It's e-asy (Posted on 2008-02-19)

	Submitted by FrankM
Rating: 4.0000 (2 votes)
Solution:	(Hide)
Approach 1: Use the generating function f(x) = exp(exp(x)) The kth derivative of f will be of the form (ex e2x .. + akk ekx)f without too much effort, we can work out formulas for the as, for instance a2k = 2k-1 - 1. So we can compute Dkf(x=0) = Σ ajk for j=1 to k But f(x) = 1 + ex + e2x/2! + e3x/3! +... Df(0) = 1 + 2/2! + 3/3! + ... = f(0) = e similarly D2f(0) = 1 + 22/2! + 33/3! ... = 2e etc. Approach 2: n/n! = 1/(n-1)! n2/n! = n/(n-1)! = 1/(n-1)! + 1/(n-2)! etc. So S1 = e S2 = 2e etc.

	Subject	Author	Date
	re: Solution	Charlie	2008-02-20 11:04:57
	re(2): Solution - question	Praneeth	2008-02-20 03:10:13
	re: Solution - question	brianjn	2008-02-20 03:03:34
	Solution	Praneeth	2008-02-20 01:22:55
	re(2): computer solution	SilverKnight	2008-02-19 16:47:01
	re: computer solution	Charlie	2008-02-19 13:03:43
	computer solution	Charlie	2008-02-19 12:42:02