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It said my IQ is 133, but I don't know if that's right. I guess it isn't... I should take a real test in the future, but yeah...I'm too lazy  :P

Edit: I forgot to say that I really liked the question "2, 3, 5, 7, 11, 13, 17, 19, ?"  :roll:

Mine is 107. Those drawings seemed really weird.

124 :mad:

I did it somewhere else and got 142.  :(

that IQ test sucks, I got a 122 and I know my IQ should be way lower than that... :P

btw, what is genius, >150?

I know average is =100

and idiot<40

I picked the 1st one for all of them and got 91.

Yes! 126, and I did it totally randomly!!!!! :lol:

:arrow: 140
[but in two or three questions I had no clue...]

169 :maybe:

 [ but on one of them I broke a nail....]

ok wtf, i did it completely random this time and got a 133...there's something very wrong....

EDIT: G Unit, you are a genius

[quote="G-UNIT"]But I broke a nail ... that's so not genius. :maybe:[/quote]

yea man it happens  :(  :(

Two points:

*IQ sucks as a measure of intelligence

*This site does not give an accurate test

[quote="Ignite168"]YOUR IQ IS SIMPLE EXPONENTIAL DECAY IF YOUR COGNITIVE AGE REMAINS THE SAME.[/quote]

DON'T TYPE IN ALL CAPS PLEASE. ;)

I have to disagree with that statement. Let $c$=cognitive age and $a$=actual age; then IQ is given by the formula

$IQ=100 \cdot \frac{c}{a}$

Holding $c$ constant and letting $a$ increase, we have a rational function (hyperbola), not exponential decay.

;)

Some got so complicated that I just guessed, it didn't really inspire me.  I got a 111, though a "real" test showed I had 150.

LOL it says my IQ is "undefi". Does that mean I'm a super-super-genius with an IQ of $\infty$?

It also gave me a bunch of weird code:

[code]<param name="allowScriptAccess" value="sameDomain" />
<param name="movie" value="http://www.iqpe.com/iq.swf" />
<param name="quality" value="high" />
<param name="bgcolor" value="#347bae" />
<PARAM NAME=FlashVars VALUE="a=213346&b=mathnerd314&c=7254333">
<embed src="http://www.iqpe.com/iq.swf" FlashVars="a=213346&b=mathnerd314&c=7254333"
quality="high" bgcolor="#347bae" width="400" height="200" name="iq" align="middle"
allowScriptAccess="sameDomain" type="application/x-shockwave-flash"

pluginspage="http://www.macromedia.com/go/getflashplayer"/>[/code]

no... that means you're an idiot with something messing with a .swf that's supposed to appear on the page =p

I took some IQ tests last year and I know its >140  :lol:

waiting for tokenadult to come and pwn this thread

Anyways, my IQ is 95

Dumbest IQ test ever, and im not just saying that because of my score. They should call it the "pattern finding/assessing ability gage" or something along those lines.

I got 107  :blush:  got bored so after a while just guessed everything :)

I did an official IQ test in grade 6 and told me its >=140 (the test only went up to 140).....although i dont know how that works.... anyway online iq tests never work... my iqs ranged from 129 to 155 on all the different tests (besides this one)...

[quote="betons42"]What you think about this IQ test?
http://www.iqpe.com

My Iq is 119[/quote]

OK, this is another IQ test:
 http://web.tickle.com/tests/uiq/index-pop.jsp?sid=&supp=&z=
 :wink:

My iq is 133 and I am a facts curator. :P

The test is wrong.  I didn't answer a single problem and got an 86. I guessed the second time and got a 189 (strangely).

That is really weird. It must be based on how long it takes you.

not accurate. it justs makes people happy.

[quote]I got 110, then did it randomly and got 86...I randomly clicked on an answer at hyperspeed... it was fun.[/quote]

I think this IQ test is very complex and thats why U got 86 on random click, because avarage people answers, except some firt questions, are close to avarage


Let $ P_1P_2\cdots P_n$ be a convex polygon in the plane. Assume that for any pair of vertices $ P_i,P_j$, there exists a vertex $ V$ of the polygon such that $ \angle P_iVP_j\equal{}\pi/3$. Show that $ n\equal{}3$.

[hide]I think this is quite straightforward, but the solution in Titu's book is very long, so I'm not sure about my solution. I'll post my solution later.[/hide]

[hide="well, can you spot any mistake here?"]Let $ A,B$ be a pair of vertices such that $ AB$ is the smallest distance between any two vertices. Let $ C$ be the vertice such that $ \angle ACB\equal{}\pi/3$. In triangle $ ABC$, if the other two angles do not equal $ \pi/3$, then there is a side which is shorter than $ AB$, which contradicts the initial assumption. So all of the angles in $ ABC$ are equal, that is $ \pi/3$, implying that $ ABC$ is equilateral. So $ AB,BC,CA$ are three pair of vertices such that the distance is the smallest. But, since the polygon is convex, the smallest distance is achieved between two consecutive vertices. So $ A,B,C,A$ are consecutive vertices, in that order. Thus $ ABC$ is the desired polygon, and $ n\equal{}3$.[/hide]

[quote="Johan Gunardi"]But, since the polygon is convex, the smallest distance is achieved between two consecutive vertices.
[/quote]
I think that it's not right. For example in this quadrilateral the smallest distance is $ BD$, and $ B$ and $ D$ are not consecutive vertices.


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Is there a function f from natural numbers to natural numbers such that f(f(n)) = 2n ?

Yes there is. There is a very simple construction. I remember this problem, solved it long ago, from some Poland olympiad. Grobber, I think, gave a solution similar to mine on the forum. Try to search for it.

Note that $2f(n)=f(f(f(n)))=f(2n)$, so $f(2^kd)=2^kf(d)$. Let us partition the odd natural numbers into two disjoint infinite sets $A$ and $B$. Let $ g:A \to B $ be any bijection with inverse $h$.

Let
$f(2^kd)= \begin{cases}
2^kg(d) & \text {if } d \in A \\
2^{k+1}h(d) & \text {if }d \in B
\end{cases}$
Where d is odd.

Each $g$ gives rise to a valid $f$, and vice versa.

Edited for typesetting 01-30-05

And one more similar problem:
Is there a function $f$ from natural numbers to natural numbers such that $f(f(n)) = n^2$?

Yes, this has also been discussed. In fact, I think there was a thread where both of them were discussed, just like this one. Anyway, we can make it multiplicative, so we only need to define the function on primes (or prime powers, or something like that :)). It can then be reduced to the previous one.

Here are two other related threads: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=2749]f(f(n)) = n + k[/url] and the [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=13743#13743]general case.[/url]

I posted something that was incorrect here. Will revise soon.

What is this topic doing in the Advanced Fields?!  :what?: I'll move it :)


Hey everyone.  My friend (Kevin) and I have created these sheets originally to train the Stuyvesant Math Team while we were there.  Since then, we have decided to let it out into the general public.

http://www.nysml.org/files/formulas.pdf

If you guys have any comments about it, you can email me at jpo1@math.jhu.edu

I am currently also writing a manual for the baltimore county math team, which will be cover a great deal of topics (probably no where close to Art of Problem Solving.. but eh.. but when I'm done, I'll probably release it here too.

BTW, Richard, you said you worked at DE Shaw?? When??

-Jack

[quote="JBL"]Jack!  Sorry to comandeer your thread -- how are you?  Also, what compelled you to the general release of the formula sheets?[/quote]

haha Joel, read my response to you on the other forum.. btw did you AIM change?  no longer as fuzzy as before?  IM me you punk!

I am just deciding on what license to open up the formula sheets.  As you can see Kevin and I have left it in disrepair since 2001, so we will probably release it soon using one of the open pub licenses and let it continue to evolve.

(main message, IM me Joel, and 9 messages a day on this forum?  at least somethingawful.com!! not a math forum!!)

[quote="JackPo"]Hey everyone.  My friend (Kevin) and I have created these sheets originally to train the Stuyvesant Math Team while we were there.  Since then, we have decided to let it out into the general public.

http://www.nysml.org/files/formulas.pdf

[/quote]

Cool! Thanks for sharing. I'm not coaching at quite that level (yet), but this will be good reference material for some of the kids I coach in Minnesota, letting them know what lies ahead.


[quote="mr. math"]For what single value of n5,3nn,672 divisible by 11? Is there some kind of shortcut to do these kinds of problems?

 :arrow:  :ninja:  :huh:[/quote]
[hide]
well there's the divisibility trick. $-n+5-3+n-n+6-7+2=-n+3$. $-n+3$ has to be divisible by 11. so $n=3$[/hide]

can u just tell me what the divisibility rule is for 11 for any number. thanks

You alternate the digits with positive and negative signs and if that sum is divisible by 11 then the original was too

So for $abcdef$ if $-a+b-c+d-e+f$ is divisible by 11 then $abcdef$ is divisible by 11.

Actually, I know it won't make a difference but I think it is a-b+c-d+e-f must be a multiple of 11, but it doesn't matter.

[quote="b-flat"]Think about it.  It will work either way, becuase the rule is that the difference of the two sets of alternating digits must be divisible by 11.

(by two sets, I mean the sum of the digits in the sets)[/quote]

I sort of said it doesn't really  matter in my post above.


AMC
USA(J)MO
USAMO
combinatorics unsolved
combinatorics

I can do a lot of combinatorics problems in USAMO, but I can't write a single proof! can anyone help?
btw, thanks for recommending me to this forum pythag1101

combinatorics is one of the hardest subjects to write a proof in...

This might be useful.

http://www.artofproblemsolving.com/Resources/AoPS_R_A_HowWrite.php


probability
symmetry
combinatorics solved
combinatorics

Euclid 97 #10
A group of n married couples arrive at a dinner party and is seated around a circular table.  The distance between a husband and wife is defined to be equal to the number of people sitting between them measured either clockwise or counter-clockwise, whichever gives the smaller result.  Consider all possible arrangements of the 2n people.
what is the average number of couples, per arrangement, where both members are sitting side by side?

I should say that this takes a lot of computations.
well, let E(i) denote the number of situations where i couples seat side by side. we want to evaluate Average=(sum iE(i))/(2n)! .
(of course I'm assuming that seats are different. otherwise the total number of cases will be (2n-1)!/2! and the final answer will be different)
let B(i) denote the number of cases where at least i couples are side by side.
clearly: E(i)=B(i)-B(i+1) where B(n+1)=0 and B(0)=(2n)!
hence Average= 1/(2n)! (sum B(i)).  1=<i<=n.
Finally B(i)= (nCi) * (2n)C(i)*2^i*(2n-2i)!

I don't know if there is a closed form for this sum, but it seems not!

I don't know why I wrte the first line.
actually as you see it doesn't envolve any computations. :D

several good ppl and also the solution booklet explained to me that you can simply take the probability of one couple sitting together (which is easy to do) and multiply by n to get our desired average (expectation)

So answer is $n\cdot \frac{4n\cdot(2n-2)!}{(2n)!}=\frac{2n}{2n-1}$.
Problem solved (?)

If you want... I did not undersand anything in the statement or in the discussion...

Pierre.

I meant that I did not understand all the stuff about the distance which can be computed clockwisely or not and so on...and the answer from Ali.
I think the problem is more easy than it looks :
Let $X_i$ be the random variable equals to $1$ if the two people from couple $i$ seat side by side, and $0$ otherwise.
We want the average of $X = X_1  + \cdots  X_n$. But using linearity, it follows that $E(X) = E(X_1) + \cdots + E(X_n)$.
In another hand, using symmetry, $E(X_1)= \cdots = E(X_n)$, thus $E(X) = nE(X_1)$.
You did compute $E(X_1)$ thus problem is closed.   :)

Pierre.

Actually, that is what bugzpodder wrote. And accordingly to this idea I computed answer.

I know, but now I understand why. :P
Let's move it!

Pierre.

I dont think ali's solution is right,because he counted some cases more than 1.
I mean in the B(i) part.he counted the at least i couples not,exactly i couples!!!!!
Imean consider j<i.so B(j) is also counted in B(i) too.


What do you guys do after yo can handle the first 5 AIME problems besides doing the old AIME problems?  Is there any book that's suitable to read for #6 to #15 on the AIME?  What other contest has similar difficulties with those problems?

Do NEW AIME problems. New ones seem to be much more difficult than old ones...

Also, do some USAMO preparation, because you have a chance of making USAMO.

What if i've done all the AIME and tried all the easy USAMO problems?

I just finished doing the AoPS series (and I mean really "doing" them, as in doing every single problem), and I think that the transition should be AoPS to AIME problems (#5 and up), not the other way around.

Well, look to see why you're not able to get the other AIME problems.
If you don't have the mathematical knowledge to do some step, read some books or look it up. In other words, if a solution uses Stewart's theorem or Vieta's, etc, look them up and work with them. This is where Volume 2 would be useful.

If you're not able to finish in time or just can't see the step or "crux move", as Zeitz calls it, I would suggest doing more problems. Hopefully, you'd be able to get faster and also see how to apply more techniques.

In my case, I have the knowledge required for most of the problems, and I *never* actually use more than half the time limit, since 3 hours is quite a lot. My problem is that I can't see the creative step at first, and then just get too tired/frustrated/bored to go on further. That can also be fixed by attempting more problems (that are challenging for you) as you'd be able to handle them better.

[quote="pythag011"][quote="cognos599"]if start doing USAMO preparation and learn as you go, AIME will become ez.[/quote]

I agree. I did practically no AIME preparation last year...[/quote]

So you made it to the USAMO last year by practicing for the USAMO?

[quote="abacadaea"][quote="mathemonster"]aops v2 then acops[/quote]

If you think you are ready, you may want to skip ACoPS and go to engel.

Where can you find olympiad problems easier than USAMO.(I already tried CMO)[/quote]

Canada, etc..

Not CMO....


all series has to be to be divergent is [i]not divergent[/i].  Since that sum doesn't converge to any single value, but rather oscillates between 0 and 1 (or 0 and -1 if the lower limit is n=0), then the series is divergent.

scratch that, the values alternate between 0 and 1 if it starts at n=0 and -1 and 0 if it starts at n=1  :blush:

[quote="Hokkage"]I've tried searching around, but I haven't found a proof. How do you prove $1-1+1-1+1 \cdots$ or $\displaystyle \sum_{n=1}^{\infty} (-1)^n$ is divergent?
Is it simply that, because it's a geometric sequence with $|r| \not< 1$ then it must diverge?[/quote]

An infinite series diverges iff the sequence of partial sums does not tend to a limit.

The sequence of partial sums here is $1, 0, 1, 0, 1, ...$ which obviously does not tend to a limit (one of the properties of a limit is that it must be [i]unique[/i]).  Hence, the sum is divergent.

what is there to [i]prove[/i] in a definition?


logarithms
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integration
real analysis
real analysis unsolved

I'd like to know what the asymptotics of the sequence $ x_{n\plus{}1}\equal{}x_n\plus{}\log x_n$ are (with, say, $ x_0>1$).  Does anyone want to tell me how I might go about doing this (or, perhaps, just tell me the answer?)

Quick empirical check: it appears to be faster than $ n\ln n$ but slower than $ n\ln n\ln\ln n.$ Although with those logarithms, who can really tell, empricially?

OK, try this: if $ x_n\equal{}n\ln n,$ then $ x_{n\plus{}1}\minus{}x_n\approx \ln n\plus{}1,$ while $ \ln x_n\approx \ln n\plus{}\ln\ln n.$ So $ n\ln n$ is a little too small.

How about the differential equation? Try $ y'\equal{}\ln y.$ The solutions to that will grow like the inverse function to $ \int\frac1{\ln y}\,dy.$

Note that $ \int_a^{x}\frac{dy}{\ln y}\approx \frac{x}{\ln x}\plus{}\frac{x}{\ln^2 x}\plus{}\cdots.$ Try inverting that.

You can use a boot-strapping argument.  First show that $ x_n \equal{} O(2^n)$.  Use that to show that $ x_n \equal{} O(n^2)$.  Use that to show that $ x_n \equal{} O(n \log n)$.

Something similar will work for lower bounds.  In the end, you will find that $ x_n$ is asymptotic to $ n \log n$.

Indeed, asymptotics is $ n\log n$. To prove this just combine what Kent has said with Schtolz theorem $ lim\limits_{n\to\infty} \frac {x_n}{n\log n} = lim\limits_{n\to\infty} \frac {\log x_n}{\log(n + 1) + 1} = 1$ because you can estimate that $ \log n < \log x_n < \log n + \log\log n + \log\log\log n + c$ (you don't even need this bound just something like $ x_n < c n(\log n)^2$) which is  easy to get)


advanced fields
advanced fields unsolved

I am looking for an example of a PDE for $f(x,y)$ in 4th order

i.e. $a_{1}\partial_{1}^{4}(f)+a_{2}\partial_{1,2}^{2}(f)(x,y)+\ldots =0$

Are there any good real life examples with physical or economical background?

[url=http://eqworld.ipmnet.ru/en/solutions/lpde/lpdetoc5.htm]Fourth-Order Partial Differential Equations[/url]

An excellent website overall.


vector
combinatorics unsolved
combinatorics

Points $ A_1,A_2,...,A_n$ and $ B_1,B_2,...,B_n$ are given on a plane. Show that the points $ B_i$ can be renumbered in such a way that the angle between vectors $ A_iA_j^{\longrightarrow}$ and $ B_iB_j^{\longrightarrow}$ is acute or right whenever $ i\neq j$.

[quote="probability1.01"][hide="big hint"]Consider $ \sum A_i \cdot B_{\sigma(i)}$. (Motivation: trying to fix angles that are not right or acute.)[/hide][/quote]

Umm... but how?


[quote="xxxyyyy"]$ 7(ab \plus{} ac \plus{} bc) < \equal{} 2 \plus{} 9abc$ when $ a \plus{} b \plus{} c \equal{} 1$[/quote]
We need prove that
$ 7(ab\plus{}bc\plus{}ca)(a\plus{}b\plus{}c) \leq 2(a\plus{}b\plus{}c)^3\plus{}9abc$
<=>$ 21abc\plus{}7\sum_{cyc} ab(a\plus{}b) \leq 2(a^3\plus{}b^3\plus{}c^3)\plus{}6(a\plus{}b)(b\plus{}c)(c\plus{}a)\plus{}9abc$
<=>$ 21abc\plus{}7\sum_{cyc} ab(a\plus{}b) \leq 2(a^3\plus{}b^3\plus{}c^3)\plus{}6\sum_{cyc} ab(a\plus{}b) \plus{}21abc$
<=>$ 2(a^3\plus{}b^3\plus{}c^3) \geq ab(a\plus{}b)\plus{}bc(b\plus{}c)\plus{}ca(c\plus{}a)$
We have:
$ a^3\plus{}b^3 \geq ab(a\plus{}b)$
$ b^3\plus{}c^3 \geq bc(b\plus{}c)$
$ c^3\plus{}a^3 \geq ca(c\plus{}a)$
Done!


calculus
integration
trigonometry
calculus computations

[b]without[/b] using  complex variables, show  :D

$ \int _ 0^\infty \; \frac{ 2\cos \; 2e x} { \cosh \; \pi x } \; \; \textbf dx \;\equal{}\; \boxed { \text{sech} \;  e  }$


inequalities
calculus
integration
geometry
3D geometry
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logarithms

Prove that if $ p\geq 2$ is a real number then \[{ \int_{-\pi}^{\pi}{|\frac{sin t}{t}|^p}dt}\leq \pi\sqrt{\frac{2}{p}}.\]
  K. Ball used this estimation to prove the following very beautiful result: any section of the unit cube by a hyperplane has measure at most $ \sqrt{2}$. Unfortunately, his proof of this inequality is a total mess, with tons of computations. Is there any neat proof of this inequality?

Maybe you mean
\[{ \int_{ - \infty}^{\infty}{|\frac {\sin t}{t}|^p}dt}\leq \pi\sqrt {\frac {2}{p}}\quad?
\]

I agree with Myth (this inequality is true). It's enough to prove Myth's  inequality  .

Oh, I'm terribly sorry, I just wrote too many $ \pi$'s. The correct form (and obviously stronger) is the one indicated by Myth.

who is k ball ?   :?:   must be he is not famous 'cause i have never heard of him before.

we have, $ \forall x$,

$ \sin \; x = x - \frac {x^3}{6} + \frac {x^5}{120} - \ldots$

$ \Rightarrow \frac {\sin\;x}{x} \le 1 - \frac {x^2}{6} + \frac {x^4}{120}$

now $ 6 < 2\pi$, so

$ \Rightarrow \frac {\sin\;x}{x} \le 1 - \frac {x^2}{2\pi} + \frac {x^4}{40\pi}$

so,
${ 2\int_{0}^{\infty}\; \bigg|\frac {\sin \;x}{x}\bigg|^p}\; dx = 2 \int_{0}^{\infty}\; \text{exp} \left( p\; \ln \bigg| \frac {\sin \;x}{x} \bigg| \right) \; dx$

$ \le 2 \int_{0}^{\infty}\; \text{exp} \left( p\; \ln\left( 1 - \frac {x^2}{2\pi} + \frac {x^4}{40\pi} \right) \; \right) \;dx < 2 \int_{0}^{\infty}\; \text{exp} \left( - \frac {p}{2\pi}\;x^2 \right) \; dx$

$ = 2\cdot \frac {1}{2} \sqrt {\frac {\pi}{\frac {p}{2\pi} }} = \pi \sqrt {\frac {2}{p} }$

where i have used

$ \int_{0}^{\infty}\; \text{exp} \left( - a\;x^2 \right) \; dx = \frac {1}{2} \sqrt {\frac {\pi}{a} }$

[quote="misan"]

so,
${ 2\int_{0}^{\infty}\; \bigg|\frac {\sin \;x}{x}\bigg|^p}\; dx = 2 \int_{0}^{\infty}\; \text{exp} \left( p\; \ln \bigg| \frac {\sin \;x}{x} \bigg| \right) \; dx$

$ \le 2 \int_{0}^{\infty}\; \text{exp} \left( p\; \ln\left( 1 - \frac {x^2}{2\pi} + \frac {x^4}{40\pi} \right) \; \right) \;dx < 2 \int_{0}^{\infty}\; \text{exp} \left( - \frac {p}{2\pi}\;x^2 \right) \; dx$

$ = 2\cdot \frac {1}{2} \sqrt {\frac {\pi}{\frac {p}{2\pi} }} = \pi \sqrt {\frac {2}{p} }$

[/quote]
$ \int_{0}^{\infty}\; \text{exp} \left( p\; \ln\left( 1 - \frac {x^2}{2\pi} + \frac {x^4}{40\pi} \right) \; \right) \;dx$ is disvergent.

Note that we actually have an equality when $ p \equal{} 2$. This shows that we have to make a careful examination.

Is this problem too easy?
I have only proved the result for large $ p$ using misan's method.
$ \int_{0}^{\pi}\equal{}\int_{0}^{\eta}\plus{}\int_{\eta}^{\pi}\equal{}\int_{0}^{\eta}\cdot(1\plus{}\frac{\int_{\eta}^{\pi}}{\int_{0}^{\eta}})\le\int_{0}^{\eta}\cdot(1\plus{}\frac{\pi(\frac{\sin{\eta}}{\eta})^{p}}{\int_{0}^{\eta/2}})$
$ \le\int_{0}^{\eta}\cdot(1\plus{}\frac{\pi(\frac{\sin{\eta}}{\eta})^{p}}{(\eta/2)\cdot(\frac{\sin(\eta/2)}{\eta/2})^{p}})\equal{}\int_{0}^{\eta}\cdot(1\plus{}\frac{2\pi}{\eta}[\cos(\eta/2)]^{p})$.
On the other hand,we have
$ \int_{0}^{\eta}\le\int_{0}^{\eta}{(1 \minus{} \frac {t^2}{6} \plus{} \frac {t^4}{120})^{p}} dt\le\int_{0}^{\eta}{\exp(p\log(1 \minus{} \frac {t^2}{6} \plus{} \frac {t^4}{120}))} dt$
$ \le\int_{0}^{\eta}{\exp(p( \minus{} \frac {t^2}{6} \plus{} \frac {t^4}{120}))} dt\le e^{\frac{p\eta^4}{120}}\cdot\int_{0}^{\eta}{e^{\minus{}\frac{pt^2}{6}}} dt\equal{}e^{\frac{p\eta^4}{120}}\cdot\sqrt{\frac{6}{p}}\cdot\int_{0}^{\eta\sqrt{\frac{p}{6}}}{e^{\minus{}t^2}} dt$
$ \le e^{\frac{p\eta^4}{120}}\cdot\sqrt{\frac{6}{p}}\cdot\frac{\sqrt{\pi}}{2}$
Letting $ \eta\equal{}\frac{1}{p^{1/4}}$,we have $ \int_{0}^{\eta}\le e^{\frac{1}{120}}\sqrt{\frac{3\pi}{2}}p^{\minus{}1/2}$.
Now it is quite easy to prove that $ \frac{2\pi}{\eta}[\cos(\eta/2)]^{p}\to 0$,when $ p\to \infty$.
So we have $ \int_{0}^{\pi}\le (1\plus{}\epsilon)e^{\frac{1}{120}}\sqrt{\frac{3\pi}{2}}p^{\minus{}1/2}$.

Indeed, there is a standart method to get exact asymptotic for large $ p$. The question is to prove the inequality for all $ p\geq 2$ and as you see we have equality for $ p\equal{}2$, so need quite accurate estimates, not just asymptotic.

Ah, that's a sweet one. Keith Ball himself gave a terrible proof. Later I and Anatoly Podkorytov found a much simpler way to show the inequality. Let's see if the mathlinkers can do even better ;). Answering Misan's question, Keith Ball is a professor from University College London. His most famous recent achievement is the proof that there are infinitely many irrational numbers among the values of the $ \zeta$-function at odd integers, but this is by no means the only thing he will be remembered for. Younger mathlinkers may know him as the author of the book "Strange curves, counting rabbits and other mathematical explorations". The inequality in question is the key step in his proof that the maximal volume of a section of the unit cube by a hyperplane is $ \sqrt 2$. :)

For even integers $ p$ we can compute the integral exactly
\[ \int_{\minus{}\infty}^{\infty}{\left(\frac{\sin t}{t}\right)^p dt}\equal{}\pi p\sum_{k\equal{}1}^{\frac{p}{2}}{(\minus{}1)^{\frac{p}{2}\plus{}k}\frac{k^{p\minus{}1}}{\left(\frac{p}{2}\plus{}k\right)!\left(\frac{p}{2}\minus{}k\right)!}}\]

So to prove the claim for even integers $ p$ is equivalent to showing
\[ \sum_{k\equal{}1}^{\frac{p}{2}}{(\minus{}1)^{\frac{p}{2}\plus{}k}\frac{k^{p\minus{}1}}{\left(\frac{p}{2}\plus{}k\right)!\left(\frac{p}{2}\minus{}k\right)!}}
\leq \sqrt{\frac{2}{p^3}}\]
Though I'm entirely unsure whether or not this is easier than the original problem.  Any ideas?

I see it's time to post full solution .

[b]lemma1.[/b]
suppose $ f,g \geq 0$ and $ f,g \in \mathbb{L(E)}$ where $ E$ is subset of $ \mathbb{R^{n}}$.
Let's define :
$ F = \lambda(E(f > t))$ and $ G = \lambda(E(g > t))$ (in measure theory it has the name ( функция распределения ))
then if
\[ \int_{E}f^{p_{0}} \geq \int_{E}g^{p_{0}}
\]
also suppose at point $ t_{0}$ function $ F - G$ change sign from $ -$ to $ +$
then we have :
\[ \int_{E}f^{p} \geq \int_{E}g^{p}
\]
for all $ p > p_{0}$
(it's supposed that $ f^{p},g^{p} \in \mathbb{L(E)})$ for $ p > 0$
Indeed since :
\[ \int_{E} f^{p} - g^{p} = p \int_{0}^{\infty}t^{p - 1}(F - G)
\]
[b]Lemma2[/b]
\[ \int_{0}^{\infty}e^{ - x^{2}p/2\pi} = \frac {1}{2} \sqrt {\frac {2\pi}{p}} \int_{0}^{\infty}t^{ - 1/2}e^{ - t} = \frac {1}{2} \sqrt {\frac {2\pi}{p}}\Gamma(1/2) = \frac {\pi}{\sqrt {2p}}
\]
and for $ x \in (0,1)$
\[ |\frac {\sin x}{x}| = \prod_{k \geq 1}\left( 1 - \frac {x^{2}}{k^{2}\pi^{2}} \right) \leq \prod _{k \geq 1} e^{ - x^{2}/(k\pi)^{2}} = e^{ - x^{2}/6}
\]
[u]proof[/u]

let's $ g = |\frac {\sin x}{x}|$ and $ f = e^{ - x^{2}/6}$
and we need to prove :
\[ \int_{(0,\infty)} g^{p} < \int_{(0,\infty)}f^{p}
\]
for $ (p > 2)$
for $ p = 2$ we have equal parts .
it's enough to show that : $ F - G$ change the sign at only one point
$ F(t) = f^{ - 1}(t) = \sqrt {2\pi \ln \frac {1}{t} }$ note that for $ t\geq 1$ we have $ G = F = 0$
let $ t_{m} = \max g$ on $ t \in (\pi m , \pi m + \pi)$ hence $ \frac {1}{\pi^{2}(m + 1/2)^{2}} < t_{m} < 1/(\pi m )^{2}$
therefore for $ t \in (t_{1},1)$ we have $ F - G > 0$ indeed :
$ G(t) \leq \sqrt { - 6\ln t } < \sqrt { - 2\pi \ln t } = F(t)$
to show that $ F - Г$ is negative on $ (0,t_{1})$ it's enough to show that :
$ |G'(t)/F'(t)| > 1$ on $ (t_{m + 1},t_{m})$
note that $ |G'(t)| = - G'(t) = \sum_{g(x) = t} |g'|^{ - 1}$  where $ x > 0$
equation $ g(x) = t$ have only one root on $ (0,\pi)$ and two root on $ (\pi k , \pi k + \pi )$ for $ k > 0$
it's easy to show that on $ (0.\pi)$ , $ |g'(x)| \leq 1/2$ and on  $ (\pi k , \pi k + \pi )$   $ |g'(x)| < (\pi k)^{ - 1}$
thus
\[ - G'(t) \geq 2 + \pi m (m + 1) > \pi (m + 3/2)
\]
hence :
\[ |G'/F'|\geq\pi (m + 3/2)\sqrt { - (2/\pi)t^{2} \ln t }
\]
last is increasing function on $ (0,t_{1})$ therefore for  $ t > t_{m + 1}$ we have
\[ \pi (m + 3/2)\sqrt { - (2/\pi)t^{2} \ln t } > 1
\]
since $ t_{m + 1} > \frac {1}{\pi (m + 3/2)}$
from the proof follows that  for $ 1 < p < 2$ we have another inequality :
\[ \int_{(0,\infty)} |\sin x/x|^{p} > \pi/ \sqrt{2p}
\]


Bonjour,

Je suis nouveau et je cherche à résoudre l'équation différentielle suivante:
xy''+y'-mxy=0
ou
my=y''+y'/x

Elle provient d'une transformé de Laplace d'une équation de diffusion de la chaleur en polaire. Pour l'instant je cale.

Merci à ceux qui ont des solutions ou des pistes.

PS: où dois-je poster cette question sur le forum global pour ne pas me tromper de rubrique?

Tu peux chercher la solution sous forme de série entière.


number theory
greatest common divisor
algebra
polynomial
factorial
number theory proposed

Find all $(x,n,k)$ natural solutions of the following equation:$3^{k}-1=x^{n}$.
Please post your solutions very fast.

it is Catalan equation so no trivial solution is
$k=2,x=2,n=3$

my proof for this case:
1)let assume that $n=2m$
of course $x=2l$
we have
$3^k=(2l)^{2m}+1=({2l}^m+i)({2l}^m-i)$
we know that $gcd({2l}^m+i,{2l}^m-i)=1$
and we have that $3$ is prime in $Z[i]$
it is easy to verify that $l=0$
and we have solution $(0,0,2m)$
2)let assume that $n=2m+1$
now we have that:
${3^k=(x+1)(x^{2m}-x^{2m-1}+....+1}$
so
$x=3^{\alpha}-1$
now it is easy to verify that only solutions is
$\alpha=1$

Why $\alpha=1$?What means $gcd(2l^{m}+i,2l^{m}-i)=1$... Complex numbers and gcd???
Please explain it.

I'm assuming by natural you mean positive...

first off, we see that n cant be even, for if it is then we get:

$-1\equiv x^n \mod 3$ but $0^n \equiv 0, 1^n\equiv 1, 2^n\equiv 1$ so there are no solutions.

Therefore n must be odd, and we get:

$3^k = (x+1)(x^{n-1}-x^{n-2} + \dots -x+1)$ so $(x+1)=3^a$, $(x^{n-1}-x^{n-2} + \dots -x+1)=3^b$, $a,b>0$

then using polynomial division we have:
$(x^{n-1}-x^{n-2} + \dots -x+1)=q(x)(x+1)+r$ and plugging in x = -1, we see r = n, and 3|n.

Therefore solving the original equation reduces to solving $3^k-1=y^3$ for some y.

then $3^k=(y+1)(y^2-y+1)$ so $(y+1)=3^m$, $(y^2-y+1)=3^n$, $m,n>0$ plugging in $y=3^m-1$ we see that:

$3(3^{2m-1}-3^m+1)=3^n$ and so n = 1, and m = 1 and so k=2, and we arrive at the solution (2,2,3).

To explain the $gcd$:
_el_doopa works in the ring $\mathbb{Z} [i] = \{ a+bi | a,b \in \mathbb{Z} \}$ which is factorial, meaning that you have a unique prime factorisation, and so you can easily define a greatest common divisor


if  (2^2005  + 2^2002) /  (2^2004  - 2^2003) is written in the form a/b where a and b are integers with no common divisors, what is a + b?

(1) 11 (2) 12 (3) 13 (4) 14 (5) None of the above

[hide="my solution"]
$ \frac{2^{2005 }\plus{} 2^{2002}}{2^{2004} \minus{} 2^{2003}}\equal{}\frac{2^3\plus{}1}{2^2\minus{}2}\equal{}\frac{9}{2}$

$ \Rightarrow a\plus{}b\equal{}9\plus{}2\equal{}11 \Rightarrow \boxed {1}$ is the correct answer[/hide]

[quote="moldovan"][hide="my solution"]
$ \frac {2^{2005 } \plus{} 2^{2002}}{2^{2004} \minus{} 2^{2003}} \equal{} \frac {2^3 \plus{} 1}{2^2 \minus{} 2} \equal{} \frac {9}{2}$

$ \Rightarrow a \plus{} b \equal{} 9 \plus{} 2 \equal{} 11 \Rightarrow \boxed {1}$ is the correct answer[/hide][/quote]

I think you mean $ \boxed{11}$; not $ \boxed{1}$.

He may have been referring to the answer choice.
[quote="sri340"](1) 11 [/quote]


modular arithmetic
number theory
greatest common divisor
number theory unsolved

Show that there exist a positive integer k such that $ k2^n\plus{}1$ is not prime for all positive integer n.

Wierd... I can only reduce it to a problem about covering systems which might have no solutions.
For example, if $ n \equiv 0 \pmod 2$ then $ k2^n \plus{} 1 \equiv k \plus{} 1 \pmod 3$ so we should take $ k \equiv 2 \pmod 3$.
If $ n \equiv 1 \pmod 3$, then $ k2^n \plus{} 1 \equiv 2k \plus{} 1 \pmod 7$, so we should take $ k \equiv 3 \pmod 7$.
By CRT, there's a solution for $ k$, for example: $ k \equal{} 17$.
The problem is that we don't know what happens when $ n \equiv 1,5 \pmod 6$.

And in general: Let $ p_i$ be a sequence of distinct odd primes, and let $ ord_{p_i}(2)$ be the order of $ 2$ modulo those primes.
Then if we can find a distinct covering system whose moduli are $ ord_{p_i}(2)$, we're done.
(in the previous case, $ p_1 \equal{} 3, ord_{3}(2) \equal{} 2, p_2 \equal{} 7, ord_{7}(2) \equal{} 3$)

The relevant sequence is the following:
2, 4, 3, 10, 12, 8, 18, 11, 28, 5, 36, 20, 14, 23, 52, 58, 60, 66, 35, 9, 39, 82, 11, 48, 100, 51, 106, 36, 28, 7, 130, 68, 138, 148, 15, 52, 162, 83, 172, 178, 180, 95, 96, 196, 99, 210, 37, 226, 76, 29, 119, 24, 50, 16,...
[url]http://www.research.att.com/~njas/sequences/A014664[/url]

Can we find a distinct covering system whose moduli are the those numbers? If we can, we're done.
(a small correction: it doesn't have to be a distinct system if some number appears more than once in the sequence
also note that we can construct a nearly covering system: $ 0(2), 0(3), 1(4), 7(12)$. We only need $ 5(6)$ but 6 doesn't appear...)

Another close covering: $ 0(2), 0(3), 3(4), 5(12), 1(36), 13(36)$. We only need 36 to appear 3 or more times in the sequence (I see 2 appearances), or alternatively, 96 to appear 3 times (I see it once)

Let $ p_{ni}$ prime divisors $ \Phi_n\equal{}2^{2^n}\plus{}1$.
Then we can chose $ k$ $ p_m|k*2^n\plus{}1$ if $ 2^{m\minus{}1}|n, 2^m\not |n$ for $ m\equal{}0,1,2,3,4$.
If $ n$ odd $ 3|k*2^n\plus{}1\to k\equal{}2\mod 2$, if $ n\equal{}2\mod 4$ $ 5|4k\plus{}1\to k\equal{}1\mod 5$,...
Obviosly $ Per_{p_m}(2)\equal{}2^m$. We had 2 prime divisors $ p_{51},p_{52}|\Phi_5$ and can chose $ p_{51}|k*2^{2^4}\plus{}1,p_{52}|k*2^5\plus{}1$. It give for any n exist prime $ q\equal{}p_0,p_1,p_2,p_3,p_4,p_{51},p_{52}$, suth that $ q|k*2^n\plus{}1$.

Found it!
The covering system: $ 0(2), 0(3), 3(4), 5(12), 13(24), 1(48), 25(48)$
Find the indices of the primes here: [url]http://www.research.att.com/~njas/sequences/b014664.txt[/url]
Let $ p_i$ be the corresponding primes, and $ a_i$ the values modulo $ ord_2(p_i)$. We need $ k$ to satisfy the following system:
$ k2^{a_i} \plus{} 1 \equiv 0 \pmod {p_i}$

Explicitly:
$ k \plus{} 1 \equiv 0 \pmod 3$
$ k \plus{} 1 \equiv 0 \pmod 7$
$ 8k \plus{} 1 \equiv 0 \pmod 5$
$ 32k \plus{} 1 \equiv 0 \pmod {13}$
$ 2^{13}k \plus{} 1 \equiv 0 \pmod {241}$
$ 2k \plus{} 1 \equiv 0 \pmod {97}$
$ 2^{25}k \plus{} 1 \equiv 0 \pmod {673}$

And then $ (k2^n \plus{} 1, 3\cdot 5 \cdot 7 \cdot 13 \cdot 97 \cdot 241 \cdot 672) > 1$ For all $ n$.
(we can take $ k$ to big enough so that this gcd won't equal one of those primes. Namely, $ k > 672$)
SUPERMAN2, do you happen to know the source? (I just like knowing the origin of problems..)

I searched the forums with the key word "Perron" and ended up with a topic which mentioned this problem.
This problem is related to the Sierpinski problem, and the good values of $ k$ are Sierpinski numbers.

[url]http://primes.utm.edu/glossary/page.php?sort=SierpinskiNumber[/url]
[url]http://en.wikipedia.org/wiki/Sierpinski_number[/url]

I was just thinking that it can make a good PEN problem  :roll:  :blush:

Select $ k$ s.t. $ k\equal{}1(modF_0F_1F_2F_3F_4)$ , $ 2^{2^{4}}k\equal{}\minus{}1(modx)$ and $ k\equal{}\minus{}1(mody)$
Where $ F_n$ is the nth Fermat number and x and y are the two prime divisors of $ F_5$

Wow you have done a nice job in definitely reading the previous posts  :roll:

Sorry ZetaX, i posted in a hurry and then read the above solutions (I was too eager to post my solution) forgive me


geometry
circumcircle
perpendicular bisector
geometry proposed

$ OX,OY$ are perpendicular. Assume that on $ OX$  we have wo fixed points $ P,P'$ on the same side of $ O$. $ I$ is a variable point that $ IP\equal{}IP'$. $ PI,P'I$ intersect $ OY$ at $ A,A'$.
a) If $ C,C'$ Prove that $ I,A,A',M$ are on a circle which is tangent to a fixed line and is tangent to a fixed circle.
b) Prove that $ IM$ passes through a fixed point.

It is clear that the circle $ \odot (IAA')$ is tangent to a fixed line and a fixed circle, but this time I am unable to guess who is $ M$ and what happens if $ C, C'.$

Since the $ \triangle IPP'$ is isosceles with $ IP \equal{} IP',$ the $ \triangle IAA'$ is also isosceles with $ IA \equal{} IA'.$ A circle $ (I)$ with center $ I$ and radius $ IP \equal{} IP'$ cuts the line $ OY$ at $ Z, Z'.$ Inversion in $ (I)$ takes the lines $ IA, IA'$ to themselves and the line $ AA' \equiv OY$ to a circle $ \odot (IZZ').$ $ IA, IA'$ cut $ \odot (IZZ')$ at the inversion images $ A^*, A'^*$ of $ A, A' \Longrightarrow$ the line $ A^*A'^*$ is image of the circle $ \odot (IAA').$ $ IA \equal{} IA' \Longrightarrow IA^*  \equal{} IA'^* \Longrightarrow A^*A'^* \parallel AA'.$ $ AA'^*$ cuts $ OX$ at $ Q.$ $ R, S$ are midpoints of the segments $ OQ$ and $ PP',$ respectively. Obviousely, the circumcircle of the isosceles $ \triangle IAA'$ is tangent to the external bisetor $ IS$ of the angle $ \angle AIA'.$ $ \overline{IA} \cdot \overline {IA^*} \equal{} IP^2$ $ \Longrightarrow$ $ \overline {SO} \cdot \overline {SQ} \equal{} SP^2$ $ \Longrightarrow$ the line $ A^*QA'^*$ is fixed for any $ I$ on the perpendicular bisector of the segment $ PP'.$ Moreover, a circle $ (R)$ with center $ R$ and diameter $ OQ$ is coaxal with the points $ P, P'$ $ \Longrightarrow$ it is perpendicular to the inversion circle $ (I)$ passing through $ P, P'$ and the inversion in $ (I)$ takes it to itself. Since the line $ A^*QA'^*$ is tangent to $ (R)$ at $ Q,$ the circle $ \odot (IAA')$ is also tangent to $ (R).$

I tried to repair the statement.

It seems that the following is true. Does anyone prove it?

$OX, OY$ are perpendicular. Assume that on $OX$ we have two fixed points $P,P'$ on the sameside of $O$. $I$ is variable point that $IP=IP'$. $PI, P'I$ intersect $OY$ at $A,A'$.

a) If $C,C'$ are circles passing through $O,P,A'$, $O,P',A$ and they intersect at $M$ other than $O$, prove that $I,A,A',M$ are on a circle which is tangent to a fixed line and is tangent to a fixed circle.

b) Prove that $IM$ passes through a fixed point.

Toshihiro, your statement is true only in case PIP` is right, but even then, M and I just coincide


Hello guys!
I am looking for the solution to the following problem(I think it is quite easy
 but I could not solve it in the last 10 minutes :D ):

[i]Two players play the  following game: there are n stones. A player can take $2^m$ stones with $m \in N$. For which n does the first player win?[/i]

[/tex]

Already thought of it but I could not see what the allowed operation becomes in the binary system. Could you write the whole  solution if you have one?

I assume: The player who takes the last stone, wins the game.

Then the losing-positions are exactly the numbers that are divisible by 3
(this can be proved by induction).

From any winning position, you can subtract 1 or 2 and leave your opponent
in a losing position.
From any losing position, you must subtract a power of 2 (and powers of 2
are never divisible by 3), and hence leave your opponent in a winning position.

Thank you very much. I had a real blackout.

Can you replace 2 by a constant k in N? The same solution modulo k+1 doesn't work now!

For $k=3$, it works $\pmod 2$ (the losing positions are the even numbers), for example, but I don't know whether for each $k$ there is a $t(k)$ s.t. we can work $\pmod{t(k)}$. The case $k=3$ seems to be a special case.

Actually, if $k$ is odd, $\pmod 2$ works like a charm :), so it's not a very interesting game in this case.

I was going to write the same now but you were faster.  :cry:

It still seems to be periodic $\pmod {k+1}$, though. This, together with the fact that if $1\le n<k$ then $n$ yields a losing game iff $n$ is even, might give the answer.


Find the sum of all positive integers $n$ for which $n^{2}-19n+99$ is a perfect square.

[hide=Solution] Let $n^2 - 19n + 99 = m^2$, where $m^2$ is a perfect square. We rewrite this as a quadratic $n^2 + n + (99 - m^2) = 0$ and use the discriminant to find acceptable values of $m^2$: $n = \frac{-(-19) \pm \sqrt{(-19)^2 - 4(1)(99 - m^2)}}{2(1)} = \frac{19 \pm \sqrt{4m^2 - 35}}{2}$. For $n$ to be an integer, $4m^2 - 35 = a^2$, where $a^2$ is an odd perfect square. ($4m^2 - 35$ is always odd). We rewrite this as $(2m - a)(2m + a) = 35$ and see we must find all ordered pairs of positive integers that multiply to $35$. These are: $(1, 35), (5, 7), (7, 5)$ and $(35, 1)$.

Our systems of equations are now: $(2m - a = 1, 2m + a = 35), (2m - a = 5, 2m + a = 7), (2m - a = 7, 2m +  a = 5)$, and $(2m - a = 35, 2m + a = 1)$. (The last two systems are duplicates of the first two: they yield the same $m$-values. The only difference is that the $1$st two yield the positive square root of $a^2$, and the other two yield the negative root). Solving them, we find that $m = 9$ or $3$, and plugging these back into $n^2 - 19n + 99 = m^2$, we find $n = 1, 9, 10$, or $18$. Thus, our sum is $1 + 9 + 10 + 18 = \boxed{38}$.

[quote=towersfreak2006][hide]
$n^2-19n+99\ge (n-10)^2$ is true for $1\le n$
$(n-9)^2\ge n^2-19n+99$ is true for $n\le18$
Thus, $n^2-19n+99$ can only be a perfect square for $1\le n\le18$.

Also, $n^2-19n+99$ is a parabola symmetrical about $n=9.5$ so for any solution $x$, the value $(19-x)$ is also a solution.

Testing from $n=1$ to $n=9$, we find that $n=1$ is a perfect square ($n^2-19n+99=81$) and $n=9$ is a perfect square ($n^2-19n+99=9$), so the expression is a perfect square for $n=1, 9, 10, 18$ which sums to be $38$.
[/hide][/quote]

I think that in $(n-9)^2\ge n^2-19n+99$ is true for $n\le18$, you want to write $n>18$


group theory
abstract algebra
induction
superior algebra
superior algebra unsolved

$ G $ is a finite group with nontrivial $ 2 $-Sylows. Show that the product of all the elements of $ G $ is never the neuter element.

What does "nontrivial" mean in this context? If it's a standard meaning, the problem is wrong.

Hi,

I made a typo in the wording. You must read "with non trivial cyclic $2$-Sylows".

Sorry again for this.

Suppose n is the order of G and n=2m, and the 2-Sylow subgroup is Cyclic. Consider the G acting on G by left multiplication. This induces a representation of G as a subgroup of permutations. The 2-sylow subgroup being cyclic means that there are odd permutations. (a generator of the 2-sylow subgroup being one).
  If m is odd, then the product of all permutations is odd so we reach a contradiction.
  If m is even note that if a permutation is odd then its inverse is different ¡an odd two!, so the product of the odd permutations is the identity. Now  the even permutation is a normal subgroup of order m, getting back to G by induction on the higher power of two dividing m the result follows.

¿How do i quote a previous post? (the set of solutions of x^m-1)

Just press "quote" ("right lower corner" of the post you want to quote).


[quote="luimichael"]Find all real solutions to the following system of equations:
$\{ \begin{array}{rcl} {16x^4+y^4+24x^2y^2}&=&4\\ {32x^3y+8xy^3}&=&1 \end{array} $[/quote]
[hide]
By adding and subtracting the two equations, we get:
$(2x+y)^4=5$ and $(2x-y)^2=3$, respectively. Then $2x+y=\pm5^{1/4}$ and $2x-y=\pm3^{1/4}$. Now u can find x and y values by solving these simul. equ.
[/hide]


[quote="test20"][quote="kunny"]Let $a,\ b,\ c,\ d$ be real numbers such that $(a-1)(b-1)(c-1)(d-1)=1$.
For $a\geq 2,\ b\geq 2,\ c\geq 2,\ d\geq 2$, prove that $\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\geq2$.[/quote]

You could also ask for a proof that under the given conditions $\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\leq2$.[/quote]
 :lol:
kunny probably has a very nice proof  :wink:

Something is not right because if we have $a,b,c,d\geq 2$ then $a-1\geq 1,b-1\geq 1,c-1\geq 1,d-1\geq 1$ so their product can'y be equal to $1$ unless $a=b=c=d=2$

[quote="test20"][quote="kunny"]Let $a,\ b,\ c,\ d$ be real numbers such that $(a-1)(b-1)(c-1)(d-1)=1$.
For $a\geq 2,\ b\geq 2,\ c\geq 2,\ d\geq 2$, prove that $\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\geq2$.[/quote]

You could also ask for a proof that under the given conditions $\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\leq2$.[/quote]
haha, because  $\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}=2$
I think in all these problems kunny has been posting either there is a mistake or they are "trick" problems...

[quote="kunny"]Let $a,\ b,\ c,\ d$ be real numbers such that $(a-1)(b-1)(c-1)(d-1)=1$.
For $a\geq 2,\ b\geq 2,\ c\geq 2,\ d\geq 2$, prove that $\ \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\geq 2$.[/quote]
Notice that if any of the inequalities $a,b,c,d\geq 2$ is strict, then $(a-1)(b-1)(c-1)(d-1)>1$.  Therefore, $a=b=c=d=2$.  In this case, $\sum_{cyc}\frac{1}{a}=2$.  I don't think you should post such trivial problems here.  I believe that the so-called "highest" forum you can post [i]this[/i] problem in is the Intermediate Forum.

[quote="kunny"]This problem seems to be very easy, but this has been made by differentiation. :wink:[/quote]
It does not seem to be very easy, [b]it really is very easy[/b]. And before using such strong methods, you may want to try the very very basic tools.
But keep investigating and discovering new things, some day you may hit gold :)


When do you think it will be safe to post the questions here (if we can) and discuss the problems. I wasn't able to partake in the event, but I would like to take a look at the problems. Also, how did you guys do?

yeah; i'm interested in seeing the questions too.  you guys better have aced that competition!

just received word... NYC swept the competition, occupying places one through four.  ohh goodness.

plus sweeping individual first place ~ third place
(I personally didnt like it tho)

congrats danny, sho and ethan!!

some people need to learn to play chinese poker and/or hold 'em!

wont mention any names now, k Geehoon?

sure sure.

I'm not on New York City Poker Team!
(As we said, it would be hilarious - thousands of people playing poker in gym hehe)


function
integration
real analysis
calculus
calculus computations

Need help!!

Let {$a_n$} be a strictly decreasing sequence in [0,1] such that lim n->$\infty$ $a_n$ = 0
Define a function f:[0,1]-->R by
f(x)= $\{$ 1 if x= $a_n$ for some n
         $\{$ 0 if x$\neq$ $a_n$ for all n

Show that f is integrable
Show that $\int_{0}^{1}$ $f$ = 0

Now I see the question again I realize that my previous answer was incomplete/incorrect.
$f=0$ a.e allows you to conclude that $f$ is Lebesgue-integrable with integral $0$, but it does not allow you to say that $f$ is Riemann-integrable. In order to do this you need to say that $f$ is bounded and continuous a.e (since $a_{n}$ is decreasing, the set of discontinuities of $f$ is actually countable, which solves the question). Now $\int_{[0,1]}{f \ dm} =1$
Thus you do not need the "strict" hypothesis, nor "$a_{n} \rightarrow 0$".

Well, if we want to work more along the statement of the question, we will show that at each $x\in f^{-1}(\{0\})$ (except $x=0$), $f(x)$ is continuous.  Since $[0,1]=f^{-1}(\{0\})\cup f^{-1}(\{1\})$ and $f^{-1}(\{1\})$ is countable, this will show that $f$ is continuous a.e.

Say that $x\in f^{-1}(\{0\})$.  Then $x\in (a_{j+1},a_j)$ for some $j$.  Take $\delta=\min\{x-a_{j+1},a_j-x\}$.  Then $(x-\delta,x+\delta)\subseteq (a_{j+1},a_j)$, so that $f(x)\equiv 0$ on $(x-\delta,x+\delta)$.  So $f(x)$ is continuous a.e (it is also discontinuous at $0$, but we don't care about this, and $f^{-1}(\{1\})$ is countable because a sequence only contains countably many points), and $f$ is Riemann integrable.

Obviously $f$ is also Lebesgue integrable, and we may omit $f^{-1}(\{1\})$ in the integration because it is a set of measure zero.  So, letting $m$ denote Lebesgue measure, we have

$\int^1_0 fdx=\int^1_0 fdm=\int_{f^{-1}(\{0\})} 0dm=0$

Well, almost.  We want to make sure that $f^{-1}(\{0\})$ is Lebesgue measurable, right?  But the Lebesgue measure on $\mathbb{R}$ is complete and contains all closed intervals, so it contains $[0,1]\setminus f^{-1}(\{1\})$ e.g, a closed interval minus a set of measure zero, so $f^{-1}(\{0\})$ is fine.


limit
function
logarithms
calculus
calculus computations

This follows from [url=http://en.wikipedia.org/wiki/Stolz-Ces%C3%A0ro_theorem]Stolz-Cesaro[/url], although it's possible the proof must be repeated for every $ x \in [0, 1)$.

Tanks alot . I have known that $ \underset {n\rightarrow\infty}{\lim}\frac {f\left( n\right) }{n} \equal{} 0$ by Stolz-Cesaro theorem which you taught

I don't understand that
 [quote="t0rajir0u"] although it's possible the proof must be repeated for every $ x \in [0, 1)$.[/quote]

Stolz-Cesaro is about sequences, not about functions!!!

It is not so easy to prove it for continuous functions.

Without Stolz-Cesaro for sequences:
If $ \underset{n\rightarrow\infty }{\lim}f\left( n \plus{} 1\right) \minus{} f\left( n\right) \equal{} 0$ then $ \underset {n\rightarrow\infty}{\lim}\frac {f\left( n\right) }{n} \equal{} 0$
Proof:

Define $ a_n \equal{} e^{f(n)}$ for all $ n$
Then the theorem becomes:
If $ \lim_{n \to \infty} \frac {a_{n \plus{} 1}}{a_n} \equal{} 1$ then $ \lim_{n\to\infty}\sqrt [n]{a_n} \equal{} 1$.
This is a well known theorem for positive sequences.

How to make " $ f$ is continuous and $ \underset {n\rightarrow\infty}{\lim}\frac {f\left( n\right) }{n} \equal{} 0$ " become " $ \underset {x\rightarrow\infty}{\lim}\frac {f\left( x\right) }{x} \equal{} 0$ " ?

[quote="Kent Merryfield"]What we actually have is this: for all $ x\in[0,1),\ \lim_{n\to\infty}\frac {f(x \plus{} n)}{n} \equal{} 0.$ Does, that, and the continuity (or uniform continuity, or boundedness) of $ f$ on $ [0,1]$ help?[/quote]

Sorry , i  don't know it.

Would you teach me step by step ?  thank you alot !

[quote="luckman168"]Suppose that $ f$ is continuous and $ \underset{x\rightarrow\infty }{\lim}f\left( x \plus{} 1\right) \minus{} f\left( x\right) \equal{} 0$ ,then $ \underset {x\rightarrow\infty}{\lim}\frac {f\left( x\right) }{x} \equal{} 0$[/quote]
This is just a thought... We define $ M$ such that $ f(x) < M \forall x \in [0;1]$, and thus $ f(x) < M \forall x \in \mathbb{R}$.
Hence $ \underset {x\rightarrow\infty}{\lim}\frac {f\left( x\right) }{x} \le \lim_{x \to \infty} \frac{M}{x} \equal{} 0$ ? The part i'm worried about is: Do such a $ M$ necessarily exist?

No, you can't claim that $ f(x)$ is bounded on $ [0,\infty).$ You can say that $ |f(x)|\le M$ for $ x\in[0,1],$ since a continuous function on a closed bounded interval is bounded and assumes its maximum. But try letting $ f(x)\equal{}\ln(x\plus{}1).$ In this case $ f(x\plus{}1)\minus{}f(x)\equal{}\ln\left(1\plus{}\frac1{x\plus{}1}\right)<\frac1{x\plus{}1},$ which goes to zero and satisfies our original hypothesis. This $ f$ is not bounded on $ [0,\infty).$

[quote="Kent Merryfield"]No, you can't claim that $ f(x)$ is bounded on $ [0,\infty).$ You can say that $ |f(x)|\le M$ for $ x\in[0,1],$ since a continuous function on a closed bounded interval is bounded and assumes its maximum. But try letting $ f(x) \equal{} \ln(x \plus{} 1).$ In this case $ f(x \plus{} 1) \minus{} f(x) \equal{} \ln\left(1 \plus{} \frac1{x \plus{} 1}\right) < \frac1{x \plus{} 1},$ which goes to zero and satisfies our original hypothesis. This $ f$ is not bounded on $ [0,\infty).$[/quote]

I knew it would be possibe unbounded  before.

[quote="Kent Merryfield"]What we actually have is this: for all $ x\in[0,1),\ \lim_{n\to\infty}\frac {f(x \plus{} n)}{n} \equal{} 0.$ Does, that, and the continuity (or uniform continuity, or boundedness) of $ f$ on $ [0,1]$ help?[/quote]

Sir ,Would you teach me by easier way ? I need your help! thank you.

hello i try to post a proof:

we will note $ p_x = [x]$ the integer part of x, that is to say : $ [x]\leqq < [x] + 1$
\[ f(x) = (f(x) - f(x - 1)) + .... + (f(x - p_x + 1) - f(p_x)) + f(x - p_x)
\]
and
\[ \frac {f(x)}{x} = \frac {p_x}{x} \frac {1}{p_x} \sum_{k = 1}^{p_x} [f(x - k + 1) - f(x - k)] + \frac {f(x - p_x)}{x}
\]
It is enough prooving that $ \lim_{x \to \infty} \frac {1}{p_x} \sum_{k = 1}^{p_x} [f(x - k + 1) - f(x - k)] = 0$

That looks like Cesaro but we can't use it. Nevertheless we can adapt the proof:

Let's $ \epsilon > 0$ and let's $ A > 0$ such that $ \forall x \geq A$ we have $ |f(x + 1) - f(x)|\leq \epsilon$

So
$ \frac {1}{p_x} \sum_{k = 1}^{p_x} [f(x - k + 1) - f(x - k)] = \frac {1}{p_x} \sum_{k = 1}^{p_x - p_A} [f(x - k + 1) - f(x - k)] + \sum_{p_x - p_A + 1}^{p_x}[f(x - k + 1) - f(x - k)]$

And then
$ |\frac {1}{p_x} \sum_{k = 1}^{p_x} [f(x - k + 1) - f(x - k)]| \leq \frac {(p_x - p_A + 1)}{p_x}\epsilon + \frac {1}{p_x}\sum_{p_x - p_A + 1}^{p_x}|f(x - k + 1) - f(x - k)|$

But when $ p_x - p_A + 1\leq k \leq p_x$ we have $ |f(x - k) - f(x-k + 1)|\leq 2 \sup_{[0,p_A + 1]} |f(x)| \leq C_A$

and then $ \frac {1}{p_x}\sum_{p_x - p_A + 1}^{p_x}|f(x - k + 1) - f(x - k)| \leq \frac {(p_A + 1)}{p_x} C_A$

So $ \limsup_{x \to \infty}|\frac {1}{p_x} \sum_{k = 1}^{p_x} [f(x - k + 1) - f(x - k)]|\leq \epsilon$ and this is true forall epsilon the result is then prooved.

@populux: Well done, the proof is (I think) correct, apart from a few minor mistakes.


Sophie-Germain Identity
quadratics
algebra
quadratic formula
special factorizations
primes
number theory

I was reading about the sophie germain identity and came across this problem (in the AOPS Wiki)

Is $ 4^{545} \plus{} 545^4$ composite?

How would you apply the identity to solve this?

[hide="hint"]
What happens when you factor out a 4 from $ 4^{545}$?

If you still don't see it:
[hide="pretty much a dead giveaway when combined with first hint"]
$ 4^{545}\equal{}2^{1190}$[/hide][/hide]

[hide="another hint"]
The sophie germain identity can be generalized to
$ x^4 \plus{} 4y^4 \equal{} \left(x^2 \minus{} 2xy \plus{} 2y^2\right)\left(x^2 \plus{} 2xy \plus{} 2y^2\right)$
EDIT: added 2's
*afterthought* my LaTeX is awfully messy
[/hide]
[hide="solution"]
The second addend is a $ 4^{th}$ power, and the first addend is 4 times a $ 4^{th}$ power.
Letting $ x \equal{} 545$ and $ y \equal{} 2^{544}$ lets us apply the sophie germain identity:
$ 4^{545} \plus{} 545^4 \equal{} (545)^4 \plus{} 4\left(2^{544}\right)^4 \equal{} \left(545^2 \minus{} 2(545)\left(2^{544}\right) \plus{} 2\left(2^{544}\right)^2\right)\left(545^2 \plus{} 2(545)\left(2^{544}\right) \plus{} 2\left(2^{544}\right)^2\right)$
and the last 2 factors are both integers greater than 1, so these are 2 non-1 and non-number factors of the given number, so the given number is composite
[/hide]

[hide="extension"]
Find all n such that $ 4^n \plus{} n^4$ is prime.
(This uses the sophie germain identity)
[/hide]

[quote="stevenmeow"]
The sophie germain identity can be generalized to
$ x^4 \plus{} 4y^4 \equal{} \left(x^2 \minus{} 2xy \plus{} y^2\right)\left(x^2 \plus{} 2xy \plus{} y^2\right)$[/quote]

That should be $ (x^2 \plus{} 2xy \plus{} 2y^2)(x^2 \minus{} 2xy \plus{} 2y^2)$.

note: I would not bother remembering this formula; everytime I need it, I just rederive it by completing the square and then factoring as a difference of squares.

[quote="stevenmeow"] Find all n such that $ 4^n \plus{} n^4$ is prime.
(This uses the sophie germain identity)[/quote]

hm....
[hide="I think this works"]
Note that $ n$ must be odd, or the expression is divisible by 2. Also note that if $ n$ is odd, then we can factor the expression as \[ 4^n \plus{} n^4 \equal{} (n^2 \plus{} n2^{\frac{n \plus{} 1}{2}} \plus{} 2^n )(n^2 \minus{} n2^{\frac{n \plus{} 1}{2} \plus{} 2^n}.\]

In order for $ 4^n \plus{} n^4$ to be prime, one of its factors in the expression above must be $ 1$; in fact, since $ n$ is positive, $ n^2 \minus{} n2^{\frac{n \plus{} 1}{2}} \plus{} 2^n$ must be the factor equal to $ 1$, so we get the quadratic $ n^2 \minus{} n2^{\frac{n \plus{} 1}{2}} \plus{} 2^n \minus{} 1 \equal{} 0.$ Using the quadratic formula, we find that \[ n \equal{} 2^{\frac{n \minus{} 1}{2}} \pm \sqrt{2^{n \minus{} 1} \minus{} 2^n \plus{} 1} \Longleftrightarrow n \equal{} 2^{\frac{n \minus{} 1}{2}} \pm \sqrt{(\minus{}1)2^{n \minus{} 1} \plus{} 1}\] from which it is clear that $ n\equal{}1$ is the only solution, since that expression above for $ n$ must be a positive integer.

note: there may be an easier solution that I cannot find[/hide]

[quote="stevenmeow"][hide=another hint]
Find all n such that $ 4^n \plus{} n^4$ is prime.
(This uses the sophie germain identity)
[/quote]

Divide into 2 cases,
1st case : if $ n$ is even number, it's clear that $ 4^n \plus{} n^4$ is also even and greater than 2, so it is definitely not a prime number.

2nd case : if $ n$ is odd number, let $ n \equal{} 2k \plus{} 1$
Then
$ 4^n \plus{} n^4 \equal{} (2k\plus{}1)^4 \plus{} 4 (2^k)^4$
The last form is in the form of $ x^4 \plus{} 4y^4$ which can be factorized by the sophie germain identity.
It is clear that $ 4^n \plus{} n^4$ is prime only when $ n \equal{} 1$


In the rectangles shown, $ \overline{EH}$ is $ 50\%$ greater than $ \overline{AD}$, and $ \overline{EF}$ is 50\% less than $ \overline{AB}$. What percent of the area of ABCD is the area of EFGH?

[asy]pair A,B,C,D,E,F,G,H;
A = (0,-2); B = (4,-2); C = (4,0);
H = (7,0); G = (9,0); E = (7,-5); F = (9,-5);
draw(A--B--C--D--A); draw(E--F--G--H--E);
label("D",D,N);label("C",C,N);label("H",H,N);label("G",G,N);
label("A",A,S);label("B",B,S);label("E",E,S);label("F",F,S);[/asy]

Let the side length of the first rectangle be x and y. Thus the area of the second square is $ \frac{3}{2}x\cdot\frac{1}{2}y=\frac{3}{4}xy$, so our answer is ${ \boxed{75\%}}$.

Actually, let AB=AD=1. Then EH=1.5 and EF=.5. The area of EFGH is .75, which is $ \boxed{75\%}$ of that of ABCD.


limit
real analysis
real analysis unsolved

If $ \sum\alpha_i\xi_i$ converges for every sequence $ \{\xi_i\}$ such that $ \xi_i\to 0$ as $ i\to\infty$, prove that $ \sum |\alpha_i |<\infty$.

I'm trying this a la Banach spaces. So I converted the terminology to: Suppose $ \sum\alpha_i\xi_i$ converges for every $ x\equal{}\{\xi_i\}\in c_0$. Show that $ \{\alpha_i\}\in \ell^1$.

Define $ \Lambda x\equal{}\sum\alpha_i\xi_i$. Then $ \Lambda$ is a linear operator. If I could get it bounded and hence $ \Lambda\in (c_0)^*$, then I'd have $ \|\{\alpha_i\}\|_1\equal{}\|\Lambda\|<\infty$ and I'd be done.

I don't really buy that it has to be bounded just because the sum converges, though, because it seems that you could construct $ x_n\in c_0$ with $ \|x_n\|_\infty\equal{}1$ such that $ \sup\Lambda x_n\equal{}\infty$. Or maybe I could invoke uniform boundedness or something?

I'm still sort of fishing around in Banach spaces, so if I've converted terminology incorrectly, or am trying to make this too difficult, please point this out.

Define $ F_n(x) \equal{} \sum_{k\equal{}1}^{n}\alpha_k \xi_k$, where $ x \equal{} (\xi_k)_k \in c_0$. Then,

$ |F_n(x)| \leq \sum_{k\equal{}1}^{n}|\alpha_k\parallel{}\xi_k| \leq \left(\sum_{k\equal{}1}^{n}|\alpha_k|\right)\|x\|$.

On the other hand, the limit $ \lim_{n \rightarrow \plus{}\infty} F_n(x) \equal{} \sum_{k\equal{}1}^{\plus{}\infty}\alpha_k \xi_k : \equal{} F(x)$ exists for each $ x \equal{} (\xi_k)_k \in c_0$.

By Banach-Steinhaus Theorem, $ F \in (c_0)' \equal{} \ell^{1}$. Hence, there is $ f \equal{} (x_k)_k \in \ell^{1}$ such that $ F \equal{} f$. In particular,

$ x_k \equal{} f(e_k) \equal{} F(e_k) \equal{} \alpha_k$.

Thus, we conclude that $ (\alpha_k)_k \in \ell^{1}$.


[quote="Lovasz"][quote="sen"]Prove that if, $a,b,c>0$
$(a^3+b^3)(b^3+c^3)(c^3+a^3)\geq a^2b^2c^2(a+b)(b+c)(c+a)$[/quote]
\[ x,y>0 \implies x^3+y^3 \ge xy(x+y)\ \ \ \left( \iff (x-y)^2(x+y) \ge 0 \right)  \][/quote]

Actually our original inequality is true for all $a,b,c\ge0$. I think that we should write the biggest set, where is an inequality true before proving it. E.G. AM-GM is true for every $n\ge1$ integer and $x_i\ge0$ for all $i=1,2,\ldots,n$.

conceive

Dividind we get: $(a^2-ab+b^2)(b^2-bc+c^2)(c^2-ca+a^2) \geq a^2b^2c^2$

This is easy to prove starting from:
$a^2+b^2 \geq 2ab$  :arrow:  $a^2-ab+b^2 \geq ab$

Similarly: $b^2-bc+c^2 \geq bc$ and $c^2-ca+a^2 \geq ca$

Multiplying we have the desired inequality.

[quote="Karth"][quote="rem"][quote="sen"]Prove that if, $a,b,c>0$

$(a^3+b^3)(b^3+c^3)(c^3+a^3)\geq a^2b^2c^2(a+b)(b+c)(c+a)$[/quote]
Expanding and simplifying, obtain:
$a^6b^3+a^6c^3+b^6c^3+b^6a^3 +c^6a^3+c^6b^3\geq a^4b^3c^2+a^4c^3b^2+ b^4a^3c^2+b^4c^3a^2+c^4a^3b^2+c^4b^3a^2$
 which is obviously true by Muirhead. :)[/quote]

What's Muirhead?  :|[/quote]
It's a really powerful ineq that can be very useful for solving some even very hard ineq.

[quote="Karth"][quote="rem"][quote="sen"]Prove that if, $a,b,c>0$

$(a^3+b^3)(b^3+c^3)(c^3+a^3)\geq a^2b^2c^2(a+b)(b+c)(c+a)$[/quote]
Expanding and simplifying, obtain:
$a^6b^3+a^6c^3+b^6c^3+b^6a^3 +c^6a^3+c^6b^3\geq a^4b^3c^2+a^4c^3b^2+ b^4a^3c^2+b^4c^3a^2+c^4a^3b^2+c^4b^3a^2$
 which is obviously true by Muirhead. :)[/quote]

What's Muirhead?  :|[/quote]

even if you don't know Muirhead, you can prove it by repeated application of Am-Gm

[i] Solution. [/i]

  The following inequality is valid: \[ 2(x^{3}+y^{3})\geq (x+y)(x^{2}+y^{2}). \]
   So, by applying this inequality, we only need to prove

                                        \[ (a^{2}+b^{2})(b^{2}+c^{2})(c^{2}+a^{2})\geq 8a^{2}b^{2}c^{2}. \]

   Bu this inequality follows immediately from $x^{2}+y^{2}\geq 2xy.$


I think that you shouldn't lock a topic where people still post topics, so I open a new topic for the same thing.

So the genius:
definition, condition, fate, utility

Write what you want.

I locked that topic because you insisted on creating a "flame" which had nothing to do with the topic. In other word the discussion we were having was off topic.

$Bump^{bump^{bump^{bump^{bump^{bump^{bump^{bump^{bump^{bump^{bump}}}}}}}}}}$

~5,200 day revive!


A point inside a square is positioned so that the distances to the four vertices are 27,21,6,x units. If x is a whole number, what is the value of x ?

[hide][b]Note[/b]
 Please , let  $ABCD$ the square and

   $MA = 27 , MB = 21 , MC = 6 , MD = x$

Applying the median theorem( consider the diagonals , which are equal)  , you must have

   $27^2 + 6^2 = 21^2 + x^2$ ,

  so $x^2 = 324$ and $x = 18$

 I hope having helped you !

[u] Babis[/u][/hide]

Yes that is the correct way to do the problem, but can you please use a hide tag so people can see the problem without the answer

Sorry , but I do not know how to do this. Can you do it for me or to tell me the way ?

 [u]Babis[/u]

Now I see , that this is a problem for grades 6-7. These pupils do not know the median theorem. You can however draw  parallels from M to the sides , name  $a , b , c , d$ the segments which we see on the sides of the square and then apply four times the pythagorian equality(theorem). If you add the first and third relation and then the second and forth ,the  second members will be the same , named
  $a^2 + b^2 + c^2 + d^2$. In this way you will receive  the same relation as in my first post.

 Babis

Wonderful solution! I was just trying to figure out a similar problem on a high school math contest!

[quote="stergiu"]Sorry , but I do not know how to do this. Can you do it for me or to tell me the way ?[/quote]

When you post, there is a line of buttons labeled [b]B[/b], [i]i[/i], [u]u[/u], Quote, etc. The last button says hide. That's how you hide your answer.

[hide](this is what it should look like) [/hide]

will be

[code][hide](this is what it should look like) [/hide][/code]

[b]To 236 factorial:[/b]

 [hide]Thank you very much ! ;)

[u]Babis[/u][/hide]

I wanted to revive this topic because I was just about to post this problem.

Does anyone have an elaborate explanation of how to do this because I didn't understand any of it (except for the theorem, but I don't know why it works).  :blush:

create a square, label the sides, and create similar diagonals to this one.
You'll see that
$a^2 +c^2 = e^2 + f^2 + g^2 + h^2$
$b^2 + d^2 = e^2 + f^2 + g^2 + h^2$
So $a^2 + c^2 = b^2 + d^2$


This does not seem right, as it will becomes false if we let one variable be very large.

Perhaps it's the other way around?

This is not true for arbitrary positive a, b, c. However, if a, b, c are the sidelengths of a triangle, the inequality is true, since, with the well-known substitution a = y + z, b = z + x, c = x + y, with positive x, y, z, we have

$6\left(a^2b^2+b^2c^2+c^2a^2\right)-\left(4abc\left(a+b+c\right)+ab\left(a^2+b^2\right)+bc\left(b^2+c^2\right)+ca\left(c^2+a^2\right)\right)$
$= 2\left(x^2\left(x-y\right)\left(x-z\right)+y^2\left(y-z\right)\left(y-x\right)+z^2\left(z-x\right)\left(z-y\right)\right)$
$+yz\left(y-z\right)^2+zx\left(z-x\right)^2+xy\left(x-y\right)^2$,

what is clearly $\geq 0$, since

$x^2\left(x-y\right)\left(x-z\right)+y^2\left(y-z\right)\left(y-x\right)+z^2\left(z-x\right)\left(z-y\right)\geq 0$

after Schur.

  Darij

Sorry I have forgotten to say that $a,b,c$ are sides of a triangle.

Thank you very much for your replies, Billzhao and Darij. :)

Manlio, you wrote Vasc as source for this inequality.
But I don't remember this inequality! :?

Oh....It is your nice ineq

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

rewritten using $a+b=z$ and similar.

[quote="manlio"]If $x,y,z$ are positive real numbers, then
\[\frac{x^{2}+2yz}{y+z}+\frac{y^{2}+2zx}{z+x}+\frac{z+2xy}{x+y}\ge \frac{3}{2}(x+y+z). \]
[/quote]
$\sum{\frac{x^{2}+2yz}{y+z}}-\frac{3}{2}\sum{x}$

$\equiv\frac{F(x,y,z)}{2(y+z)(z+x)(x+y)}$.

Without losing generality, we may assuming $x\leq y\leq z$.

$F(x,y,z)=F(x,x+s,x+s+t)$

$=4(s^{2}+st+t^{2})x^{2}+(2s+t)(s^{2}+st+6t^{2})x+t^{2}(7s^{2}+7st+2t^{2})\geq0$.
[hide="Link."]http://www.mathlinks.ro/Forum/viewtopic.php?t=24555[/hide]


Let $ S\equal{}x\plus{}y\plus{}u\plus{}v$, knowing that:
\[ x\plus{}y\equal{}3\]
\[ xu\plus{}yv\equal{}\minus{}1\]
\[ xu^2\plus{}yv^2\equal{}3\]
\[ xu^3\plus{}yv^3\equal{}\minus{}1\]

Calculate $ S$

One obvious solution is $ S\equal{}3$, with $ x\equal{}1, y\equal{}2, u\equal{}1$ and $ v\equal{}\minus{}1$ and the question seems to imply that the solution is unique.

hello, we have two solutions
$ x\equal{}1,y\equal{}2,u\equal{}1,v\equal{}\minus{}1$
or
$ x\equal{}2,y\equal{}1,u\equal{}\minus{}1,v\equal{}1$
and our sum is $ 3$.
Sonnhard.

Assume $ u \equal{} 1$ and $ v \equal{} \minus{}1$ (From a guess based on the pattern of alternating values.

We then have

$ x \plus{} y \equal{} 3$
$ x \minus{} y \equal{} \minus{}1$

The other equations reduce to the same two equations.

So $ x \equal{} 1$ and $ y \equal{} 2$, and the sum is $ x \plus{} y \plus{} u \plus{} v \equal{} 1 \plus{} 2 \plus{} \minus{}1 \plus{} 1 \equal{} 3$

$ \minus{}(xu^i\plus{}y v^i)(uv)\plus{}(x u^{i\plus{}1}\plus{}y v^{i\plus{}1})(u\plus{}v)\equal{}(x u^{i\plus{}2}\plus{}y v^{i\plus{}2})$

In case anyone was wondering, that identity has to do with characteristic equations of linear recurrences: a very straightforward idea, even though it may seem unnecessarily abstract/general in this context.
To find solutions to recurrences, one constructs closed form solutions that look like \[ xu^n \plus{} y v^n\]
This particular solution solves a second order recursion whose characteristic polynomial has roots u,v.

For example:
$ a_{n\plus{}2}\plus{}ba_{n\plus{}1}\plus{}ca_n\equal{}0$ has char. poly. $ x^2\plus{}bx\plus{}c\equal{}0$. We want this polynomial to have roots $ u,v$, so we let $ b\equal{}\minus{}(u\plus{}v)$ and $ c\equal{}uv$. Then the recurrence
\[ a_{n\plus{}2}\plus{}uva_n\equal{}(u\plus{}v)a_{n\plus{}1}\] is solved, in general, by a series with closed form like the one given by balan. From here, we can go from term to term using the recursive relation (or, identically, by using Altheman's identity)


Given an acute-angled triangle $ABC$, let points $A' , B' , C'$ be located as follows:  $A'$ is the point where altitude from $A$ on $BC$ meets the outwards-facing semicircle on $BC$ as diameter. Points $B', C'$ are located similarly.
Prove that $A[BCA']^2 + A[CAB']^2 + A[ABC']^2 = A[ABC]^2$ where $A[ABC]$ is the area of triangle $ABC$.

$(\dfrac{\triangle{BA'C}}{\triangle{BAC}})^2=(\dfrac{BA' \cdot CA'}{AD \cdot BC})^2=\dfrac{{BA'}^2 \cdot {CA'}^2}{AD^2 \cdot BC^2}$

Since: ${BA'}^2=BD \cdot BC$, ${CA'}^2=CD \cdot CB$, put it in we get:
\[ (\dfrac{\triangle{BA'C}}{\triangle{BAC}})^2=\dfrac{BD\cdot CD}{AD^2} \]

Because: $\dfrac{BD}{AD}=\cot{B}$, $\dfrac{CD}{AD}=\cot{C}$, so the equation is transformed into:
\[ \sum \cot{B}\cot{C}=1 \]
Use $\cot{C}=\dfrac{1}{\tan{C}}=-\dfrac{1}{\tan{(A+B)}}=\dfrac{\tan{A}\tan{B}-1}{\tan{A}+\tan{B}}$

Therefore: $\cot{C}(\cot{A}+\cot{B})=\dfrac{\tan{A}\tan{B}-1}{\tan{A}\tan{B}}$

$\cot{A}\cot{B}+\cot{C}(\cot{A}+\cot{B})=\dfrac{1}{\tan{A}\tan{B}}+1-\dfrac{1}{\tan{A}\tan{B}}=1$

So the area equation is also true.
Hence the problem is solved.

I got essentially the same answer as shobber ....But I prefer pure geometrical solutions....
Can anyone give a pure geometrical solution to this one?

[quote=cmtappu96][geogebra]2f19e7b549f6b0374f982b5015869cc47bdfae33[/geogebra]

Let $[ABC] = \Delta, [HBC] = \Delta_1, [HCA] = \Delta_2, [HAB] = \Delta_3$

Lemma: $DB \cdot DC = DH \cdot DA$
This can be easily proved as $\triangle HBD ~ \triangle CAD \Rightarrow \frac{DH}{DC} = \frac{DB}{DA} \Rightarrow DB \cdot DC = DH \cdot DA$

$A'D^2 = DB \cdot DC = DH \cdot DA$
$\frac{1}{4} \cdot A'D^2 \cdot BC^2 = \frac{1}{4} \cdot DH \cdot DA \cdot BC^2$
$\left(\frac{1}{2} \cdot BC \cdot A'D \right)^2 = \left(\frac{1}{2}\cdot BC\cdot DA\right)\left(\frac{1}{2}\cdot BC\cdot DH\right)$
$[A'BC]^2 = [ABC][HBC] = \Delta \Delta_1$
Similarly $[B'CA]^2 = \Delta \Delta_2 \text{ and } [C'AB]^2 = \Delta \Delta_3$
Adding we get $[A'BC]^2 + [B'CA]^2 + [C'AB]^2 = \Delta (\Delta_1 + \Delta_2 + \Delta_3) = \Delta \cdot \Delta = [ABC]^2$[/quote]

Please attach the figure

We know that,
$\frac{\triangle A'BC^2}{\triangle ABC^2}  =A'D^2/AD^2$
$\frac{\triangle AB'C^2}{\triangle ABC^2}  =B'E^2/BE^2$
$\frac{\triangle ABC'^2}{\triangle ABC^2}  =C'F^2/CF^2$
So,
$A'D^2/AD^2+B'E^2/BE^2+C'F^2/CF^2=1$,should be true
$A'D^2/AD^2+B'E^2/BE^2+C'F^2/CF^2=1$.... (i)
Now,
Let the altitudes from $A,B,C$  meet $BC,AC,AB$ at $ D,E,F$  respectively.
$\triangle A'DC^2$ and $\triangle A'BD^2$ are two similar triangles.
$\triangle AC'F^2$ and $\triangle C'FB^2$ are two similar triangles.
$\triangle AB'E^2$ and $\triangle B'EC^2$ are two similar triangles.
So,
we can say,
$C'F/AF=BF/C'F$
$C'F^2=AF.BF$
Similarly,
$B'E^2=AE.EC$
$A'C^2=AD.CD$
Plugging the values in (i),we get,
$BD.CD/AD^2+AE.EC/BE^2+AF.BF/CF^2=1$
Now,
$cotB=BD/AD$
$cotC= CD/AD$
$cot A=AE/BE$

So,
$cotBcotC+cotAcotC+cotAcotB= 1$
$tanA+tanB+tanC=tanAtanBtanC$,which is true .
If $A+B+C=180^{\circ{}}$,
then ,
$tanA+tanB+tanC=tanAtanBtanC$ holds true.


For b>a>0, maximize (ab-a^2)/(ba+b^2-a^2).

[hide]This problem was best approached from an analytical point of view.  Since you have (ab - a :^2: ) in the top and bottom, you can rearrange it so that it looks like this:

(ab - a :^2: )/((ab - a :^2: )+b :^2: )

This can also be described as X/(X+b :^2: ).  This function approaches its maximum value as b :^2: approaches zero.  (Anything larger than a zero will result in a value of less than 1, but a zero will give you a 1.)

Therefore, the function is maximized as b and a approach zero, given the initial parameters.

[/hide]

Um... Your first step was good, but ur final answer is not correct (pretty much because there is a numeric answer to this).

One problem with what you're saying is that X is not a constant, X is a function X(b).  So changing b changes X as well, which means that sending b closer and closer to zero might not in fact result in larger values for the whole expression.

[hide](ab-a^2)/(ab+b^2-a^2)

= ab/(ab+b^2-a^2) - a^2/(ab+b^2-a^2)

= 1/(1 + b/a - a/b) - 1/(b/a + (b/a)^2 - 1)

Let x = b/a, and the things becomes a single-variable expression. You can now use calculus.[/hide]

I know next to nothing about inequalities, but with (semi) random manipulation i get

[hide] 1/5 [/hide]

Calculus?  You can find something nicer than that, can't you?

I chose the steps I did because no matter what values there are for a and b, what I equated as "X" will equal the same thing on both the top and the bottom.  On the bottom the b :^2: term is added onto it, hence my reasoning.

Coder:  By your reasoning, no matter what X is, as long as it's the same, the expression will get bigger when b goes to 0, right?  Here's a counter-example.  Let X(1) = 1 and let X(1/2) = 1/20.  Then we get
X(1)/(X(1) + 1) = 1/2
X(1/2)/(X(1/2) + 1/4) = (1/20)/(1/20 + 1/4) = 1/6 < 1/2.  So making b go to zero [i]could[/i] result in a smaller value for the whole thing, depending on what happens to X.  X = X isn't the only important fact -- how X acts as we change b is also important.

Geomar -- Post a solution!!

Geomar is correct.... btw, this is a number 8 on arml from 99... well this wasnt the entire thing, first it was geometric, and i got this with about 3 mins left, but wasnt able to maximize it. Does anyone know a site with more than just 99 and 00 for ind armls?

Well, i figured out what i did... but it was in an annoying round-about way

[hide]

i substituted b=ka and made the thing a function of k, and then i minimized that...

however for some reason i had to take 2 reciprocals and stuff untill i got that it was at a min at k=2 ... thats the part im trying to see how to make it much easier and evident.

[/hide]

[hide]We can just do the reciprocal

(ab-a :^2: +b :^2:)/(ab-a :^2: )=1+b/a+b/(b-a)

Note that b/a+b/(b-a)=b :^2: /a(b-a)=(b/a)(b/b-a)

let x=b/a, y=b/(b-a)

x+y-xy=0, so (x-1)(y-1)=1

all the solutions are (x, x/(x-1))

We can show that the minimum of x(x/x-1) :ge: 4

because (x-2) :^2:  :ge: 0

 or x :^2: -4x+4 :ge: 0

x :^2:  :ge: 4(x-1)

so x :^2: /(x-1) :ge: 4

we want the reciprocal to be small, so we want b/a+b/(b-a) to be 4. the maximum is therefore 1/5[/hide]

[hide]

much like beta's solution, plugging in b=ka (for a k>1)

and taking the reciprocal,

1/z=1+k^2/(k-1)

and you gotta minimize k^2/(k-1) which i did in another way heh...taking the reciprocal again !

so um that part is

1/k-1/k^2 = y-y^2,

so the min is y=1/2 or k =2 and then 1/z = 5

so mine = 1/5

But my problem with that is that its not pretty, and all math must be pretty right? anyone got a quick way to minimize or do all take time like this.

[/hide]

Hmm...   I see your point about considering how the X portion changes in relationship to B.  A different approach would be to try to maximize ab - a :^2: since there there are no lower extrema to that part.

ab - a :^2: = y
dy/da = b - 2a = 0
b = 2a

So, this would change the original equation to:

(2a * a - a :^2: ) / (2a * a - a :^2:  + 4a :^2: )

simplified to....

(a :^2: ) / (5a :^2: )

simplified to a final answer of

1/5

I guess my original method amounts to about as much as that humorous proof to make 1=2.


topology
function
real analysis
real analysis unsolved

k this is a qn from my book and i cant do it. let (An) be a nested sequence of non empty compact sets. ie A1 contains A2 contains....  . Let A= intersection of all An's. the diameter of each An>= k for some k>0. Is it necessary that diam A>=k

how do i approach this problem? please give me hints only, dont post a complete solution. i want to do it myself

$>$? No. This should be pretty obvious.

$\ge$? Yes. Consider a pair of sequences $x_n$ and $y_n$ with the distance between $x_n$ and $y_n$ at least $k$. Look for convergent subsequences- can you find such sequences with the same indices for $x$ and $y$?

hmmm ok, i edited my original post, its >= , not just > .i'll write whatever i'm thinking now. since diam of each of the An's >=k, we can find xn,yn such that d(xn,yn)>=k , for each n. where xn,yn belong to An. so we have a pair of sequences each of which is in A1. since A1 is compact, there is a sub seq xni which converges to p in A1. the sub seq yni has a sub seq ynij which converges  to q in A1 and xnij continues to converge to p. so we have a convergent sub seq ( xnij,ynij) converging to (p,q) in A1. but p also belongs to A2 since except probably for the first term of the seq xnij, rest all lie in A2. similarly p lies in all An. and q the same way too. so we have d(p,q)>=k, because p,q lie in each An. correct? i'm not very sure

That's basically right, but we should clear up why $d(p,q)\ge k$. Jet $x_n,y_n$ be the $n$-th terms of the subsequences converging to $p,q$ respectively. By construction, $d(x_n,y_n)\ge k$.

Choose $\epsilon>0$. There exist $N_1$ and $N_2$ such that for $n>N_1$, $d(x_n,p)<\frac{\epsilon}2$ and for $n>N_2$, $d(y_n,q)<\frac{\epsilon}2$. Let $N=\max\{N_1,N_2\}$.

For $n>N$, $d(p,q)\ge d(x_n,y_n)-d(p,x_n)-d(q,y_n)>k-\epsilon$. Letting $\epsilon$ tend to zero, $d(p,q)\ge k$.

It's got nothing to do with being in the intersection and everything to do with the fact that we picked the sequences far apart to begin with.

There was no need to post twice asking for help four hours apart- I was asleep then anyway.

Why can we find $x_n,y_n$ such that $d(x_n,y_n)\ge k$?  If $diam A_N=k$ for some $N$ and $diam A_N=\sup_{p,q\in A_N} d(p,q)$, then $d(x_n,y_n)\le \sup_{p,q\in A_N} d(p,q) = k$, but what if it is less than the $sup$?

I have a feeling what I said makes no sense  :blush:

Compactness. That's a sup of a continuous function on a compact set, so it's a maximum.

On the other hand, the whole thing would work just fine with distances $>k-\frac1n$ or something similar.


Let the period be $ y$, then $ |e^{x \plus{} y} \minus{} e^{x}| \to 0$ when $ x \to \minus{} \infty$, impossible, so there is no such period.

I was thinking like this:
Let $ p$ be a period.
then$ sin(e^{x\plus{}p})\equal{}sin(e^x)$
$ \implies$ $ e^{x\plus{}p}\equal{}n\pi\plus{}(\minus{}1)^ne^x$ for all real $ x$
The above is true for $ x\equal{}0,\minus{}p$
Putting $ x\equal{}0$
$ e^{p}\equal{}n\pi\plus{}(\minus{}1)^n$
Putting $ x\equal{}\minus{}p$,
$ 1\equal{}n\pi\plus{}(\minus{}1)^ne^{\minus{}p}$
$ \implies$ $ e^{\minus{}p}\equal{}\frac{(1\minus{}n\pi)}{(\minus{}1)^n}$

Therefore
$ 1\equal{}e^{\minus{}p}.e^{p}\equal{}\frac{(n\pi\plus{}(\minus{}1)^n)(1\minus{}n\pi)}{(\minus{}1)^n}
\equal{}((\minus{}1)^nn\pi\plus{}1)(1\minus{}n\pi)\equal{}(\minus{}1)^nn\pi\plus{}(\minus{}1)^{n\plus{}1}n\pi\minus{}n\pi\plus{}1$
$ \implies$ $ (\minus{}1)^nn\pi\plus{}(\minus{}1)^{n\plus{}1}n^2\pi^2\minus{}n\pi\equal{}0$
$ \implies$ $ n\pi\equal{}(\minus{}1)^{n\plus{}1}\plus{}1$
And this is impossible for an integer $ n$
Hence by contradiction , we prove $ sin(e^x)$ is not periodic

[quote="Xaenir"]I was thinking like this:
Let $ p$ be a period.
then$ sin(e^{x \plus{} p}) \equal{} sin(e^x)$
$ \implies$ $ e^{x \plus{} p} \equal{} n\pi \plus{} ( \minus{} 1)^ne^x$ for all real $ x$
The above is true for $ x \equal{} 0, \minus{} p$
Putting $ x \equal{} 0$
$ e^{p} \equal{} n\pi \plus{} ( \minus{} 1)^n$
Putting $ x \equal{} \minus{} p$,
$ 1 \equal{} n\pi \plus{} ( \minus{} 1)^ne^{ \minus{} p}$
$ \implies$ $ e^{ \minus{} p} \equal{} \frac {(1 \minus{} n\pi)}{( \minus{} 1)^n}$

Therefore
$ 1 \equal{} e^{ \minus{} p}.e^{p} \equal{} \frac {(n\pi \plus{} ( \minus{} 1)^n)(1 \minus{} n\pi)}{( \minus{} 1)^n} \equal{} (( \minus{} 1)^nn\pi \plus{} 1)(1 \minus{} n\pi) \equal{} ( \minus{} 1)^nn\pi \plus{} ( \minus{} 1)^{n \plus{} 1}n\pi \minus{} n\pi \plus{} 1$
$ \implies$ $ ( \minus{} 1)^nn\pi \plus{} ( \minus{} 1)^{n \plus{} 1}n^2\pi^2 \minus{} n\pi \equal{} 0$
$ \implies$ $ n\pi \equal{} ( \minus{} 1)^{n \plus{} 1} \plus{} 1$
And this is impossible for an integer $ n$
Hence by contradiction , we prove $ sin(e^x)$ is not periodic[/quote]

As $ n$ can be $ 0$, I think you should state that when $ n\equal{}0$, $ e^p\equal{}1$, impossible. There is no mistake but a small typo (the 5th row from the bottom) in your solution except this .

Actually there are many different ways to prove this weak result.

For example, let the period (if exist) be $ y$ and $ e^y \equal{} k \neq 1$ , then it leads to $ sin(kx) \equal{} sin x$ for all $ x$, which is easy to check to be impossible because of their different first derivatives.


calculus
integration
trigonometry
function
complex analysis
real analysis
real analysis unsolved

using complex variables (especially for sylow_theory  :rotfl: ) or otherwise, show :D

for $ 0 < m\plus{}1 < n$ and $ m,\,n\in\mathbb N^{\plus{}}$, and $ a\in\mathbb R$,

$ \textbf{(i)}\;\;\int_{0}^{\infty}\;\;\frac{x^{m}}{ x^{n}\plus{}a^{n}}\;\;\textbf dx\;\;\equal{}\;\;\boxed{\frac{\pi\cdot a^{m\plus{}1\minus{}n}}{n\cdot\sin\Big(\frac{m\plus{}1}{n}\pi\Big)}}\;\;\;\quad\text{(probably has been shown before)}$

and, for $ 0 < m\plus{}1 < n\cdot p$ and $ m,\,n,\,p\in\mathbb N^{\plus{}}$, and $ a\in\mathbb R$,

$ \textbf{(ii)}\;\;\int_{0}^{\infty}\frac{x^{m}}{(x^{n}\plus{}a^{n})^{p}}\;\;\textbf dx\;\;\equal{}\;\;\boxed{\frac{(\minus{}1)^{p\minus{}1}\cdot\pi\cdot a^{m\plus{}1\minus{}n\cdot p}\cdot\Gamma\Big(\frac{m\plus{}1}{n}\Big)}{n\cdot(p\minus{}1)!\cdot\sin\Big(\frac{m\plus{}1}{n}\pi\Big)\cdot\Gamma\Big(\frac{m\plus{}1}{n\minus{}p\plus{}1}\Big)}}$

here is solution. (without complex ...)

http://www.mathlinks.ro/Forum/viewtopic.php?t=137064
http://www.mathlinks.ro/Forum/viewtopic.php?t=146820

[quote="misan"]using complex variables (especially for sylow_theory  :rotfl: ) or otherwise, show :D

for $ 0 < m\plus{}1 < n$ and $ m,\,n\in\mathbb N^{\plus{}}$, and $ a\in\mathbb R$,

$ \textbf{(i)}\;\;\int_{0}^{\infty}\;\;\frac{x^{m}}{ x^{n}\plus{}a^{n}}\;\;\textbf dx\;\; \equal{}\;\;\boxed{\frac{\pi\cdot a^{m\plus{}1\minus{}n}}{n\cdot\sin\Big(\frac{m\plus{}1}{n}\pi\Big)}}\;\;\;\quad\text{(probably has been shown before)}$
[/quote]
(Thanks for having so much faith in me. :))

1)First it doesn't work if $ a\equal{}0$ so I will do it by assuming you meant to say $ a\in\mathbb{R}^{\plus{}}$.

2)Let $ \gamma$ be the following contour. 1)It is a horizontal line on the real line from $ 0$ to $ R$. 2)It turns in a contor-clockwise manner along $ Re^{i\theta}$ but only by $ \frac{2\pi}{n}$ radians, i.e. $ 0\leq\theta\leq\frac{2\pi}{n}$. 3)It return back to were it started along the line $ te^{2\pi i/n}$.

3)Before we start using complex analysis notice that $ \int_{0}^{\infty}\frac{x^{m}dx}{x^{n}\plus{}a^{n}}\equal{}\frac{1}{a^{n}}\int_{0}^{\infty}\frac{x^{m}dx}{\left(\frac{x}{a}\right)^{n}\plus{}1}$.
Use the substitution $ \mu \equal{}\frac{x}{a}$ and we get: $ a^{m\minus{}n\plus{}1}\int_{0}^{\infty}\frac{\mu^{m}}{\mu^{n}\plus{}1}d\mu$.
So our strategy is simple, instead find $ \int_{0}^{\infty}\frac{x^{m}}{x^{n}\plus{}1}dx$ by complex analysis and then multiply by $ a^{m\minus{}n\plus{}1}$.

4)Define the meromorphic function $ f(z) \equal{}\frac{z^{m}}{z^{n}\plus{}1}$. Now by the way we chose this contour the only pole occurs at $ z \equal{} e^{\pi i/n}$ which is simple. Now to find the residue note that $ f(z) \equal{}\frac{g(z)}{h(z)}$ and the $ h'(z)\not \equal{} 0$ at $ z\equal{}e^{\pi i/n}$ so $ \mbox{res}(f,e^{\pi i/n}) \equal{}\minus{}\frac{1}{n}e^{\pi i(m\plus{}1)/n}$.

5)By the residue theorem we find that,
$ \int_{0}^{R}\frac{x^{m}}{x^{n}\plus{}1}dx\plus{}\int_{0}^{2\pi/n}\frac{R^{m\plus{}1}ie^{(m\plus{}1)i\theta}d\theta}{R^{n}e^{ni\theta}\plus{}1}\minus{}\int_{0}^{R}\frac{t^{m}e^{2\pi i(m\plus{}1)/n}dt}{t^{n}\plus{}1}\equal{}\minus{}\frac{2\pi i}{n}e^{\pi i(m\plus{}1)/n}$

6)The middle term goes to zero cuz $ m\plus{}1<n$ and so:
$ \int_{0}^{\infty}\frac{x^{m}dx}{x^{n}\plus{}1}\left( 1\minus{}e^{2\pi i(m\plus{}1)/n}\right) \equal{}\minus{}\frac{2\pi i}{n}e^{\pi i(m\plus{}1)/n}$.

7)Divide through,
$ \int_{0}^{\infty}\frac{x^{m}dx}{x^{n}\plus{}1}\left(\frac{e^{\pi i(m\plus{}1)/n}\minus{}e^{\minus{}\pi i(m\plus{}1)/n}}{2}\right) \equal{}\frac{\pi i}{n}$

8)Some hyperbolic identity stuff,
$ \int_{0}^{\infty}\frac{x^{m}dx}{x^{n}\plus{}1}\cdot\sinh\left(\frac{\pi i(m\plus{}1)}{n}\right) \equal{}\frac{\pi i}{n}$.

9)Some more hyperbolic identities,
$ \int_{0}^{\infty}\frac{x^{m}}{x^{n}\plus{}1}dx\cdot i\sin\left(\frac{\pi (m\plus{}1)}{n}\right) \equal{}\frac{\pi i}{n}$.

10)Da taa,
$ \int_{0}^{\infty}\frac{x^{m}}{x^{n}\plus{}1}dx \equal{}\frac{\pi}{n}\cdot\csc\left(\frac{\pi (m\plus{}1)}{n}\right)$

[quote="misan"]
$ \textbf{(ii)}\;\;\int_{0}^{\infty}\frac{x^{m}}{(x^{n}\plus{}a^{n})^{p}}\;\;\textbf dx\;\; \equal{}\;\;\boxed{\frac{(\minus{}1)^{p\minus{}1}\cdot\pi\cdot a^{m\plus{}1\minus{}n\cdot p}\cdot\Gamma\Big(\frac{m\plus{}1}{n}\Big)}{n\cdot(p\minus{}1)!\cdot\sin\Big(\frac{m\plus{}1}{n}\pi\Big)\cdot\Gamma\Big(\frac{m\plus{}1}{n\minus{}p\plus{}1}\Big)}}$[/quote]

Can you do this integral? I don't think the contour I used for $ p\equal{}1$ generalizes.

Assuming that $ a > 0$, let $ x = a\tan^{\frac{2}{n}}\theta$. Using this substitution, you can use beta function identity to have:

\begin{eqnarray*}\int_{0}^{\infty}\frac{x^{m}\, dx}{(x^{n}+a^{n})^{p}}& = &\int_{0}^{\frac{\pi}{2}}\frac{\left(a^{m}\tan^{\frac{2m}{n}}\theta\right)\left(\tfrac{2a}{n}\tan^{\frac{2}{n}-1}\sec^{2}\theta\, d\theta\right)}{(a^{n}\tan^{2}\theta+a^{n})^{p}}\\ & = &\frac{2a^{m+1-np}}{n}\int_{0}^{\frac{\pi}{2}}\sin^{\frac{2(m+1)}{n}-1}\theta\cos^{2p-\frac{2(m+1)}{n}-1}\theta\, d\theta\\ & = &\frac{a^{m+1-np}}{n}B\left(\frac{m+1}{n}, p-\frac{m+1}{n}\right)\\ & = &\frac{a^{m+1-np}}{n}\frac{\Gamma\left(\frac{m+1}{n}\right)\Gamma\left(p-\frac{m+1}{n}\right) }{\Gamma(p)}.\end{eqnarray*}

I think this is not a rigorous one, but maybe someone can justify it as long as he/she is willing to do... and it would agree with misan's result for positive integer $ p$.


abstract algebra
group theory
superior algebra
superior algebra theorems

Let $ B \equal{} B_1 \oplus B_2$ be a finite abelian group and let $ \mu_1 : B_1 \rightarrow B$ , $ \mu_2 : B_2 \rightarrow B$

be injective homomorhisms. If $ \mu_1(B_1) \cap \mu_2(B_2) \equal{} 0$, then the mapping $ \mu : B \rightarrow B$ defined by

$ \mu(b_1 \plus{} b_2 ) \equal{} \mu_1 (b_1) \plus{} \mu_2 (b_2)$, $ b_1 \in B_1$, $ b_2 \in B_2$ is an automorphism of B

Hint: Why is it sufficient to show that $ \mu$ is injective? Then show that $ \mu$ is indeed injective.


Prove that a network $ G \equal{} (V,A)$, with a finite number of vertices $ V$ and set of arcs $ A$ has infinite flow if and only if there exists a forward path from the source $ s$ to sink $ t$ with all its arcs having infinite capacity.

Seems pretty obvious  :maybe: That's why I think my solution is wrong  :P
[hide]Let's assume that the network doesn't have a forward path from s to t with all its arc having infinite capacity. Then every path from s to t would have a finite flow. Thus the flow of the network will be finite. So only if we have at least one infinite path, can we have an infinite flow.

Now, let's assume that we have a forward path in G with infinite capacity. Since this "infinite path" would have the maximum flow the flow of the network is infinite.[/hide]


integration
limit
real analysis
real analysis unsolved

Let $X=(C[0,1],\parallel \cdot\parallel ), \parallel x\parallel =max_{t\in [0,1]}|x(t)|$ and $U: X\to X$,
$U(x)(s)=\int_{0}^{1}{stx(t)}dt,s\in [0,1].$  Compute or prove that is true or false:
a) $\parallel U\parallel =?$

b) U is compact

c) U is an injection

d) $\frac{1}{3}$ is an eigenvalue for U

e) the spectrum of U=?

f) $\lim_{n\to \infty}U^{n}(x)=?$

g) $\lim_{n\to \infty}\sqrt[n]{U^{n}(x)}=?$

h) $x-U(x)=y$ has an unic solution for every $y\in C_{[0,1]}$

i) the ecuetion $3U(x)=x$ has solution different from 0.

If I understand U correctly :) it is just a linear functional $f\colon x\mapsto \int_{0}^{1}tx(t)\,dt$ followed by an isometric embedding of $\mathbb R$ into $C[0,1]$ via $\alpha\mapsto \alpha s$. Of course, $U$ is compact and is not an injection. Its norm is equal to the norm of $f$, which is $\int_{0}^{1}t\,dt=1/2$. I'll leave d)-i) for someone else.


algebra
polynomial
superior algebra
superior algebra solved

$K$ field characteristic $\neq 2$
$P \in K[X,Y]$ polynomial symetric $P(X,Y)=P(Y,X)$
Suppose that $P$ is divisible by $X-Y$
Prove that $P$ is divisible by $(X-Y)^2$

The condition that $K$ doesn't have characteristic $2$ simply says $-1\ne 1$. Let $P(X,Y)=(X-Y)Q(X,Y)$. We get $Q(Y,X)=-Q(X,Y)$, so we must have $Q(X,Y)=\sum_{i\le j}a_{ij}(X^iY^j-X^jY^i)$ (where finitely many of the coefficients are non-$0$, of course, in order for it to be a polynomial). Each of the summands is divisible by $X-Y$, so $Q$ must be divisible by $X-Y$ too.

However, I think it works when the characteristic of $K$ is $2$ as well, because $Q$ is symmetric, and every symmetric polynomial over such a field is divisible by $X-Y$ (which is the same as $X+Y$).

What am I doing wrong?

[quote="grobber"]... $Q(Y,X)=-Q(X,Y)$...[/quote]

$Q$ is antisymetric $Q(X,Y)+Q(Y,X)=0$ so $2Q(X,X)=0$ characteristic is $\neq 2$
$Q(X,X)=0$ , $Q(X,Y)=(X-Y)R(X,Y)$

About characteristic 2: $X+Y=X-Y$ is symmetric, divisible by $X-Y$, and not divisible by $(X-Y)^2.$ Or, if you don't like first-order examples, how about $X^3+Y^3=(X+Y)(X^2+XY+Y^2)$ which is also divisible by $X-Y$ but not $(X-Y)^2.$ Does that answer your question, grobber?

Yup.. Thanks. That was actually rather hard to miss :). I didn't think about monomials of the form $X^iY^i$ for some reason (or first order examples, for that matter). Thanks.


Simplify:  $\sin25\sin35\sin85$

Bah, I propose we start posting some real Math on this forum.  Perhaps this teaser will be a nice start.  Anyone up for a problem of the day thing?

preferably simple enough to be on comc or euclid

Okay, here's a homemade problem that I think, is a little bit more approachable.

If $\frac{\tan 22.5+\sin^{2}22.5}{\tan 15+\cos^{2}15}$

=

$\frac{a\sqrt{6}+b\sqrt{2}+c\sqrt{3}+d}{e}$

where a, b, c, d, and e are integers, then calculate $a+b+c+d$.

[hide]

Hint:  This is everyone's favourite number![/hide]

Ah, 42 is a good guess.  But perhaps I should specify that this is everyone's favourite number when thinking emo thoughts. :wink:

Yes, I would love if we all pitched in together on a problem of the day type thing but maybe it would be better if it was a 2 or 3 problems of the day of varying difficulty... For the noobs like myself! (cut us some slack... I haven't prepared for a math contest since the 8th grade I'm a little rusty  :P).

Using half angle formulas, you get $\frac{\tan 22.5+\sin^{2}22.5}{\tan 15+\cos^{2}15}=\frac{9\sqrt{6}-6\sqrt{3}+30\sqrt{2}-20}{73}\implies\boxed{13}$
Masoud Zargar

[quote="lowest_reynolds"]$\sin25\sin35\sin85$[/quote]

$\frac{1}{4}(\sin75+\sin95-\sin25-\sin145)$?
Not sure if that's simpler...

[quote="lowest_reynolds"]Simplify:  $\sin25\sin35\sin85$[/quote]I got it down to
$\left(\cos^{2}5^\circ-\frac{3}{4}\right)\cos5^\circ$.

[hide="how"]by expanding
$\sin(30^\circ-5^\circ)\sin(30^\circ+5^\circ)\cos5^\circ$[/hide]

Is there better?

lol. First time to use this type setting stuff.
$\frac{\sin75^\circ}{4}$

[hide]
$sin25^\circ\sin(60^\circ-25^\circ)sin(60^\circ+25^\circ)$[/hide]

Even simpler, $\frac{1}{8}\sqrt{2+\sqrt{3}}$.

I'll post the next Problem:

[b]Problem:[/b] Prove the following equality:
\[\sin 25^{\circ}\sin 35^{\circ}\sin 85^{\circ}=\frac{1}{8}\sqrt{\cot 5^{\circ}\cot 55^{\circ}\cot 65^{\circ}}\]
Masoud Zargar

I think this is how you do it
EDIT: To the most simplified answer (hopefully) as proposed by andy_tok..the solution before didnt make sense either

[hide]
continuing from where andy_tok left off

$= \frac{\sin75^\circ}{4}$
$= \frac{\sin(30^\circ+45^\circ)}{4}$
$= \frac{\sin(30^\circ)\cos(45^\circ)+sin(45^\circ)\cos(30^\circ)}{4}$
$= \frac{1}{4}({\frac{1}{2}\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}})$
$= \frac{1}{4}(\frac{1+\sqrt{3}}{2\sqrt{2}})$
$= \frac{1+\sqrt{3}}{8\sqrt{2}}$
$= \sqrt{2}\frac{1+\sqrt{3}}{16}$
$= \frac{\sqrt{2}+\sqrt{6}}{16}$

[/hide]

Ahhh, great job guys!  Now I'll have to think of another one...

The radical we have so far can be simplified further though  :wink: .

lol, if u want to simplify more:
$\frac{\sqrt{6}+\sqrt{2}}{16}$

[quote="max_tm"]boxedexe would you mind posting a couple midsteps you used to get that solution?? thx![/quote]
I think it was clear enough.  $\sin 75=\frac{1}{2}\sqrt{2+\sqrt{3}}$.

[quote] $\sin 75=\frac{1}{2}\sqrt{2+\sqrt{3}}$.[/quote]

how did you get $\sin 75=\frac{1}{2}\sqrt{2+\sqrt{3}}$ .. like i mean in that form i thought i got it in that form but then i quickly realized that it didnt make sense from what i was doing...

You use a half angle formula on sin150, which is equivalent to sin30.  Hope that helps! :)

Yup i got it in that form..thx lowest_reynolds

[quote="boxedexe"]
[b]Problem:[/b] Prove the following equality:
\[\sin 25^{\circ}\sin 35^{\circ}\sin 85^{\circ}=\frac{1}{8}\sqrt{\cot 5^{\circ}\cot 55^{\circ}\cot 65^{\circ}}\]
Masoud Zargar[/quote]

lol, use a generalization
[hide="click text"]
$sina^{\circ}\sin(60^{\circ}-a^{\circ})\sin(60^{\circ}+a^{\circ})=\frac{sin(3a^{\circ})}{4}$
$cosa^{\circ}\cos(60^{\circ}-a^{\circ})\cos(60^{\circ}+a^{\circ})=\frac{cos(3a^{\circ})}{4}$
[/hide]

nice generalization the problem is pretty basic now i guess

I simplified it down but i still have like 22 sin terms left..how is this actually solved??

Show me what you have got.
PS:My solution has 3 main steps

[quote="andy_tok"]Ya, it is.
Try this one:

$\sin(1^\circ)\sin(2^\circ)\sin(3^\circ)...\sin(89^\circ)$[/quote]

I know this is not very strong....
But the problem asks what the product of the sines is if the interval (0,pi/2) is subdivided into n terms (in the case of the problem, 89).  I'm looking for a pattern from small values of n. When n=1, sin(pi/4)=(sqrt2)/2
When n=2, sin(pi/6)sin(pi/3)=(sqrt3)/4
When n=3, sin(pi/8)sin(pi/4)sin(3pi/8)=(sqrt4)/8
When n=4, sin(pi/10)sin(pi/5)sin(3pi/10)sin(2pi/5)=(sqrt5)/16

So without bothering to do the induction, I'm guessing that when n=89, sin(1)sin(2)sin(3)...sin(89)=3(sqrt10)/(2^89).

It's not a real solution, but if this was an AIME problem, that's not a bad way to get it.
Unless I'm wrong, of course. :maybe:

ya, your answer is right.
A little Typo here:
[quote]
When n=3, sin(pi/8)sin(pi/4)sin(3pi/4)=(sqrt4)/8[/quote]
3pi/8

Thanks, I fixed it.

To do it the real way, you would start

sin(1)sin(2)sin(3)...
=cos(1)cos(89)cos(2)cos(88)
And since cosAcosB=(cos(A+B)+cos(A-B))/2
cos(1)cos(89)...
=sqrt2/(2^45)cos2cos4cos8...cos88
=sqrt2/(2^67)cos2cos6cos10...cos86
And I'm stuck.

After Andy TOk gve me that tip i did it exactly the same way as he did above....

hmmmm....
here's a nice one, but requires some strong insights and not just trig identities...

Evaluate the sum:
\[\tan^{2}1^\circ+\tan^{2}3^\circ+\tan^{2}5^\circ+...+\tan^{2}89^\circ \]

[hide="hint1"]the sum is an integer
[/hide]
[hide="hint2"]use demoiver's theorem and vieta's theorem
[/hide]


Given $ a,b,c>0$ and$ a\plus{}b\plus{}c\equal{}3$.Prove that:
$ \frac{2b\plus{}c}{\sqrt{a^2\plus{}2b\plus{}c}}\plus{}\frac{2c\plus{}a}{\sqrt{b^2\plus{}2c\plus{}a}}\plus{}\frac{2a\plus{}b}{\sqrt{c^2\plus{}2a\plus{}b}} \le \frac{9}{2}$ :)

We only need to prove:
$ \sum\ \frac{a^2}{a^2\plus{}2b\plus{}c} \geq\ \frac{3}{4}$ (obviously trues by CauchySchwarts and Vasc's result)  :)

[quote="nguoivn"]We only need to prove:
$ \sum\ \frac {a^2}{a^2 \plus{} 2b \plus{} c} \geq\ \frac {3}{4}$ (obviously trues by CauchySchwarts and Vasc's result)  :)[/quote]
Yes :lol:

[quote="nguoivn"]We only need to prove:
$ \sum\ \frac {a^2}{a^2 \plus{} 2b \plus{} c} \geq\ \frac {3}{4}$ (obviously trues by CauchySchwarts and Vasc's result)  :)[/quote]

I also solved it using Cauchy-Schwarz this way

$ \sum a^2(a^2\plus{}2b\plus{}c) \sum\ \frac {a^2}{a^2 \plus{} 2b \plus{} c}\geq (\sum a^2)^2$

and then using Vasc ineq

$ (\sum a^2)^2 \geq 3\sum a^3b$

[quote="manlio"][quote="nguoivn"]We only need to prove:
$ \sum\ \frac {a^2}{a^2 \plus{} 2b \plus{} c} \geq\ \frac {3}{4}$ (obviously trues by CauchySchwarts and Vasc's result)  :)[/quote]

I also solved it using Cauchy-Schwarz this way

$ \sum a^2(a^2 \plus{} 2b \plus{} c) \sum\ \frac {a^2}{a^2 \plus{} 2b \plus{} c}\geq (\sum a^2)^2$

and then using Vasc ineq

$ (\sum a^2)^2 \geq 3\sum a^3b$[/quote]
Nice,manlio :lol:


[quote="April"]In a circle of diameter $ 1$, some chords are drawn. The sum of their lengths is greater than $ 19$. Prove that there is a diameter intersecting at least $ 7$ chords.[/quote]
It's quite funny this problem :) Mostly because $ 7$ is the wrong number, because the person who made the problem made an error in calculation. (All the team leaders did too, so it wasn't discovered, when we got it at the competition...!)

Anyway divide the circle in $ n$ points $ P_1,P_2,..,P_n$ such that the lie on the circle an $ |P_1P_2| \equal{} |P_2P_3| \equal{} ... \equal{} |P_nP_1|$. And let $ n \to \infty$. Give every point $ P_i$ a value and let it be $ a_i$. Initially let it be $ a_1 \equal{} a_2 \equal{} ... \equal{} a_n \equal{} 0$. Then every time a chord is drawn between two points on a circle, then increment $ a_i$ by $ 1$ for every point $ P_i$ which's diameter intersect the chord. Then we easily obtain $ \frac{\sum_{i\equal{}1}^n a_i}{n} > 2\frac{19}{\pi} \equal{} \frac{38}{\pi} > 12$. Since $ \pi$ is the perimeter. So by dirichlets principle there exists $ a_i$ such that $ a_i > 12$, and thus $ a_i \ge 13$. So the diameter through $ P_i$ intersects at least $ 13$ chords!!


We are given 5040 balls in k different colors, where the number of balls of each color is the same. The balls are put into 2520 bags so that each bag contains two balls
of different colors. Find the smallest k such that, however the balls are distributed into the bags, we can arrange the bags around a circle so that no two balls of the same
color are in two neighboring bags.

$ 2520 = \text{lcm}(2, 3, ... 10)$.  I suspect that the answer is $ 11$, but I don't have time to give an argument now.

i think "however" should be "whenever"    i will come back with a solution if i figure that out; nice problem!!!  :idea:


geometry rhombus rotation limit trigonometry

[hide="are you sure that"] $\theta=\displaystyle\frac{n}{2\pi}$?
cause $\displaystyle\lim_{n\to\infty}\theta=\infty$...
[/hide]
[hide="if it is a typo"]
$\theta=\displaystyle\frac{2\pi}{n}$
So there are $n$ rhombi with angle $\theta$, with side length $1$
The area of one rhombus is $A=\displaystyle\frac{(2\cos{\frac{2\pi}{2n}})(2\sin{\frac{2\pi}{2n}})}{2}=\displaystyle\frac{\text{(length of longer diagonal)}\text{(length of shorter diagonal)}}{2}$
$A=2\cos{\frac{2\pi}{2n}}\sin{\frac{2\pi}{2n}}=\sin{\frac{2\pi}{n}}$ by double angle formula for sin
Area of the whole star is $nA=n\sin{\frac{2\pi}{n}}$
\begin{eqnarray*} \displaystyle\lim_{n\to\infty}n\sin{\frac{2\pi}{n}}&=&\lim_{n\to\infty}\frac{\sin{\frac{2\pi}{n}}}{\frac{1}{n}}\\ &=&\lim_{n\to\infty}\frac{(\cos{\frac{2\pi}{n}})(\frac{-2\pi}{n^2})}{\frac{-1}{n^2}}\text{ by LHopital}\\ &=&\lim_{n\to\infty}2\pi\cos{\frac{2\pi}{n}}\\ &=&2\pi \end{eqnarray*}
[/hide]

[quote="Hexaditidom"]A rhombstar is formed by taking a rhombus with sidelength 1 and acute angle A = theta = n/360 (where n is an int >= 5). Then an image of ABCD is rotated about A, theta degrees. Then another image of ABCD is rotated about A, 2theta degrees. This process is done n-1 times. The end result is an n-pointed star.

As n approaches infinity, what does the area of the figure approach?[/quote]
How can $n\rightarrow \infty$? ($\theta$ must be less than $90$ becuase its acute). I think you meant to say $\theta=\frac{360}{n}$?

i think he is trying to get a regular polygon, draw the diagonals of the individual rhombi, the area bounded by the diagonals is clearly half of the total figure, and also approaches a circle , so $2\pi$ is the total area

[hide]Model a circle as many infinitely thin isoceles triangles. Compare this to the rhombstar:
[img]http://img.photobucket.com/albums/v194/mukmasterxps/circle.gif[/img]

We see that the area of the rhombstar is half the area of the circle.

As theta gets smaller, the radius of the circle approaches two, and the area is half the area of the circle, or $\frac{1}{2}\pi 2^2=\boxed{2\pi}$
[/hide]

You're all correct, you got the idea :yup: