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Math Help - Approximation for large Catalan Numbers

  1. #1
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    Approximation for large Catalan Numbers

    Show that for a very large m, C(m) \approx \frac{4^m}{m^{\frac{3}{2}} \times \sqrt{\pi}}<br />
    where C(m) denotes the mth catalan number.

    So I did:

    C(m) = \frac{1}{m+1} \binom{2m}{m} = \frac{(2m)!}{(m+1)!m!} = \frac{(2m)(2m-1)(2m-2)}{m!} after some canceling.

    Now \lim_{m \to \infty} \frac{(2m)(2m-1)(2m-2)}{m!} and I'm stuck here... we are given Stirling's approximation formula n! \approx \sqrt{2\pi n} \times \left(\frac{n}{e}\right)^n for n >>>>1 but I do not know how to apply that here to my final step...

    Any help would be greatly appreciated!

    Thanks heaps!
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  2. #2
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    I don't agree that \frac{(2m)!}{(m+1)!m!} = \frac{(2m)(2m-1)(2m-2)}{m!} unless m = 4.

    I would recommend not to simplify (2m)!/((m+1)m!m!) but instead to apply Stirling's formula to replace each factorial.
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  3. #3
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    Yup actually I figured it out after I posted haha, thanks anyway.
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  4. #4
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    Actually I think I am still a bit stuck. I did this: (After a bit of simplification)

    <br />
C(m) \approx \frac{4^me^{-2m}}{\sqrt{\pi}m^{-m}(m+1)^{m+\frac{3}{2}}e^{-2m-1}} \approx \frac{4^me}{\sqrt{\pi}m^{-m}(m+1)^{m+\frac{3}{2}}}<br />

    But in the end the e does not cancel and there is an extra \sqrt{m+1}

    I've checked all my simplifications, they are all correct, how do I get rid of the e and \sqrt{m+1} to match what's required?

    Thanks!
    Last edited by usagi_killer; September 14th 2010 at 04:12 AM.
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  5. #5
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    Represent C(m) as (2m)!/((m+1)(m!)^2) and don't modify m+1. Then, after replacing the factorials by their Stirling approximations, note that (m+1)\sqrt{m}\sim m^{3/2}. Here f(n)\sim g(n) means \lim_{n\to\infty}f(n)/g(n)=1.
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  6. #6
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    Thanks again, actually I noticed another cool method, is this way also acceptable?

    \displaystyle{\approx \frac{4^me}{\sqrt{\pi}m^{-m}(m+1)^{m+\frac{3}{2}}}}

    \displaystyle{\approx \frac{4^me}{\sqrt{\pi} m^{\frac{3}{2}}m^{-\left(m+\frac{3}{2}\right)}(m+1)^{m+\frac{3}{2}}}}

    \displaystyle{\approx \frac{4^me}{\sqrt{\pi} m^{\frac{3}{2}} \left(\frac{m+1}{m}\right)^{m+\frac{3}{2}}}}

    \displaystyle{\approx \frac{4^me}{\sqrt{\pi}m^{\frac{3}{2}}\left(1+\frac  {1}{m}\right)^m\left(1+\frac{1}{m}\right)^{\frac{3  }{2}}}}

    Now note as m gets large \displaystyle{\left(1+\frac{1}{m}\right)^m} approachs e. Formally we can write, \displaystyle{\lim_{m \to \infty} \left(1+\frac{1}{m}\right)^m = e}

    Also note as m gets large \displaystyle{\left(1+\frac{1}{m}\right)^{\frac{3}  {2}}} approaches 1. Formally we can write, \displaystyle{\lim_{m \to \infty} \left(1+\frac{1}{m}\right)^{\frac{3}{2}} = 1}

    So substituting these back in yields:

    \displaystyle{\approx \frac{4^me}{\sqrt{\pi}m^{\frac{3}{2}}e(1)}}

    \displaystyle{\approx \frac{4^m}{\sqrt{\pi} \times (m)^{\frac{3}{2}}}} as required.
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  7. #7
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    Yes, this works as well.
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