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Math Help - A Problem involving the Intermediate Value Theorem

  1. #1
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    A Problem involving the Intermediate Value Theorem

    Greets,

    Anyone out there . . . please help ! I'm stuck on this problem:

    Suppose f is a continuous function on [0,1] with f(0)=f(1). Prove that for any positive integer n, there exists some x in the interval [0,1-(1/n)] such that
    f(x+(1/n))=f(x).

    Thanks in advance.
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  2. #2
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    Quote Originally Posted by Ian Moore View Post
    Greets,

    Anyone out there . . . please help ! I'm stuck on this problem:

    Suppose f is a continuous function on [0,1] with f(0)=f(1). Prove that for any positive integer n, there exists some x in the interval [0,1-(1/n)] such that
    f(x+(1/n))=f(x).

    Thanks in advance.
    Here is a start. I will assume that f (x) \geq 0 for x\in [0,1]. Note, that if f is a function having the above property then f+k will also have the same property where k is a real number. Thus, it is safe to assume that f(0)=f(1)=0. Now define g(y) = f(y+1/n) - f(y) for y\in [0, 1-1/n] for a particular integer n\geq 2. Then g(0) = f(1/n) - f(0) = f(1/n)\geq 0 and g(1-1/n) = f(1) - f(1-1/n) = f(1-1/n)\leq 0 by IVT there is an x\in [0,1-1/n] such that g(x) = 0 and so f(x+1/n) = f(x).

    Note: I did not do the general case case when f can be positive and negative valued. Are you sure you did not forget this condition?
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  3. #3
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    Thanks !

    I've just looked at your reply. I have typed out the question exactly as I have it in front of me. There are no additional conditions on f such as you suggest. This is what makes it hard to prove. If it were just f>0 on [0,1] then it is straightforward as you suggest.

    Thankyou for responding to this question.

    Incidentally, what software are you using to provide Math type notation ?
    Is it easy to learn?
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  4. #4
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    It's called Latex and isn't really that tough to learn. Here's a link for you to check out some of the basics...

    http://www.mathhelpforum.com/math-he...-tutorial.html
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    More on the Intermediate Value Theorem

    I have tried the original problem for some time now and still have made no headway. I agree that the problem is straightforward enough with f > 0 (or f< 0).

    A further problem is this :

    Suppose 0<a<1, but a is not equal to 1/n for any natural number n. Find a
    function, f, which is continuous on [0,1] and which satisfies f(0)=f(1),
    but which does not satisfy f(x)=f(x+a) for any x.

    Again I have made no headway with this one either.
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  6. #6
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    Thanks for your response. However, I am afraid it is not very clear to me what you are saying. Would it be possible to elaborate ?
    Thanking you in advance.
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  7. #7
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    Quote Originally Posted by Ian Moore View Post
    Suppose f is a continuous function on [0,1] with f(0)=f(1). Prove that for any positive integer n, there exists some x in the interval [0,1-(1/n)] such that f(x+(1/n))=f(x).
    Here's a sketch of a proof.

    Let g(x) = f(x+(1/n)) f(x) (for 0 ≤ x ≤ 1(1/n)). Suppose that g(x) is always strictly positive. Then f(x)<f(x+(1/n)), and so
    f(0) < f(1/n) < f(2/n) < f(3/n) < ... < f(1).
    This contradicts the fact that f(0) = f(1). Therefore g is not always positive. A similar argument shows that g is not always negative. Therefore g takes both positive and negative values. It then follows from the intermediate value theorem that g must be zero at some point x. In other words, f(x+(1/n))=f(x).
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  8. #8
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    Thankyou for your suggested proof. I think I now have a proof for any f on [0,1]. Perhaps you might like to correct me. Here goes:

    If f is constant then f(x)=f(x+1/n) for every natural number n and every x in [0,1]. This disposes the case f constant.

    For f not constant, let g(w) = f(w)-f(w+(1/n)) for w in [0,1-1/n]. Then

    g(0) = f(0) - f(1/n)
    g(1/n) = f(1/n) -f(2/n)
    g(2/n) = f(2/n) -f(3/n)
    ...
    ...
    ...
    g(1-2/n) = f(1-2/n)-f(1-1/n)
    g(1-1/n)=f(1-1/n)-f(1)

    Adding, and noting that f(0)=f(1), we obtain

    g(0)+g(1/n)+g(2/n)+...+g(1-2/n)+g(1-1/n)=0.

    So not all terms are of the same sign and not all terms are zero because f is assumed to be not a constant. Thus there is an x in [0,1-1/n] such that g(x)=0. That is, there is an x such that f(x)=f(x+1/n).

    A further problem, which still remains unresolved is this :

    Suppose 0<a<1, but a is not equal to 1/n for any natural number n. Find a
    function, f, which is continuous on [0,1] and which satisfies f(0)=f(1),
    but which does not satisfy f(x)=f(x+a) for any x.
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  9. #9
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    Quote Originally Posted by Ian Moore View Post
    A further problem, which still remains unresolved is this :

    Suppose 0<a<1, but a is not equal to 1/n for any natural number n. Find a
    function, f, which is continuous on [0,1] and which satisfies f(0)=f(1),
    but which does not satisfy f(x)=f(x+a) for any x.
    If a is not of the form 1/n then 1/a is not an integer, and therefore sin(π/a)≠0. Take a look at the function f(x) = \sin^2(\pi x/a) - x\sin^2(\pi/a).
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