◆ dfzero()

Solution of a single general nonlinear equation

Purpose: dfzero searches for a zero of a function f(x) between the given values b and c until the width of the interval [b, c] has collapsed to within a tolerance specified by the stopping criterion, abs(b - c) <= 2*(re*abs(b) + ae).
It is designed primarily for problems where f(b) and f(c) have opposite signs. The method used is an efficient combination of bisection and the secant rule.

Returns: (x, info)

x (float):
The final approximation to a zero.

info (int):
= 0: Normal return
= -1: The argument f is invalid
= 1: f(x) = 0, however the interval may not have collapsed to the requested tolerance
= 2: b may be near a singular point of f(x)
= 3: No change in sign of f(x) was found although the interval
= 4: Too many (> 500) function evaluations used

Parameters

[in]	f	User supplied function which evaluates f(x) for the equation f(x) = 0 defined as follows: _CODE def f(x): return computed function value f(x) _ENDCODE
[in]	b	Lower endpoint of the initial interval.
[in]	c	Upper endpoint of the initial interval.
[in]	r	A (better) guess of a zero of f which could help in speeding up convergence. When no better guess is known, it is recommended that r be set to b or c.
[in]	re	(Optional) Relative error used for the stopping criterion. If the requested re is less than machine precision, then it is set to approximately machine precision. (default = 1.0e-10)
[in]	ae	(Optional) Absolute error used in the stopping criterion. If the given interval [b, c] contains the origin, then a nonzero value should be chosen for ae. (default = 1.0e-10)

Reference: D. Kahaner, C. Moler, S. Nash, "Numerical Methods and Software", Prentice-Hall (1989)

Example Program: Find the root of the following equation in the interval [1, 3].
x^3 - 2x - 5 = 0

def f(x):

return (x*x - 2.0)*x - 5.0

def TestDfzero():

b = 1.0

c = 3.0

r = b

print(b, c)

x, info = dfzero(f, b, c, r)

print(x, info)

dfzero
def dfzero(f, b, c, r, re=1.0e-10, ae=1.0e-10)
Solution of a single general nonlinear equation