Difference between revisions of "User talk:Baijiangchen"

Revision as of 23:40, 21 July 2012

If:

$W(0):=1$

$W(n):=\sum_{i=0}^{n-1}({n-1 \choose i}W(i)(x-i-1)!(2^{x-i-1}))$

Then:

$W(n)=(2n-1)!!$

Let $W(n)=\sum_{i=1}^{n}(\binom{i-1}{n-1}W(i-1)(n-i)!(2^{n-i}))$

Assume that for some integer $x$ , $W(x)=(2x-1)!!$ . We intend to show that $W(x+1)=(2(x+1)-1)!!=(2x+1)!!$ .

$W(x+1)=\sum_{i=1}^{x+1}(\binom{i-1}{x}W(i-1)(x-i+1)!(2^{x-i+1}))$

$=\sum_{i=1}^{x+1}(\binom{i-1}{x-1}(\frac{x-i+1}{x})W(i-1)(x-i)!(x-i+1)(2^{x-i})(2))$

$=\sum_{i=1}^{x+1}(\binom{i-1}{x-1}W(i-1)(x-i)!(x-i+1)(2^{x-i})(2x))$

$=2x\sum_{i=1}^{x+1}(\binom{i-1}{x-1}W(i-1)(x-i)!(x-i+1)(2^{x-i}))$

$=2x[W(n)=\sum_{i=1}^{x}(\binom{i-1}{x-1}W(i-1)(x-i)!(2^{x-i}))+\binom{x}{x-1}W(x)(1)!(2)]$

$=2x[(2x-1)!!+\binom{x}{x-1}W(x)(1)!(2)]$

And this is where I'm stuck.

Return!

Revision as of 23:40, 21 July 2012 (view source) Saprmarks (talk \| contribs) ← Older edit		Revision as of 23:40, 21 July 2012 (view source) Baijiangchen (talk \| contribs) (→‎Sam's stuff) Newer edit →
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	And this is where I'm stuck.		And this is where I'm stuck.
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		+	Return!