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Sterling approximation of Beta Function

  1. Nov 1, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-11-1_15-37-21.png
    2. Relevant equations
    upload_2016-11-1_15-37-31.png
    upload_2016-11-1_15-37-41.png
    3. The attempt at a solution
    I think this problem is probably a lot simpler than I am making it out to be. However, when I apply sterling's approx., I get a very nasty equation that does not simplify easily.
    upload_2016-11-1_15-39-48.png

    One of the biggest problems I have though is dealing with that n^x term. it seems to me that as n approaches infinity, the outcome depends heavily on whether x is a positive or negative integer, or if its zero.
     
  2. jcsd
  3. Nov 1, 2016 #2

    stevendaryl

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    Staff Emeritus
    Science Advisor

    Rearrange it to:

    [itex]n^x B(x,n) =\Gamma(x) n^x \frac{\Gamma(n)}{\Gamma(n+x)}[/itex]

    So what you're really interested in is the limit of

    [itex]n^x \frac{\Gamma(n)}{\Gamma(n+x)}[/itex]

    Use Sterling's approximation:

    [itex]\Gamma(n) \approx n^{n+\frac{1}{2}} \sqrt{2 \pi} e^{-n}[/itex]
    [itex]\Gamma(n+x) \approx (n+x)^{n+x+\frac{1}{2}} \sqrt{2 \pi} e^{-(n+x)}[/itex]

    Write [itex](n+x)^{n+x+\frac{1}{2}}[/itex] as [itex](n(1+\frac{x}{n}))^{n+x +\frac{1}{2}} = n^{n+x+\frac{1}{2}} (1+\frac{x}{n})^{n+x+\frac{1}{2}}[/itex]
    [itex]= n^{n+\frac{1}{2}} n^x (1+\frac{x}{n})^{n} (1+\frac{x}{n})^{x+\frac{1}{2}}[/itex]

    So
    [itex]\Gamma(n+x) \approx n^{n+\frac{1}{2}} n^x (1+\frac{x}{n})^{n} (1+\frac{x}{n})^{x+\frac{1}{2}} \sqrt{2 \pi} e^{-(n+x)}[/itex]

    Now, you can just use a fact about [itex]e[/itex]: [itex]e^x = lim_{n \rightarrow \infty} (1+\frac{x}{n})^n[/itex]
     
  4. Nov 1, 2016 #3

    Charles Link

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    Homework Helper

    Suggest using the simplified Stirling formula without the ## \sqrt{2 \pi n} ##.Then the result follows immediately with ## ln(n+x-1)=ln(n) ## in the limit that n approaches infinity... editing...You can include the ## \sqrt{2 \pi n} ## but in the limit that n approaches infinity, this puts the same factor in the numerator that it does in the denominator.
     
    Last edited: Nov 1, 2016
  5. Nov 1, 2016 #4
    Thanks to both of you! The key to solving that problem noticing the what Steven showed, that
    Then the entire problem simplifies quite nicely.
     
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