[math-fun] Analytic continuation software?
Speaking of math software, can anyone tell me if there exists analytic continuation software? In particular, given a subroutine f_0 to evaluate an analytic function f on a disk D_0 in C centered at z_0 and of radius R_0, and a curve a: [0,L] -> C with a(0) = z_0 . . . . . . then -- the software would output a set of curve parameters 0 = t_0 < t_1 < t_2 < . . . < t_n = L and radii R_k, 1 <= k <= n such that the original function f_0 element can be successively continued along curve a(t) to the function elements f_k, via the disks D_k of radius R_k about z_k = a(t_k), for 1 <= k <= n (where of course f_(k-1) must agree with f_k on the overlap of D_(k-1) and D_k). The software would also output a routine that will evaluate the function f_k inside the disk D_k. (Or a statement that the function cannot be continued along a(t), 0 <= t <= L.) . . . . . . and all this to a pre-specified level of numerical accuracy. NOTE: The hardest part here is probably ensuring the numerical accuracy of the f_k's. --Dan _____________________________________________________________________ "It don't mean a thing if it ain't got that certain je ne sais quoi." --Peter Schickele
On Sun, Jun 8, 2008 at 5:05 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Speaking of math software, can anyone tell me if there exists analytic continuation software?
In particular, given a subroutine f_0 to evaluate an analytic function f on a disk D_0 in C centered at z_0 and of radius R_0, and a curve a: [0,L] -> C with a(0) = z_0 . . .
. . . then -- the software would output a set of curve parameters
0 = t_0 < t_1 < t_2 < . . . < t_n = L
and radii R_k, 1 <= k <= n such that the original function f_0 element can be successively continued along curve a(t) to the function elements f_k, via the disks D_k of radius R_k about z_k = a(t_k), for 1 <= k <= n (where of course f_(k-1) must agree with f_k on the overlap of D_(k-1) and D_k).
The software would also output a routine that will evaluate the function f_k inside the disk D_k.
(Or a statement that the function cannot be continued along a(t), 0 <= t <= L.) . . .
. . . and all this to a pre-specified level of numerical accuracy.
I don't think this is possible. In finite tie the software can only evaluate your original function f at a finite number of points. And there are analytic functions that have exactly the given value at that finite number of points, but have any desired value at any other point. Think of the same task where f is constrained to be a polynomial. If you don't have the coefficients of f, or any information about its degree, but only the values of f at a finite number of points, you can't say anytyhing sensible about its value at other points, unless you make an assumption that f is of low degree compared to the number of points sampled. If you only have the assumption that f is analytic, that's even weaker than "f is polynomial, of arbitrarily high degree". -- Andy.Latto@pobox.com
On Monday 09 June 2008, Andy Latto wrote:
On Sun, Jun 8, 2008 at 5:05 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Speaking of math software, can anyone tell me if there exists analytic continuation software? ... . . . and all this to a pre-specified level of numerical accuracy.
I don't think this is possible. In finite tie the software can only evaluate your original function f at a finite number of points. And there are analytic functions that have exactly the given value at that finite number of points, but have any desired value at any other point.
The same argument proves that numerical integration and differentiation are impossible. And yet, if you pick a function you're interested in and ask Maple or Mathematica or whatever to integrate it for you over a given range to a given level of accuracy, they can generally do it pretty well. (That doesn't mean that I know of software to do what Dan wants, or that I have a proposal for how to do it so that it works well in practice. I wonder whether it might be more appropriate for what you feed the analytic continuation routine to be a black box that computes not only f but also any requested number of its derivatives.) -- g
On Mon, Jun 9, 2008 at 3:03 AM, Gareth McCaughan <gareth.mccaughan@pobox.com> wrote:
On Monday 09 June 2008, Andy Latto wrote:
On Sun, Jun 8, 2008 at 5:05 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Speaking of math software, can anyone tell me if there exists analytic continuation software? ... . . . and all this to a pre-specified level of numerical accuracy.
I don't think this is possible. In finite tie the software can only evaluate your original function f at a finite number of points. And there are analytic functions that have exactly the given value at that finite number of points, but have any desired value at any other point.
The same argument proves that numerical integration and differentiation are impossible. And yet, if you pick a function you're interested in and ask Maple or Mathematica or whatever to integrate it for you over a given range to a given level of accuracy, they can generally do it pretty well.
Though the above argument shows that there exist pathological functions for which these routines will give very wrong answers. So for any given method of numerical integration or numerical analytic continuation that know a function only by evaluating it at a finite number of poitns, there are implicit assumptions about the nature of the function. A simple trapezoidal rule will work well for numerical integration if the second derivative of the function doesn't get too big. So what I meant to be indirectly suggesting is that a first step towards working out how to do this sort of numerical analytic continuation would be considering what implicit assumptions one wishes to make on the function being continued.
(That doesn't mean that I know of software to do what Dan wants, or that I have a proposal for how to do it so that it works well in practice. I wonder whether it might be more appropriate for what you feed the analytic continuation routine to be a black box that computes not only f but also any requested number of its derivatives.)
That sounds right; it would allow you to expand f as a power series, which is what I think you want for analytic continuation. -- Andy.Latto@pobox.com
participants (3)
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Andy Latto -
Dan Asimov -
Gareth McCaughan