I'd like to solve SymPy expressions to high (arbitrary) precision. I tried
`nsolve` with `mpmath.mp.dps = 128`, and filed #8564,
as nsolve is currently limited to double precision.
My current idea is to convert my expression to an lambda involving only
mpmath objects. Then call `mpmath.mp.findroot`. This is essentially what
nsolve does anyway (except for the part about "only mpmath" objects).
How do I do that? `lambdify(x, expr, module=mpmath)` is limited to
double-precision (filed #8818, but not sure if this is by design or a bug).
`mpmath.mpmathify` works for individual floats...
Best I have so far is something like:
````
d = 128 % precision desired.
g = sqrt(2) - x % I want to solve g == 0 for x
% here's my approach
h = g.evalf(d)
f = lambda meh: h.subs(x, meh) % yuck
q = mpmath.mp.findroot(f, 1.0)
q*q % should be 2 to 128 digits, looks good:
%
mpf('2.000000000000000000000000000000000000000000000000000000000000000038')
````
thanks,
Colin
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