Adaptive numerical limit (and residue) estimation

バージョン 1.1.0.0 (50.5 KB) 作成者: John D'Errico
Numerical extrapolation of a limit (with an error estimate) from only function values
ダウンロード: 2.9K
更新 2015/3/13

ライセンスの表示

LIMEST will find the limit of a general function (specified only for evaluation) at a given point. You might think of limest like quad, but for limits.
While to me this seems to appear more often as a homework exercise than anything else, it was an interesting problem to solve as robustly as possible for a general case.

As an example, I'll use a moderately difficult one that is simple to analyze, but more difficult to deal with numerically.

fun = @(x) (exp(x)-1-x)./x.^2;

This function cannot be evaluated in MATLAB at x = 0, returning a NaN. While a Taylor series expansion shows the limit to be 1/2, the brute force evaluation of fun anywhere near zero results in numerical trash because of the two orders of cancellation.

fun(0)
ans =
NaN

fun(1e-15)
ans =
110223024625156

fun(1e-10)
ans =
827.403709626582

fun(1e-5)
ans =
0.500000696482408

fun(1e-2)
ans =
0.501670841679489

Limest computes the limit, also returning an approximate error estimate.

[lim,err] = limest(fun,0)
lim =
0.499999999681485
err =
2.20308196660258e-09

I've now added the residueEst tool, for computation of the residue of a function at a known pole. For example, here is a function with a first order pole at z = 0

[r,e]=residueEst(@(z) 1./(1-exp(2*z)),0)
r =
-0.5
e =
4.5382e-12

Again, both an estimate of the residue, as well as an uncertainty around that estimate are provided. Next, consider a function with a second order pole around z = pi.

[r,e]=residueEst(@(z) 1./(sin(z).^2),pi,'poleorder',2)

r =
1
e =
2.6336e-11

See the included demos for many other examples of use.

引用

John D'Errico (2024). Adaptive numerical limit (and residue) estimation (https://www.mathworks.com/matlabcentral/fileexchange/20058-adaptive-numerical-limit-and-residue-estimation), MATLAB Central File Exchange. に取得済み.

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