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January 31st, 2012, 02:54 PM  #1 
Senior Member Joined: Dec 2011 From: Argentina Posts: 216 Thanks: 0  Wallis and Stirling and Poisson: a relation.
It isn't hard to prove that and thus that for odd for even Morever, since we are in the first quadrant, the following inequation holds. Thus: Arranging the inequation properly produces: Thus we conclude that and that with which is a variation of Wallis infinite product. Let's focus on Poisson's integral It is easily proven by Taylor series that Thus Upon integration we have Similar to the integration in Wallis' identity we can prove that for odd for even and with appropiate change of variables, namely and we conclude that Thus: And with (but keep the ) Let's set and thus we have, using Wallis identity: But if we get and thus which yields I'll wait for some feedback untill I get to Stirling's approximation. I learned this proofs from a Spanish Analisis book, by Rey Pastor (a true diamond if you have to ask). 

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poisson, relation, stirling, wallis 
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