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Math Help - Not sure where this should exactly go, but...

  1. #1
    Senior Member Pinkk's Avatar
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    Not sure where this should exactly go, but...

    ...For my proofwriting class, I have to show that  n! > a^{n} when  n \geq A , with a,n \in \mathbb{N}. I know the eventual proof will be by induction, but I am not sure where to determine what the value of A. Messing around with the numbers 1-9, I was able to generalize it to two cases: If a is even, then A = \frac{5a-2}{2} and if a is odd, then A = \frac{5a-1}{2} , but is there a way to generalize what A must equal for ANY natural number a?

    Edit: Looking at  a > 9, my generalization doesn't seem to hold. Now I am definitely confused.

    Edit2: And just to clarify, I want A to be the least possible value.
    Last edited by Pinkk; October 19th 2009 at 07:02 PM.
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  2. #2
    MHF Contributor
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    I'm going to use the perks of my Admin status and bump this question. I'm stumped right now but I'm sure some of our other members could help you out.

    EDIT: I've been thinking about this more and more and the only thing I could come with was to use Stirling's Approximation to help choose a generalized A.

    n! \approx \sqrt{2 \pi}n^{n+.5}e^{-n}
    Last edited by Jameson; October 26th 2009 at 04:05 PM.
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  3. #3
    Senior Member Pinkk's Avatar
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    Well, I went ahead with an induction proof using A=\frac{5a-1}{2} when A is odd and A=\frac{5a-2}{2} when A is even (I rechecked and those two cases of A always hold). I showed it to my professor and she said it was good.

    Someone also showed me a neat way of doing it for even numbers

    n \ge 2a^{2} where n=2m, so m \ge a^{2}

    It follows then that:

    n!=(2m)(2m-1)\cdot \cdot \cdot (m+1)(m)\cdot \cdot \cdot(2)(1) > (2m)(2m-1)\cdot \cdot \cdot(m+1)  > a^{2}\cdot \cdot \cdot a^{2} = (a^{2})^{m} = a^{2m} = a^{n}
    Last edited by Pinkk; October 28th 2009 at 07:06 AM.
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