Problemas Teoremas

Maio 6, 2012

Two Mathematics Stack Exchange Questions and Answers on the Gamma Function

Filed under: Cálculo,Função Gama,Funções Especiais,Integrais,Integrais impróprios,Matemática,Math,Mathematics Stack Exchange — Américo Tavares @ 11:49 pm
Tags: , ,

I post here together two answers of mine on MSE both on the gamma function.

Question by pomme. Definition of the gamma function

“I know that the Gamma function with argument (-\dfrac{1}{ 2}) — in other words \Gamma(-\dfrac{1}{2}) is equal to -2\pi^{1/2}. However, the definition of

\Gamma(k)=\displaystyle\int_0^\infty t^{k-1}e^{-t}dt

but how can \Gamma(-\frac{1}{2}) be obtained from the definition? WA says it does not converge…”

My answer

Your doubt makes sense if for k=-1/2 you try using the definition of the gamma function by the integral you have written, because it diverges at k=-1/2 as you stated. I will try to clarify if as follows. This integral representation of the gamma function (I use x instead of k)

\Gamma (x)=\displaystyle\int_{0}^{\infty }t^{x-1}e^{-t}dt\qquad(0)

holds in the reals if and only if x>0. Using integration by parts we can show that

\Gamma (x+1)=x\Gamma (x).\qquad(1)

For x<0 we can define \Gamma (x) for all negative values of x<0 except -1,-2,-3,\cdots not by the integral (0), rather by means of the functional equation (1) in the form

\Gamma (x)=\dfrac{\Gamma(x+1)}{x}.\qquad(2)

Then x+1>0 [and] \Gamma (x+1) is convergent. In your example x=-1/2, so x+1=1/2 and \Gamma (1+x)=\Gamma (1/2). Hence we obtain

\Gamma (-1/2)=\dfrac{\Gamma(1/2)}{-1/2}=-2\Gamma (1/2)=-2\sqrt{\pi },\qquad(3)

where we use the known value of the integral \Gamma(1/2)=\sqrt{\pi }, which can be evaluated, e.g. from the equality

\displaystyle\int_{0}^{\infty }\displaystyle\int_{0}^{\infty }e^{-x^{2}-y^{2}}dxdy=\dfrac{\pi }{4}=\left( \displaystyle\int_{0}^{\infty }e^{-x^{2}}dx\right) ^{2}.\qquad(4)

Similarly, for x=-3/2 by (2) we find \Gamma (-3/2)=-\frac{2}{3}\Gamma (-1/2), using (3) twice. This process is called analytic continuation, but its true understanding requires knowledge of complex analysis.

    \text{Plot of }y=\Gamma(x)\quad -5<x<5

Question by Amitabh Udayiman. Convergence of this integral

“My statistics text book prescribed by my school states that the integral

\Gamma(n)=\displaystyle\int_{0}^{\infty}e^{-x}x^{n-1}dx

is convergent for n>0.It does not prove the statement. So can anyone please help me prove it? Thanks again!”

My answer. I assume that n is a real number. Split the gamma improper integral

\Gamma(n)=\displaystyle\int_{0}^{\infty}e^{-x}x^{n-1}dx\qquad(0)

into I_1+I_2, where

I_1=\displaystyle\int_{0}^{1}e^{-x}x^{n-1}dx\qquad(1)

and

I_2=\displaystyle\int_{1}^{\infty}e^{-x}x^{n-1}dx\qquad(2)

1. To prove that the integral I_2 is always convergent use the fact that for any real number \alpha the integral

\int_{1}^{\infty }e^{-x}x^{\alpha }dx\qquad(3)

is convergent, by the limit comparison test

\displaystyle\lim_{x\rightarrow \infty }\dfrac{e^{-x}x^{\alpha }}{x^{-2}}=0\qquad(4)

with the convergent integral

\displaystyle\int_{1}^{\infty }\dfrac{dx}{x^{2}}.\qquad(5)

2. As for I_1 consider two cases. (a) If n\geq 1 observe that \lim_{x\rightarrow 0}e^{-x}x^{n-1}=0, so I_1 is a proper integral. (b) If 0<n<1, the integrand e^{-x}x^{n-1} behaves like x^{n-1} near n=0, because e^{-x}\rightarrow 1 as x\rightarrow 0. Since

\displaystyle\int_{0}^{1}\dfrac{dx}{x^{1-n}}\qquad(6)

is convergent if and only if 1-n<1, i.e. n>0, so is I_1. It follows that \Gamma(n)=I_1+I_2 is convergent for n>0.

Março 23, 2012

Evaluating zeta function at 6, ζ(6)

Filed under: Análise de Fourier,Matemática,Math,Mathematics Stack Exchange,Séries — Américo Tavares @ 12:08 pm
Tags: ,

Here is my answer to this Mathematics Stach Exchange question  by Chon  on how to compute \zeta(6).

I have posted here in Portuguese a recursive method based on the computation of the Fourier trigonometric series expansion for the function defined in \left[ -\pi ,\pi \right] by f(x)=x^{2p} and extended to all of {\mathbb R} periodically with period 2\pi. This is a shorter description than the original. In this reply I outline the case \zeta(4). For p=3 the expansion is

x^{6}=\dfrac{\pi ^{6}}{7}+2\displaystyle\sum_{n\ge 1}^{}\left( \left( \dfrac{6}{n^{2}}\pi ^{4}-\dfrac{120}{n^{4}}\pi ^{2}+\dfrac{720 }{n^{6}}\right)\cos n\pi \right)\cos nx.\qquad (1)

The computation is as follows:

f(x)=x^{2p}=\dfrac{a_{0,2p}}{2}+\displaystyle\sum_{n=1}^{\infty }\left( a_{n,2p}\cos nx+b_{n,2p}\sin nx\right),

where the coefficients are given by the following integrals

\begin{aligned}a_{0,2p}&=\dfrac{1}{\pi }\displaystyle\int_{-\pi }^{\pi }x^{2p}\;\mathrm{d}x=\dfrac{2\pi ^{2p}}{2p+1}\\a_{n,2p}&=\dfrac{1}{\pi }\displaystyle\int_{-\pi }^{\pi }x^{2p}\cos nx\;\mathrm{d}x=\dfrac{2}{\pi }\displaystyle\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x,\\b_{n,2p}&=\dfrac{1}{\pi }\displaystyle\int_{-\pi }^{\pi }x^{2p}\sin nx\;\mathrm{d}x=0.\end{aligned}

The series expansion is thus

x^{2p}=\dfrac{\pi ^{2p}}{2p+1}+\dfrac{2}{\pi }\sum_{n=1}^{\infty }\cos nx\displaystyle\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x.\qquad(2)

For f(\pi )=\pi ^{2p} we obtain

\begin{aligned}\pi ^{2p}=\dfrac{\pi ^{2p}}{2p+1}+\dfrac{2}{\pi }\displaystyle\sum_{n=1}^{\infty }\cos n\pi\displaystyle\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x, \end{aligned}

where the integral

\begin{aligned}I_{n,2p}:=\displaystyle\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x \end{aligned}

satisfies the following recurrence, as can be shown by integration by parts

\begin{aligned} I_{n,2p}=\dfrac{2p}{n^{2}}\pi ^{2p-1}\cos n\pi -\dfrac{2p(2p-1)}{n^{2}} I_{n,2\left( p-1\right) },\qquad I_{n,0}=0.\qquad(3) \end{aligned}

For p=1, we get

\begin{aligned} I_{n,2}=\frac{2p}{n^{2}}\pi ^{2p-1}\cos n\pi \end{aligned}

and

\begin{aligned} \pi ^{2} &=\dfrac{\pi ^{2}}{3}+\dfrac{2}{\pi }\displaystyle\sum_{n=1}^{\infty }\cos n\pi \cdot I_{n,2}\\&=\dfrac{\pi ^{2}}{3}+\dfrac{2}{\pi }\displaystyle\sum_{n=1}^{\infty }\cos n\pi \left(\dfrac{2}{n^{2}}\pi \cos n\pi \right)\\&=\dfrac{\pi ^{2}}{3}+4\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{n^{2}}\\&\Rightarrow\zeta (2)=\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{n^{2}}=\dfrac{\pi ^{2}}{6}.\end{aligned}

For p=2, we get

\begin{aligned}I_{n,4}=\left( \dfrac{4\pi ^{3}}{n^{2}}-\dfrac{24\pi }{n^{4}}\right) \cos n\pi\end{aligned}

and

\begin{aligned}\pi ^{4}&=\dfrac{\pi ^{4}}{5}+\dfrac{2}{\pi }\displaystyle\sum_{n=1}^{\infty }\cos n\pi\cdot I_{n,4}=\dfrac{\pi ^{4}}{5}+\dfrac{4\pi ^{4}}{3}-48\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{n^{4}}\\ &\Rightarrow\zeta (4)=\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{n^{4}}=\dfrac{\pi ^{4}}{ 90}.\end{aligned}

Finally for p=3, we get

\begin{aligned}I_{n,6}=\left( \dfrac{6\pi ^{5}}{n^{2}}-\dfrac{120\pi ^{3}}{n^{4}}+\dfrac{720}{n^{6}}\right) \cos n\pi\end{aligned}

and

\begin{aligned}\pi ^{6}=\dfrac{\pi ^{6}}{7}+2\displaystyle\sum_{n=1}^{\infty }\left( \dfrac{6\pi ^{4}}{n^{2}}-\dfrac{120\pi ^{2}}{n^{4}}+\dfrac{720}{n^{6}}\right),\end{aligned}

from which the result follows

\zeta(6)=\begin{aligned}\displaystyle\sum_{n=1}^{\infty }\dfrac{1}{n^{6}}=\dfrac{\pi ^{6}}{945}.\end{aligned}

Plots of the periodic function defined in \left[ -\pi ,\pi \right] by f(x)=x^{6} (blue curve) and of the partial sum with the first 10 terms of its Fourier trigonometric series (red curve).

This method generates recursively the sequence (\zeta(2p))_{p\ge 1}.

Julho 18, 2011

Tall fraction puzzle, MSE

Filed under: Matemática,Math,Mathematics Stack Exchange — Américo Tavares @ 11:19 am
Tags: , , ,

Question by Ross Millikan:

“ I was given this problem 30 years ago by a coworker, posted it 15 years ago to rec.puzzles, and got a solution from Barry Wolk, but have never seen it again.  Consider the series:

1, \dfrac{1}{2},\dfrac{\dfrac{\ 1\ }{2 }}{\dfrac{3}{ 4}},\dfrac{\ \dfrac{\ \dfrac{\ 1\ }{2}\ }{ \dfrac{\ 3\ }{4}}\ }{\ \dfrac{\ \dfrac{\ 5\ }{6}\ }{\dfrac{\ 7\ }{8}}\ },\ldots

I don’t know how to format it to show the larger fraction bars, but you can guess, particularly with what follows.  Each fraction keeps its large bars while being put atop a similar structure. [Improved formatting AT]

This can also be represented as \dfrac{1\cdot 4 \cdot 6 \cdot 7 \dots}{2 \cdot 3 \cdot 5 \cdot 8 \dots} terminating at 2^n for some n, where it is much closer to the limit than elsewhere.

The challenge:

1)Find the limit, not too hard by experiment

2)In the last expression, find a simple, nonrecursive, expression to say whether n is in the numerator or denominator

3)Prove the limit is correct-this is the hard one. “

My answer:

“ This problem (E 2692) was proposed by D. Woods in Americ. Math. Monthly 85, No.1, p.48, (link needs a subscription) and a solution by E. Robbins was published in Americ. Math. Monthly 86, No. 5, p.394f, (link needs a subscription) in 1979. A solution from 1987 by Jean-Paul Allouche is given in Proposition 5 of Jean-Paul Allouche and Jeffrey Shallit’s paper The ubiquitous Prouhet-Thue-Morse sequence (or here slides 24-28).

In 3. apart from a sketch of Allouche and Shallit’s proof of Proposition 5, I give my interpretation why the limit can be expressed as the infinite product \displaystyle\prod_{m=0}^{\infty }\left( \dfrac{2m+1}{2m+2}\right)^{(-1)^{t_{m}}}, where \left( t_{m}\right) _{m\geq 0} is the Prouhet-Thue-Morse sequence. This product is the starting point of their proof.

1. The first few terms of this sequence are

\begin{aligned}\left( f_{n}\right) _{n\geq 0}=\left( 1,\dfrac{1}{2},\dfrac{2}{3},\dfrac{7}{10},\dfrac{286}{405},\dfrac{144\,305}{204\,102},\dfrac{276\,620\,298\,878}{391\,202\,754\,597},\ldots \right)\end{aligned}

These numerical values suggest that \left( f_{n}^{2}\right) _{n\geq 0} converges relatively fast to \dfrac{1}{2}, and thus f_{n} to \dfrac{\sqrt{2}}{2}:

\begin{aligned} \left( f_{n}^{2}\right) _{n\geq 0}=\left(1, 0.25,0.444\,44,0.49,0.498\,68,0.499\,88,0.499\,99,\ldots \right) \end{aligned}

2. The Prouhet-Thue-Morse sequence (A010060) OEIS page gives the closed form formula (already  in Eelvex’s answer) by Benoit Cloitre (benoit7848c(AT)orange.fr), May 09 2004.

3. The term f_{n} can be written as the product of integers 1\leq k\leq 2^{n} raised to e_{k}\in \left\{ -1,+1\right\}. For instance,

\begin{aligned}f_{3} &=\dfrac{\ \dfrac{1}{2}/\dfrac{3}{4}\ }{\dfrac{5}{6}/\dfrac{7}{8}}=\dfrac{1}{2}\left( \dfrac{3}{4}\right) ^{-1}\left( \dfrac{5}{6}\left( \dfrac{7}{8}\right)^{-1}\right) ^{-1}=1\cdot 2^{-1}\cdot 3^{-1}4\cdot 5^{-1}\cdot 6\cdot 7\cdot 8^{-1}\\&=\prod_{k=1}^{2^{3}}k^{e_{k}}=\prod_{k=1}^{2^{3}}k^{(-1)^{t_{k-1}}}\end{aligned},

where \left( t_{k}\right) _{k\geq 0}=\left( 0,1,1,0,1,0,0,1,\ldots \right) is the binary sequence known as the Prouhet-Thue-Morse sequence (A010060), which has several equivalent definitions. One that is related directly to the way the numbers k exchange between numerators and denominators, in other words, whether the exponent e_{k}=(-1)^{t_{k-1}} is +1 or -1, is the following. Let A_{k} be a sequence of strings of 0′s and 1′s with length 2^{k}, with A_{0}=0. For k\geq 0, A_{k+1}=A_{k}\overline{A}_{k}, where \overline{A}_{k} is obtained from A_{k} by interchanging 0′s and 1′s. Then \left( t_{k}\right) _{k\geq 0} is the infinite sequence generated by A_{k} as k\rightarrow \infty . It has the following property: t_{2m}=t_{m} and t_{2m+1}=1-t_{m} for m\geq 0. Thus t_{2m}+t_{2m+1}=1 and since t_{k}\in \left\{ 0,1\right\} , one of t_{2m+2}, t_{2m+1} is 0 and the other is 1. In terms of the exponents we have e_{2m+1}=(-1)^{t_{2m}}=(-1)^{t_{m}} and e_{2m+2}\ e_{2m+1}=(-1)^{t_{2m}+t_{2m+1}}=-1. This means that one of the integers 2m+1 and 2m+2 is in the numerator and the other in the denominator, which is in accordance with the way how the tall fraction is constructed from fractions \frac{1}{2},\frac{2}{3},\frac{4}{5},\ldots . Similarly, we have in general (when k runs from 1 to 2^{n}, m varies from 0 to 2^{n-1}-1.)

\begin{aligned}f_{n} &=\prod_{k=1}^{2^{n}}k^{e_{k}}=\prod_{k=1}^{2^{n}}k^{(-1)^{t_{k-1}}}\\&=\prod_{m=0}^{2^{n-1}-1}\left( 2m+1\right)^{(-1)^{t_{2m}}}\left( 2m+2\right)^{(-1)^{t_{2m+1}}}=\prod_{m=0}^{2^{n-1}-1}\left( \frac{2m+1}{2m+2}\right)^{(-1)^{t_{m}}} \end{aligned}

and we want to evaluate the limit of the sequence f_{n}

\begin{aligned}\underset{n\rightarrow \infty }{\lim }f_{n}=\prod_{m=0}^{\infty }\left(\dfrac{2m+1}{2m+2}\right) ^{(-1)^{t_{m}}}.\qquad(\ast )\end{aligned}

In Proposition 5 of the mentioned paper, the authors show that

\begin{aligned}\underset{n\rightarrow \infty }{\lim }f_{n}=\prod_{m=0}^{\infty }\left(\dfrac{2m+1}{2m+2}\right) ^{(-1)^{t_{m}}}=\dfrac{1}{2}\prod_{m=0}^{\infty }\left(\dfrac{2m+1}{2m+2}\right) ^{(-1)^{t_{2m+1}}} \end{aligned}

and, since (-1)^{t_{2m+1}}=-(-1)^{t_{2m}}=-(-1)^{t_{m}}, they get

\begin{aligned} \underset{n\rightarrow \infty }{\lim }f_{n}=\dfrac{1}{2\ \underset{n\rightarrow\infty }{\lim }f_{n}},\end{aligned}

thus proving that \underset{n\rightarrow \infty }{\lim }f_{n}^{2}=\dfrac{1}{2}. The trick they use is to multiply both sides of \left( \ast \right) by the auxiliary product

\begin{aligned}\prod_{m=1}^{\infty }\left( \frac{2m}{2m+1}\right) ^{(-1)^{t_{m}}}\qquad(\ast \ast ) \end{aligned}

pretty much as in Moron’s answer. Concerning the issue of the convergence of the infinitive products,  namely \left( \ast \right) and (\ast\ast ) the authors state that they “are convergent by Abel’s theorem”, but I must confess I have no idea which theorem is this.* “

* See this answer by Plop to this question of mine.

Outubro 26, 2010

Novo design de Mathematics Stack Exchange

Filed under: Divulgação,Matemática,Math,Mathematics Stack Exchange — Américo Tavares @ 12:15 am
Tags: , , ,

Terminada a fase beta  foi lançado o novo design de Mathematics SE – divulgado anteriormente neste blogue. Pode ver o anúncio feito, no site,  neste post.

Outubro 16, 2010

Problema do mês :: Problem of the month #7

Filed under: Combinatória,Combinatorics,Matemática,Matemática Discreta,Math,Problem,Problem Of The Month,Problema do mês,Problemas — Américo Tavares @ 6:11 pm
Tags: ,

pdmpom20101017


Mostre que :: Show that

\displaystyle\sum_{n=k+1}^{N}\dfrac{(-1)^{n+k}n}{\displaystyle\binom{n}{k}\displaystyle\binom{n+k}{k}}-\displaystyle\sum_{n=k+1}^{N}\dfrac{(-1)^{n+k}n}{\displaystyle\binom{n-1}{k}\displaystyle\binom{n-1+k}{k}}

=\dfrac{k}{\displaystyle\binom{2k}{k}}+\dfrac{(-1)^{N+k-1}k}{\displaystyle\binom{N}{k}\displaystyle\binom{N+k}{k}}

 

Soluções: até 8 Novembro 2010, via acltavares@sapo.pt ou caixa de comentários.

Solutions: until November 8, 2010, via acltavares@sapo.pt or comment box.

Outubro 11, 2010

Tradução do problema U151 de Mathematical Reflections e da minha solução

Filed under: Matemática,Math,Mathematical Reflections,Problem,Problemas — Américo Tavares @ 4:20 pm

Issue 2, 2010 — Mathematical Reflections — U151

« Seja n um número positivo e seja

f(x)=x^{n+8}-10x^{n+6}+2x^{n+4}-10x^{n+2}+x^{n}+x^{3}-10x+1.

Determine f(\sqrt{2}+\sqrt{3}). »

Proposto por Dorin Andrica,

Babeş-Bolyai University, Cluj-Napoca, Roménia

Enunciado original:

” Let n be a positive integer and let

f(x)=x^{n+8}-10x^{n+6}+2x^{n+4}-10x^{n+2}+x^{n}+x^{3}-10x+1.

Evaluate f(\sqrt{2}+\sqrt{3}). “

Proposed by Dorin Andrica,

Babeş-Bolyai University, Cluj-Napoca, Romania

Tradução da minha resolução (aceite):

Resolução: Se a=\sqrt{2}+\sqrt{3}, b=a^{8}-10a^{6}+2a^{4}-10a^{2}+1 e c=a^{3}-10a+1, então f(\sqrt{2}+\sqrt{3})=f(a)=ba^{n}+c.

Dado que

-10a^{2}=-10\left( \sqrt{2}+\sqrt{3}\right) ^{2}=-20\sqrt{6}-50

2a^{4}=2\left( \sqrt{2}+\sqrt{3}\right) ^{4}=40\sqrt{6}+98

-10a^{6}=-10\left( \sqrt{2}+\sqrt{3}\right) ^{6}=-1980\sqrt{6}-4850

a^{8}=\left( \sqrt{2}+\sqrt{3}\right) ^{8}=1960\sqrt{6}+4801

e

b=a^{8}-10a^{6}+2a^{4}-10a^{2}+1

=\left( 4801-4850+98-50+1\right) +\left( 1960-1980+40-20\right) \sqrt{6}

=0+0\cdot \sqrt{6}

=0,

temos

f(\sqrt{2}+\sqrt{3})=f(a)=ba^{n}+c=c.

Agora calculamos c

c=a^{3}-10a+1

=\left( \sqrt{2}+\sqrt{3}\right) ^{3}-10\left( \sqrt{2}+\sqrt{3}\right) +1

=\sqrt{2}-\sqrt{3}+1.

Logo

f(\sqrt{2}+\sqrt{3})=\sqrt{2}-\sqrt{3}+1.

Outubro 2, 2010

Problema do mês :: Problem of the month #6. (Círculos :: Circles). Resolução :: Solution

Filed under: Geometria,Matemática,Matemática-Secundário,Math,Problem,Problem Of The Month,Problema do mês,Problemas — Américo Tavares @ 11:59 am
Tags: , , , , ,

pdmpom20101003

Problema: Os quatro círculos têm o mesmo raio. Determine-o no caso do triângulo medir 1 m².

Resolução de Jacques Glorieux (minha tradução):

Seja r o raio dos círculos. Visto que DE=EF=2r tem-se \measuredangle DEF=\pi /4. Como DF é parlalelo  a AB, o ângulo \measuredangle CAB=\pi /4. Assim, o ângulo \measuredangle ABG=\pi /2. Logo HBIF é um quadrado de lado r. Tem-se

BF=r\sqrt{2}.

BC=BF+FC=r\sqrt{2}+3r=r\left( 3+\sqrt{2}\right) .

AG=2\times BC=2r\left( 3+\sqrt{2}\right) .

A área S de ABG é, portanto,  r^{2}\left( 3+\sqrt{2}\right) ^{2}. Mas esta área S=1. Por este motivo

r=\dfrac{1}{3+\sqrt{2}} =\dfrac{3-\sqrt{2}}{7}

Outras resoluções por: josejuan , Prof. Paulo Sérgio.


Problem: The four circles have equal radius. Find it if the size of the triangle is 1 m².


Solution by Jacques Glorieux:

Let r be the radius of the circles. We have DE=EF=2r thus \measuredangle DEF=\pi /4. As DF is parallel to AB, angle \measuredangle CAB=\pi /4. Thus angle \measuredangle ABG=\pi /2. Thus HBIF is a square of side r. We have

BF=r\sqrt{2}.

BC=BF+FC=r\sqrt{2}+3r=r\left( 3+\sqrt{2}\right) .

AG=2\times BC=2r\left( 3+\sqrt{2}\right) .

The area S of ABG is thus r^{2}\left( 3+\sqrt{2}\right) ^{2}. But this area S=1. Thus

r=\dfrac{1}{3+\sqrt{2}} =\dfrac{3-\sqrt{2}}{7}

Other solvers:  josejuan , Prof. Paulo Sérgio.

Setembro 19, 2010

Um problema de juros compostos de uma série não uniforme da Universidade de Purdue

Filed under: Cálculo financeiro,Matemática,Problem,Problemas,Purdue University,Séries,Sucessões — Américo Tavares @ 11:49 pm
Tags: , ,

Esta é a tradução do problema Problem No. 2 (Fall 2010 Series) e da minha resolução aceite pela Universidade de Purdue.

« Qual é o montante mais pequeno que deverá investir-se à taxa de juro de i\%, composta anualmente, de maneira a poder levantar-se 1^{2},2^{2}, 3^{2},\ldots dólares no final do ano 1,2,3,\ldots , perpetuamente? (Para i=10, a resposta é 2310 dólares.) »

Transcrição do original

“ What is the smallest amount that may be invested at interest rate i\%, compounded annually, in order that we may withdraw 1^{2},2^{2}, 3^{2},\ldots dollars at the end of the 1st, 2nd, 3rd, … year, in perpetuity? (For i=10, the answer is 2310 dollars.) “

Resolução: O principal resultado que usaremos é o cálculo da soma da série \displaystyle\sum_{n=1}^{\infty }n^{2}x^{n}.

Proposição: se -1<x<1, a série \displaystyle\sum_{n=1}^{\infty }n^{2}x^{n} converge para \dfrac{x\left( 1+x\right) }{\left( 1-x\right) ^{3}}.

Demonstração: Tomemos a seguinte série geométrica, que é convergente para |x|<1:

\displaystyle\sum_{n=1}^{\infty }x^{n}=\dfrac{x}{1-x}\qquad (1)

e diferenciemos ambos os membros \displaystyle\sum_{n=1}^{\infty }nx^{n-1}=\left( 1-x\right) ^{-2}. Agora multipliquemo-los por x: x\displaystyle \sum_{n=1}^{\infty}nx^{n-1}=\sum_{n=1}^{\infty }nx^{n}=x\left( 1-x\right) ^{-2}. Diferenciando novamente, obtemos \displaystyle\sum_{n=1}^{\infty }n^{2}x^{n-1}=\left( 1+x\right) \left( 1-x\right) ^{-3}. Multipliquemos ambos os membros por x e completaremos a demonstração da Proposição:

x\displaystyle\sum_{n=1}^{\infty }n^{2}x^{n-1} =\displaystyle\sum_{n=1}^{\infty}n^{2}x^{n}=\dfrac{x\left( 1+x\right) }{\left( 1-x\right) ^{3}}\qquad (2)

Pondo x=1/c obtemos na forma alternativa, válida para |c|>1,

x\displaystyle\sum_{n=1}^{\infty }\dfrac{n^2}{c^n} =\dfrac{c(c+1)}{(c-1)^{3}}\qquad (3)

Designemos por P o valor actual total da série de levantamentos 1^{2},2^{2}, 3^{2},\ldots dólares, no fim do ano 1, 2, 3,\ldots . O levantamento n^{2} no final do ano n contribui para P no valor de n^{2}/(1+i/100)^{n}, em que i é a taxa de juro (em percentagem) composta anualmente. Sumando todas as contribuições desde n=1 a \infty dá P (no princípio do ano 1),  que é o montante A mais pequeno que é necessário investir-se para equilibrar (A-P=0) os levantamentos como enunciado no problema: P=\sum_{n=1}^{\infty}n^{2}/\left( 1+i/100\right) ^{n}.

Usando (3) com c=1+i/100>1 obtemos o valor actual P(i)=A(i), em dólares, em função da taxa de juro i em percentagem:

A(i)=P(i)=\dfrac{\left( 1+i/100\right) (2+i/100)}{(i/100)^{3}}\qquad (4)

Para i=10, confirmamos que A(10)=P(10)=2310.

Cópia do Texto original

[Correcção gramatical: "alternative form" em vez de "alternatively form"]

* * *

Comentário: Ao iniciar este problema não fazia a mínima ideia de como o iria resolver na prática. De repente consegui associar dois conceitos diferentes: um proveniente da Cálculo financeiro e o outro das Séries, que consegui concretizar na resolução apresentada.

Setembro 15, 2010

Divulgando o Mathematics Stack Exchange

Filed under: Divulgação,Matemática,Math,Mathematics Stack Exchange — Américo Tavares @ 12:13 pm
Tags: , , ,

Exemplos do que respondi e perguntei, neste novo site, em dois casos, num como autor de uma resposta e no outro como autor de uma pergunta.

Pergunta de jkottnauer

“ Adding powers of  i

I’ve been struggling with figuring out how to add powers of i.

For example, the result of i^{3}+i^{4}+i^{5} is 1. But how do I get the result of i^{3}+i^{4}+\ldots +i^{50}? (…) “

Minha resposta

“ Observing that i^{3}+i^{4}+\ldots +i^{50} is a geometric progression with ratio i,  first term i^3 and 50-3+1=48 terms, we have

i^{3}+i^{4}+\ldots +i^{50}=i^{3}\times \dfrac{1-i^{50-3+1}}{1-i}= =i^{3}\times\dfrac{1-i^{48}}{1-i}=i^{2}i\times \dfrac{1-(i^{2})^{24}}{1-i}=

=-i\dfrac{1-(-1)^{24}}{1-i}=-i\dfrac{1-1}{1-i}=0 “

Minha pergunta

“ Generalizing \sum_{n=1}^{\infty }n^{2}/x^{n} to \sum_{n=1}^{\infty }n^{p}/x^{n}

For computing the present worth of an infinite sequence of equally spaced payments (n^{2}) I had the need to evaluate

\displaystyle\sum_{n=1}^{\infty}\frac{n^{2}}{x^{n}}=\dfrac{x(x+1)}{(x-1)^{3}}\qquad x>1.

The method I used was based on the geometric series \displaystyle\sum_{n=1}^{\infty}x^{n}=\dfrac{x}{1-x} differentiating each side followed by a multiplication by x, differentiating a second time and multiplying again by x. There is at least a second (more difficult) method that is to compute the series partial sums and letting n go to infinity.

Question: Is there a closed form for

\displaystyle\sum_{n=1}^{\infty }\dfrac{n^{p}}{x^{n}}\qquad x>1,p\in\mathbb{Z}^{+}?

What is the sketch of its proof in case it exists? “

Uma das respostas: esta de Qiaochu Yuan

Adenda de 17/9/2010: Uma pergunta de Wade

Aunt and Uncle’s fuel oil tank dip stick problem

” (…)

The problem is usually stated in the form of a letter from an Aunt and Uncle:
Dear niece/nephew, How are things going for you and your folks? We hear you are doing quite well it school. Keep it up! Given this success, we were hoping you could help us figure out a little dilemma. As you know, our home is heated by fuel oil, and we have a big tank buried in the side yard. The tank is a cylinder, 20 feet long and 10 feet in diameter, lying on its side five feet deep, with a narrow tube coming to a fill cap at ground level. Your uncle has a 15 foot length of old pipe that we’d like to utilize as a dip stick in order to know when we are getting close to needing a fill-up. We know that 0 feet is empty, 5 feet is half full, and 10 feet is completely full. Trouble is, we don’t know how to mark any other points. We are pretty sure they will not be uniformly spaced. What we really want is to know, within the nearest 0.01 foot, where to mark the dip stick for every multiple of 10% from 0% to 100%. Can you figure this out for us? Of course, we will want to see details of your solution and check it ourselves, and it would especially help if you could draw us a scale model of the dip stick. Love, Auntie Flo and Uncle Jim (…) “

Minha resposta:  “


Upper figure: cylinder oil tank cross-section perpendicular to its horizontal axe. The vertical coordinate is the oil level in percentage.

Lower figure: graph of oil volume/max. volume (in %) versus oil level l (in feet). The horizontal straight lines represent the area/volume ratio A(l)/A(10)=V(l)/V(10) (in %) for every multiple of 10% from 0% to 100%.

Since the tank radius is 5, the oil level with respect to the bottom of the tank is given by l=5-5\cos \dfrac{\theta }{2}, where \theta is the central angle as shown in the figure. The area of the tank cross section filled with oil is A(\theta )=\dfrac{25}{2}\theta -\dfrac{25}{2}\sin \theta

or

A(l)=25\arccos (\dfrac{5-l}{5})-\dfrac{25}{2}\sin (2\arccos (\dfrac{5-l}{5}))

The area ratio A(l)/A(10)=V(l)/V(10) where V(l) is the oil volume.

Let f(l) denote this area ratio in percentage:

f(l)=\dfrac{100}{\pi }\arccos \left( 1-\dfrac{1}{5}l\right) -\dfrac{50}{\pi }\sin \left( 2\arccos \left( 1-\dfrac{1}{5}l\right) \right)

Here are is the the sequence of f(l) for l=0,1,2,\ldots ,10. The graph of f(l) is shown above.

f(0)=0, f(1)=5.2044, f(2)=14.238, f(3)=25.232, f(4)=37.353, f(5)=50,

f(6)=62.647, f(7)=74.768, f(8)=85.762, f(9)=94.796, f(10)=100

(…)

The oil level marks (in feet) should be placed at

0,1.57,2.54,3.40,4.21,

5,5.79,6.60,7.46,8.44,10

corresponding to the oil volume percentage of

0,10,20,30,40,

50,60,70,80,90,100.

This calculation was based on the following f function values:

f(0)=0.0, f(1.5648)=10.0, f(2.5407)=20.0, f(3.40155)=30.0, f(4.21135)=40.0

f(5)=50.0, f(5.7887)=60.0, f(6.59845)=70.000, f(7.4593)=80.000,

f(8.4352)=90.000, f(10)=100.0

Figure of marks:

”

Setembro 12, 2010

Problema do mês :: Problem of the month #6

Filed under: Calculus,Cálculo,Geometria,Matemática,Matemática-Secundário,Math,Problem,Problem Of The Month,Problema do mês,Problemas,Trigonometria — Américo Tavares @ 11:42 pm
Tags: , , , , , ,

pdmpom20100913

Enunciado do Problema

Os quatro círculos têm o mesmo raio. Determine-o no caso do triângulo medir 1 m².

  • Serão bem-vindas soluções até ao fim do mês, comentando ou por email: acltavares@sapo.pt

Problem Statement

The four circles have equal radius. Find it if the size of the triangle is 1 m².

  • Solutions till the end of the month will be welcome, in the comments
    box or via e-mail:     acltavares@sapo.pt

Setembro 8, 2010

Série de termo geral n²/xⁿ :: Series of general term n²/xⁿ

Filed under: Análise Matemática,Calculus,Cálculo,Exercícios Matemáticos,Matemática,Math,Problem,Problemas,Séries — Américo Tavares @ 9:37 pm
Tags: , , ,

Demostre que para x>1, x\in\mathbb{R} a função f(x)=\displaystyle\sum_{n=1}^{\infty }\dfrac{n^{2}}{x^{n}} é convergente e determine, justificando, a sua expressão analítica, na forma de fracção racional f(x)=\dfrac{P(x)}{Q(x)}.

Sugestão: utilize a série geométrica de razão x e 1.º termo 1, diferencie e multiplique por x duas vezes.

Prove that for x>1, x\in\mathbb{R} the function f(x)=\displaystyle\sum_{n=1}^{\infty }\dfrac{n^{2}}{x^{n}} converges and find, with proof, its analytical expression in the form of a rational function f(x)=\dfrac{P(x)}{Q(x)}.

Hint: use the geometrical series with ratio x and first term 1, differentiate and multiply by x twice.

Agosto 25, 2010

Solução do Puzzle trigonométrico :: Trigonometric Puzzle Solution

Filed under: Matemática,Matemática-Secundário,Math,Trigonometria — Américo Tavares @ 6:39 pm
Tags: , , ,

Puzzle trigonométrico :: Trigonometric Puzzle

Sejam n um inteiro positivo e f(x) uma função trigonométrica. Determine n e f(x) tais que:

\displaystyle 2f\left( \frac{\pi }{n}\right) =\sqrt{1-\frac{1}{2}\sqrt{2}}+\sqrt{1+\frac{1}{2}\sqrt{2}}

Let n be a positive integer and f(x) some trigonometric function. Find n and f(x) such that:

\displaystyle 2f\left( \frac{\pi }{n}\right) =\sqrt{1-\frac{1}{2}\sqrt{2}}+\sqrt{1+\frac{1}{2}\sqrt{2}}.

 

Solução de Jacques Glorieux ::  Solution by Jaques Glorieux:

\displaystyle\sqrt{1-\frac{1}{2}\sqrt{2}}+\sqrt{1+\frac{1}{2}\sqrt{2}}

=\displaystyle\sqrt{\cos 0-\cos \frac{\pi }{4}}+\sqrt{\cos 0-\cos \frac{\pi }{4}}

=\displaystyle\sqrt{-2\sin \left( \frac{1}{2}\left( 0+\frac{\pi }{4}\right) \right) \sin\left( \frac{1}{2}\left( 0-\frac{\pi }{4}\right) \right) } +\displaystyle\sqrt{2\cos \left( \frac{1}{2}\left( 0+\frac{\pi }{4}\right) \right) \cos\left( \frac{1}{2}\left( 0-\frac{\pi }{4}\right) \right) }

=\displaystyle\sqrt{2\sin ^{2}\frac{\pi }{8}}+\sqrt{2\cos ^{2}\frac{\pi }{8}}

=\displaystyle\sqrt{2}\left( \sin \frac{\pi }{8}+\cos \frac{\pi }{8}\right)

=\displaystyle2\left( \frac{\sqrt{2}}{2}\sin \frac{\pi }{8}+\frac{\sqrt{2}}{2}\cos \frac{\pi }{8}\right)

=\displaystyle2\left( \sin \frac{\pi }{4}\sin \frac{\pi }{8}+\cos \frac{\pi }{4}\cos\frac{\pi }{8}\right)

=\displaystyle2\cos \left( \frac{\pi }{4}-\frac{\pi }{8}\right)

=\displaystyle2\cos \left( \frac{\pi }{8}\right)

\displaystyle\implies f(x)=\cos x \wedge n=8

Página Seguinte »

Tema: Rubric. Blog em WordPress.com.