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Math Help - Leibnitz Formula.

  1. #1
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    Leibnitz Formula.

    Okay I'm not sure if this is University level or below, but anyway

    Apply the Leibnitz theorem to the function y=x^{2n}

    and hence prove that,

    1 + (n^2)/1 + ((n^2)(n-1)^2)/(1^2)(2^2) + ........ = (2n!)/(n!)^2


    Now I can easily prove this using the Binomial theorem as the terms on the LHS are merely squares of the binomial coefficients in the expansion (1+x)^n
    And the RHS is the coefficient of x^n in the expansion of
    (1+x)^{2n}

    Using this data the above relation can be proved easily.

    But I am not able to prove it using the method the question demands, that is, by using the Lebnitz formula for derivative of a function.

    Can anyone help?
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  2. #2
    Super Member Matt Westwood's Avatar
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    Is the Leibniz Formula that thing which is used to determine the n'th derivative of a product?

    In that case, see what happens if you treat it as x^2n = x^n x^n.

    From my limited knowledge:

    D^n fg = \sum_{j=0}^n \binom n j D^j f D^{n-j} g

    You find that D^j f D^{n-j} g gives you the appropriate factorial stuff in it and it looks as though everything works. Haven't actually tried it as I'm too lazy.
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  3. #3
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    In that case, see what happens if you treat it as .
    This should do the trick. Thanks. I wonder why I didn't think of this myself! :-/
    Maybe I'm too sleepy at the moment.
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  4. #4
    Super Member Random Variable's Avatar
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     \frac{d^{n}}{dx^{n}} x^{k} = \frac{k!}{(k-n)!} x^{k-n}

     \frac{d^{n}}{dx^{n}} x^{2n} = \frac{(2n)!}{n!} x^{n}


     \frac{d^{n}}{dx^{n}} x^{2n} = \frac{d^{n}}{dx^{n}} (x^{n}\cdot x^{n}) = \sum_{r=0}^{n} \binom{n}{r} \frac{d^{r}}{dx^{r}} x^{n} \frac{d^{n-r}}{dx^{n-r}} x^{n}

     = x^{n}n!x^{0} + nnx^{n-1}n!x^{1} + \frac{n(n-1)}{2!}n(n-1)x^{n-2}\frac{n!}{2!} x^{2} +  \frac{n(n-1)(n-2)}{3!}n(n-1)(n-2)x^{n-3}\frac{n!}{3!}x^{3} + ... n!x^{0} x^{n}= \frac{(2n)!}{n!} x^{n}

     1 + \frac{n^{2}}{(1!)^{2}} + \frac{n^{2}(n-1)^{2}}{(2!)^{2}} + \frac{n^{2}(n-1)^{2}(n-2)^{2}}{(3!)^{2}} + ... + 1 = \frac{(2n)!}{(n!)^{2}}
    Last edited by Random Variable; August 21st 2009 at 12:51 PM.
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