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How would I solve this? (real analysis I think)

  1. Jul 16, 2013 #1
    1. The problem statement, all variables and given/known data

    f is a function defined on natural numbers such that

    2f(n) * f(2n+1) = f(2n) *[2f(n) + 1] and

    8f(n) > f(2n) > 4f(n)

    Find value of f(12)

    3. The attempt at a solution

    [itex]f(n)=\frac{f(2n)}{2f(2n+1)-2f(2n)}[/itex]

    [itex]f(12)=\frac{f(24)}{2f(25)-2f(24)}\frac{}{}[/itex]

    2f(n) > f(n) so f(n) > 0 and f(2n) > 0

    Because [itex]\frac{f(2n)}{2f(2n+1)-2f(2n)} > 0[/itex] and f(2n) > 0 we know that

    f(2n+1) > f(2n) so [itex]f'(n) > 0[/itex]

    Other than that, I have no idea. Math level: ODEs
     
  2. jcsd
  3. Jul 16, 2013 #2
    Do you mean that the function f is given by ##f:\mathbb{N}\rightarrow\mathbb{R}##? If so, why are we using derivatives?
     
  4. Jul 16, 2013 #3

    haruspex

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    The problem is underspecified. I have a family of solutions with two partly arbitrary constants. Requiring f:N→N fixes one of them; knowing f(1) fixes the other. Whether there are other solutions, I don't know.
     
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