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Finding function, simplifying the summation

  1. Jul 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Let ## n \geq 2## be a fixed integer. ##f(x)## be a bounded function defined in ##f:(0,a) \rightarrow R## satisfying
    [tex]f(x)=\frac{1}{n^2}\sum_{r=0}^{(n-1)a} f\left(\frac{x+r}{n}\right)[/tex]
    then ##f(x)## =
    a)-f(x)
    b)2f(x)
    c)f(2x)
    d)nf(x)


    2. Relevant equations



    3. The attempt at a solution
    I see no way of simplifying the summation. I need a few hints to start with.

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. Jul 9, 2013 #2
    I don't see any way to simplify it at the first glance. If I spot anything, I will edit this post.

    Edit: Don't try to simplify the sum, because it won't turn out so nicely anyway. Instead, inspect the functional equations given to you in the options and try substituting them back into the original sum formula.
     
    Last edited: Jul 9, 2013
  4. Jul 9, 2013 #3
    That gives a), b) and d), what about the c) option?
     
  5. Jul 9, 2013 #4

    haruspex

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    (a), (b) and (d) are equivalent to f(x)=0, no?
    What does the range limit mean when (n-1)a is not an integer?
     
  6. Jul 9, 2013 #5
    I figured out a), b) and d). :)

    I can't understand what you ask me here.
     
  7. Jul 9, 2013 #6

    haruspex

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    The OP doesn't say whether a is an integer. If it isn't, (n-1)a need not be either. In that case I'm not sure how to interpret the sum. Does it mean the sum up to the greatest integer <= (n-1)a?
    Anyway, assuming a is an integer, I would start by seeing what a constant solution would look like. Since f is not given as continuous, you could then see what changing f at only a countable set of points looks like (e.g. at a2-m, m=1, 2...). That still satisfies (c). Admittedly that's backwards; just because it satisfies (c) does not mean it should satisfy the given condition, but it might yield some insight.
     
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