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Math Forum / Mathematics / Undergraduate Math / October 2008Caracterización de una función racional por su des arrollo en serie de potencias
Excusem me for using Spanaish instead English. Nevertheless, I'll try to translate this question to English if anybody want. Thanks. Es sabido que si f : C--->C es una función racional de variable compleja, su desarrollo en serie de potencias en un punto z=a de su dominio tiene la propiedad de que los coeficientes de este desarrollo constituyen una sucesión linealmente recurrente. ¿Es cierto el recíproco? Es decir, si el desarrollo de la función es SUM (a(n) (z-a)^n), de n=0 a N=inf, verifica que a(n) es linealmente recurrente, ¿Puede afirmarse que la función es racional? Como caso particular, propongo obtener la suma de la serie SUM (a(n) (z-a)^n), de n=0 a N=inf, donde a(n) es la sucesión de Fibbonacci: a(n)=(1,1,2,3,5,8,...) In article <2d294f8e-2a08-4382-8660-69641ae800b5@t65g2000hsf.googlegroups.com>, J Antonio Torné <jatorne@gmail.com> wrote: Perdóneme para usar el inglés en vez del español. > Excusem me for using Spanaish instead English. Nevertheless, I'll try > to translate this question to English if anybody want. Thanks. [quoted text clipped - 7 lines] > SUM (a(n) (z-a)^n), de n=0 a N=inf, verifica que a(n) es linealmente > recurrente, ¿Puede afirmarse que la función es racional? Yes, I think so. I do not have a reference however. You will see it mentioned at <http://en.wikipedia.org/wiki/Rational_function>. > Como caso particular, propongo obtener la suma de la serie SUM (a(n) > (z-a)^n), de n=0 a N=inf, donde a(n) es la sucesión de Fibbonacci: > a(n)=(1,1,2,3,5,8,...) I do not know what rational function is represented by the power series with the Fibonacci numbers as coefficients - it looks difficult. If you do a Google search on "rational functions fibonacci" you will get many references. I wish I could be of more help.  SignaturePaul Sperry Columbia, SC (USA) On Mon, 06 Oct 2008 02:26:13 -0400, Paul Sperry <plsperry@sc.rr.com> wrote in <news:061020080226135129%plsperry@sc.rr.com> in alt.math.undergrad: > In article > <2d294f8e-2a08-4382-8660-69641ae800b5@t65g2000hsf.googlegroups.com>, J > Antonio Torné <jatorne@gmail.com> wrote: > Perdóneme para usar el inglés en vez del español. >> Excusem me for using Spanaish instead English. >> Nevertheless, I'll try to translate this question to >> English if anybody want. Thanks. [...] >> Como caso particular, propongo obtener la suma de la >> serie SUM (a(n) (z-a)^n), de n=0 a N=inf, donde a(n) es >> la sucesión de Fibbonacci: a(n)=(1,1,2,3,5,8,...) > I do not know what rational function is represented by the > power series with the Fibonacci numbers as coefficients - > it looks difficult. If you do a Google search on > "rational functions fibonacci" you will get many > references. > I wish I could be of more help. Let F(z) = SUM[n >= 0; a(n) (z - a)^n] = 1 + (z - a) + 2 (z - a)^2 + 3(z - a)^3 + 5(z - a)^4 + ... . Then F(z) + (z - a) F(z) = SUM[n >= 0; a(n) (z - a)^n] + SUM[n >= 0; a(n) (z - a)^(n+1)] = SUM[n >= 0; a(n) (z - a)^n] + SUM[n >= 1; a(n-1) (z - a)^n] = 1 + SUM[n >= 1; (a(n) + a(n-1) (z - a)^n] = 1 + SUM[n >= 1; a(n+1) (z - a)^n] Assume that a(n) = 0 for n < 0. Then a(n) = a(n-1) + a(n-2) for all n except n = 0. Let c(0) = 1 and c(n) = 0 if n != 0. Then a(n) = a(n-1) + a(n-2) + c(n) for all n. Let F(z) = SUM[a(n) (z - a)^n], taken over all n. Then F(z) = SUM[a(n) (z - a)^n] = SUM[(a(n-1) + a(n-2) + c(n)) (z - a)^n] = SUM[a(n-1) (z - a)^n] + SUM[a(n-2) (z - a)^n] + 1. Now SUM[a(n-1) (z - a)^n] = (z - a) SUM[a(n-1) (z - a)^(n-1)] = (z - a)F(z), and similarly SUM[a(n-2) (z - a)^n] = (z - a)^2 F(z), so F(z) = (z - a) F(z) + (z - a)^2 F(z) + 1, and we can solve for F(z) to get F(z) = 1/[1 - (z - a) - (z - a)^2] (if I haven't made any computational errors -- I'm not awake yet). Brian > On Mon, 06 Oct 2008 02:26:13 -0400, Paul Sperry > <plspe...@sc.rr.com> wrote in [quoted text clipped - 52 lines] > > Brian Hello, Brian. Brave! Very interesting your method. I say you the same thing I said Paul: Thank you, for your answer. If you want, I can send you a work in which I worked about linear recurrence sequences, where I study this question, and others relationed whit it. Excuse my bad English. I tried this question in another way. The Fibonacci sequence a(n)=(1,1,2,3,..) verifies a(n+2)=a(n+1)+a(n), or a(n+2)-a(n+1)-a(n)=0. Then, I asociate it the polinomial x^2-x-1, ( Let´s say x is the sequence operator x[a(n)]=a(n+1), and then x^2 is the operator x[ a(n) ]=a(n+2) ) , which anulates the Fibonacci sequence. Calling its coefficients c(0)=-1, c(1)=-1, c(2)=1, it can be prove that: SUM [ n>=0, a(n) z^n ] = [ (a(0)c(1)+a(1)c(2)) z +a(0) c(2) ] / ( c(2)+c(1) z+ c(0) z^2 )= 1/ (1-z-z^2) Now, we can substitute z by z-a. The result is valid for every linear sequence as coefficients. All we need is an anulator polinomial for the suquence. So, you were right. Regards, JA Torné > In article > <2d294f8e-2a08-4382-8660-69641ae80...@t65g2000hsf.googlegroups.com>, J [quoted text clipped - 32 lines] > Paul Sperry > Columbia, SC (USA) Thank you, Paul, for your answer. If you want, I can send you a work in which I worked about linear recurrence sequences, where I study this question, and others relationed whit it. Excuse my bad English. Regards, JA Torné |
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