1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    Zurtex

    User Avatar
    Science Advisor
    Homework Helper

    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

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Is there a function such that
Loading...