Find roots of a system of equations to an arbitrary decimal precision

Question

Given an initial guess for an array of values x, I am trying to find the root of a system that is closest to x. If you are familiar with finding roots of a system, you will understand that finding a root for a system of equations f satisfies:

0 = f_1(x)
0 = f_2(x)
....
0 = f_n(x)

Where f_i is one particular function within f

There is a package within scipy that will do this exactly: scipy.optimize.newton_krylov. For example:

import scipy.optimize as sp

def f(x):
    f0 = (x[0]**2) + (3*(x[1]**3)) - 2
    f1 = x[0] * (x[1]**2)
    return [f0, f1]
# Nearest root is [sqrt(2), 0]
print sp.newton_krylov(f, [2, .01], iter=100, f_tol=Dc('1e-15')) 

>>> [  1.41421356e+00   3.49544535e-10] # Close enough!

However, I am using the decimal package within python because I am doing extremely precise work. decimal offers more than normal decimal precision. scipy.optimize.newton_krylov returns float-precision values. Is there a way to get my answer at an arbitrarily precise decimal precision?

score 2 · Answer 1 · answered Apr 06 '17 at 19:21

You could try copying the code in and referenced by scipy.optimize.newton_krylov then modifying it to use decimal values rather than floating point values. This may be difficult and time-consuming, of course.

I have done the equivalent for other situations, but never quite this.

score 1 · Accepted Answer · answered Apr 06 '17 at 20:00

I have found the mpmath module, which contains mpmath.findroot. mpmath uses arbitrary decimal-point precision for all of its numbers. mpmath.findroot will find the nearest root within tolerance. Here is an example of using mpmath for the same problem, to a higher precision:

import scipy.optimize as sp
import mpmath
from mpmath import mpf
mpmath.mp.dps = 15

def mp_f(x1, x2):
    f1 = (x1**2) + (3*(x2**3)) - 2
    f2 = x1 * (x2**2)
    return f1, f2

def f(x):
    f0 = (x[0]**2) + (3*(x[1]**3)) - 2
    f1 = x[0] * (x[1]**2)
    return [f0, f1]

tmp_solution = sp.newton_krylov(f, [2, .01], f_tol=Dc('1e-10'))
print tmp_solution

>>> [  1.41421356e+00   4.87315249e-06]

for _ in range(8):
    tmp_solution = mpmath.findroot(mp_f, (tmp_solution[0], tmp_solution[1]))
    print tmp_solution
    mpmath.mp.dps += 10 # Increase precision

>>> [    1.4142135623731]
[4.76620313173184e-9]
>>> [    1.414213562373095048801689]
[4.654573673348783724565804e-12]
>>> [    1.4142135623730950488016887242096981]
[4.5454827012374811707063801808968925e-15]
>>> [    1.41421356237309504880168872420969807856967188]
[4.43894795688326535096068850443292395286770757e-18]
>>> [    1.414213562373095048801688724209698078569671875376948073]
[4.334910114213471839327827177504976152074382061299675453e-21]
>>> [     1.414213562373095048801688724209698078569671875376948073176679738]
[4.2333106584123451747941381835420647823192649980317402073699554127e-24]
>>> [    1.41421356237309504880168872420969807856967187537694807317667973799073247846]
[4.1340924398558139440207202654766836515453497962889870471467483995909717197e-27]
>>> [     1.41421356237309504880168872420969807856967187537694807317667973799073247846210703885]
[4.037199648296693366576484784520203892002447351324378380584214947262318103197216393589e-30]

The precision can be raised arbitrarily.

Find roots of a system of equations to an arbitrary decimal precision

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