Steffensen's method: Difference between revisions
Python example
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(Python example) |
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Output same as C
=={{header|Python}}==
{{trans|C}}
<syntaxhighlight lang="python">from math import nan, isnan
from numpy import arange
def aitken(f, p0):
""" Aitken's extrapolation """
p1 = f(p0)
p2 = f(p1)
return p0 - (p1 - p0)**2 / (p2 - 2 * p1 + p0)
def steffensen_aitken(f, pinit, tol, maxiter):
""" Steffensen's method using Aitken """
p0 = pinit
p = aitken(f, p0)
iter = 1
while abs(p - p0) > tol and iter < maxiter:
p0 = p
p = aitken(f, p0)
iter += 1
return nan if abs(p - p0) > tol else p
def deCasteljau(c0, c1, c2, t):
""" deCasteljau function """
s = 1.0 - t
return s * (s * c0 + t * c1) + t * (s * c1 + t * c2)
def xConvexLeftParabola(t): return deCasteljau(2, -8, 2, t)
def yConvexRightParabola(t): return deCasteljau(1, 2, 3, t)
def implicit_equation(x, y): return 5 * x**2 + y - 5
def f(t):
""" Outside of NumPy arithmetic may return NoneType on overflow """
if type(t) == type(None):
return nan
return implicit_equation(xConvexLeftParabola(t), yConvexRightParabola(t)) + t
def test_steffensen(tol=0.00000001, iters=1000, stepsize=0.1):
""" test the example """
for t0 in arange(0, 1, stepsize):
print(f't0 = {t0:0.1f} : ', end='')
t = steffensen_aitken(f, t0, tol, iters)
if isnan(t):
print('no answer')
else:
x = xConvexLeftParabola(t)
y = yConvexRightParabola(t)
if abs(implicit_equation(x, y)) <= tol:
print(f'intersection at ({x}, {y})')
else:
print('spurious solution')
return 0
if __name__ == '__main__':
test_steffensen()
</syntaxhighlight>{{out}} Output is the same as C.
=={{header|RATFOR}}==
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