Results 1 to 3 of 3

Math Help - Differentiate n times

  1. #1
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7

    Differentiate n times

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

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

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

    My attempt:

    Differentiating each side of the equation once,

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

    I don't seem to be getting anywhere.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Sep 2011
    Posts
    58

    Re: Differentiate n times

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

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

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

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

    So all of your terms follow the pattern perfectly... now just keep differentiating
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,074
    Thanks
    7

    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 (1-x^2)v^{(3)} corresponds to the (1-x^2)v^{n+2}

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

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

    And now 2nv' = 2v' which corresponds to 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,

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

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

    Differentiating again,

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

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

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

    Differentiating again,

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

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

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

    Claim: After m differentiations, the equation becomes

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

    Substituting m=n, we get the required result.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Simplify then differentiate times 3
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 5th 2010, 11:26 AM
  2. Replies: 1
    Last Post: February 7th 2010, 01:18 AM
  3. How do I differentiate a vector times vector?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 13th 2009, 11:12 PM
  4. Replies: 4
    Last Post: February 8th 2009, 09:30 AM
  5. Replies: 2
    Last Post: December 9th 2007, 03:33 PM

Search Tags


/mathhelpforum @mathhelpforum