From: Alexei Sheplyakov Date: Sat, 21 Aug 2010 16:13:29 +0000 (+0300) Subject: fsolve: avoid useless numerical evaluation of the function X-Git-Tag: release_1-6-0~44 X-Git-Url: https://www.ginac.de/ginac.git//ginac.git?p=ginac.git;a=commitdiff_plain;h=beeb0818e9cdb1b5de0ba2754286ad7bb2a9d032;hp=515171f0bcd42099c266713c3d605cd92cedd2e2 fsolve: avoid useless numerical evaluation of the function Don't compute f(x) if new x is outside of the interval. We don't need that value anyway, and the function might be difficult to compute numerically or even ill defined outside the interval. As a result fsolve is able to find root(s) of some weird functions. For example fsolve((1/(sqrt(2*Pi)))*integral(t, 0, x, exp(-1/2*t^2)) == 0.5, x, 0, 100) actually works now. --- diff --git a/ginac/inifcns.cpp b/ginac/inifcns.cpp index f366e779..cc020920 100644 --- a/ginac/inifcns.cpp +++ b/ginac/inifcns.cpp @@ -1011,13 +1011,20 @@ fsolve(const ex& f_in, const symbol& x, const numeric& x1, const numeric& x2) if (!is_a(dx_)) throw std::runtime_error("fsolve(): function derivative does not evaluate numerically"); xx[side] += ex_to(dx_); - - ex f_x = f.subs(x == xx[side]).evalf(); - if (!is_a(f_x)) - throw std::runtime_error("fsolve(): function does not evaluate numerically"); - fx[side] = ex_to(f_x); - - if ((side==0 && xx[0]xxprev) || xx[0]>xx[1]) { + // Now check if Newton-Raphson method shot out of the interval + bool bad_shot = (side == 0 && xx[0] < xxprev) || + (side == 1 && xx[1] > xxprev) || xx[0] > xx[1]; + if (!bad_shot) { + // Compute f(x) only if new x is inside the interval. + // The function might be difficult to compute numerically + // or even ill defined outside the interval. Also it's + // a small optimization. + ex f_x = f.subs(x == xx[side]).evalf(); + if (!is_a(f_x)) + throw std::runtime_error("fsolve(): function does not evaluate numerically"); + fx[side] = ex_to(f_x); + } + if (bad_shot) { // Oops, Newton-Raphson method shot out of the interval. // Restore, and try again with the other side instead! xx[side] = xxprev;