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.
(cherry picked from commit beeb0818e9cdb1b5de0ba2754286ad7bb2a9d032)

diff --git a/ginac/inifcns.cpp b/ginac/inifcns.cpp
index f366e779e70fbdb18a64e0447b8b21263f3704ee..cc0209208da959329d60fd3dbd9d1e1076d5d477 100644 (file)
--- 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<numeric>(dx_))
                        throw std::runtime_error("fsolve(): function derivative does not evaluate numerically");
                xx[side] += ex_to<numeric>(dx_);
                if (!is_a<numeric>(dx_))
                        throw std::runtime_error("fsolve(): function derivative does not evaluate numerically");
                xx[side] += ex_to<numeric>(dx_);

-
-               ex f_x = f.subs(x == xx[side]).evalf();
-               if (!is_a<numeric>(f_x))
-                       throw std::runtime_error("fsolve(): function does not evaluate numerically");
-               fx[side] = ex_to<numeric>(f_x);
-
-               if ((side==0 && xx[0]<xxprev) || (side==1 && xx[1]>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<numeric>(f_x))
+                               throw std::runtime_error("fsolve(): function does not evaluate numerically");
+                       fx[side] = ex_to<numeric>(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;
                        // Oops, Newton-Raphson method shot out of the interval.
                        // Restore, and try again with the other side instead!
                        xx[side] = xxprev;