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B Is it impossible to prove that some functions are periodic?

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  1. Dec 25, 2016 #1
    Heres an example.
    AyDL3ws.jpg
    Let G(s) be the moving average of all previous values of f(s).

    G(s) and F(s) intersect at multiple points. Is it possible to prove that the intersections happen periodically?
     
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  3. Dec 25, 2016 #2

    pwsnafu

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    Could you explain what the 1 over infinity is supposed to be?
     
  4. Dec 25, 2016 #3
    That's the summation term from 1 to infinity. The software garbled it.
     
  5. Dec 25, 2016 #4

    mfb

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    Like this?$$F(s) = \sum_{n=1}^{\infty} \frac{1}{n^{s i +2}}$$
    What do you mean by "moving average over all previous values", and does f(s) mean F(s)?
    Do you have a proof that they intersect at all? With complex numbers this is not trivial.
     
  6. Dec 25, 2016 #5
    I just graphed the functions. If you dont know what a moving average is then look it up.

    As I suspected this problem seems very difficult or impossible to solve.
     
  7. Dec 26, 2016 #6

    mfb

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    I know what a moving average is, but the definition is not unique (average over which range?), and your description doesn't make sense.

    Did you plot both the real and imaginary part?

    Currently the problem is not even well-defined.
     
  8. Dec 26, 2016 #7
    I plotted both the real and imaginary parts.

    The range is over the entire function, or if your software does not support that, then try from the origin to S.
     
  9. Dec 27, 2016 #8

    mfb

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    That doesn't make sense either.
    How did you define G(s), can you show the formula?
    And they had intersections at the same point?
     
  10. Dec 27, 2016 #9
    I don't have the formula handy. You just average all the previous points.

    They intersected at different points. I want to know if either or both of the graphs are periodic.

    Im not too concerned about the specifics. If you can answer anything related to this question then that would be interesting. I am just asking if its always possible to prove whether a function is periodic.
     
  11. Dec 28, 2016 #10

    mfb

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    That would work as description if there would be a finite set of those points, but you don't have that. Average from 0 to s? Is your function defined for negative s? Make some limit procedure to "average" from -infinity to s? Something else?
    Which means the functions do not intersect. Only their real and imaginary parts.
    A clear no. You can construct functions that are periodic if and only if some turing machine halts, and there is no algorithm that can determine this for all turing machines (halting problem). Those function definitions will look weird, however (like "f(x)=sin(x) if this turing machine halts, f(x)=x if it does not").
     
  12. Dec 28, 2016 #11
    Yes

    Yes

    Is that a proof? Are you sure they never intersect?

    You seem to be claiming I claimed they always intersect at different points. No, I said they intersected at some points and the parts intersected at others.

    Anyway you basically answered my questions to a good enough level with your last paragraph.
     
    Last edited: Dec 28, 2016
  13. Dec 28, 2016 #12

    mfb

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    No, but you claimed they would intersect, and I don't see evidence for that so far.
    An intersection of two functions requires that the function values are identical - for complex numbers, this means both real and imaginary part have to be identical.
     
  14. Dec 28, 2016 #13
    I really don't see how that helps. If you have any more questions about the function let me know, but it seems pretty well defined for me. Cant you just use the moving average feature in matlab? I really don't see where the ambiguity was, I am able to graph the function fine.
     
  15. Dec 29, 2016 #14

    mfb

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    It is well-defined now, after 10 posts with multiple rounds of questions how it is defined.
    The range of definition and the range you average over.

    In addition, the first post had the (probably wrong, but certainly not backed by evidence) claim that the functions intersect.
     
  16. Dec 29, 2016 #15
    So basically, the example problem is well defined and unanswerable.
     
  17. Dec 29, 2016 #16

    jbriggs444

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    Nothing in this thread has indicated that the example problem (as clarified in #11) is unanswerable. Only that it is unanswered and that it might be unanswerable.
     
  18. Dec 30, 2016 #17
    So its unanswerable, as far as anyone knows.
     
  19. Dec 30, 2016 #18

    jbriggs444

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    And it's answerable, as far as anyone knows. Might be answerable, might not be. We do not know. Trying to spin it one way or the other does not change that.
     
    Last edited: Dec 31, 2016
  20. Dec 31, 2016 #19

    Mark44

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    Please dial back the attitude.
    People give help here out of the kindness of their hearts -- none of us gets paid for this. Also, I guarantee that @mfb knows what a moving average is, but that's not what he asked:
    Your question was unclear from the start, and it took 11 posts to figure out what you were asking. It's not the responsibility of responders to go "look it up" -- it's your responsibility to ask a question that makes sense, and with as much information as you can provide.
     
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