of sine
   Let A>B>C . Then
    $NA+NC=NB <=>sin \measuredangle{BDN} = sin\measuredangle{ADN} + sin\measuredangle{CDN} <=> \frac{1}{2} BD.DPsin \measuredangle{BDN} = 1/2 DC.DP sin\measuredangle{CDN}+1/2 IM.MPsin\measuredangle{CDN}
<=> S[BDP] = S[CDQ]+S[IPD] <=> S[BNE] = S[CNE] + S[INE] <=> S[BNC] = S[INC]$
(obvious)
    where E is the intersection of DP and BC.[/quote]

<=> S[BDP] = S[CDP]+S[IPD] <=> is obvious


vector
induction
superior algebra
superior algebra unsolved

For which $n$ there exists a partition of $\mathbb{R}$ in $n$ disjoint additive semigroups.

Using axiom of choice:

Take a basis $B$ of $\mathbb{R}$ over $\mathbb{Q}$ and split $B$ into disjoint sets $A_k$ for $k=1,2,...,n-1$.
Then there are sub-semi-groups $<A_k>: = \{ \sum a_i q_i | a_i \in A_k , q_i \in \mathbb{Q}_0^+ , \text{ as good as all but not all } q_i=0 \}$. Now $\mathbb{R} \backslash \left( \bigcup_k <A_k> \right) = \{ \sum b_i q_i | b_i \in B , q_i \in \mathbb{Q}_0^- , \text{ as good as all } q_i=0 \}$ is closed under addition, too.
So we have a division into $n$ semigroups.

[quote="ZetaX"]
Then there are sub-semi-groups $<A_k>: = \{ \sum a_i q_i | a_i \in A_k , q_i \in \mathbb{Q}_0^+ , \text{ as good as all but not all } q_i=0 \}$. Now $\mathbb{R} \backslash ( \bigcup_k <A_k> ) = \{ \sum b_i q_i | b_i \in B , q_i \in \mathbb{Q}_0^- , \text{ as good as all } q_i=0 \}$ is closed under addition, too.
[/quote]

Thank you, ZetaX. I don't get why $\mathbb{R} \backslash ( \bigcup_k <A_k> ) = \{ \sum b_i q_i | b_i \in B , q_i \in \mathbb{Q}_0^- , \text{ as good as all } q_i=0 \}$ ?
If we have for some real $x$ the representation $x=\sum_I a_i q_i^{+}+\sum_J b_i q_i^{-}$, (by $q^{+}$ and $q^{-}$ i meant the nonegative and nonpositive rationals), if there is not a set $A_k$ from you partition such that all $a_i \in A_k, , i\in I$ i don't understand why $\sum_I a_i q_i^{+} \in \bigcup_k <A_k> $.

And could you please explain  the meaning of the words "as good as all but not all" and "as good as all"- it seems i don't quite understand it.

My construction above is nonsense... (as good as all meant: all but finetely many)

But another way:

Lemma 1:
A nontrivial vector space over an ordered field $F$ can be written as disjoint union of two additive semigroups $A_1 , A_2$.
Proof:
There is at least one element $a_1$ in the basis $\{a_i\}$, so lets consider $A_1=\{ \sum a_i q_i | q_i \in F , q_1 \geq 0 \}$ and $A_2=\{ \sum a_i q_i | q_i \in F , q_1 < 0 \}$. This clearly works.

Lemma 2:
A nontrivial vector space over an ordered field $F$ can be written as disjoint union of three additive semigroups $A_1 , A_2 , A$ where $A$ is a (sub)vector space again. When the vector space is $\infty$ dimensional, then $A$ is that, too.
Proof:
As above: There is at least one element $a$ in the basis $\{a_i\}$, so lets consider $A_1=\{ \sum a_i q_i | q_i \in F , q_1 > 0 \}$ and $A_2=\{ \sum a_i q_i | q_i \in F , q_1 < 0 \}$ and $A=\{ \sum a_i q_i | q_i \in F , q_1 = 0 \}$. It's clear that $A$ is a vector space over $F$ with the desired property.

Theorem:
When $F$ is an ordered field and $V$ is some $\infty$ dimensional vector space over $F$, then $V$ is the union of $n$ semigroups, where $n$ is arbitrary.
Proof by induction:
$n=1$: trivial
$n=2$: Lemma 1
$n \geq 3$:
Write $V$ as union of $A_1 , A_2 , A$ as in Lemma $3$. Then it was already proved that $A$ instead of $V$ has the required property for $n-2$, thus we are done.

This seems like it can be generalised to arbitrary cardinalities smaller than the cardinality of $V$.

But I hope this time it's correct  :?


The relevant formulas are $ a(1 \plus{} r \plus{} r^2 \plus{} \cdots \plus{} r^n) \equal{} \frac {a(1 \minus{} r^{n \plus{} 1})}{1 \minus{} r}$ and $ a(1 \plus{} r \plus{} r^2 \plus{} \cdots) \equal{} \frac a{1 \minus{} r}$.

In this problem, $ \frac23 \plus{} \frac29 \plus{} \frac2{27} \plus{} \frac2{81} \plus{} \frac2{243}$

$ \equal{} \frac23\left(1 \plus{} \frac13 \plus{} \cdots \plus{} \frac1{3^4}\right) \equal{} \frac23\left(\frac {1 \minus{} \left(\frac13\right)^5}{1 \minus{} \frac13}\right) \equal{} 1 \minus{} \left(\frac13\right)^5 \equal{} \boxed{\frac {242}{243}}$.


Hi all, Here's the problem.

[b] Problem [/b]

In how many ways can you distribute 8 different computers to 5 different institutes such that each institute gets atleast a single (I mean, one) computer?

[b] P.S.: [/b] I am seeking a precise, elaborate solution to this problem. One of my peers regarded this problem as being very abstruse. (This claim is "relative", it may vary from person to person). I would like to know if there exists any other way of solving this problem other than treating it as an instance of  "Inclusion-Exclusion" principle.

Thanks!

This is a balls-and-urns problem; see if you can find the relevant part of the [url=http://en.wikipedia.org/wiki/Twelvefold_way]twelvefold way[/url].

I'm actually seeking a  solution to the problem.

Which you would have if you read the link.  The answer is the number of partitions of an $ 8$-element set into $ 5$ nonempty partitions.  The answer is one of [url=http://en.wikipedia.org/wiki/Twelvefold_way#Formulas]these[/url].  Which one?


In how many ways can you color the six sides of a cube in black or white? (Do note that the cube is unchanged when rotated?)

A. 7
B. 10
C. 20
D. 30
E. 36

[hide="Case Work"][b]Case 1:[/b] 6 black sides; 0 white sides. $ 1$ way.

[b]Case 2:[/b] 5 black sides; 1 white side. $ 1$ way.

[b]Case 3:[/b] 4 black sides; 2 white sides. $ 2$ ways: either the white sides are next to each other, or they are on opposite faces of the cube.

[b]Case 4:[/b] 3 black sides; 3 white sides. $ 2$ ways: either three same-colored sides have a common corner, or they don't.

The remaining cases are logically equivalent to the first three, so they need not be written out.

$ 2\cdot(1 \plus{} 1 \plus{} 2) \plus{} 2 \equal{} 10\quad\Rightarrow\quad\boxed{\mathrm{B}}$[/hide]

While two coloring a cube is easy enough that casework can knock it out, for problems like these Burnside's lemma is generally very useful. If you don't know what it is you may want to look it up before reading below.

[hide]

There are 2^6 different fixed cubes, and we consider how the symmetry group of the cube acts upon these:

the identity element fixes everything, rotate any of the sides one turn and there are 2^3 different fixed cubes, rotate a side 180 degrees and that fixes 2^4 cubes, (a 270 degree turn is a 90 degree turn on the other side), rotate around a corner 120 degrees (a 240 degree turn is the same as a 120 degree turn on the opposite corner) and 2^2 cubes are fixed, and just flip an edge (opposite edges do the same thing though) and 2^3cubes are fixed.

so by burnsides lemma we get

1/24 * (2^6+6*8+3*16+8*4+ 6*8) = 10 different cubes.[/hide]


function
induction
algebra unsolved
algebra

[color=darkred]Find all functions $ f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that

$ f(m\plus{}n)\plus{}f(m)f(n)\equal{}f(mn\plus{}1)$
for all integers $ m,n$.[/color] :?:

[quote="baleanu"]Put $ m \equal{} n \equal{} 0$, we have $ f(0) \plus{} (f(0))^2 \equal{} f(1)$ ...............................................(1;
Put $ m \equal{} 1$ , $ n \equal{} 0$ , we have $ f(1) \plus{} f(1)f(0) \equal{} f(1)$ $ \equal{} >$ $ f(1)f(0) \equal{} 0$.............(2);
  From (1) and (2) we have:
CASE 1)   $ f(0) \equal{} 0$ $ \equal{} >$  $ f(1) \equal{} 1$ (in 1).
Now for $ n \equal{} 0$ we obtain $ f(n) \equal{} f(1) \equal{} f(0) \equal{} 0$ .
CASE 2) $ f(1) \equal{} 0$ $ \equal{} >$ for $ n \equal{} 0$ : $ f(m) \plus{} f(m)f(0) \equal{} f(m \plus{} 1)$ and by induction on $ Z$  we obtain $ f(n) \equal{} 0$ (Obs: for $ f(0) \equal{} \minus{} 1$ we have contradiction with $ n \equal{} 0$ and $ m \equal{} \minus{} 1$ .[/quote]

your analysis in case 2 is wrong.
And i must say the answer set is [size=150]largely[/size] incomplete. :wink:  :wink:  :cool:

[quote="ith_power"]And i must say the answer set is [size=150]largely[/size] incomplete. :wink:  :wink:  :cool:[/quote]
Yeah. There are at least:

$ f(n)\equal{}0$
$ f(n)\equal{}n\minus{}1$
$ f(n)\equal{}0$ when $ n$ odd and $ f(n)\equal{}\minus{}1$ when $ n$ even.

Yes I see now my mistake for case2. Sorry.

Let $ n \equal{} 0$, we have $ f(m)(1 \plus{} f(0)) \equal{} f(1)$.
If $ f(0)\ne \minus{} 1$, we get first solution $ f(m) \equal{} 0$.
Now we assume $ f(0) \equal{} \minus{} 1,f(1) \equal{} 0$.
$ P(2, \minus{} 1): f( \minus{} 1)f(2) \equal{} f( \minus{} 1)$. So we have either $ f( \minus{} 1) \equal{} 0$ or $ f(2) \equal{} 1$.
Case 1) $ f( \minus{} 1) \equal{} 0$
$ P(m, \minus{} 1): f(m \minus{} 1) \equal{} f(1 \minus{} m)$. So $ f$ is even
Let $ a \equal{} f(2)$, we have
$ f(2m \plus{} 1) \equal{} af(m) \plus{} f(m \plus{} 2)$
or
$ f( \minus{} 2m \plus{} 1) \equal{} f(2m \minus{} 1) \equal{} af(m) \plus{} f(m \minus{} 2)$.
or
$ f(2m \minus{} 1) \equal{} af(m \minus{} 1) \plus{} f(m \plus{} 1)$.
Compare the above two, we get
$ f(m \plus{} 1) \equal{} af(m) \minus{} af(m \minus{} 1) \plus{} f(m \minus{} 2)$.
The characteristic function is $ x^3 \minus{} ax^2 \plus{} ax \minus{} 1 \equal{} (x \minus{} 1)(x^2 \plus{} (1 \minus{} a)x \plus{} 1)$.
If this function has root $ r$ with $ |r| > 1$, then we can write $ g(m) \equal{} c_1 \plus{} c_2r^m \plus{} c_3r^{ \minus{} m},\forall m$,
where $ c_1,c_2\ne0,c_3\ne0$ are constants.
Now we look at the original f.e, we see the two sides have different exponential growth rate, contradiction!
Hence, we have $ |1 \minus{} a|\le2$, or $ a \equal{} \minus{} 1,0,1,2,3$.
When $ a \equal{} \minus{} 1$, we get second solution $ f(m) \equal{} \minus{} 1,\forall 2|m;f(m) \equal{} 0,\forall 2\not|m$.
When $ a \equal{} 0$, we get third solution $ f(m) \equal{} \minus{} 1,\forall 3|m;f(m) \equal{} 0,\forall 3\not|m$.
When $ a \equal{} 1$, we get fourth solution  $ f(m) \equal{} \minus{} 1,\forall 4|m;f(m) \equal{} 1,\forall m \equal{} 2\mod4;f(m) \equal{} 0,\forall 2\not|m$.
When $ a \equal{} 2$, there is no solution.
When $ a \equal{} 3$, we get fifth solution $ f(m) \equal{} m^2 \minus{} 1$.

Case 2) $ f(2) \equal{} 1,f( \minus{} 1)\ne0$
Let $ a \equal{} f( \minus{} 1)\ne0$, we have
$ P(m, \minus{} 1): f(m \minus{} 1) \plus{} af(m) \equal{} f( \minus{} m \plus{} 1)$ (*).
$ P( \minus{} m \plus{} 2, \minus{} 1): f( \minus{} m \plus{} 1) \plus{} af( \minus{} m \plus{} 2) \equal{} f(m \minus{} 1)$
Compare the above two, we get
$ af( \minus{} m \plus{} 2) \equal{} \minus{} af(m)\Rightarrow f( \minus{} m \plus{} 2) \equal{} \minus{} f(m)$
or $ f( \minus{} m \plus{} 1) \equal{} \minus{} f(m \plus{} 1)$, plug this in (*), we get
$ f(m \plus{} 1) \equal{} \minus{} af(m) \minus{} f(m \minus{} 1)$.
The characteristic function is $ x^2 \minus{} ax \plus{} 1$, which can not have root with greater-than-1 modulus.
Hence, $ a \equal{} \minus{} 2, \minus{} 1,1,2$.
When $ a \equal{} \minus{} 2$, we get sixth solution $ f(m) \equal{} m \minus{} 1$.
When $ a \equal{} \minus{} 1$, there is no solution
When $ a \equal{} 1$, we get seventh solution $ f$ is periodic $ \minus{} 1,0,1, \minus{} 1,0,1,\cdots$
When $ a \equal{} 2$, there is no solution.

Conclusion, there are totally seven solutions
1) $ f(m) \equal{} 0$
2) $ f$ is periodic with period $ 2$,
$ \minus{} 1,0, \minus{} 1,0,\cdots$.
3) $ f$ is periodic with period $ 3$,
$ \minus{} 1,0,0, \minus{} 1,0,0,\cdots$
4) $ f$ is periodic with period $ 4$,
$ \minus{} 1,0,1,0, \minus{} 1,0,1,0,\cdots$
5) $ f(m) \equal{} m^2 \minus{} 1$
6) $ f(m) \equal{} m \minus{} 1$
7) $ f$ is periodic with period $ 3$,
$ \minus{} 1,0,1, \minus{} 1,0,1,\cdots$

[quote="xxp2000"]Conclusion, there are totally seven solutions
1) $ f(m) \equal{} 0$
2) $ f$ is periodic with period $ 2$,
$ \minus{} 1,0, \minus{} 1,0,\cdots$.
3) $ f$ is periodic with period $ 3$,
$ \minus{} 1,0,0, \minus{} 1,0,0,\cdots$
4) $ f$ is periodic with period $ 4$,
$ \minus{} 1,0,1,0, \minus{} 1,0,1,0,\cdots$
5) $ f(m) \equal{} m^2 \minus{} 1$
6) $ f(m) \equal{} m \minus{} 1$
7) $ f$ is periodic with period $ 3$,
$ \minus{} 1,0,1, \minus{} 1,0,1,\cdots$[/quote]
 :spider:

[quote="xxp2000"]Let $ n \equal{} 0$, we have $ f(m)(1 \plus{} f(0)) \equal{} f(1)$.
If $ f(0)\ne \minus{} 1$, we get first solution $ f(m) \equal{} 0$.
Now we assume $ f(0) \equal{} \minus{} 1,f(1) \equal{} 0$.
$ P(2, \minus{} 1): f( \minus{} 1)f(2) \equal{} f( \minus{} 1)$. So we have either $ f( \minus{} 1) \equal{} 0$ or $ f(2) \equal{} 1$.
Case 1) $ f( \minus{} 1) \equal{} 0$
$ P(m, \minus{} 1): f(m \minus{} 1) \equal{} f(1 \minus{} m)$. So $ f$ is even
Let $ a \equal{} f(2)$, we have
$ f(2m \plus{} 1) \equal{} af(m) \plus{} f(m \plus{} 2)$
or
$ f( \minus{} 2m \plus{} 1) \equal{} f(2m \minus{} 1) \equal{} af(m) \plus{} f(m \minus{} 2)$.
or
$ f(2m \minus{} 1) \equal{} af(m \minus{} 1) \plus{} f(m \plus{} 1)$.
Compare the above two, we get
$ f(m \plus{} 1) \equal{} af(m) \minus{} af(m \minus{} 1) \plus{} f(m \minus{} 2)$.
The characteristic function is $ x^3 \minus{} ax^2 \plus{} ax \minus{} 1 \equal{} (x \minus{} 1)(x^2 \plus{} (1 \minus{} a)x \plus{} 1)$.

Oooooh! Thanks. :)  :D
If this function has root $ r$ with $ |r| > 1$, then we can write $ g(m) \equal{} c_1 \plus{} c_2r^m \plus{} c_3r^{ \minus{} m},\forall m$,
where $ c_1,c_2\ne0,c_3\ne0$ are constants.
Now we look at the original f.e, we see the two sides have different exponential growth rate, contradiction!
Hence, we have $ |1 \minus{} a|\le2$, or $ a \equal{} \minus{} 1,0,1,2,3$.
When $ a \equal{} \minus{} 1$, we get second solution $ f(m) \equal{} \minus{} 1,\forall 2|m;f(m) \equal{} 0,\forall 2\not|m$.
When $ a \equal{} 0$, we get third solution $ f(m) \equal{} \minus{} 1,\forall 3|m;f(m) \equal{} 0,\forall 3\not|m$.
When $ a \equal{} 1$, we get fourth solution  $ f(m) \equal{} \minus{} 1,\forall 4|m;f(m) \equal{} 1,\forall m \equal{} 2\mod4;f(m) \equal{} 0,\forall 2\not|m$.
When $ a \equal{} 2$, there is no solution.
When $ a \equal{} 3$, we get fifth solution $ f(m) \equal{} m^2 \minus{} 1$.

Case 2) $ f(2) \equal{} 1,f( \minus{} 1)\ne0$
Let $ a \equal{} f( \minus{} 1)\ne0$, we have
$ P(m, \minus{} 1): f(m \minus{} 1) \plus{} af(m) \equal{} f( \minus{} m \plus{} 1)$ (*).
$ P( \minus{} m \plus{} 2, \minus{} 1): f( \minus{} m \plus{} 1) \plus{} af( \minus{} m \plus{} 2) \equal{} f(m \minus{} 1)$
Compare the above two, we get
$ af( \minus{} m \plus{} 2) \equal{} \minus{} af(m)\Rightarrow f( \minus{} m \plus{} 2) \equal{} \minus{} f(m)$
or $ f( \minus{} m \plus{} 1) \equal{} \minus{} f(m \plus{} 1)$, plug this in (*), we get
$ f(m \plus{} 1) \equal{} \minus{} af(m) \minus{} f(m \minus{} 1)$.
The characteristic function is $ x^2 \minus{} ax \plus{} 1$, which can not have root with greater-than-1 modulus.
Hence, $ a \equal{} \minus{} 2, \minus{} 1,1,2$.
When $ a \equal{} \minus{} 2$, we get sixth solution $ f(m) \equal{} m \minus{} 1$.
When $ a \equal{} \minus{} 1$, there is no solution
When $ a \equal{} 1$, we get seventh solution $ f$ is periodic $ \minus{} 1,0,1, \minus{} 1,0,1,\cdots$
When $ a \equal{} 2$, there is no solution.

Conclusion, there are totally seven solutions
1) $ f(m) \equal{} 0$
2) $ f$ is periodic with period $ 2$,
$ \minus{} 1,0, \minus{} 1,0,\cdots$.
3) $ f$ is periodic with period $ 3$,
$ \minus{} 1,0,0, \minus{} 1,0,0,\cdots$
4) $ f$ is periodic with period $ 4$,
$ \minus{} 1,0,1,0, \minus{} 1,0,1,0,\cdots$
5) $ f(m) \equal{} m^2 \minus{} 1$
6) $ f(m) \equal{} m \minus{} 1$
7) $ f$ is periodic with period $ 3$,
$ \minus{} 1,0,1, \minus{} 1,0,1,\cdots$[/quote]


Find all fuctions f:R->R satisfying: $ f(f(x))=x+a$ (a=const)

There are far too many. We can choose $f(0)$ arbitrarily, as long as $f(0) \neq na$ for any integer $n$. Then, for any $b$ not equal to an integer multiple of $a$ or $f(0)$ plus an integer multiple of $a$, we can choose $f(b)$ almost arbitrarily.

Here is an example, considerably nicer than most:
For $0<x<\frac{a}{2}$, $f(x)=\frac{a}{2}+\sqrt{ax-x^2}$. For $\frac{a}{2}<x<a$, $f(x)=\frac{3a}{2}-\sqrt{ax-x^2}$. Also $f(0)=\frac{a}{2}$ and $f(\frac{a}{2})=a$. We then extend this function by $f(x+a)=f(x)+a$.

That example is even continuous- I'm not sure what you were expecting from this problem.


Prove that if $ a>b>0$ and $ a^5\plus{}b^5\equal{}a\minus{}b$ then $ a^4\plus{}b^4<1$.

[quote="moldovan"]Prove that if $ a > b > 0$ and $ a^5 \plus{} b^5 \equal{} a \minus{} b$ then $ a^4 \plus{} b^4 < 1$.[/quote]

We have
$ a \minus{} b \equal{} a^5  \plus{} b^5  > a^5  \minus{} b^5  \equal{} (a \minus{} b)(a^4  \plus{} a^3 b \plus{} a^2 b^2  \plus{} ab^3  \plus{} b^4 )$
So
$ a^4  \plus{} a^3 b \plus{} a^2 b^2  \plus{} ab^3  \plus{} b^4  < 1$
$ \Rightarrow a^4  \plus{} b^4  < 1$


Let $ a,\ b,\ c$ be side lengths of the triangle such that $ (a \plus{} b): (b \plus{} c): (c \plus{} a) \equal{} 4: 5: 6$ and the area of $ 15\sqrt{3}$.

(1) Find the ratio of $ a: b: c$.

(2) Find the values of $ \cos C$.

(3)  Find the  exradius $ R$ of $ \triangle{ABC}$.

(4) Find the inradius $ r$ of $ \triangle{ABC}$.

[hide="my solution"]

Let $ a\plus{}b\equal{}4x,b\plus{}c\equal{}5x,c\plus{}a\equal{}6x$, so $ a\plus{}b\plus{}c\equal{}\frac{15x}{2}$
$ a\equal{}5x/2,b\equal{}3x/2,c\equal{}7x/2$
So, $ a: b: c\equal{}5: 3: 7$ (a)

From heron formula
$ 15\sqrt{3}\equal{}\sqrt{\frac{15x.5x.9x.x}{4^4}}$
$ 225.3.4^4\equal{}15.5.9.x^4$
$ x\equal{}4$

$ a\equal{}10,b\equal{}6,c\equal{}14$
$ \frac{1}{2}10.6.sinC\equal{}15\sqrt{3}$
$ sinC\equal{}\frac{1}{2}\sqrt{3} \rightarrow \angle C\equal{}60^0$
$ Cos C\equal{}\frac{1}{2}$ (b)

$ R\equal{}\frac{abc}{4.15\sqrt{3}}\equal{}\frac{14}{3}\sqrt{3}$ (c)

$ r\equal{}\frac{15\sqrt{3}}{15}\equal{}\sqrt{3}$ (d)

[/hide]

Your answer is correct, except the values of $ \cos C$, but why is $ \angle{C} \equal{} 60^\circ$?


En un trapecio ABCD de base mayor AB, base menor DC y lados no
    paralelos BC y DA, sea K el punto del lado BC tal que BK = 1/3 BC. Se
    traza por K la recta paralela a DA que corta a AB en L. Si BL = CD y el
    área del trapecio ABCD es 20, calcular el área del triángulo ADL.

:?  no se, la verdad no me motiva a publicr la solucion, peor la respuesta es [b]12[/b]

$PD.$ $colgar$ $problemas$ $mas$ $dificiles$

[quote="hucht"]$PD.$ $colgar$ $problemas$ $mas$ $dificiles$[/quote]

Algo así?? http://www.mathlinks.ro/Forum/viewtopic.php?t=76357

Supongo que si, no lo se en realidad porque a mi no me gusta la aritmetica, pero es bueno que los problemas tengan buen nivel si quieren puedo colgar problemas de geometria de buen nivel.... :D

:D  ya que mi novia no entra al msn colgare problemas, que tal?

[color=blue][b]Problema 1. [/b]Dado el triangulo $ABC$ recto en $C,$ sea $D$ un punto en $AB$ tal que los triangulo $ACD$ y $DCB$ tienen el mismo inradio, demostrar que $[ABC]=CD^2$ ($[ABC]$ significa area dle triangulo $ABC$)

[b]Problema 2.[/b] Dado un triangulo, demostrar que sus puntos de Fermat, el centro de su circunferencia de Euler y el circuncentro estan en una circunferencia. [/color]

(vean en:
$\square$ [url]http://http://centros5.pntic.mec.es/~marque12/matem/fermat.htm[/url],
$\square$ http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=fermat+%5C%5Ctriangle&t=47955,
 $\square$ [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?highlight=triangles+constructed++the+outside+%28on+the+inside&t=26139]www.artofproblemsolving.com/Forum/viewtopic.php?highlight=triangles+constructed++the+outside+%28on+the+inside&t=26139[/url])

[color=blue][b]Problema 3. [/b]En un trapecio se conoce que la suma de ambas diagonales es 4 y que el área es 2. Según lo anterior se pide calcular la altura.

[b]Problema 4.[/b] Sea $ABCDE$ un pentágono convexo de área $1$ (las diagonales quedan dentro del pentágono). Sean $P,$ $Q,$ $R$ y $S$ los baricentros de los triángulos $ABE,$ $BCE,$ $CDE$ y $DAE,$ respectivamente. Indicar la naturaleza del cuadrilátero $PQRS$ y dar su área.

[b]Problema 5.[/b] La circunferencia inscrita en $\triangle ABC$ tiene centro $O$ y es tangente a los lados $BC,$ $AC$ y $AB$ en los puntos $X,$ $Y$ y $Z,$ respectivamente. Las rectas $BO$ y $CO$ intersecan a la recta $YZ$ en los puntos $P$ y $Q,$ respectivamente. Si  $XP$ y $XQ$ tienen la misma longitud, Indicar la naturaleza del triangulo $ABC.$

[b]Problema 6.[/b] Los vértices de un triangulo $ABC$ pertenecen a la hipérbola $xy=1.$ Demostrar que el ortocentro también pertenece a la hipérbola.[/color]

(Sug. Usar tangentes)

[color=blue][b]Problema 7.[/b] En $ABC,$ sean $AD,BE,CF$ cevianas concurrentes en $M.$ Si los triángulos $BDM,$ $CME$ y $AMF$ tienen la misma área y el mismo perímetro, entonces $ABC$ es equilátero.[/color]

[color=red][b]Problema 8.[/b] (Dedicado a Tipe) Los números enteros del 1 al 2002, ambos inclusive, se escriben en una pizarra en orden creciente 1, 2, . . . , 2001, 2002. Luego, se borran los que ocupan el primer lugar, cuarto lugar, séptimo lugar, etc., es decir, los que ocupan los lugares de la forma 3k + 1. En la nueva lista se borran los números que están en los lugares de la forma 3k +1. Se repite este proceso hasta que se borran todos los números de la lista. ¿Cuál fue el último número que se borró?  :dry: $(detesto$ $la$ $aritmetica)$
Espero manden sus soluciones prontito  ;) [/color]

Cuidense

$Jose^/ Carlos$


geometry
3D geometry
sphere
function
trigonometry
integration
real analysis

Prove that

$\sum_{n=-\infty}^\infty e^{-(x+n)^{2}/2}= \sqrt{2\pi}$

The Poisson summation formula could be useful  :)

I was trying to come up with some unusual solution. Consider the sphere of radius $\sqrt{N}$ in the $N$-dimensional space. Let $\mu_{N}$ be its normalized surface measure. Then the gaussian measure is the weak* limit of orthogonal projections of $\mu_{N}$ onto a line.
  Now take an infinite sequence of equally spaced spheres and try to imagine that as $N\to\infty$, the total surface somehow converges to an (infinite-dimensional) cylinder. :ewpu:

Poisson summation formula is useful, indeed, but it immediately shows that the problem is wrong as stated :P. The series does [b]not[/b] represent a constant function!

[quote]oh, fedja... appearence of the expression can be decieving and in fact, this formula is correct.[/quote]

ah, i'm sorry. i had a mistake in calculating fourier coefficient...  :blush:

here, it is almost 3 am and i often make critical mistakes with sleepy head...

let $f(x) = \sum_{n=-\infty}^\infty e^{-(x+n)^{2}/2}$, then one can show that

$f(x)=\sqrt{2\pi}\sum_{k=-\infty}^{\infty}e^{-2\pi^{2}k^{2}}e^{2\pi ikx}$.

obviously $f(0)$ and $f(1/2)$ have different values, because

$f(0)=\sqrt{2\pi}\sum_{k=-\infty}^{\infty}e^{-2\pi^{2}k^{2}}$

and

$f(1/2)=\sqrt{2\pi}\sum_{k=-\infty}^{\infty}e^{-2\pi^{2}k^{2}}(-1)^{k}$.

fedja is right, and this is not constant.

On the one hand, I tried it out numerically, summing it from $n=-10$ to $n=10,$ for $x$ ranging from $-0.5$  to $0.5,$ and it does appear to be accurate to within about one part in $10^{8}.$

But on the other hand:

We have the function $f(x)=e^{-x^{2}/2}.$ I know its Fourier transform.

We let $g(x)=\sum_{n=-\infty}^{\infty}f(x+n).$ We should have that $g$ is periodic of period $1,$ and we should be able to find the Fourier coefficients of $g$ by discretely sampling the Fourier transform of $f.$ In fact, we should have

$g(x)=\sqrt{2\pi}\sum_{k=-\infty}^{\infty}e^{-2\pi^{2}k^{2}}e^{2\pi^{2}ikx}.$

To see how much this deviates from being constant, look at the size of the $k=1$ term of this sum. That Fourier coefficient is $e^{-2\pi^{2}}\approx 3\times 10^{-9}.$ That matches up pretty well with what I'm seeing numerically.

[quote]The graph would look better with $\sqrt{2\pi}$ subtracted.[/quote]

ok, re-plotted. numerical calculation with mathematica 5.1 shows that the graph of $f(x)-\sqrt{2\pi}$ is

[img]http://myhome.shinbiro.com/~sosQED/nsol083.GIF[/img]

He-he... You do not need mathematica to sketch the graph of $\cos 2\pi x$ (the difference between the actual function with $\sqrt{2\pi}$ subtracted and a multiple of $\cos 2\pi x$ is less than the thickness of the curve representing the graph, so you'll never see it on the computer screen). Returning to Xevarion's wrong formula, here comes a problem:

Does there exist an entire function $f: \mathbb C\to \mathbb C$ such that $f$ has no zeroes on the complex plane, $f$ is non-negative on $\mathbb R$ and $\sum_{n\in \mathbb Z}f(x+n)=\text{\,const}$ ?

Why, yes, there is such an $f.$ There are many such functions. Here's one:

$f(x)=\int_{0}^{1}(1-\xi)^{2}\cos(2\pi\xi x)\,d\xi= \frac1{2\pi^{2}x^{2}}\left(1-\frac{\sin(2\pi x)}{2\pi x}\right).$

On second thought ... you wanted no zeros in the complex plane? My function has no zeros in $\mathbb{R},$ but I'm not so sure about elsewhere.

Yes, I wanted no zeroes on the entire complex plane and Kent's function has plenty of complex zeroes ;). The request for no zeroes on the plane may seem a bit strange but it eliminates simple examples and agrees well with the title of the topic.

$f(x)=\int_{x-\frac12}^{x+\frac12}e^{-\pi t^{2}}\,dt$ has some attractive properties: it satisfies the sum identity, is entire, and is postive on the real axis. But I'm afraid it likely also has zeros in the complex plane.

The general principle: any entire function satisfying the sum identity is the convolution of an entire function (with certain decay properties along the real axis) and a "box". The hard part is avoiding complex zeros.


The book comes out sometime around the 27th of this month.

[quote="brady00n"]When's the AoPS Intro Number Theory coming out?  What are some other number theory books that start at the very basics - no knowledge of modular arithmetic?

Thanks![/quote]

If you have no knowledge, like I did, just review AoPS 1. You have it don't you?


function
algebra
system of equations
algebra proposed

[quote="nkht-tk14"][u]

  [i]Problem [/i][/u]

  [i]  Solve the following system of equations of n unknowns

   $ \sqrt {x_1} \plus{} \sqrt {x_2} \plus{} .... \plus{} \sqrt {x_n} \equal{} n$

    $ \sqrt {x_1 \plus{} 8} \plus{} \sqrt {x_2 \plus{} 8} \plus{} .... \plus{} \sqrt {x_n \plus{} 8} \equal{} 3n$

    ( $ n$ is a given positive integer )

   Generalize the problem [/i][/quote]

$ \sqrt {x_1 \plus{} 8} \plus{} \sqrt {x_2 \plus{} 8} \plus{} .... \plus{} \sqrt {x_n \plus{} 8} \equal{} 3n$ ---> $ (\frac{3}{2})(\sqrt {x_1 \plus{} 8} \plus{} \sqrt {x_2 \plus{} 8} \plus{} .... \plus{} \sqrt {x_n \plus{} 8}) \equal{} \frac{9n}{2}$  --->  $ \frac{9n}{2}\equal{}\sqrt{\frac{1}{4}\plus{}2}*\sqrt{x_{1}\plus{}8}\plus{}\sqrt{\frac{1}{4}\plus{}2}*\sqrt{x_{2}\plus{}8}\plus{}...\plus{}\sqrt{\frac{1}{4}\plus{}2}*\sqrt{x_{n}\plus{}8}\geq (\frac{\sqrt{x_{1}}}{2}\plus{}4)\plus{}(\frac{\sqrt{x_{2}}}{2}\plus{}4)\plus{}...\plus{}(\frac{\sqrt{x_{n}}}{2}\plus{}4)\equal{}\frac{9n}{2}$

 ---->

$ \frac{x_{1}}{\frac{1}{4}}\equal{}\frac{8}{2}\equal{}4$
$ \frac{x_{2}}{\frac{1}{4}}\equal{}\frac{8}{2}\equal{}4$
...........
$ \frac{x_{n}}{\frac{1}{4}}\equal{}\frac{8}{2}\equal{}4$

$ x_{1}\equal{}x_{2}\equal{}...\equal{}x_{n}\equal{}1$

An other method ..  $ f(x)\equal{}\sqrt x , g(x) \equal{} \sqrt {8\plus{}x}\ ; \ f(x_1)\plus{}...\plus{}f(x_n)\equal{}nf(1)$$ \  , \  g(x_1)\plus{}...\plus{}g(x_n)\equal{}ng(1)$
Let's define $ F(x_1,...,x_n)\equal{} f(x_1)\plus{}...\plus{}f(x_n)\ , \ G(x_1,...,x_n)\equal{}g(x_1)\plus{}...\plus{}g(x_n)$ .
With the Lagrange multipliers you  obtain that the extremum of $ G$ holds iff $ G'_{x_k}\minus{}\lambda F'_{x_k}\equal{} g'(x_k)\minus{}\lambda f'(x_k)\equal{}0$, so $ x_1\equal{}..\equal{}x_n\equal{}1$
So with the condition $ F(x_1,...,x_n)\equal{}nf(1)$ the extremum is $ G_{min}\equal{} ng(1)$ and it holds only for   $ x_1\equal{}..\equal{}x_n\equal{}1$ ... and that's the solution ..
You may generalize with other convenient $ f,g$  functions ..
 :cool:


Good problem
[hide]
First, we need to simplify the bases in the logs.
$121_x = x^2 + 2x + 1 = (x + 1)^2$
Doing the same on the bottom gives us $1331_x = (x + 1)^3$

So our equation becomes
$\prod_{x=4}^{1331}\left(\frac{\log_x{(x+1)^2}}{\log_x{(x+1)^3}}\right)$
(much thanks to elemennop for the latex)

Anyway, this is pretty easy to work with.  The fraction simplifies to
$\left(\frac{2\log_x{(x+1)}}{3\log_x{(x+1)}}\right)$
$= \frac{2}{3}$
Therefore, our solution is $(\frac{2}{3})^{1328}$
Which I don't think simplifies... this is twice today I've gotten a large exponent answer and couldn't simplify it.  The first one was simplified, and I think this one can be too... it's frustrating.
[/hide]

To the previous poster, that would be

$\prod_{x=4}^{1331}\displaystyle\left(\frac{\log_x{(x+1)^2}}{\log_x{(x+1)^3}}\right)$

That's what I got, but I solved it with just some properties of logs. What's that giant pi mean?

Have you seen a big sigma to mean sum?  This is a big pi to mean product.


geometry
circumcircle
ratio
symmetry
perpendicular bisector

$ (O)$ is a semicircle and $ AB$ is a diameter. $ (O')$ is the semicircle with diameter $ AO$ (in the same half-plane than $ (0)$).
$ C\in AO$, $ C$ variable. The perpendicular to $ AB$ at $ C$ cuts $ (O')$ at $ D$ and $ (O)$ at $ F$. $ AD$ intercepts $ (O)$ at $ E$. $ (I)$ is the circumcircle of triangle $ DEF$.
[b](i)[/b] Prove that $ AF$ is tangent to $ (I)$.

[b](iii)[/b] Prove that $ I$ belongs to the circle with center $ O$ and radius $ \sqrt{2}/2\cdot AO$.

I made part (i). I'm missing (ii).

[hide]I first proved using similarity that $ AF/AE\equal{}\sqrt{2}$ . Then $ AD\cdot AE \equal{} 1/2\cdot AE^2 \equal{} AF^2$.
Therefore, power of $ A$ with respect to $ (I)$ implies $ AF$ is tangent to it.[/hide]

Do you think I should move this problem to the Olympiad Section?

The original problem was only part (i) and I found that solution. Others are welcome.

But then I asked myself what is the locus of I, so I made that conjecture.

 :)

Well, the 'conjecture' is true.
Let's prove all from beginning: let's complete the two circles $(O), (O')$ and note $\{A;M\}\in AF\cap\odot(AOD)$, $\{D; N\}\in FD\cap\odot (AOD)$, $\{F; P\}\in FD\cap\odot (ABF)$.
Obviously, $\widehat{AFC}=\widehat{ABF}=\widehat{AEF}$ so $AF$ is tangent to the circle $\odot(DEF)$.
As the circles $(O), (O')$ are homothetic of ratio $2$ and center $A :$ $M, D$ are midpoints of $AF, AE$ respectively.
Consequently $FM=AM$ and $DE=AD$.
From power of $D$ w.r.t. $(O): DE\cdot AD= DF\cdot DP\ (\ 1\ )$, but $PD=FN$ by symmetry, so $AE^2=2\cdot FD\cdot FN\ (\ 2\ )$. From the power of $F$ w.r.t. $(O'): FD\cdot FN=FM\cdot FA=\frac{AF^2}{2}\ (\ 3\ )$. From the relations $(2), (3)$ we get $AE^2=2\cdot AF^2\implies AD=AM\sqrt{2}\ (\ 4\ )$.
Next we shall prove that $\triangle BOI\sim\triangle NAM$. Indeed, $\angle BOE=2\angle BAE=\angle DAN\ (\ 5\ )$, $OI$ is perpendicular bisector og $EF$ so $\angle EOI=\frac{\angle FOE}{2}=\angle FAE\ (\ 6\ )$; consequently $\angle BOI=\angle BOE+\angle EOI=\angle DAN+\angle DAM=\angle MAN\ (\ 7\ )$.
Since $OM\parallel BF\implies\angle IBO=\angle AOM=\angle ANM\ (\ 8\ )$; from $(7), (8)$ our claim is proven.
But $AN=AD$ and, with $(4)$ we are done $($i.e. $\frac{OB}{OI}=\frac{AD}{AM}=\sqrt{2})$.

Best regards,
sunken rock


algebra
polynomial
Ring Theory
superior algebra
superior algebra solved

Let I denote the set of all polynomials in Z[x] that have an even number as the constant term.  Prove that I is an ideal of Z[x]. Also prove that I is not a principal ideal of Z[x].

The first one is simple to check. For the second one, assume to the contrary that it is principal and generated by some $a$. But then $a | 2$, and the rest is easy.


Does anybody have any links for some past amc 12 problems?

Try the AMC's practice problem archive:

http://www.unl.edu/amc/mathclub/04,1-problems.html

thanx alot that really helped try my challange problem in the pre olympiad section :lol:


function
limit
integration
calculus
real analysis
real analysis unsolved

Let $f: [0,\infty)\to\mathbb{R}$ be a continuous and bounded function such that $f(0)=0$.

     a) Prove that \[ \lim_{n\to\infty}\int_{0}^{1}f(nx^{n})dx=0.  \]

     b) Find \[ \lim_{n\to\infty}\int_{0}^{1}\sqrt{1+n^{2}x^{2n}}dx.  \]

For (a): pick $\epsilon > 0$. Suppose $f$ is bounded by $M$. Obviously for any fixed $x < 1$, $nx^n \to 0$ as $n \to \infty$. So pick $N$ large enough that $n > N \Rightarrow n(1-\epsilon)^n < \epsilon$. Then the integral from 0 to $1-\epsilon$ is arbitrarily small for large enough $n$ (by continuity of $f$), and then the rest of the integral is less than $M\epsilon$. But $\epsilon$ is arbitrarily small, so we win.

(b) should be 1 for similiar reasons, but since the function isn't uniformly bounded for $x \in [0,1)$ it is not so obvious to me how we should estimate the small end part of the integral.

Actually, $\sqrt{1+n^{2}x^{2n}}$ blows up near $1$ and you can't control it like you wanted to.

But since it's pretty obvious that we must use (a), let's do just that.

Let $f : \left[ 0, \infty \right) \to \mathbb R, \, f(x) = \sqrt{1+x^{2}}-x-1$. We have that \[\sqrt{1+n^{2}x^{2n}}= \left( \sqrt{1+n^{2}x^{2n}}-n x^{n}-1 \right)+n x^{n}+1 = f \left( n x^{n}\right)+n x^{n}+1 .\]
We have that $f(0) = 0$ and that
\[| f(x) | = \left| \frac1{\sqrt{1+x^{2}}+x}-1 \right| \leq 1 ,\]
so by (a) we have that $\lim_{n \to \infty}\int_{0}^{1}f \left( n x^{n}\right) \, dx = 0$.

Hence, $\lim_{n \to \infty}\int_{0}^{1}\sqrt{1+n^{2}x^{2n}}\, dx = \lim_{n \to \infty}\left( n x^{n}+1 \right) = \lim_{n \to \infty}\frac{2n+1}{n+1}= 2$.


inequalities
Euler
induction
function
inequalities unsolved
109

Find the biggest $k$ such that
\[\sum_\text{cyc}a(a-b)(a-c)\ge k\sum_\text{cyc}\frac{ab}{a+b}(a-b)^{2}\]
for all $a,b,c>0$.
[hide="maybe"]I think $k_{max}=\frac12$. :wink: [/hide]

[quote="fjwxcsl"]Let c=1,b=1,a->0,then
        f= a*(a-b)*(a-c)+b*(b-c)*(b-a)+c*(c-a)*(c-b)-k*(a*b/(a+b)
           *(a-b)^2+b*c/(b+c)*(b-c)^2+c*a/(c+a)*(c-a)^2)
        =a*(a-1)^2*(a+1-2*k)/(a+1)>=0
        =>k<=1/2,
   and if k=1/2,then
      numer( f)=F[6][0,2]+1/2*F[6][0,3]+2*F[6][1,1]>=0,
   in whch
      $F[6][0,2]=\sum(a^{3}*(b+c)*(a-b)*(a-c))\geq 0;$
$F[6][0,3]=(a-b)^{2}*(b-c)^{2}*(c-a)^{2}\geq 0;$
$F[6][1,1]=\sum(a^{2}*b*c*(b+c)*(a-b)*(a-c))\geq 0.$[/quote]
Your last term has degree 7 not 6. An error by your computer?  :D Please check  :D

Anyway I guess it's essentially Vornicu-Schur and a bit of $(a-b)^{2}(b-c)^{2}(c-a)^{2}$.
Now I doubt there are solutions without expanding, given that $p=\frac14$ yields Iran96. :)

yes, definitely 1/2.
[hide]
it's equivalent to
\[\sum_\text{cyc}\left( a+b-c-\frac{ab}{a+b}\right) (a-b)^{2}\geq 0\]
this can be proven by setting $a\geq b\geq c$ and then writing $(a-c)^{2}=(a-b)^{2}+(b-c)^{2}+2(a-b)(b-c)$ so that the above is
\[a^{2}\left( \frac{1}{a+b}+\frac{1}{a+c}\right) (a-b)^{2}+c^{2}\left( \frac{1}{a+c}+\frac{1}{b+c}\right) (b-c)^{2}\]
\[+2\left( (a-b)+\frac{c^{2}}{a+c}\right) (a-b)(b-c)\]
so it's nonnegative.

When $k>\frac{1}{2}$ it fails for $a=b=1$ and $c$ close enough to 0.
[/hide]

I’ve been looking how to make the ineq even sharper by adding HOT (=higher order terms) to the RHS and searching for maximal possible coefficients. Here is the result: :D
For $a,b,c>0$ and $N\in\mathbb N$, the following holds:
\[\sum_{cyc}a(a-b)(a-c)\ge \sum_{cyc}\Biggl[\frac{ab}{a+b}\cdot\frac{(a-b)^{2}}2 \sum_{k=0}^{N}E_{k}\frac{(ab)^{k}}{(a+b)^{2k}}\Biggr],\]
where $E_{k}=\frac1{k+1}{2k\choose k}$ (I think these are called the Euler numbers): $(E_{0},E_{1},...)=(1,1,2,5,14,42,...)$
Note that $N=0$ is the original problem (with coefficient $\frac12$).

My proof of this is funny and amazingly simple. It uses power series :D Let’s start with a

[b]Lemma[/b]: $\sum_{cyc}a(a-b)(a-c)\ge \sum_{cyc}\frac{(a-b)^{2}}4(a+b-|a-b|).$
[b]Proof of Lemma[/b]: assume wlog $a\ge b\ge c$. Then $LHS-RHS=\frac12(a-b)^{2}(2a+b-3c)\ge0$. :)

And that’s it by virtue of the identity $\boxed{\frac{(b-c)^{2}}4(b+c-|b-c|)= \frac{ (b-c)^{2}}2\frac{bc}{b+c}\sum_{k=0}^\infty E_{k}\Bigl(\frac{bc}{(b+c)^{2}}\Bigr)^{k}.}$ :D Just sum it up cyclically.
[hide="proof of the identity"]

Put $x=\frac{bc}{(b+c)^{2}}$. By AM-GM, $x\le\frac14$, and we have $\sqrt{1-4x}=\frac{|b-c|}{b+c}$.
Using ${{\frac12}\choose k+1}=(-1)^{k}\frac{1\cdot3\cdots(2k-1)}{2^{k+1}(k+1)!}=\frac{1}{2\cdot(-4)^{k}(k+1)}{2k \choose k}$, we have
\[\sqrt{1-4x}= \sum_{k=0}^\infty{{\frac12}\choose k}(-4x)^{k}=1-\sum_{k=0}^\infty\frac2{k+1}{2k \choose k}x^{k+1},\]
thus $\frac{1-\sqrt{1-4x}}4 = \frac12\sum_{k=0}^\infty E_{k}x^{k+1}=\frac12(x+x^{2}+2x^{3}+5x^{4}+...)$

So we get
$\frac{b+c-|b-c|}4 = \frac{b+c}4(1-\frac{|b-c|}{b+c})$ $= (b+c)\frac{1-\sqrt{1-4x}}4 = (b+c) \frac12\sum_{k=0}^\infty E_{k}x^{k+1}= \frac12\frac{bc}{b+c}\sum_{k=0}^\infty E_{k}x^{k}.$
[/hide]

The funny thing is that this strong (thus difficult) inequality follows from an easy lemma, and I don’t really understand why. But I can’t find a flaw in my proof. All terms on the RHS are non-negative, and there is no problem with convergence, as the series for $\sqrt{1-4x}$ converges for all $0\le x\le\frac14$.
My explanation is the following: note that the lemma has many equality cases, as there is equality wherever two of $a,b,c$ are equal and $\ge$ the third one. So it is (at least in a certain sense ;) ) very sharp. And almost trivial, too :lol:

I wonder if this HOT method (sort of) can be applied elsewhere...

Just for fun the same with general Schur:
For $a,b,c>0$ and $n,N\in\mathbb N$, the following holds:
\[\sum_{cyc}a^{n}(a-b)(a-c)\ge \sum_{cyc}\Biggl[\bigl(\frac{ab}{a+b}\bigr)^{n}\cdot\frac{(a-b)^{2}}2 \sum_{k=0}^{N}E_{k,n}\frac{(ab)^{k}}{(a+b)^{2k}}\Biggr], \]
where $E_{k,n}=\frac n{k+n}{2k+n-1\choose k}=\frac n{2k+n}{2k+n\choose k}=\frac{n(2k+n-1)!}{k!(k+n) !}$. In fact, the $E_{k,1}$ are not the Euler numbers, but the Catalan numbers. :blush:
The lemma can be written in an easy way:
$\sum_{cyc}a^{n}(a-b)(a-c)\ge\frac12\sum_{cyc}(b-c)^{2}\min(a,b)^{n}$ and this holds because of
$LHS-RHS=(a-b)^{2}\bigl(\frac{b^{n}-c^{n}}2+(b-c)\sum_{k=1}^{n-1}a^{n-k}b^{k-1}+a^{n}-b^{n-1}c\bigr)\ge0.$ :)
To prove the corresponding identity $\boxed{\min(b,c)^{n}=\bigl(\frac{bc}{b+c}\bigr)^{n}\sum_{k=0}^\infty E_{k,n}x^{k}}$  (*), we can use induction over $n$:
If we define the generating function $f_{n}(x): =\sum_{k=0}^\infty E_{k,n}x^{k}$, then $f_{0}=1, f_{1}=\frac{1-\sqrt{1-4x}}{2x}$, and because of $E_{k,n}= E_{k,n-1}+E_{k-1,n+1}$, we have $f_{n}=\frac1x(f_{n-1}-f_{n-2})$. It is easy to prove that a corresponding recursion holds for both sides of the identity (*) ;)

All this is true for all real $n>0$. To see this, here is a better way to prove the lemma. (there was a little typo, of course the lemma must be
$\sum_{cyc}a^{n}(a-b)(a-c)\ge\frac12\sum_{cyc}(b-c)^{2}\min(b,c)^{n}$.)
If $a\ge b\ge c$, this is equivalent to
$(a-b)^{2}\frac{b^{n}-c^{n}}2+(a-b)(a^{n}-b^{n})(a-c)\ge0$. :)
Now further note that $\frac{b+c}{bc}\min(b,c)= \frac{1-\sqrt{1-4x}}{2x}$ where again $x=\frac{bc}{(b+c)^{2}}$. No need for induction. :D

Note that the sign flips if $n\le0$.
 From the lemma, we easily get the following:
For real $t\ge1$,
\[\boxed{\sum_{cyc}\frac{(b-c)^{2}}{2\min^{t}(b,c) }\ge \sum_{cyc}a^{-t}(a-b)(a-c)\ge\sum_{cyc}\frac{a(b-c)^{2}}{2bc\max^{t-1}(b,c)}}\]
The first ineq even holds for $t\ge0$ :D

[quote]Find the biggest $ k$ such that
\[ \sum_\text{cyc}a(a \minus{} b)(a \minus{} c)\ge k\sum_\text{cyc}\frac {ab}{a \plus{} b}(a \minus{} b)^{2}
\]
for all $ a,b,c > 0$.[/quote]

[quote="spanferkel"] given that $ k \equal{} \frac14$ yields Iran96. :)[/quote]
This was wrong. But I have found a neat way of writing Iran 96 as a refinement of Schur’s, see [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=1316509]here[/url].


abstract algebra
group theory
number theory
relatively prime
superior algebra
superior algebra unsolved

Let $ G$ be a finite abelian group with $ a,b\in G$.  Suppose the order of $ a$ is $ m$ and the order of $ b$ is $ n$.  Prove that the order of $ ab$ is $ mn$.

Clearly $ \left(ab\right)^{mn}\equal{}e$.  So we suppose there exists $ k<mn$ such that $ \left(ab\right)^{k}\equal{}e$.  Since $ G$ is abelian, $ \left(ab\right)^{k}\equal{}a^{k}b^{k}$.  How do we proceed from here?

Yes, sorry, $ m$ and $ n$ are relatively prime.

$ a^kb^k\equal{}e\equal{}a^m$
$ a^{m\minus{}k}\equal{}b^k$
$ a^{n(m\minus{}k)}\equal{}b^{nk}\equal{}e$
$ m|n(m\minus{}k)$
$ m|nk$
Since $ m$ and $ n$ are relatively prime, $ m|k$. Similarly, $ n|k$, and thus $ mn|k$.

Suppose $ a^k b^k \equal{} e$ and that $ k < mn$. Since $ (m,n) \equal{} 1$, $ k$ cannot be a multiple of both $ m$ and $ n$. WLOG it is not a multiple of $ m$. Then $ (a^k b^k)^n \equal{} e^n \equal{} e$. But $ (a^k b^k)^n \equal{} a^{kn} b^{kn} \equal{} a^{kn} e \equal{} a^{kn}$. So $ a^{kn} \equal{} e$. But $ (m,n) \equal{} 1$ and $ k$ is not a multiple of $ m$, so $ m \nmid kn$. This contradicts the fact that the order of $ a$ is $ m$.

EDIT: Beaten? This post didn't even take that long to write.  :(


calculus
integration
LaTeX
Ring Theory
advanced fields
advanced fields unsolved

Consider this equation in the integers :

$y^2+d=x^3$

d is a squarefree integer greater than 1

Formulate a general theorem for its integral solutions (x,y)

It is allowed and encouraged to use factorisation.

Yes, this is a homework problem, and that is not my style, but I'm really stuck on it.  I do not expect a complete solution.  Some opinions on the feasibility of such a theorem would even help me.
I understand that unique  factorisation is only usable in a very few limited cases.  So is there any factorisation-hope for the general case?

Perhaps a nice theorem would be that I only have a finite amount of integral solutions? :oops:

As a side note, it is also asked to discuss the rational solutions.  Is that doable?  I am familiar with some elliptic curve (basic ) theory (well I know what it is and that it is a group), and Mordell's theorem, but I don't know any applications.

So, does anyone have any ideas?  Or do you recognise it?

Solutions I have seen for some special cases make me think that the general case should quite hard...  ;)

Pierre.

At least the case $d = -7$ is very nice and quite easy :)

And I believe the case $d = 1$ or $d = -1$ was a Mathlinks contest problem.

However, this seems a bit hard to me for an homework assignment ;) ...

because this is the famous Mordell equation, here is a link:

[url]http://mathworld.wolfram.com/MordellCurve.html[/url]

Thanks for the replies

But Arne, that Mordell equation seems to involve negative d in my case

I am speaking of equations like $y^2+10=x^3$

On that site, it says Mordell's equation never has infinitely many solutions.

Is the same true for my question?

And Arne, $y^2-7=x^3$ was done by 'more elementary methods' too in our course,
 by rewriting $y^2+1=(x+2)(x^2-2x+4)$

If you have (or can link me to) (elementary ) solutions of $y^2+/-1=x^3$, although not asked in my problem, I would be very interested :|

Btw : as you can see , I am finally trying to live up to mathlinks standards and write in LaTeX , any comments on my latex behaviour is welcome :ninja:

[quote="fredbel6"]But Arne, that Mordell equation seems to involve negative d in my case[/quote]

Oops, sorry :)

[quote="fredbel6"]If you have (or can link me to) (elementary ) solutions of $y^2 \pm 1=x^3$, although not asked in my problem, I would be very interested :|[/quote]

Have a look here:

http://www.mathlinks.ro/Forum/contest/2nd%20Edition/2nd%20edition%20-%20round%207%20solutions.pdf

[quote="fredbel6"]
Btw: as you can see , I am finally trying to live up to mathlinks standards and write in LaTeX, any comments on my latex behaviour is welcome :ninja:[/quote]

That's great :D

Thanks, that looks interesting

However, what about $x^3=y^2-1$ (although that is not part of my original question either, but my shame that I can't do it is too bog  :P )

With Catalan it is easy but such a theorem is overkill in this case I guess

Any other thoughs on my original question?

Can you show us how to do it properly then? ;)

yes, enlighten us  :idea:

but when you speak of elliptic curves, you probably want to solve the 'rational solutions" second part of the question

Can you be more specific about what happens in that ring?

You get something like
$(y+\sqrt(-d))(y-\sqrt(-d)) / x^ 3$
sometimes this does help you to find all solutions? sometimes... :mad:

[quote="cimabue"]you guys are messing around... with the ELLIPTIC <snipped>!!!![/quote]Some of us [b]can[/b] read Romanian, even the subtle ones! :mad: Don't [b]ever[/b] do that again!

[quote="cimabue"]valentin: beauty is in the eye of the beholder. now that i think about it, the word "elliptic" poses a problem as well (even subtler). [/quote]I strongly suggest you stop with the "smart guy" attitude :!:. This will be the last post in this topic about this.

Yes I have heard of factorisation in into ideals.

However, if you denote by $(Z)(a)$ the ring of algebraic integers over $(Z)$ of $(Q)(\sqrt(-d))$, I thought the positive squarefree d's for which $Z(a)$ is UFD were listed as
 1,2,3,7,11,19,43,67,163

Or am I being confused again

About the finiteness, indeed Mordell Siegel is of use here about the integral solutions, it seems

But for the rationals, I don't understand your $|x|<d^(1/3)$ argument, when we are talking about $y^2=x^3+d$ with d greater than 1 , integer, and squarefree?

Hmm, it has come to my attention that even when I have UFD
it is not that easy

Perhaps a decent question that can be answered explicitly by users here : what about the integral solutions of
$y^2+7=x^3$

$(2,1)$ and $(32,181)$ seem to be solutions, where the latter is not so obvious :o

The tricks that come up in my mind seem to be full of plotholes, I would write
$(y+\sqrt(-7))(y-\sqrt(-7))=x^3$ and then do some coprime argument of the two factors in the LHS, but that's
exactly something that I don't have

Any ideas?

Sorry for trying to keep this topic alive, but I felt stupid starting a new topic, so I choose for the lesser of two evils :ninja:

Please delete this message, I made a correction but I cannot remove this one?

So uhm

I am still stuck on this

As I said I found some solutions

My professor told me that $y+\sqrt(-d)$ and $y-\sqrt(-d)$ are coprime for all squarefree d greater than one , regardless of $d=3$ mod 4

But for $d=7$ that is not true because when we observe the solution $(x,y)=(2,1)$ we find that
$1+\sqrt(-7)$ and $1-\sqrt(-7)$ are both divisible by $\frac^{1}_{2}$

Anyone can enlighten me how to solve in the integrals $y^2+7=x^3$ ?

So uhm
I am still stuck on this

As I said I found some solutions

My professor told me that $y+\sqrt(-d)$ and $y-\sqrt(-d)$ are coprime for all squarefree d greater than one , regardless of $d=3$ mod 4

But for $d=7$ that is not true because when we observe the solution $(x,y)=(2,1)$ we find that
$1+\sqrt(-7)$ and $1-\sqrt(-7)$ are both divisible by $\frac{1+\sqrt(-7)}{2}$

Anyone can enlighten me how to solve in the integrals $y^2+7=x^3$ ?

[quote="fredbel6"]My professor told me that y+\sqrt(-d) and $y-\sqrt(-d)$ are coprime[/quote]

A "small" LaTeX hint: when typing square roots, use {} instead of (). You are using (), so you are taking the square root of the bracket "(" but not of the rest of the expression. If you use \sqrt{}, no brackets will appear and the root will look much nicer.

cimabue, I don't understand, what exactly are you saying now, that the conclusion
that $y+\sqrt{-d}$ is a third power in our extension ring is still valid?
And why would you call this case easier?  I really do not understand at all

Thanks for the tip Arne

sorry for bringing it up again
what is up with cimabue's reaction, all those flatliners?

anyway everyone i know is still puzzled about the evil $y^2+7=x^3$  equation

On the other hand, I have found that one rational solution of $y^2+d=x^3$ with $d>1$ and squarefree implies infinitely many solutions over the rationals, as the torsion group is definitely trivial

however, i was wondering if it is possible to have rational solutions without integer solutions


inequalities
vector
linear algebra
linear algebra unsolved

Suppose $ A\in\mathbb{R}^{n\times n}$ is an isometry of $ \ell_p$, how can I show that $ A^t$ is an isometry of $ \ell_q$, where $ \frac{1}{p}\plus{}\frac{1}{q}\equal{}1$?

We need to know two theorems. The first is Hölder's inequality:

$ |y^Tx|\le\|y\|_q\|x\|_p$ for $ x,y\in\mathbb{R}^n$

But the second is that for any given $ x$ we can choose $ y$ so as to have equality in Hölder's inequality. Following up on that, we have

$ \|x\|_p \equal{} \sup_{\|y\|_q \equal{} 1}y^Tx.$

Now suppose $ \|Ax\|_p \equal{} \|x\|_p$ for all $ x.$

$ \|A^Ty\|_q \equal{} \sup_{\|x\|_p \equal{} 1}x^TA^Ty \equal{} \sup_{\|x\|_p \equal{} 1}y^TAx\le \|y\|_q$ since $ \|Ax\|_p \equal{} \|x\|_p \equal{} 1.$

We're not quite there yet. Now comes the part of the argument for which we do need that these are finite dimensional vector spaces. Since $ A$ is an isometry, its null space is trivial, and since its null space is trivial, it has an inverse. For a given $ y\ne0,$ there is a vector $ z$ with $ \|z\|_p \equal{} 1$ such that $ y^Tz \equal{} \|y\|_q.$ But since $ A$ is invertible, there is an $ x$ such that $ Ax \equal{} z,$ and since $ A$ is an isometry, $ \|x\|_p \equal{} 1.$ Then by the computations above,

$ x^TA^Ty \equal{} z^Ty \equal{} \|y\|_q.$

This completes the argument; $ \|A^Ty\|_q \equal{} \|y\|_q$ for all $ y.$