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Analysis Function Question

  1. Oct 15, 2013 #1
    1. The problem statement, all variables and given/known data
    Suppose f is continous on [0,1] and f(0)=f(1). Let n be a natural number. Prove that there is some number x, such that f(x)=f(x+1/n).


    2. Relevant equations
    The hints says to consider g(x)=f(x)-f(x+1/n)

    3. The attempt at a solution
    I've tried to consider the function g(x), but I haven't gotten anything useful from it. When I've tried various values of n, like 1/2, 1/3...I've noticed that there are repeating terms and I can manipulate the terms a bit to get like g(0)=-g(1/2) for n=1/2 and the like, but I am not sure where to go with this
     
  2. jcsd
  3. Oct 15, 2013 #2
    What's wrong with x =0 and n=1?
     
  4. Oct 15, 2013 #3

    jbunniii

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    You don't get to choose ##n##. It is chosen for you. In other words, you have to show there is a solution for ANY ##n##.
     
  5. Oct 15, 2013 #4

    jbunniii

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    If ##g(0) = 0## then you can simply take ##x = 0##. If ##g(0) > 0##, then I claim there must be some other point ##x## such that ##g(x) < 0##, and then you can apply the intermediate value theorem. Can you prove this claim? Hint: consider ##g(0) + g(1/n) + \ldots + g((n-1)/n)##.
     
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