Exercise « Problemas Teoremas

Março 29, 2010

Solução do Desafio — área entre círculos :: Challenge solution — area between circles

Filed under: Exercícios Matemáticos,Exercise,Geometria,Matemática,Matemática-Secundário,Math,Trigonometria — Américo Tavares @ 1:34 pm
Tags: Desafios, Exercícios, Matemática, Matemática-Secundário

Enunciado do desafio :: Challenge Statement

Tem cinco moedas iguais, cada uma a tocar nas duas adjacentes. No espaço que fica no meio coloca uma moeda tangente às cinco.

Qual é a área total entre as moedas em função do diâmetro $d$ das maiores?

Five equal coins are placed in such a way that each of them is tangent to the two adjacents ones. Another coin tangent to those five is put in the centre.

Find the total area between the coins as a function of the diameter $d$ of each one of the outer five.

* * *

Transcrição com diferente formatação (em LaTeX) da Resolução de RSCS publicada no blogue Simplesmente porque sim . . . : [corrigido autor da solução]

Transcription in a different format (using LaTeX) and translation by myself of the Solution by RSCS posted in the blog Simplesmente porque sim . . . : [solver corrected]

Área do pentágono / Area of the pentagon

$A_{P}=\dfrac{d^{2}}{4}\sqrt{25+10\sqrt{5}}$

Área do círculo / Area of the circle

$A_{C}=\pi\dfrac{d^{2}}{4}$

[Soma dos] ângulos internos do pentágono: $540^\circ=3\pi$ / [Sum of the] pentagon internal angles: $540^\circ=3\pi$

Raio da moeda mais pequena / Radius of the small coin

$\dfrac{d}{2}\cdot\dfrac{1}{\cos 54^\circ}-\dfrac{d}{2}$

$[$ Área do círculo pequeno / Area of the small circle

$A_{c}=\pi\left( \dfrac{d}{2}\cdot\dfrac{1}{\cos 54^\circ}-\dfrac{d}{2}\right) ^{2}\qquad$ $]$

Área Final / Final Area

$A_{P}-5A_{C}\times\dfrac{108}{360}-A_{c}$

* * *

Uma nota minha: Substituindo os resultados parciais na área final indicada na resolução, obtive, em função de $d^2$ , para a área pedida:

A note of mine: Replacing the parcial results into the final area indicated in the above solution, I got the requested area as a function of $d^2$ :

$A=\dfrac{1}{4}\left( \sqrt{25+10\sqrt{5}}-\left( \dfrac{8}{5-\sqrt{5}}-\dfrac{4\sqrt{2}}{\sqrt{5-\sqrt{5}}}+\dfrac{5}{2}\right) \pi \right) d^{2}$

ou seja / i. e.

$\dfrac{A}{d^{2}}=\dfrac{\sqrt{25+10\sqrt{5}}}{4}-\left( \dfrac{2}{5-\sqrt{5}}-\dfrac{\sqrt{2}}{\sqrt{5-\sqrt{5}}}+\dfrac{5}{8}\right) \pi$

Usei, entre outras relações, esta / Among others I used this identity:

$\cos 54^\circ=\sin (90^\circ-54^\circ)=\sin 36^\circ=\sin (\pi /5)$

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Março 5, 2010

Duas Questões de Exame de Introdução à Análise Complexa: Contribuição do Prof. Paulo Sérgio

Filed under: Análise Complexa,Cálculo,Exames,Exercícios Matemáticos,Exercise,Integrais,Integrais impróprios,Matemática,Math,Problem,Problemas,Transformadas de Laplace — Américo Tavares @ 1:25 pm
Tags: Exames, Exercícios, Integração, Matemática, Problema

Publico a resolução apresentada pelo Prof. Paulo Sérgio (nestes dois comentários) às duas questões seguintes.

Duas questões de Análise Complexa (listadas também aqui)

« Nota de 27-5-2009: o blogue echoone deixou de estar disponível.

Passagem do blogue

http://echoone.wordpress.com/, entrada Introductory Complex Analysis Final

(tradução e adaptação do inglês). »

« (…) Demonstre que as equações de Cauchy-Riemann se escrevem em coordenadas polares

$u_r=\dfrac{v_\theta}{r}$

$v_r=-\dfrac{u_\theta}{r}$

(…) Determine o valor de

$\displaystyle\int_{0}^{\infty}\dfrac{\sin^{2}x}{x^2}dx$ (…) »

Resolução :

Seja $f(z)=u+iv$ , satisfazendo $u_{x}=v_{y}$ e $u_{y}=-v_{x}$ .

Sendo

$u=u(x,y)$ , $v=v(x,y)$ ,

$x=x(r,\theta )=r\cos \theta$ e $y=y(r,\theta )=r\sin \theta$ .

Assim,

$u_{r}=u_{x}x_{r}+u_{y}y_{r}=u_{x}\cos \theta +u_{y}\sin \theta$

$=v_{y}\cos \theta -v_{x}\sin \theta =\dfrac{v_{y}r\cos \theta +v_{x}\left( -r\sin \theta \right) }{r}$

$=\dfrac{v_{\theta }}{r}$

A outra é análoga.

A resolução da integral está neste link.

http://img63.imageshack.us/img63/7049/integrald.png

–

Transcrição:

$\displaystyle\int_{0}^{\infty }\dfrac{\sin ^{2}x}{x^2}dx=\dfrac{\pi }{2}$ .

De fato,

$\dfrac{1}{s^{2}}=\mathcal{L}\{x\}=\displaystyle\int_{0}^{\infty }xe^{-sx}dx$ .

Assim,

$\displaystyle\int_{0}^{\infty }\dfrac{\sin ^{2}s}{s^{2}}ds=$ $\displaystyle\int_{0}^{\infty}\displaystyle\int_{0}^{\infty }e^{-sx}x\sin ^2 s\;dx\,ds$ $=\displaystyle\int_{0}^{\infty }x\left( \displaystyle\int_{0}^{\infty }e^{-sx}\sin ^{2}s\;ds\right) dx$

Sendo

$\sin ^{2}s=\dfrac{1-\cos 2s}{2}$ ,

temos:

$\displaystyle\int_{0}^{\infty }\dfrac{\sin ^{2}s}{s^{2}}ds=\dfrac{1}{2}\displaystyle\int_{0}^{\infty}x\left[ \mathcal{L}\{1\}-\mathcal{L}\{\cos \left( 2s\right) \}\right] dx$ $=\dfrac{1}{2}\displaystyle\int_{0}^{\infty }x\left( \dfrac{1}{x}-\dfrac{x}{x^{2}+4}\right) dx$

Logo,

$\displaystyle\int_{0}^{\infty }\dfrac{\sin ^{2}s}{s^{2}}ds=\dfrac{1}{2}\displaystyle\int_{0}^{\infty}\left( \dfrac{4}{x^{2}+4}\right) dx=$ $\arctan \left. \left( \dfrac{x}{2}\right) \right\vert _{0}^{\infty }=\dfrac{\pi }{2}$

* * *

5.04.10 – Notas (de Américo Tavares) sobre a notação aqui utilizada

i. derivadas parciais

Exemplo: $u_{x}=\dfrac{\partial u(x,y)}{\partial x}$

ii. transformada de Laplace

A transformada de Laplace de uma função $F\left( t\right)$ , definida para $t>0$ , é o integral

$\mathcal{L}\left\{ F\left( t\right) \right\} =f\left( s\right) =\displaystyle\int_{0}^{\infty }e^{-st}F\left( t\right) \;dt$

Comentários (2)

Agosto 26, 2009

Three gamma function identities

Filed under: Calculus,Cálculo,Exercícios Matemáticos,Exercise,Função Gama,Funções Especiais,Identidade matemática,Matemática,Math,Problem,Problemas — Américo Tavares @ 1:35 pm
Tags: Matemática, Math, Problema

Let $n=1,2,\ldots$ . Show that

$\sqrt{\pi}\;\Gamma (2n+1)=2^{2n}\Gamma\left( n+\dfrac{1}{2}\right) \Gamma (n+1)\qquad\left( 1\right)$

and

$\sqrt{\pi}\;\Gamma (2n)=2^{2n-1}\Gamma (n)\Gamma\left( n+\dfrac{1}{2}\right)\qquad \left( 2\right)$ .

Let $x\in\mathbb{R}$ . If $x>0$ , show that

$\sqrt{\pi}\;\Gamma (2x)=2^{2x-1}\Gamma (x)\Gamma\left( x+\dfrac{1}{2}\right)\qquad \left( 3\right)$ .

Hints: for the first two identities use the formula proved here. As for the last one evaluate the beta function value B $(x,x)$ and by means of an appropriate change of variable find a relation between B $(x,x)$ and B $\left(x,\dfrac{1}{2}\right)$ .

PS. Listed in the Carnival of Mathematics #56. See pingback in the 1st comment.

Comentários (2)

Março 9, 2008

Putnam problem of the day (by the HMD dated March 1, 2008)

Filed under: Caderno,Exercícios Matemáticos,Exercise,Fracções Contínuas,Matemática,Math,Problem,Problemas,Putnam — Américo Tavares @ 11:33 am
Tags: Caderno, Matemática, Math, Problem, Problema

pdf: included in Caderno (see “caderno” page)

On March 1st, 2008, the Putnam problem of the day displayed on the Harvard’s Math Department site was stated as follows:

“ Evaluate

$\displaystyle\sqrt[8]{2207-\dfrac{1}{2207-\dfrac{1}{2207-\cdots }}}$

Express your answer in the form

$\dfrac{a+b\sqrt{c}}{d}$ ,

where $a,b,c,d$ are integers. “

$\bigskip$

Solution

To evaluate the radicand I start by seeing that the continued fraction

$x=\dfrac{1}{2207-\dfrac{1}{2207-\cdots}}$

satisfies

$x=\dfrac{1}{2207-x}$ .

Thus, since $\dfrac{1}{2}\left( 2207+\sqrt{2207^2-4}\right) \approx 2207,$ the only solution left is

$x=\dfrac{2207-\sqrt{2207^2-4}}{2}$ .

A few algebraic manipulations give

$2207-x=\dfrac{2207+987\sqrt{5}}{2};$

hence

$\displaystyle\sqrt[8]{2207-\dfrac{1}{2207-\dfrac{1}{2207-\cdots}}}=\sqrt[8]{\dfrac{2207+987\sqrt{5}}{2}}.$

In order to have

$\sqrt[8]{\dfrac{2207+987\sqrt{5}}{2}}=\dfrac{a+b\sqrt{c}}{d}$

or equivalently,

$\dfrac{d^8}{2}\left( 2207+987\sqrt{5}\right) =\left( a+b\sqrt{c}\right) ^8,$

with $a,b,c$ integers, $d^8/2$ should also be an integer; therefore $d$ should be even. I assume that $d=2;$ On the other hand $c$ should be $5.$ Thus,

$2^7\left( 2207+987\sqrt{5}\right) =126\,336\sqrt{5}+282\,496=\left( a+b\sqrt{5}\right) ^8$

$\bigskip$

$\displaystyle a+b\sqrt{5}=\sqrt[8]{2^{7}\left( 2207+987\sqrt{5}\right) }$

$\bigskip$

$\displaystyle a=\sqrt[8]{2^{7}\left( 2207+987\sqrt{5}\right) }-b\sqrt{5}.$

$\bigskip$

Since, for $b=2$

$\bigskip$

$\displaystyle a=\sqrt[8]{2^{7}\left( 2207+987\sqrt{5}\right) }-2\sqrt{5}<1$ ,

this possibility is excluded. It remains $b=1$

$\bigskip$

$\displaystyle a=\sqrt[8]{2^{7}\left( 2207+987\sqrt{5}\right) }-\sqrt{5}\approx 5,\,236\,1-2,\,236\,1=3,\,000$

$\bigskip$

Now I confirm

$\displaystyle \left( 3+\sqrt{5}\right) ^{8}=126\,336\sqrt{5}+282\,496.$

$\bigskip$

So, the solution I came was

$\bigskip$

$\displaystyle\sqrt[8]{2207-\dfrac{1}{2207-\dfrac{1}{2207-\cdots }}}=\dfrac{3+\sqrt{5}}{2}.$

$\bigskip$

Remark: The calculation of $\displaystyle \left( 3+\sqrt{5}\right) ^{8}=126\,336\sqrt{5}+282\,496$

can be done by hand as follows

$\displaystyle\left( 3+\sqrt{5}\right) ^{2}=6\sqrt{5}+14$

$\bigskip$

$\displaystyle\left( 3+\sqrt{5}\right) ^{4}=\left( 6\sqrt{5}+14\right) ^{2}=168\sqrt{5}+376$

$\bigskip$

$\displaystyle\left( 3+\sqrt{5}\right) ^{8}=\left( 168\sqrt{5}+376\right)^{2}=126\,336\sqrt{5}+282\,496$

Update March, 20: you can compare with this solution (Putnam 1995, Problem B5 )

Addendum of March 7, 2009: Comment/Proof of the convergence of the continued fraction by Vishal Lama (comment dated March 7, 2009)

“ In the solution presented in your post, $x$ denotes an expression that we can’t assume, beforehand, is a finite number. $x$ may perhaps be infinite! Therefore, the way to go about computing the expression (the infinite continued fraction) given in the problem is as follows.

The infinite continued fraction is defined as the limit of the sequence $(a_n)$ , where $a_0 = 2207$ and $a_n = 2207 - 1/a_{n-1}$ for all $n \geq 1$ . Then, we show that the sequence $a_n$ is bounded from below ( $a_n>2206$ for all $n \geq 0$ , which can be shown by a simple induction) and that it is also strictly decreasing (which can be shown using induction, again). Now, we invoke the Monotone Convergence Theorem to conclude that the sequence does indeed have a (finite) limit, which we can now denote by $x$ . Once we establish that $x$ (which is the infinite continued fraction!) is finite, we can compute $x$ the way you did in your solution. Basically, we have to go through all that trouble just to prove that the given infinite continued fraction is indeed finite! Only after that can the computation begin! “

Part of my reply was: “I did not prove the convergence of the continued fraction. Thanks for doing it.
Basically I assumed that convergence based on a certain numerical evidence, but of course this evidence proves nothing.”

ADDENDUM of May 5, 2010: Alternatively we can prove the convergence applying the Śleszýnki-Pringsheim Theorem (a reference here, wikipedia): if for all naturals $j$ , $\left\vert b_{j}\right\vert \geq \left\vert a_{j}\right\vert +1$ , then the continued fraction $\mathcal{K}_{1}^{\infty }\left( a_{j}/b_{j}\right)$ converges. Since $a_j=-1$ and $b_j=2207$ , the theorem hypothesis is satisfied:

$\left\vert b_{j}\right\vert=2207 \geq 1+1=\left\vert a_{j}\right\vert +1$ .

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