why is there this TypeError?

Question

i am trying to find the square root a number through the function sqrt(a). fixedPoint(f, epsilon) is a helper function. the problem is that i get a this TypeError: 'float' object is not callable. i am new to programming, so if anybody can help and find were is the bug and explain what does this error mean ??

def fixedPoint(f, epsilon):
    """
    f: a function of one argument that returns a float
    epsilon: a small float

    returns the best guess when that guess is less than epsilon 
    away from f(guess) or after 100 trials, whichever comes first.
    """
    guess = 1.0
    for i in range(100):
        if abs(f(guess) - guess) < epsilon:
            return guess
        else:
            guess = f(guess)
    return guess


def sqrt(a):
    def tryit(x):
        return 0.5 * (a/x + x)
    return fixedPoint(tryit(a), 0.0001)

Answer 1

In sqrt function, the code is passing the return value of the tryit (which is a float value), not tryit itself.

Passing the function itself will solve the problem.

def sqrt(a):
    def tryit(x):
        return 0.5 * (a/x + x)
    return fixedPoint(tryit, 0.0001)