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A question about a function

  1. Dec 6, 2011 #1
    Suppose fn:[0,p/q] → ℝ be a function defined by :
    [itex]f_{n}(x) = \frac{x^{n}(1-qx)^{n}}{n!}[/itex] where p,n and q are natural numbers .

    Is that true that [itex]f_{n}^{(2n)}[/itex] is always an integer for any natural number n .

    Thanks .
  2. jcsd
  3. Dec 6, 2011 #2


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    The function is a polynomial with highest order term qnx2n/n! Taking 2n derivatives, the lower order terms all = 0. This highest term ends up as qn(2n)!/n!, so the answer is yes.
  4. Dec 7, 2011 #3

    ok that is was I tried to show but the step I was stuck in that if we differentiate x^n , ntimes we will have n! , so I don't know if I have proved it in the right way , I had done this by induction , where if n=1 , we will have (x^1)' = 1 = 1! , and hence we will assume that this true for any k , and then we will prove that this true for k+1 as following

    [itex]\frac{d^{k+1}}{dx^{k+1}} x^{k+1} = \frac{d^{k}}{dx^{k}} ( \frac{d}{dx} x^{k+1} ) = \frac{d^{k}}{dx^{k}} (k+1)(x) = (k+1) \frac{d^{k}}{dx^{k}} x^{k} = (k+1).k! = (k+1)![/itex]

    And hence we have the result is true . Afterthat I use the binomial theorem to expand the polynomial fn and differentiate each term 2n times all will be zero except the leading term and then our result follows . is these arguments are true
  5. Dec 7, 2011 #4


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    As far as I can tell you are saying the same thing I did.
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