Re: Rational approximation of exponential function

From: Rob Johnson (rob_at_trash.whim.org)
Date: 06/15/04 

Date: Tue, 15 Jun 2004 02:32:06 +0000 (UTC)

In article <cal4kb$gel$1@newslocal.mitre.org>,
lewis@PROBE.mitre.org (Keith A. Lewis) wrote:
>"Stephen J. Herschkorn" <herschko@rutcor.rutgers.edu> writes in article <40CE0381.1070704@rutcor.rutgers.edu> dated Mon, 14 Jun 2004 19:56:53 GMT:
>>>>In article <caetaq$7d2$1@newslocal.mitre.org>,
>>>>lewis@PROBE.mitre.org (Keith A. Lewis) wrote:
>
>>>>>kariwala@ualberta.ca (Vinay Kariwala) writes in article
>>>>><200406112211.i5BMBDa10382@proapp.mathforum.org> dated Fri, 11 Jun 2004
>>>>>23:33:31 +0000 (UTC):
>>>>>>This problem would be trivial for most of you. I am looking for an
>>>>>>approximation of the exponential function, e^(-x) with the following
>>>>>>properties:
>>>>>>
>>>>>>1. The nth order approximation f(x,n) should be rational, i.e. it
>>>>>>should be possible to write f(x,n) as ratio of polynomials in x. f
>>>>>>(x,n) = g(x,n)/h(x,n), where g(x,n) has less number of zeros than h(x,n).
>>>>>>
>>>>>>2. As n approaches infinity, f(x,n) should converge back to the
>>>>>>exponential function.
>
>>>>>Use the Taylor series. The derivative of e^(-x) is -e^(-x), and e^0=1.
>
>>A assume you mean to let g be the nth order Taylor approximation of x
>>|-> exp(-x) and h be indentically 1. Then g has more zeroes than h.
>
>Yes, that's what I meant. And g does indeed have more zeroes in x, although
>I'm having a hard time seeing why you'd even want zeroes in the denominator
>of an approximation for an exponentially decreasing function.

There are times when a rational approximation is nice. For example,
last month, in response to a post by David Cantrell, I used the rational
approximation e^x = (2+x)/(2-x) while estimating n! to get a refinement
of Stirling's formula

                        n n 1/(12n)
    n! ~ sqrt(2 pi n) ( - ) e
                        e

                        n n 24n+1
       ~ sqrt(2 pi n) ( - ) -----
                        e 24n-1

and a refinement of Burnside's formula

                     n+1/2 n+1/2 -1/(24n+12)
    n! ~ sqrt(2pi) ( ----- ) e
                       e

                     n+1/2 n+1/2 48n+23
    n! ~ sqrt(2pi) ( ----- ) ------
                       e 48n+25

If you are interested, see

    http://groups.google.com/groups?threadm=20040504.032925@whim.org

Rob Johnson <rob@trash.whim.org>
take out the trash before replying

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