Periodic functions

  • Thread starter Bipolarity
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  • #1
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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
 

Answers and Replies

  • #2
pwsnafu
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Can a function have two periods? If so, which is the fundamental period?
Yes, and the smallest positive number is the fundamental.

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.
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.
 
  • #3
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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.
Ahh, I see, my mistake.
Are there functions for which 2 and 3 are both periods?

BiP
 
  • #4
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Ahh, I see, my mistake.
Are there functions for which 2 and 3 are both periods?

BiP
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.
 
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  • #5
jbriggs444
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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.
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.
 
  • #6
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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.
 
  • #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##.
 
  • #8
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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##.
What would be considered the period of a constant function?

BiP
 
  • #9
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What would be considered the period of a constant function?

BiP
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.
 
  • #10
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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.
I see, but what would be the fundamental period? 0?

BiP
 
  • #11
jbriggs444
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I see, but what would be the fundamental period? 0?
What is the definition of a fundamental period? If you apply that definition, would a constant function have a fundamental period?
 
  • #12
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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?
No, because the set of periods is unbounded below?

BiP
 
  • #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.
 
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  • #14
jbriggs444
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No, because the set of periods is unbounded below?
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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