Status: Valid
Owner: ----
Labels: Type-Defect Priority-Medium
New issue 3770 by smi...@gmail.com: nsolve could employ scaling while
searching for solution
http://code.google.com/p/sympy/issues/detail?id=3770
mp is a polynomial with large coefficients
print filldedent(mp)
(x - 15)**16 - 22216*(x - 15)**14 + 184230412*(x - 15)**12 -
689044484248*(x - 15)**10 + 1067007063010870*(x - 15)**8 -
385661200155268792*(x - 15)**6 + 35979069569359907692*(x - 15)**4 -
1171725834641905675624*(x - 15)**2 + 11624829278428085430481
It is known to have a root between 100 and 101:
Poly(mp).intervals()[-1][0]
(100, 101)
Although this root can be refined with refine_root,
Poly(mp,x).refine_root(100,101,eps=1e-3)
(9145/91, 9346/93)
we can imagine that mp is an arbitrary function, not necessarily a
polynomial. When the bisect method is used with nsolve, and error is raised
nsolve(mp, (100, 101), solver='bisect')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "sympy\solvers\solvers.py", line 2250, in nsolve
return findroot(f, x0, **kwargs)
File "sympy\mpmath\calculus\optimization.py", line 972, in findroot
% (norm(f(*xl))**2, tol))
ValueError: Could not find root within given tolerance. (4.38593e+18 >
2.1684e-19)
Try another starting point or tweak arguments.
If we simply scale by a characteristic magnitude near the root of interest,
the solution is found
nsolve(mp/abs(mp.subs(x,100)), (100, 101), solver='bisect')
mpf('100.49450956304526')
==> perhaps nsolve could use some sort of scaling heuristic or (if bisect
is only using a function norm, perhaps it could use a norm in x so that if
either the value of the function or the uncertainty in the x-value falls
below a given value then the solution is returned.
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