Comment #5 on issue 886 by Vinzent.Steinberg: simplify for general
functions and derivatives fails
http://code.google.com/p/sympy/issues/detail?id=886
>>> from sympy import *
>>> var('x')
x
>>> f=Function('f')
>>> f(x)*f(x).diff(x,2)/f(x).diff(x)**2
f(x)*D(f(x), x, x)/D(f(x), x)**2
>>> z=f(x)*f(x).diff(x,2)/f(x).diff(x)**2
>>> z.diff(x)
D(f(x), x)*D(f(x), x, x)/D(f(x), x)**2 + f(x)*D(f(x), x, x, x)/D(f(x),
x)**2 -
2*f(x)*D(f(x), x, x)*D(f(x), x, x)/(D(f(x), x)**2*D(f(x), x))
>>> simplify(z.diff(x))
(D(f(x), x)**6*D(f(x), x, x) - 2*f(x)*D(f(x), x, x)**2*D(f(x), x)**4 +
f(x)*D(f(x),
x)**3*D(f(x), x, x, x)*D(f(x), x)**2)/D(f(x), x)**7
It does not raise an exception, but it's still not what I want (as you can
see, it
can be further simplified).
Here is the problem I wanted to solve:
Cubic convergence of Halley's method
------------------------------------
Let
phi(x) = x - f(x) / f'(x) * G(z) with z = f(x)*f''(x) / f'(x)**2
and
G(0) = 1, G'(0) = 0.5.
Assuming that phi(x*) = x* (x* is a fix point of phi) and f(x*) = 0 (the
iteration
x_{k+1} = phi(x_k) converges towards a root of f), prove the cubic
convergence rate
of this fix point iteration.
The aim is to show that the first two derivatives of phi in x* are 0. (This
proves
that convergence is at least cubic, to prove that it's exactly cubic you'll
need to
show that the third derivative is not 0).
You can do this by hand, but this is the kind of mathematical work you want
to use a
computer algebra system for.
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