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Proving d^l[(x-1)^l]/dx^l = l!

  1. Dec 20, 2011 #1
    1. The problem statement, all variables and given/known data

    Sorry I can't do this in latex, the PC I'm on doesn't display it and I can't confidently reel it off without checking so I'll have to do this in an ugly way! :/

    Prove:
    d^n/dx^n [(x-1)^n] = n!
    For integer n

    2. Relevant equations

    I'm trying induction but it's never been my strong point, I'm not convinced I actually understand how induction works, whenever I try and get my head around it, it seems circular..

    3. The attempt at a solution

    Testing when n=0 gives (x-1)^0 = 1 = 0! so it holds when n=0

    Now assuming true for n=k gives:
    d^k/dx^k [(x-1)^k] = k!

    If this holds for n=k, that it holds when n=k+1 must be implied by n=k being true:
    d^(k+1)/dx^(k+1) [x-1]^(k+1) = d^k/dx^k(d/dx((x-1)^(k+1))) separating the differentiation operator
    = d^k/dx^k((k+1)(x-1)^k) evaluating the inner derivative
    = (k+1)d^k/dx^k((x-1)^k) bringing the constant (k+1) outside the derivative
    = (k+1) k! using the result I'm trying to prove
    = (k+1)!

    Sorry, that's hideous.. is it OK?
     
  2. jcsd
  3. Dec 20, 2011 #2
    Yes, it is hideous, but (unless my eyes glazed over something important), your induction works just fine!
     
  4. Dec 20, 2011 #3
    OK thanks :)

    A further question:
    This proof is actually part of my solution to Q20 part C on this paper:
    http://www.maths.cam.ac.uk/undergrad/nst-pastpapers/2009/PaperNST_IA_2.pdf [Broken]

    I've managed to do the proof OK and get the required result but it seems a bit odd to me. I'm not sure why P(1) = 1 when if we sub x=0 into v, we get 0^l for all values of l greater than 0.

    If v(1) = 0 for all positive l, then surely P(1) = 0 as well since P(1) is simply the derivative of v(1) l times multiplied by a constant factor?
    Thanks

    Edit: That was dumb of me... just realised we're taking the 'l'th derivative of the function v at x=1.. just because v(1) = 0, doesn't mean its derivatives are.

    Thanks again
     
    Last edited by a moderator: May 5, 2017
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