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    I think I have the answer to this, but am not 100%.

    Let f:[0,1]-->[0,1] be continuous such that f(0)=f(1)=0.
    Show that for any p such that 0<p<1, there exists a c such that 0<c<1-p and f(c)=f(c+p).

    Here's my answer:

    Given f(0)=f(1)=0. Rolle's Theorem states that for a continuous function where f(a)=f(b), there exists at least one point in (a,b) where the derivative is zero.

    Since 0<p<1, there exists a point 0<c<1-p, or p<c+p<1.

    Hence, 0<c<c+p<1. We know by Rolle's Theorem that there exists at least one point such that the derivative is zero.

    This gives f(c)=f(c+p) for some 0<p<1 and 0<c<1-p.



    Thanks in advance
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    If all you know is that f is continuous, then you can't assume it's differentiable, so you can't possibly use Rolle.
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    (Original post by Don John)
    I think I have the answer to this, but am not 100%.

    Let f:[0,1]-->[0,1] be continuous such that f(0)=f(1)=0.
    Show that for any p such that 0<p<1, there exists a c such that 0<c<1-p and f(c)=f(c+p).

    Here's my answer:

    Given f(0)=f(1)=0. Rolle's Theorem states that for a continuous function where f(a)=f(b), there exists at least one point in (a,b) where the derivative is zero.

    Since 0<p<1, there exists a point 0<c<1-p, or p<c+p<1.

    Hence, 0<c<c+p<1. We know by Rolle's Theorem that there exists at least one point such that the derivative is zero.

    This gives f(c)=f(c+p) for some 0<p<1 and 0<c<1-p.



    Thanks in advance
    you have assumed that the continuous function is differentiable which isn't necessarily true - look up Weirstrass' function or something like that
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    Intermediate Value Theorem is the way to go here.
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    (Original post by DFranklin)
    Intermediate Value Theorem is the way to go here.
    You thinking
    Spoiler:
    Show
    for 0 \leq x \leq 1-p define g(x) = f(x+p)-f(x), then g(0) \geq 0, g(1-p)\leq 0. You can use the IVT at this point, but then it maybe only an end of the interval for which g(x) is zero, thus not satisfying the conditions that g(x) needs to be zero for some 0<x<1-p
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    Edit: Didn't read the question properly.

    -----

    Here's my counter-example so that people can point out if I'm being stupid:

    Spoiler:
    Show
    Statement we're contradicting:
    "Let f:[0,1]-->[0,1] be continuous such that f(0)=f(1)=0.
    Then for any p such that 0<p<1, there exists a c such that 0<c<1-p and f(c)=f(c+p)."

    Let f(x)=sin(2pi*x) and p=3/4. Then for any 0<c<1/4, f(c)>0 and f(c+p)<0.

    sin(x) is differentiable, so even when Rolle applies it doesn't imply the result.
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    (Original post by harr)
    Here's my counter-example so that people can point out if I'm being stupid:

    Spoiler:
    Show
    Statement we're contradicting:
    "Let f:[0,1]-->[0,1] be continuous such that f(0)=f(1)=0.
    Then for any p such that 0<p<1, there exists a c such that 0<c<1-p and f(c)=f(c+p)."

    Let f(x)=sin(2pi*x) and p=3/4. Then for any 0<c<1/4, f(c)>0 and f(c+p)<0.

    sin(x) is differentiable, so even when Rolle applies it doesn't imply the result.
    sin goes negative, function range stipulates otherwise.
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    (Original post by elixir)
    sin goes negative, function range stipulates otherwise.
    So it does. You'd have thought I'd have got the hang of reading the question after ~14 years at school and a couple at university...
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    (Original post by harr)
    So it does. You'd have thought I'd have got the hang of reading the question after ~14 years at school and a few at university...
    happens all the time

    I admit I thought exactly the same thing s you at first (same function), then thought they wouldn't ask a question with such an obvious counter example, and looked again.

    Then I thought IVT, as DFranklin also suggested, but Im not sure about that, as I say in the spoiler.
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    harr says something stupid (Mk.II)
    f=/\_
    p=3/4
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    (Original post by harr)
    harr says something stupid (Mk.II)
    f=/\_
    p=3/4
    if the interval I for which f(x) > 0 in your example was such that the length of 1/4&lt;\|I\| &lt; 3/4, then that seems to be a counter example. If the OP relaxed his condition to 0\leq c\leq 1-p, then it wouldn't be a counter example, since then c=0 would do.
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    (Original post by elixir)
    if the interval I for which f(x) > 0 in your example was such that the length of , then that seems to be a counter example. If the OP relaxed his condition to , then it wouldn't be a counter example, since then c=0 would do.
    I was going for f(x) being something like x (0<x<1/4), 0.5-x (1/4<x<1/2), 0 (x>1/2). So we relax the conditions and conclude that IVT is the way to go.
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    For info: I posted a bit off the cuff because I'd seen a version of this question where IVT worked; of course, it's much easier to remember the general 'idea' than it is the exact details on the conditions.

    My suspicion here is that this is supposed to be "easy" and the examiners missed the fact that c could end up as an endpoint of the range.

    (The version I remember didn't have the f>=0 condition, and I think you had that p <=1/2. It's actually a bit harder than I think this Q was supposed to be).

    N.B. I posted this yesterday but TSR seemed to have a DB error at the same time - thought it had gone through but it obviously didn't!
 
 
 
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