From: Gareth.McCaughan@pobox.com (Gareth McCaughan)
Subject: Re: any good approximation of this??
Date: Mon, 28 Aug 2000 23:43:49 +0100
Newsgroups: sci.math.num-analysis
Summary: [missing]

Tom Wood wrote:

> is there a good approximation of the following product?
> 
> a   a+1   a+2 	      a+K+1
> - * --- * --- *... * -------
> b   b+1   b+2 	      b+K+1
> 
> in terms of a,b,and K?  here a,b are real and b>a>0, and K is a positive 
> integer.

The Euler-Maclaurin summation formula is your friend.

sum from p..q-1 of f(x) = integral from p..q of f(x) dx
                          + sum from 1..m of Bk/k! (f^{k-1}(q)-f^{k-1}(p))
                          + remainder

where f^r is the r'th derivative of f, and Bk is the k'th Bernoulli number.
So, stick in f(x)=log((a+x)/(b+x)), p=0, q=K+2.

The remainder unfortunately *doesn't* -> 0 in your case as m->oo.
But for fixed m, it is well-behaved as K->oo. And for moderate m
and sensible values of the other parameters, it's pretty small.

Here are some sample results. I hope they're self-explanatory.

a=5.7d0 b=5.8d0 n=100

    g0 0.7473743714034967d0
    g1 0.741253867655732d0
    g2 0.7410675978263289d0
    g4 0.7410677861867584d0
    g6 0.7410677845585493d0
    g8 0.741067784595498d0

    f  0.7410681550121796d0

a=4.3d0 b=16.8d0 n=100

    g0 7.780560904231502d-14
    g1 4.1655251478539496d-14
    g2 4.1062423206120274d-14
    g4 4.1062893387141985d-14
    g6 4.1062888958308613d-14
    g8 4.106288908675789d-14

    f  4.1063812458291915d-14

a=5.3d0 b=67.7d0 n=10000

    g0 1.9657301945829872d-157
    g1 5.517185479384265d-158
    g2 5.437805372218202d-158
    g4 5.437839175994781d-158
    g6 5.437838969596783d-158
    g8 5.43783897353324d-158

    f  5.437905778567048d-158

a=5.3d0 b=5.4d0 n=10000

    g0 0.470772967914622d0
    g1 0.46639592003372576d0
    g2 0.46626013856019854d0
    g4 0.4662602966985406d0
    g6 0.466260295119473d0
    g8 0.4662602951608676d0

    f  0.4662606053617043d0

-- 
Gareth McCaughan  Gareth.McCaughan@pobox.com
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