摘要:
#-*- 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 __ 阅读全文
posted @ 2011-03-03 19:41
一碗豆腐
阅读(4153)
评论(0)
推荐(0)