Newton-Raphson method(python)

qwe88

Implement Newton-Raphson by completing the following skeleton:
def findRoot(n, c):
      """
      Return an approximate root of the equation (x^n - c = 0). The value of x
      that is returned should be such that (x^n - c) differs from 0 by no more
      than 0.001.
      This function also prints a message showing the number of iterations needed

      to find the root.

      """

      # Your code goes here.

Newton-Raphson requires us to know the first derivative of f(x) = xn - c, which is n xn-1. Recall that in Python, xn
is written x**n.
The solution that Newton-Raphson gives may depend on the initial guess you used. In your code, please use c/2
as your first guess.
Here is a sample use of findRoot:
>>> findRoot(2, 25) # Computes sqrt(25)

(Number of iterations: 4)

5.0000129530486843

>>> findRoot(2, 2) # Computes sqrt(2)

(Number of iterations: 3)

1.4142156862745099

>>> findRoot(4, 143) # Computes fourth root of 143

(Number of iterations: 14)

3.458071842417946

It's fine if your solutions differ slightly from these sample outputs.
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我的答案

def findRoot(n, c):
    xZero = float (c/2)
    xFirst = xZero - ((xZero**n - c) / (n*xZero**n-1))
    xFirstOld = xFirst
    xZero = xFirst
    xFirst = xZero - ((xZero**n - c) / (n*xZero**n-1))
    count = 0

    while xFirstOld != xFirst:
        xFirstOld = xFirst
        xZero = xFirst
        xFirst = xZero - ((xZero**n - c) / (n*xZero**n-1))
        count += 1

    print xFirst
    print count

findRoot(4, 143)
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这个是python的题目
请问我做得对吗？
及，我的findRoot(4,143)用了343个loop才算到，那个例子用14个就算到了，我的算法有问题？
谢谢。