HP50G Bisection methode

[Albert Chan]

Hi, acser.

Welcome to the forum.
Your code had a few bugs. Example, what happens if F(C) = 0 ?

Also, we can reduce 4 functions calls per loop down to 1
Bonus: this avoided underflow issue, when F(A)*F(C) = 0.0, losing sign info.

Code:

def solve_bisect(f,a,b, eps=1e-8, verbal=True):
    fa, fb = f(a), f(b)
    if fa==0: return a
    if fb==0: return b
    if fa > fb: a, fa, b, fb = b, fb, a, fa
    assert fa < 0 and fb > 0, "bad interval"
    while abs(a-b) > eps:
        if verbal: print a, b
        c = (a+b)/2
        fc = f(c)
        if fc <= 0: a=c
        if fc >= 0: b=c
    return (a+b)/2

>>> f = lambda x: x**5 - 5
>>> solve_bisect(f,0,1)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 6, in solve_bisect
AssertionError: bad interval

>>> solve_bisect(f,1,2)
1 2
1 1.5
1.25 1.5
1.375 1.5
1.375 1.4375
1.375 1.40625
1.375 1.390625
1.375 1.3828125
1.37890625 1.3828125
...