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Thread: Differentiate n times

  1. #1
    MHF Contributor alexmahone's Avatar
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    Differentiate n times

    $\displaystyle (1-x^2)v''+2(n-1)xv'+2nv=0$

    Differentiate each side of the equation $\displaystyle n$ times in succession to obtain

    $\displaystyle (1-x^2)v^{(n+2)}-2xv^{(n+1)}+n(n+1)v^{(n)}=0$

    My attempt:

    Differentiating each side of the equation once,

    $\displaystyle (1-x^2)v^{(3)}-2xv''+2(n-1)(xv''+v')+2nv'=0$

    I don't seem to be getting anywhere.
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  2. #2
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    Re: Differentiate n times

    actually you're doing just fine =). After differentiating once we have n=1. So the $\displaystyle (1-x^2)v^{(3)}$ corresponds to the $\displaystyle (1-x^2)v^{n+2}$

    Then $\displaystyle -2xv''$ corresponds to $\displaystyle -2xv^{n+1}$

    Now the part you prolly didnt realize, since $\displaystyle n=1$, we have $\displaystyle 2(n-1)(xv''+v')=2(1-1)(xv''+v')=0$

    And now $\displaystyle 2nv' = 2v'$ which corresponds to $\displaystyle n(n+1)v^{(n)} = 1(1+1)v^{(1)}=2v'$

    So all of your terms follow the pattern perfectly... now just keep differentiating
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Re: Differentiate n times

    Quote Originally Posted by MonroeYoder View Post
    actually you're doing just fine =). After differentiating once we have n=1. So the $\displaystyle (1-x^2)v^{(3)}$ corresponds to the $\displaystyle (1-x^2)v^{n+2}$

    Then $\displaystyle -2xv''$ corresponds to $\displaystyle -2xv^{n+1}$

    Now the part you prolly didnt realize, since $\displaystyle n=1$, we have $\displaystyle 2(n-1)(xv''+v')=2(1-1)(xv''+v')=0$

    And now $\displaystyle 2nv' = 2v'$ which corresponds to $\displaystyle n(n+1)v^{(n)} = 1(1+1)v^{(1)}=2v'$

    So all of your terms follow the pattern perfectly... now just keep differentiating
    Differentiating each side of the equation once,

    $\displaystyle (1-x^2)v^{(3)}-2xv''+2(n-1)(xv''+v')+2nv'=0$

    $\displaystyle (1-x^2)v^{(3)}+2(n-2)xv''+2(2n-1)v'=0$

    Differentiating again,

    $\displaystyle (1-x^2)v^{(4)}-2xv^{(3)}+2(n-2)(xv^{(3)}+v'')+2(2n-1)v''=0$

    $\displaystyle (1-x^2)v^{(4)}+2(n-3)xv^{(3)}+6(n-1)v''=0$

    $\displaystyle (1-x^2)v^{(4)}+2(n-3)xv^{(3)}+3(2n-2)v''=0$

    Differentiating again,

    $\displaystyle (1-x^2)v^{(5)}-2xv^{(4)}+2(n-3)(xv^{(4)}+v^{(3)})+6(n-1)v^{(3)}=0$

    $\displaystyle (1-x^2)v^{(5)}+2(n-4)xv^{(4)}+(8n-12)v^{(3)}=0$

    $\displaystyle (1-x^2)v^{(5)}+2(n-4)xv^{(4)}+4(2n-3)v^{(3)}=0$

    Claim: After $\displaystyle m$ differentiations, the equation becomes

    $\displaystyle (1-x^2)v^{(m+2)}+2(n-m-1)xv^{(m+1)}+(m+1)(2n-m)v^{(m)}=0$

    Substituting $\displaystyle m=n$, we get the required result.
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