python - Solve this equation with fixed point iteration -

how can solve equation

x³ + x - 1 = 0

using fixed point iteration?

is there fixed-point iteration code (especially in python) can find online?

using scipy.optimize.fixed_point:

import scipy.optimize optimize  def func(x):     return -x**3+1  # finds value of x such func(x) = x, is, # -x**3 + 1 = x print(optimize.fixed_point(func,0)) # 0.682327803828 

the python code defining fixed_point in scipy/optimize/minpack.py. exact location depends on scipy installed. can find out typing

in [63]: import scipy.optimize  in [64]: scipy.optimize out[64]: <module 'scipy.optimize' '/usr/lib/python2.6/dist-packages/scipy/optimize/__init__.pyc'> 

here code in scipy 0.7.0:

def fixed_point(func, x0, args=(), xtol=1e-8, maxiter=500):     """find point func(x) == x      given function of 1 or more variables , starting point, find     fixed-point of function: i.e. func(x)=x.      uses steffensen's method using aitken's del^2 convergence acceleration.     see burden, faires, "numerical analysis", 5th edition, pg. 80      example     -------     >>> numpy import sqrt, array     >>> scipy.optimize import fixed_point     >>> def func(x, c1, c2):             return sqrt(c1/(x+c2))     >>> c1 = array([10,12.])     >>> c2 = array([3, 5.])     >>> fixed_point(func, [1.2, 1.3], args=(c1,c2))     array([ 1.4920333 ,  1.37228132])      see also:        fmin, fmin_powell, fmin_cg,              fmin_bfgs, fmin_ncg -- multivariate local optimizers       leastsq -- nonlinear least squares minimizer        fmin_l_bfgs_b, fmin_tnc,              fmin_cobyla -- constrained multivariate optimizers        anneal, brute -- global optimizers        fminbound, brent, golden, bracket -- local scalar minimizers        fsolve -- n-dimenstional root-finding        brentq, brenth, ridder, bisect, newton -- one-dimensional root-finding      """     if not isscalar(x0):         x0 = asarray(x0)         p0 = x0         iter in range(maxiter):             p1 = func(p0, *args)             p2 = func(p1, *args)             d = p2 - 2.0 * p1 + p0             p = where(d == 0, p2, p0 - (p1 - p0)*(p1-p0) / d)             relerr = where(p0 == 0, p, (p-p0)/p0)             if all(relerr < xtol):                 return p             p0 = p     else:         p0 = x0         iter in range(maxiter):             p1 = func(p0, *args)             p2 = func(p1, *args)             d = p2 - 2.0 * p1 + p0             if d == 0.0:                 return p2             else:                 p = p0 - (p1 - p0)*(p1-p0) / d             if p0 == 0:                 relerr = p             else:                 relerr = (p-p0)/p0             if relerr < xtol:                 return p             p0 = p     raise runtimeerror, "failed converge after %d iterations, value %s" % (maxiter,p)