Function with two periods

Office_Shredder

Suppose a function f has two periods a, b with a =/= n*b or vice versa. Then is f necessarily a constant? What if a is not a rational multiple of b? Answers involving whether f is continuous or not are appreciated. Basically, here's now I started:

WLOG, suppose b > a. Then b-a < b, and either a < b-a or a > b-a. Either way, we've found a third period which does not satisfy the integral (or rational) multiplicity requirement. Since one of these periods is smaller than the other, we can subtract the smaller from the larger to get another smaller period. This process should be repeatable infinite times to get infinite discrete periods which tend to zero. But what is this enough for?

EX:

Suppose p is Pi and e is, well, e. Then p-e is a period, and e-(p-e) = 2e-p is a period. 2e-p - (p-e) = 3e-2p is a period, and so is 3e -2p + (2e-p) 5e-3p. So you can make a whole bunch of periods.... I don't know which of the periods I have listed here is the smallest (I'm too lazy to get a calculator out and do this for more than a couple runs), but I think you can see the process of shrinking the period. So you'll get periods of the form k_i*a - m_i*b, where, I believe, k_i/m_i approaches b/a.

Thoughts?

morphism

http://en.wikipedia.org/wiki/Doubly-periodic_function

Gib Z

Ahh Im not too sure about anything, but i was wondering how a function could have to periods, and then I realised the most obvious answer of extending in more diemensions :)

uart

Actually I think the OP was referring to one dimentional functions something like y=sin(x) + sin(Pi x) for example.

Rather than "having two period" (as far as I can see) such a function as aperiodic.

NateTG

There are also fractal functions. For example, consider the characteristic function of the rationals (f(x)=1 if x is rational, and 0 otherwise) which has a periods equal to any rational number. Or, if you prefer non-rational ratios, you could go with the characteristic function of the algebraic numbers (restricted to the real line).

Office_Shredder

I was referring to one dimensional functions. I'm fairly certain that a function can't have two periods that aren't multiples of each other, but am wondering how to prove that

matt grime

Did you not read Nate's post?

It is elementary to show that any continuous periodic function on R with no smallest period is constant, and that anything with a smallest period has periods that are integer mutliples of this smallest period.

However, it is easy to find noncontinuous functions on R that have infinitely many periods that are not integer multiples of each other and are not constant. Just pick the indicator function on any additive (divisible) subgroup of R, such as Q.

HallsofIvy

Another example- if f(x) is a non-continuous function satisfying f(x+ y)= f(x)+ f(y), then it has every positive rational number as period.

Office_Shredder

Sorry to bump this, but I was looking at the problem again and was wondering if there was an official name for functions like the indicator function of an additive sub-group (or particularly a dense one) of R. I tried searching for stuff on fractal functions, but not surprisingly got a ton of stuff about Mandelbrot and Julia sets, then remembered about this post I made in times of Yore

Thanks for the help