Question

I understand that this has been solved in C/C++, but I am not comfortable enough with those languages to be able to convert it to python. I am trying to create this in python. The closest I was able to come was this:

#This is meant to work for functions of the form x^a + b = 0

def secant(base, exp=2, it=20):
    def f(x):
        return x**exp - base
    x1 = base / float(exp**2)
    xnm1 = x1 - 5
    xnm2 = x1 + 5
    xn = 0
    for n in range(it):
        q = (xnm1-xnm2)/float(f(xnm1)-f(xnm2))
        xn = xnm1 - (f(xnm1)*q)
        xnm1, xnm2 = xn, xnm1
    return xn

print secant(2, 2)

This returns the error:

Traceback (most recent call last):
  File "/Users/Joe/Desktop/secant.py", line 16, in <module>
    print secant(2, 2)
  File "/Users/Joe/Desktop/secant.py", line 11, in secant
    q = (xnm1-xnm2)/float(f(xnm1)-f(xnm2))
ZeroDivisionError: float division by zero

I was able, however, to program the Newton method, which I based this code off of. If it helps, here it is:

def newton(base, exp=2, it=20):
    def f(x):
        return x**exp - base
    def df(x):
        return exp*(x**(exp-1))
    x1 = base / float(exp**2)
    xnp = x1
    xn = 0
    for n in range(it):
        xn = xnp - ((f(xnp)/df(xnp)))
        xnp = xn
    return xn

The following method gives an answer with 12 digits of accuracy after 20 iterations. Any help would be appreciated.

Answer 1

You are getting a division by zero error because the algorithm converged to the answer within the precision of Python floats. In addition to iterating a maximum number of times (to avoid an infinite loop), you should also check if the last two guesses are “close enough”.