ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Mon, 09 Mar 2020 10:19:57 +0100How to solve numericaly with arbitrary precisionhttps://ask.sagemath.org/question/50197/how-to-solve-numericaly-with-arbitrary-precision/Hello,
I want to solve a numerical equation for which I can only access by a lambda function, the M matrix is to big for the determinant to be computed on the symbolic ring.
f = lambda om: RRR(M.subs({omega:om}).change_ring(RRR).det())
But , find-root solve in the built in float type of Python which is lacking the precision I need. Is there a way in sage to solve numericaly with arbitrary precision?
Thank you
Regards
*(edited with example more in line with what I am trying to achieve)*
sage: x=var('x')
sage: M= Matrix(SR,[[cos(x),cosh(x)],[sin(x),sinh(x)]])
sage: RRR = RealField(200)
sage: f = lambda om: M.subs({x:om}).change_ring(RRR).det()
sage: find_root(f,1,2)JonasAMon, 09 Mar 2020 10:19:57 +0100https://ask.sagemath.org/question/50197/