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Periodic functions

  1. Mar 6, 2014 #1
    Can a function have two periods? If so, which is the fundamental period?

    Consider the following function, $$ f : \mathbb{N} → \mathbb{R} $$, defined by
    f[n] = 1 if n is a multiple of 2 or 3, and 0 otherwise.
    Then it is clear that 2 and 3 are both periods of this function, since translation of the input by either 2 or 3 renders the function's value invariant.

    6, being the least common multiple of 2 and 3, is also "a period" of this function. But which is the fundamental period?

    Thanks to anyone who can clarify this confusion!

    BiP
     
  2. jcsd
  3. Mar 6, 2014 #2

    pwsnafu

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    Yes, and the smallest positive number is the fundamental.

    Well

    f(1) = 0
    f(2) = 1
    f(3) = 1
    f(4) = 1
    f(5) = 0
    f(6) = 1

    But ##1 = f(3) \neq f(3+2) = f(5) = 0## so 2 isn't a period.
     
  4. Mar 6, 2014 #3
    Ahh, I see, my mistake.
    Are there functions for which 2 and 3 are both periods?

    BiP
     
  5. Mar 6, 2014 #4

    Mark44

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    pwsnafu already answered this question by his example in post #2. f(2) = f(4) = f(6) = 1, and f(3) = f(6) = 1.

    (Embarrassed mod note): Disregard what I wrote: 2 is NOT a period of this function.
     
    Last edited: Mar 6, 2014
  6. Mar 6, 2014 #5

    jbriggs444

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    pwsnafu answered in the negative. This f is not periodic with period 2. Nor is it periodic with period 3.

    If any f() is periodic with both period x and with period y then it is clear that it is also periodic with every period that is a non-zero sum of integer multiples of x and y. In particular, it must be periodic with period (y-x).

    3-2 = 1. It folllows that in order for f() to be periodic with period 3 and with period 2 that it must then also be periodic with period 1.
     
  7. Mar 6, 2014 #6

    Mark44

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    jbriggs, yes you are correct. I was lulled into thinking that since f(2) = f(4) = f(6) = 1, that 2 was a period. Not so. Thanks for the correction.
     
  8. Mar 6, 2014 #7

    LCKurtz

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    On this subject, you might notice that the constant function ##f(x)\equiv C## has period ##r## for any real number ##r##.
     
  9. Mar 6, 2014 #8
    What would be considered the period of a constant function?

    BiP
     
  10. Mar 7, 2014 #9

    Mark44

    Staff: Mentor

    Pick any number a, let b be an arbitrary real number. Then, since C = f(a) = f(a + b) = C, the period is b. That's pretty much what LCKurtz was saying.
     
  11. Mar 7, 2014 #10
    I see, but what would be the fundamental period? 0?

    BiP
     
  12. Mar 7, 2014 #11

    jbriggs444

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    What is the definition of a fundamental period? If you apply that definition, would a constant function have a fundamental period?
     
  13. Mar 7, 2014 #12
    No, because the set of periods is unbounded below?

    BiP
     
  14. Mar 7, 2014 #13

    LCKurtz

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    You might find this short article in the American Mathematical Monthly interesting:

    R. H. Cox and L. C. Kurtz. 1966. Real periodic functions. Am.Math.Mon.,73,761

    Here's a link to the article:

    http://www.jstor.org/stable/2313992

    If you are logged in to a university account, you should just be able to open it. It's only about 1 page long in the "classroom notes" section.
     
    Last edited: Mar 7, 2014
  15. Mar 7, 2014 #14

    jbriggs444

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    Yes. Though I might quibble that the set of [positive] periods is bounded below (by zero). But that lower bound is not member of the set. Accordingly, the set has no minimum element.
     
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