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January 31st, 2012, 02:54 PM   #1
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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.


Arranging the inequation properly produces:

Thus we conclude that

and that


which is a variation of Wallis infinite product.

Let's focus on Poisson's integral

It is easily proven by Taylor series that


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


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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