牛顿二项式与 e 级数

复习一下数学, 找一下回忆.

先是从二项式平方开始:

其实展开是这样的:

再看立方:

通过排列组合的方式标记, 于是:

通过数学归纳法可以拓展:

使用求和简写可得:

e 级数

数学常数 e (The Constant e – NDE/NDT Resource Center) 的定义爲下列极限值:

使用二项式定理能得出

第 k 项之总和为

因为 n → ∞，右边的表达式趋近1。 因此

由于序列的极限可以相加, 所以 e 可以表示为:

计算情况:

e = 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + …

As an example, here is the computation of e to 22 decimal places:

1/0!	=	1/1	=	1.0000000000000000000000000
1/1!	=	1/1	=	1.0000000000000000000000000
1/2!	=	1/2	=	0.5000000000000000000000000
1/3!	=	1/6	=	0.1666666666666666666666667
1/4!	=	1/24	=	0.0416666666666666666666667
1/5!	=	1/120	=	0.0083333333333333333333333
1/6!	=	1/720	=	0.0013888888888888888888889
1/7!	=	1/5040	=	0.0001984126984126984126984
1/8!	=	1/40320	=	0.0000248015873015873015873
1/9!	=	1/362880	=	0.0000027557319223985890653
1/10!	=	1/3628800	=	0.0000002755731922398589065
1/11!	=	1/39916800	=	0.0000000250521083854417188
1/12!	=	1/479001600	=	0.0000000020876756987868099
1/13!	=	1/6227020800	=	0.0000000001605904383682161
1/14!	=	1/87178291200	=	0.0000000000114707455977297
1/15!	=	1/1307674368000	=	0.0000000000007647163731820
1/16!	=	1/20922789887989	=	0.0000000000000477947733239
1/17!	=	1/355687428101759	=	0.0000000000000028114572543
1/18!	=	1/6402373705148490	=	0.0000000000000001561920697
1/19!	=	1/121645101098757000	=	0.0000000000000000082206352
1/20!	=	1/2432901785214670000	=	0.0000000000000000004110318
1/21!	=	1/51091049359062800000	=	0.0000000000000000000195729
1/22!	=	1/1123974373384290000000	=	0.0000000000000000000008897
1/23!	=	1/25839793281653700000000	=	0.0000000000000000000000387
1/24!	=	1/625000000000000000000000	=	0.0000000000000000000000016
1/25!	=	1/10000000000000000000000000	=	0.0000000000000000000000001

For more information on e, visit the the math forum at mathforum.org
The sum of the values in the right column is 2.7182818284590452353602875 which is “e.”

Reference: The mathforum.org