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Using Stirling's Formula

  1. Jan 27, 2009 #1
    1. The problem statement, all variables and given/known data
    Ok, my teacher wants us to compute:
    (2n choose n)^2/(4n choose 2n) using Stirling's formula.

    2. Relevant equations

    Stirling's formula is sqrt(2pi) e^-n * n^(n+.5)

    3. The attempt at a solution

    Ok, I'm just confused about how you compute something with n in it....I mean, I get how to use Stirling's formula for factorials, but how do you compute that probability that is requested when there are ns in the (2n choose n)^2/(4n choose 2n)....what do I plug into stirling's formula? I don't know how to really attempt a sensible solution when I don't know what to put for n. Thanks.
  2. jcsd
  3. Jan 28, 2009 #2

    Gib Z

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    Okay well my first step would be to replace the "choose"s by their factorial definition;

    [tex] ^nC_r = \frac{n!}{r! (n-r)!} [/tex]

    Then using the properties of factorials, simplify the expression. Only once you have done all of that, everywhere you see the expression n!, replace it with Stirlings Formula.
  4. Jan 28, 2009 #3
    Ok, so I have (2n! 2n! 2n! 2n!)/(n! n! n! n! 4n!).

    Then, I cancelled n! terms to get 16/(4n!)=16/(24)n!=(2/3)*(1/n!)

    Now if I plug stirling's, this is like..... (2/3)* (1/(sqrt(2 pi)*e^-n*n^(n+0.5))

    I don't really see how this simplifies to sqrt(2/pi*n) which is the answer my teacher said we should get...what am I doing wrong/not seeing? Thanks.
  5. Jan 28, 2009 #4


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    You are using cancellation rules on the factorials that don't exist. You can't 'cancel' anything in there. Use Stirling on each factor.
  6. Jan 28, 2009 #5
    Ok so 2n!/n! is not = 2? This seems like it will get extremely messy, extremely fast since there are 9 instances of n! and how do I adjust stirling's formula for 2n!, just plug in 2n for each n in the formula?
  7. Jan 28, 2009 #6


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    2n!/n! is definitely not 2. Try some examples. 4!/2!=12. And the number of factors isn't that bad. You've got (2n!)^4/((n!)^4*4n!). Collect them into powers. And yes, to get 2n! substitute 2n for n in Stirling.
  8. Jan 28, 2009 #7

    I have ((sqrt(2pi)*e^(-2n)*2n^(2n+.5))^4/((sqrt(2pi)*e^(-4n)*(4n^(4n+2))*((sqrt(2pi)*e^(-n)*(n+.5))^4

    I cancelled out sqrt (2pi)^4 on the top and bottom and expanded a bit to get

    (e^(-16n)*16n^(8n+4))/((sqrt 2pi)*4n^(4n+.5)*e^(-8n)*n^(4n+2))=

    e^(-8n)*(4n^1.5)/(sqrt 2 pi)

    but this isn't sqrt(2/pi*n)

    I'm sorry, but my algebra is a bit rusty, so you see what I am doing wrong?
  9. Jan 28, 2009 #8
    ok, did it again and keep getting 2*(sqrt(2/pi*n) which is pretty much the right answer but an extra factor of 2...hmm where did it come from?
  10. Jan 28, 2009 #9


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    Don't know. I get sqrt(2/(pi*n)). You are pretty close. You'll have to post your whole solution if you want someone to check.
  11. Jan 28, 2009 #10
    Ok, got it, thanks
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