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Math Help - Find a function not satisfying f(x) = f(x+a)

  1. #1
    Junior Member nimon's Avatar
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    Find a function not satisfying f(x) = f(x+a)

    So I'm trying to find a solution to this question. In part (a), I have shown that for a function f which is continuous on [0,1] satisfying f(0) = f(1) then for any number n\in\mathbb{N} there is some x\in[0,1-1/n] for which f(x) = f(x+1/n). I did this by considering the function g(x) = f(x) - f(x+1/n) and assuming g(x) \neq 0 for any x. We can then show that either f(1) > f(0) or f(1) < f(0) by noting that, since g is continuous we must have g(x)<0 or g(x) > 0 for all x. So, in the latter case for example, we have  g(x) > 0 \quad \Rightarrow \quad f(0) > f(1/n) > f(2/n) > \ldots > f(1).


    This clearly relies on the fact that n \cdot \frac{1}{n} = 1 with n\in\mathbb{N}.


    Now for part (b), which has me totally stumped. For 0<a<1 with a \neq 1/n for any n\in\mathbb{N}, find a function f continuous on [0,1] satisfying f(1) = f(0) for which f(x) \neq f(x + a) for any x \in [0,1-a]. I just can't see how that could work. Imagine any function, and connect the points (x,f(x)) and (x+a,f(x+a)) with a straight line - surely if we have f(0) = f(1) this line must be horizontal at some point!

    How do I find such an f?
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  2. #2
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    Re: Find a function not satisfying f(x) = f(x+a)

    What about a constant function?
    If f(x)=c , where c is any constant then we can't find any a for which f(x)=f(x+a)
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  3. #3
    Junior Member nimon's Avatar
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    Re: Find a function not satisfying f(x) = f(x+a)

    On the contrary, we would then have f(x) = f(x+a) = c for all x\in[0,1-a].
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    Junior Member nimon's Avatar
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    Re: Find a function not satisfying f(x) = f(x+a)

    Ok, I found the solution online. The graph of f can be a saw function starting at f(0) and with f(a)>0which gradually climbs. Where 1/(n+1) < a < 1/n it has local minima of -kf(a) at 1-ka for k = 0 \ldots n and local maxima of f(1-ka)+(n+1)f(a) at ka. Now you can just connect the minima and maxima with straight lines. If you sketch it, you'll see that f(x+a) -f(a) = f(a) > 0 for all appropriate x and that f(1) = 0.

    Now I've seen a solution, I think this was probably easier than I was making out. Did get me stuck for a bit though
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    Re: Find a function not satisfying f(x) = f(x+a)

    It's perfect
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    Re: Find a function not satisfying f(x) = f(x+a)

    I found this to be an interesting challenging problem. Anyway, I thought up a solution similar to the one you describe, but thought I'd post it anyway. The first image is of such a function f (a is .3) and the dotted red graph is f(x+a):

    Find a function not satisfying f(x) = f(x+a)-mhfcalc.png

    The entire description is in two image files (I find it much easier to use my own math editor and paste images):

    Find a function not satisfying f(x) = f(x+a)-mhfcalc1.png

    Find a function not satisfying f(x) = f(x+a)-mhfcalc2.png
    Thanks from nimon
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  7. #7
    Junior Member nimon's Avatar
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    Re: Find a function not satisfying f(x) = f(x+a)

    Thank you for that johng,

    I'm really impressed (and a bit jealous!) that you achieved this on your own! Apologies for not putting a spoiler alert on the thread, I didn't realise people were working on it. My own image is drawn below with a bit of labelling in case anyone is confused by my sloppy description above!

    The function from my previous post.

    In case you're interested, the problem is question 19.(b) of chapter 7 from Calculus by Spivak. This book has lots of problems that make you think like this and I highly recommend it.
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