From: israel@math.ubc.ca (Robert Israel)
Subject: Re: power series for ln(x+1) for x>1
Date: 26 Jan 1999 00:05:38 GMT
Newsgroups: sci.math
Keywords: Example of Chebyshev polynomial expansion
In article <78gare$92c$1@eagle.cs.unc.edu>, denelsbe@cs.unc.edu (Kevin Denelsbeck) writes:
|> I apologize if this is the wrong newsgroup for this question.
|>
|> I've been trying to find an approximation for log(x+1) for x in the
|> range [0..19]. My first thought is to use the first few terms of a
|> Taylor series. The Taylor series I've found in the usual texts are
|> for ln(x+1) [so far so good] but are restricted to |x| < 1 [darned
|> radius of convergence!]. My calculus is a bit rusty and I feel like
|> I'm missing the "obvious way" of creating a power series that works
|> for bigger values of x. Do I need to use some logarithm tricks?
The Taylor series for log(x+1) in powers of x-9.5 converges for
|x - 9.5| < 10.5. Note that if y = (x - 9.5)/10.5,
log(x+1) = log(10.5 y + 10.5) = log(10.5) + log(y + 1)
so knowing the series for log(y+1) in powers of y will do the trick.
However, if you really want a good polynomial approximation for a
function on a closed interval, try Chebyshev series rather than Taylor
series.
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
==============================================================================
From: kovarik@mcmail.cis.McMaster.CA (Zdislav V. Kovarik)
Subject: Re: power series for ln(x+1) for x>1
Date: 26 Jan 1999 03:01:22 -0500
Newsgroups: sci.math
In article <78gare$92c$1@eagle.cs.unc.edu>,
Kevin Denelsbeck wrote:
>I apologize if this is the wrong newsgroup for this question.
>
Not wrong but one would find more informed advice in
sci.math.num-analysis.
>I've been trying to find an approximation for log(x+1) for x in the
>range [0..19]. My first thought is to use the first few terms of a
>Taylor series. The Taylor series I've found in the usual texts are
>for ln(x+1) [so far so good] but are restricted to |x| < 1 [darned
>radius of convergence!]. My calculus is a bit rusty and I feel like
>I'm missing the "obvious way" of creating a power series that works
>for bigger values of x. Do I need to use some logarithm tricks?
>
>Thanks for any assistance.
>
>Kevin Denelsbeck
Let me first change the variable to what it probably was meant to be.
Define parameters k, q by
k = (20^(1/2) - 1) / (20^(1/2) + 1)
q = (20^(1/4) - 1) / (20^(1/4) + 1)
Given any y in [1, 20], calculate
t = (1/k) * (y - 20^(1/2)) / (y + 20^(1/2))
so that -1 <= t <= 1
Then Chebyshev expansion reads
ln y = (1/2)*ln(20)
+ 4 * sum[n=0 to infinity] (1/(2*n+1)) * q^(2*n+1) * T_(2*n+1)(t)
where T_k is the k-th Chebyshev polynomial:
T_k(t) = cos(k*arccos(t))
and the odd-indexed polynomials have a recursion
T_(-1)(t) =t, T_1(t) = t,
T_(2*k+1)(t) = (4*t^2-2) * T_(2*n-1)(t) - T_(2*n-3)(t).
Note that the series is uniformly dominated by a geometric series with
ratio q^2, which is about 0.1281. Hence it is easy to find an upper bound
for the error of truncating the series after N terms.
For efficient evaluation, a version of Clenshaw Scheme (three-term
analogue of Horner's scheme) can be developed.
Good luck, ZVK(Slavek).