Re: Truncation of infinite series

From: Virgil (ITSnetNOTcom#virgil_at_COMCAST.com)
Date: 10/08/04 

Date: Thu, 07 Oct 2004 23:09:29 -0600

In article <snn9d.88311$He1.62739@attbi_s01>,
 Anonymous <nospam@noISP.com> wrote:

> (this was posted in sci.math.num-analysis a while ago, with no useful
> responses)
>
> I am doing many computations of the infinite series
>
> S = SUM { i^(2/3) * x^i } i=1,... and 0 < x < 1
>
> To evaluate this series, I want to truncate after some number of terms, n. n
> is to be
> chosen so the error, E(n), in replacing the infinite sum with the finite sum
> is less than
> a fixed fraction of the infinite sum. In other words, E(n)/S < eps, where eps
> = constant.
>
> The ratio between successive terms of the series is r = (i+1/i)^(2/3) * x, so
> the absolute
> error of truncating at the n-th term, u_n, is
>
> E(n) <= (r * u_n ) / (1-r)
>
> How can I determine n to limit the relative error? There must be something
> simple I am
> missing here.
>
> sherNOwoodSPAM@computer.org (remove caps to get e-mail)

The relative error is the absolute error divided by the correct sum.

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