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Nathan Urban science forum beginner
Joined: 16 Jul 2006
Posts: 2

Posted: Thu Jul 20, 2006 12:48 am Post subject:
Function from Taylor series?



I have a series that looks like this:
g(x) = a_0 f(0) + a_1 f''(0) x^2 + a_2 f^(4)(0) x^4 + ...
where f(x) is an arbitrary even function and a_i are coefficients.
This looks like a Taylor series expansion of f(x) about x=0, except
that the a_i are not the usual 1/n! for Taylor series.
I would like to know if, given the a_i, one can find a closedform
expression for g(x), expressed somehow in terms of the unknown
function f(x), for which the above series is the Taylor expansion?
For concreteness, the actual coefficients I'm looking at are a_i = 1/S_i,
where S_i is the ith term in Series A110468 from the Encyclopedia
of Integer Sequences, "Convolution of (1)^n*n! and n!". 

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Robert B. Israel science forum Guru
Joined: 24 Mar 2005
Posts: 2151

Posted: Thu Jul 20, 2006 9:45 pm Post subject:
Re: Function from Taylor series?



In article <e9mjtd$5pc$1@dizzy.math.ohiostate.edu>,
Nathan Urban <nurban@psu.edu> wrote:
Quote: 
I have a series that looks like this:
g(x) = a_0 f(0) + a_1 f''(0) x^2 + a_2 f^(4)(0) x^4 + ...
where f(x) is an arbitrary even function and a_i are coefficients.
This looks like a Taylor series expansion of f(x) about x=0, except
that the a_i are not the usual 1/n! for Taylor series.
I would like to know if, given the a_i, one can find a closedform
expression for g(x), expressed somehow in terms of the unknown
function f(x), for which the above series is the Taylor expansion?

Obviously it will depend on the a_i.
We have f(z) = sum_{j=0}^infty f^(2j)(0)/(2j)! z^(2j).
If A(z) = sum_{j=0}^infty a_j (2j)! z^{2j}, and both of these
series have positive radii of convergence r_f and r_A respectively,
then you can write g as a contour integral
g(z) = (2 pi i)^(1) int_{C_r} A(w) f(z/w) dw/w
where C_r is a circle of radius r centred at the origin and oriented
counterclockwise, r < r_A and z < r_A r_f.
Quote:  For concreteness, the actual coefficients I'm looking at are a_i = 1/S_i,
where S_i is the ith term in Series A110468 from the Encyclopedia >of
Integer Sequences, "Convolution of (1)^n*n! and n!". 
In this case a_i = (i+1)/(2i+1)!, so
A(z) = sum_{j=0}^infty (j+1)/(2j+1) z^(2j)
= (arctanh(z)/z + 1/(1  z^2))/2
with radius of convergence 1.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada 

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