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Math Help - Derivative Problem

  1. #1
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    Derivative Problem

    Challenge problem: Suppose [LaTeX ERROR: Convert failed] , and denote [LaTeX ERROR: Convert failed] by [LaTeX ERROR: Convert failed] .

    For what [LaTeX ERROR: Convert failed] is it true that [LaTeX ERROR: Convert failed] ?


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    Last edited by CaptainBlack; November 4th 2010 at 02:10 PM.
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  2. #2
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    Spoiler:
    By laboriously differentiating y twice, you can check that (1-x^2)y^{(2)} - xy^{(1)} + y^{(0)} + 2x = 0. Differentiate that to get (1-x^2)y^{(3)} - 3xy^{(2)} + 0y^{(1)} + 2 = 0. One more differentiation shows that (1-x^2)y^{(4)} - 5xy^{(3)} - 3y^{(0)} = 0. But that is the given relation for n=2.

    Next, if you differentiate the given relation, assuming it is true for n, you find that it also holds for n+1. Therefore by induction it holds for all n\geqslant2.
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    Quote Originally Posted by Opalg View Post
    Spoiler:
    By laboriously differentiating y twice, you can check that (1-x^2)y^{(2)} - xy^{(1)} + y^{(0)} + 2x = 0. Differentiate that to get (1-x^2)y^{(3)} - 3xy^{(2)} + 0y^{(1)} + 2 = 0. One more differentiation shows that (1-x^2)y^{(4)} - 5xy^{(3)} - 3y^{(0)} = 0. But that is the given relation for n=2.

    Next, if you differentiate the given relation, assuming it is true for n, you find that it also holds for n+1. Therefore by induction it holds for all n\geqslant2.
    Very nice - I didn't think of induction.
    Spoiler:
    After we have the relation (1-x^2)y^{(2)} - xy^{(1)} + y^{(0)} + 2x = 0

    We can write (1-x^2)y^{(2)} - xy^{(1)} + y^{(0)} = -2x

    Applying Leibniz's rule (f \cdot g)^{(n)}=\sum_{k=0}^n {n \choose k} f^{(k)} g^{(n-k)} to the LHS gives:

    (1-x^2)y^{n+2}-2nxy^{n+1}-xy^{n+1}-n(n-1)y^{n}-ny^n+y^{n}

    = (1-x^2)y^{n+2}-(2n+1)xy^{n+1}-(n^2-1)y^{n}

    For the RHS, if u = -2x, then u_{1} = -2 and u_{n} = 0, whenever n\ge 2.

    Thus it holds for n\ge 2.
    Last edited by TheCoffeeMachine; November 6th 2010 at 11:35 AM.
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