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Interesting problem from my analysis class

  1. Nov 1, 2005 #1
    Let n be a positive integer and suppose [tex]f[/tex] is continuous on [tex][0,1][/tex] and [tex]f(0) = f(1)[/tex]. Prove that the graph of [tex]f[/tex] has a horizontal chord of length [tex]1/n[/tex]. In other words, prove there exists [tex]x \in [0,(n - 1)/n][/tex] such that [tex]f(x+1/n) = f(x)[/tex]
    Last edited: Nov 1, 2005
  2. jcsd
  3. Nov 2, 2005 #2
    no one even wants to try?
  4. Nov 7, 2005 #3
    hint: write f(1) - f(0) as a telescoping sum.
  5. Nov 8, 2005 #4

    Tom Mattson

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    I would hazard a guess that people are suspicious that this is your homework.
  6. Nov 13, 2005 #5


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    Is f differentiable, or just continuous?
  7. Nov 13, 2005 #6


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    seems trivial. indeed trivial for all numbers less than or equal to 1, not just 1/n.

    after thinking about an actual proof for a few minutes let me rephrase that as "intuitively plausible", rather than "trivial".

    it apparently follows from the intermediate value theorem but the details seem tedious, even elusive. cute problem.
    Last edited: Nov 13, 2005
  8. Nov 14, 2005 #7


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    assume f(x) non negative, then what?
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