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Thread: Derivative Problem

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    Derivative Problem

    Challenge problem: Suppose $\displaystyle \displaystyle y = \sqrt{1-x^2}\arcsin{x}$, and denote $\displaystyle \displaystyle \frac{d^ky}{dx^k}$ by $\displaystyle \displaystyle y_{k}$.

    For what $\displaystyle \displaystyle n$ is it true that $\displaystyle \displaystyle (1-x^2)y_{n+2}-(2n+1)xy_{n+1}-(n^2-1)y_{n} = 0$?


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    Last edited by CaptainBlack; Nov 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 $\displaystyle (1-x^2)y^{(2)} - xy^{(1)} + y^{(0)} + 2x = 0$. Differentiate that to get $\displaystyle (1-x^2)y^{(3)} - 3xy^{(2)} + 0y^{(1)} + 2 = 0$. One more differentiation shows that $\displaystyle (1-x^2)y^{(4)} - 5xy^{(3)} - 3y^{(0)} = 0$. 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 $\displaystyle n\geqslant2$.
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    Quote Originally Posted by Opalg View Post
    Spoiler:
    By laboriously differentiating y twice, you can check that $\displaystyle (1-x^2)y^{(2)} - xy^{(1)} + y^{(0)} + 2x = 0$. Differentiate that to get $\displaystyle (1-x^2)y^{(3)} - 3xy^{(2)} + 0y^{(1)} + 2 = 0$. One more differentiation shows that $\displaystyle (1-x^2)y^{(4)} - 5xy^{(3)} - 3y^{(0)} = 0$. 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 $\displaystyle n\geqslant2$.
    Very nice - I didn't think of induction.
    Spoiler:
    After we have the relation $\displaystyle (1-x^2)y^{(2)} - xy^{(1)} + y^{(0)} + 2x = 0$

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

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

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

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

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

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