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Wallis Product into Stirlings Formula

  1. Dec 3, 2007 #1
    Hey all, I'm having a difficult time seeing why the approximation of the Wallis Product equals the square root of 2pi.

    I know for a fact that:


    How then does this produce:

  2. jcsd
  3. Dec 3, 2007 #2


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    The standard Stirlings Formula is
    I think
  4. Dec 3, 2007 #3
    I know, that's what I'm getting to later on in the proof, however getting through this step is a little bit of a mess to me. I already know that [tex]\sqrt{2\pi}\approx(\frac{e}{n})^n\frac{n!}{\sqrt{n }}[/tex]

    I'm trying to get from [tex]\pi\approx\left(\frac{2*4*6*8*\ldots*\left(2n\righ t)}{1*3*5*7*\ldots*\left(2n-1\right)}\right)\frac{1}{n}[/tex]

    [tex]\frac{\left(n!\right)^{2}\left(2^{2n}\right)}{\left(2n)!}[/tex] = [tex]\sqrt{2\pi}[/tex]

    and I'm not sure how to do that.
    Last edited: Dec 3, 2007
  5. Dec 3, 2007 #4
    You're not going to, because it should be [tex]2\pi[/tex] rather than [tex]\sqrt{2\pi}[/tex]. To do that, try:
    (1) Let the expression in the numerator be E. Multiply by E/E. What's the denominator now?
    (2) factor a 2 out of every term in the numerator. What does that look like now?
    (3) Win!
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