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Math Help - Asymptotic Methods

  1. #1
    Super Member Deadstar's Avatar
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    Asymptotic Methods

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    1^1 2^2 3^3 \cdots n^n \sim Cn^{\tfrac{1}{2}n^2 + \tfrac{1}{2}n +  \tfrac{1}{12}}e^{-\tfrac{1}{4}n^2} \qquad \textrm{ as } n \to  \infty



    Moderator edit: Approved Challenge question.
    Last edited by mr fantastic; May 19th 2010 at 09:22 PM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Deadstar View Post
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    1^1 2^2 3^3 \cdots n^n \sim Cn^{\tfrac{1}{2}n^2 + \tfrac{1}{2}n +  \tfrac{1}{12}}e^{-\tfrac{1}{4}n^2} \qquad \textrm{ as } n \to  \infty
    I don't want to give the solution because I have seen this before. You do realize this is "commonly" known? The value of C is usually denoted A and is known as the Glaisher-Kinkelin constant.
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  3. #3
    Super Member Deadstar's Avatar
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    Quote Originally Posted by Drexel28 View Post
    You do realize this is "commonly" known?
    Yeah totally that's why I posted it in a maths challenge section.
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  4. #4
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    Spoiler:

    It wouldn't be a very cool proof, but you can let u_n=\log(1^12^2\cdots n^n)-\frac 12 n^2\log n-\frac 12 n\log n+\frac 14 n^2 and check that u_n-u_{n-1}=\frac{1}{12n}+O\left(\frac{1}{n^2}\right) (very simple but lengthy, or Maple can do it for you). Since \sum_{k=2}^n\frac{1}{k} = \log n -\gamma+o(1) and \sum_k\frac{1}{k^2} converges, we deduce u_n=\sum_{k=2}^n (u_k-u_{k-1})=\frac{1}{12}\log n-C+o(1). Taking exponentials gives the result...
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