python 求自然对数的底数 e

#-*- coding: UTF-8 -*- 
'''
Created on 2011-3-3

@author: lihex
求自然对数的底数 e
'''
#求阶乘
def factorial(number):
    if number==1:
        return 1
    return number*factorial(number-1)

#根据e^x 幂级数张开式，当x=1时,n为精度参数
def sum_toE(n):
    retv=2.0
    for x in range(2,n+1):
        retv += 1/float(factorial(x))
    return retv   
        
if __name__ == '__main__':
    retv = sum_toE(6)
    print retv

转载于:https://www.cnblogs.com/lihex/archive/2011/03/03/1970226.html