Problemas Teoremas

Julho 24, 2008

Problem of the Week #2 [from Walking Randomly blog] – Submission

Filed under: Calculus,Cálculo,Demonstração,Integrais,Matemática,Math,Problem,Problemas,Proof — Américo Tavares @ 11:56 am
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I’ve just submitted the following solution for the

« Integral of the Week #2

The second Integral Of The Week (IOTW) is rather different from the first in that I am going to give you the evaluation. Your task is to prove it.

\displaystyle\int_{-\infty}^{\infty}e^{-x^2}\; dx=\sqrt{\pi}

But WAIT! Almost every time I have seen this integral evaluated, it has been done by squaring it and converting to polar co-ordinates and that’s the one method of evaluation you can’t use for this particular challenge. I am looking for more ‘interesting’ proofs. Have fun.

Solutions can be posted in the comments section or sent to me by email (obtaining my email address is another puzzle for you to solve) and will be discussed in a future post. Feel free to send your solution in just about any format you like – plain text, uncompiled Latex, PDF, postscript, Mathematica, ODF, even Microsoft Word. When I get around to posting the solutions I will attempt to standardize them (to PDF probably).

By the way – you still have time to submit a solution for the first IOTW.

July 22nd, 2008 »

Solution

Firstly I show that

\displaystyle\int_{0}^{\infty}xe^{-x^{2}y^{2}}\; dy=I,

where

I=\displaystyle\int_{0}^{\infty}e^{-x^2}\; dx.

Then I evaluate I^2 to derive the value of I by reversing the order of integration.

The second Integral Of The Week equals 2I:

\displaystyle\int_{-\infty}^{\infty}e^{-x^2}\; dx=2I.

If x>0, then

\displaystyle\int_{0}^{\infty}xe^{-x^{2}y^{2}}\; dy=x\displaystyle\int_{0}^{\infty}e^{-{(xy)}^2}\; dy=x\displaystyle\int_{0}^{\infty}e^{-u^2}\dfrac{du}{x}=\displaystyle\int_{0}^{\infty}e^{-u^2}\; du=I.

Thus

I^2=\displaystyle\int_{0}^{\infty}e^{-x^2}\; dx\displaystyle\int_{0}^{\infty}xe^{-x^{2}y^{2}}\; dy =\displaystyle\int_{0}^{\infty}dy\displaystyle\int_{0}^{\infty}xe^{-x^2}e^{-x^{2}y^{2}}\; dx =\displaystyle\int_{0}^{\infty}dy\displaystyle\int_{0}^{\infty}xe^{-x^{2}(1+y^2)}\; dx =\displaystyle\int_{0}^{\infty}dy\dfrac{1}{2\left( 1+y^{2}\right) }\left[ -e^{-x^{2}\left( 1+y^{2}\right) }\right] _{x=0}^{\infty } =\displaystyle\int_{0}^{\infty }\dfrac{-e^{-\underset{x\rightarrow \infty}{\lim }x}+e^{0}}{2\left( 1+y^{2}\right) }\; dy =\displaystyle\int_{0}^{\infty }\dfrac{1}{2\left( 1+y^{2}\right) }\; dy

=\dfrac{1}{2}\left[ \tan^{-1}y\right] _{y=0}^{\infty } =\dfrac{1}{2}\left( \dfrac{\pi}{2}-0\right) =\dfrac{\pi}{4}.

So

I=\dfrac{\sqrt{\pi}}{2}

and

\displaystyle\int_{-\infty}^{\infty}e^{-x^2}\; dx=2I=\sqrt{\pi}.

\square

[Posted in my auxiliary blog Cálculos Auxiliares

http://calcauxprobteor.wordpress.com/2008/07/24/integral-of-the-week-2-walking-randomly/]

[updated at 22:30 GMT: two typos in integrals corrected and explanation improved]

[update of July 26, 2008: \square added]

- : – : – leia o resto »

Abril 24, 2008

Apéry Numbers

Filed under: Demonstração,Matemática,Math,Problem,Proof — Américo Tavares @ 9:22 am
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This is a translation of a previous Portuguese post of mine. 

Show that for all integers n\geq 0, the following identity holds

\displaystyle\sum_{i=0}^{n}\dbinom{n}{i}^{2}\sum_{k=0}^{i}\dbinom{i}{k}\dbinom{n}{k}\dbinom{2n-k}{k}=\sum_{k=0}^{n}\dbinom{n}{k}^{2}\dbinom{n+k}{k}^{2}.

 

Solution

In order to have a short notation and because these numbers are the Apéry Numbers, let´s denote  them A_{n},

A_{n}=\displaystyle\sum_{i=0}^{n}\dbinom{n}{i}^{2}\sum_{k=0}^{i}\dbinom{i}{k}\dbinom{n}{k}\dbinom{2n-k}{k}=\displaystyle\sum_{i=0}^{n}\dbinom{n}{i}^{2}\sum_{k=0}^{n}\dbinom{i}{k}\dbinom{n}{k}\dbinom{2n-k}{k}

 

Since

\displaystyle\sum_{k=0}^{i}\dbinom{i}{k}\dbinom{n}{k}\dbinom{2n-k}{k}=\sum_{k=0}^{n}\dbinom{i}{k}\dbinom{n}{k}\dbinom{2n-k}{k},

 

because for i+1\leq k\leq n every term equals zero and also \dbinom{i}{k}=0.  

Multiplying both sides of the identity proved in my post “Uma proposição da análise combinatória” (here) [see "caderno" in the PDFs page] by

\dbinom{n}{k}\dbinom{2n-k}{k},

 

we have

\displaystyle\binom{n}{k}^{2}\binom{2n-k}{n}^{2}=\binom{n}{k}\binom{2n-k}{k}\sum_{i=k}^{n}\binom{n}{i}^{2}\binom{i}{k}.

 

Now summing in k , we get successively

\displaystyle\sum_{k=0}^{n}\binom{n}{k}^{2}\binom{2n-k}{n}^{2}=\sum_{k=0}^{n}\binom{n}{k}\binom{2n-k}{k}\sum_{i=0}^{n}\binom{n}{i}^{2}\binom{i}{k}\displaystyle=\sum_{k=0}^{n}\sum_{i=0}^{n}\binom{n}{k}\binom{2n-k}{k}\binom{n}{i}^{2}\binom{i}{k}\displaystyle=\sum_{i=0}^{n}\sum_{k=0}^{n}\binom{n}{k}\binom{2n-k}{k}\binom{n}{i}^{2}\binom{i}{k}\displaystyle=\sum_{i=0}^{n}\binom{n}{i}^{2}\sum_{k=0}^{n}\binom{i}{k}\binom{n}{k}\binom{2n-k}{k}

 

But

\displaystyle\sum_{k=0}^{n}\binom{n}{k}^{2}\binom{2n-k}{n}^{2}=\sum_{k=0}^{n}\binom{n}{k}^{2}\binom{2n-k}{n-k}^{2} \displaystyle=\sum_{k=0}^{n}\binom{n}{k}^{2}\binom{n+k}{k}^{2}.

 

Therefore,

\displaystyle A_{n}=\sum_{k=0}^{n}\dbinom{n}{k}^{2}\dbinom{n+k}{k}^{2}

 

which proves the proposition. \qquad\blacktriangleleft

Dezembro 1, 2007

A note on a combinatorial identity related to the Apéry’s constant ζ(3)

Filed under: Caderno,Combinatória,Combinatorics,Demonstração,Matemática,Math,Proof — Américo Tavares @ 10:50 am
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pdf: included in Caderno (see “caderno” page)

Abstract. A purely combinatorial proof of an identity (that can be used to justify one step of the Apéry´s sequences formulas) is established. Equivalent formulas for the Apéry’s double sequences are also presented, as well as the numerical values of the first four lines of these sequences.

Tema: Rubric. Blog em WordPress.com.