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Finding the nth+1 derivative via Leibniz

  1. Feb 14, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm told to find the nth derivative of a function via Leibniz.


    2. Relevant equations

    Then I'm asked to show that f(n+1)=f(n)+3n(n-1)f(n-2) evaluated at x=0 when n>1


    3. The attempt at a solution

    I can solve this just fine when plugging in a constant for n. For example, when n=2, f '" = f " + 6 f, no problem. My question though is whether this is enough to prove that the above equation balances for all n>1?

    If not, how would I show this? Would I need to use the general form of f(n) I found via the Leibniz formula? How would this work? I ask this because it is of the form f(n) = Ʃk=0n (n!/k!(n-k!) * (dk/dxk*f) * (dn-k/dxn-k*g)) and am not sure how I would transform that into f(n+1) or f(n-2) with the summation and combination.

    Thank you in advance!
     
  2. jcsd
  3. Feb 14, 2012 #2

    vela

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    Are you using f to stand for more than one function here? It appears on both the left and in the summation. How does g fit into all of this?
     
  4. Feb 14, 2012 #3

    SammyS

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    It looks like a straight up proof by induction. You claim to now how to do the base step: that's showing it's true when n=2 .

    For the inductive step:
    Assume the statement is true when n = k .

    From that assumption (plus your mathematical skill) show that the statement is true for n = k+1.​

    In other words:

    Assume that the following is true for k > 1 :
    f(k+1) = f(k)+3k(k-1)f(k-2) evaluated at x=0 .​

    Show that the following follows from this assumption :
    f((k+1)+1) = f(k+1)+3(k+1)((k+1)-1)f((k+1)-2)
    which simplifies to:

    f(k+2) = f(k+1)+(3k+3)(k)f(k-1)
    All evaluated at x=0 .
     
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