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Is there a function such that

  1. Nov 16, 2004 #1
    ... f(0)=f(1) but for all x in [0,3/5] f(x) does not equal f(x+2/5) ???
  2. jcsd
  3. Nov 16, 2004 #2


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    I would say probably yes. Try splitting up the interval into (lots of) groups of points that differ by 2/5 and see if you can assign values to each point.

    Now, have you stated the whole question? Is f supposed to be, say, continuous?
  4. Nov 16, 2004 #3


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    Surely there are an infinite amount of functions that satisfy those conditions?
  5. Nov 16, 2004 #4
    Function is continuous

    Yes, unfortunately, the function is continuous.
  6. Nov 16, 2004 #5


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    Well, you want to make sure that f(x) does not equal f(x+2/5)...

    A common trick is to rewrite this constraint so that a part of it is constant: you want to make sure that f(x) - f(x+2/5) does not equal 0.
  7. Nov 17, 2004 #6
    so, i've gotten to the point at which g(x) = f(x) - f(x+2/5) does nto equal zero for any x in [0, 3/5].... finding a form for g(x) is not difficult, but the constraint f(0) = f(1), i have no idea how to include it.
  8. Nov 26, 2004 #7
    If f(x) is continuous, I'd say there's no such func. Here's the reason:

    1.The simplest case, that f(x) have no zero points in (0,1), so the curve is like bridge. For instance, let f(x)=sin(4*pi*x), and g(x)=f(x)-f(x+2/5), then
    g(0)<0, g(3/5)>0, and since g(x) is continuous, there should be at least one zero point in (0,3/5).

    2.Then if f(x) have zero points in (0,1), however, the result is the same. Suppose f(x)=sin(2*pi*x), and in part of (0,1)--in this case, (0,1/2),we can find zero points in (0,1/2-2/5), the reason is the same as in case 1.
    Last edited: Nov 27, 2004
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