My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Reply
 
LinkBack Thread Tools Display Modes
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).
Weiler is offline  
 
Reply

  My Math Forum > College Math Forum > Calculus

Tags
poisson, relation, stirling, wallis



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Help Stirling's formula mathmari Real Analysis 4 March 31st, 2013 07:43 AM
What the Wallis product really reveals Eureka Algebra 14 September 4th, 2011 07:30 AM
Stirling formula mathdude Applied Math 2 December 19th, 2010 02:21 PM
Stirling numbers madziulka_lap Number Theory 1 September 14th, 2010 07:47 AM
Help Stirling's formula mathmari Algebra 4 December 31st, 1969 04:00 PM





Copyright © 2019 My Math Forum. All rights reserved.