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Article #33Re: tre serie (tres series)
From: Elio Fabri
Date: Mon, 07 Nov 2011 19:51
Date: Mon, 07 Nov 2011 19:51
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1820 bytes
superpollo ha scritto:
> A = sum(1/((2n-1)(3n-1)),n,1,inf)
>
> B = sum(1/((2n-1)(4n-1)),n,1,inf)
>
> C = sum(1/((3n-1)(4n-1)),n,1,inf)
Altra soluzione per B.
B = f(1) dove
f(x) = sum(x^(4n-1)/((2n-1)(4n-1)),n,1,inf)
f(x) = 0.
f'(x) = sum(x^(4n-2)/(2n-1)),n,1,inf) =
(1/2) log(1 + x^2) + (1/2) log(1 - x^2)
(valida per 0<a<1).
f(a) = (1/2) int((log(1+x^2) + log(1-x^2)),0,a).
L'integrale si fa elementarmente (per parti):
f(a) = (1/2) a log(1+a^2) + arctg(a) - (1/2)(1+a) log(1+a) +
(1/2)(1-a) log(1-a).
Passando al limite per a --> 1:
f(1) = pi/4 - (1/2) log(2).
A e C si calcolano in modo simile, anche se gli integrali sono piu'
complicati.
--
Elio Fabri
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From: Elio Fabri <elio.fabri@tiscali.it>
Newsgroups: it.scienza.matematica,es.ciencia.matematicas
Subject: Re: tre serie (tres series)
Date: Mon, 07 Nov 2011 20:51:16 +0100
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