Can Multiple Periods Determine a Fundamental Period in Functions?

[Bipolarity]

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

 

[pwsnafu]

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.

 

[Bipolarity]

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

 

[Mark44]

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.

 

[jbriggs444]

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.

 

[Mark44]

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.

 

[LCKurtz]

On this subject, you might notice that the constant function $f(x)\equiv C$ has period $r$ for any real number $r$.

 

[Bipolarity]

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

 

[Mark44]

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.

 

[Bipolarity]

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

 

[jbriggs444]

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?

 

[Bipolarity]

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

 

[LCKurtz]

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 into a university account, you should just be able to open it. It's only about 1 page long in the "classroom notes" section.

 

[jbriggs444]

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.