题目：

用Python求一个数的平方根.

At each iteration (loop) of the algorithm,the approximation x is replaced by the average of x and s divided by x.Written as an assignment statement,it looks like this:

x = (x + s/x)/2.0.There are various criteria for deciding when to stop improving the answer of a square root algorithm.In your implementation,I want you keep looping until the square of x is very close to the value of s.In other words,you want to reduce the error of your approximation x until it is acceptably small.You can measure the relative error by comparing the absolute difference of x squared and s.This is written mathematically as:

error = | x * x – s |.Use the math.fabs() function to compute absolute value.

So keep looping and calculating better and better values for x until the error is less than some small constant.For

your implementation of heron(s),stop when error < 0.0000001 which is a small discrepancy.Add a print

statement inside your algorithm loop which prints out both x and x*x at each step so you can see how fast it converges to a good answer.It is very interesting to watch it work.Try different starting guess values for x and see if it

makes any difference to the number of loops required (always use a positive guess value)

解答：

下面代码定义一个函数heron(s)用迭代的方法取得平方根,其中x=s/2可以使用x=s/3,s/5之类的多个值实验一下,看分别需要多少步.

以s=500为例,

x=s时需要9步

x=s/2时需要8步

x=s/3时需要7步

x=s/5时需要5步

”’

Created on 2011-10-26

@author:legendxx

”’

import math

def heron(s):

x=s/2

count=0

sqr=x*x

while math.fabs(sqr – s)>=0.0000001:

count+=1

x = (x + s/x)/2.0

sqr=x*x

print count,”:”,x,sqr

print count,”steps needed”

if __name__ == ‘__main__’:

s=float(raw_input(“input a number”))

heron(s)