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Working with the definition of periodic function

Physics Asked by hcp on January 28, 2021

I’m exploring periodic conditions and I’m stuck in a problem. Given that $mcdot L_1 = n cdot L_2$, we have two periodic functions $f_1$ and $f_2$ with periods $L_1$ and $L_2$ where we can find, for example that $f_1(x+mcdot L_1) = f_1(x)$. What can I say about the period of the sum of these two functions? If I use the defition of a periodic funcion $f(x+L) = f(x)$ then i have $f_1(x+mcdot L_1)$ + $f_2(x+ncdot L_2)$ = $f_1(x)$ + $f_1(x)$. But, what about the new period?

One Answer

If you know that $f_1$ and $f_2$ are periodic functions with periods $L_1$ and $L_2$, then, by definition, $f_1(x+L_1) = f_1(x)$ and $f_2(x+L_2) = f_2(x)$. If you've expressed yourself clearly, you're interested in the periodicity properties of $f(x) = f_1(x)+f_2(x)$ for the case where there exist integers $m$ and $n$ such that $mL_1 = n_L2$.

For this case, if you define $L=mL_1=nL_2$, then both components will satisfy $$ f_1(x+L) = f_1 (x+m L_1) = f_1(x) $$ (and similarly for $f_2(x)$), so you'll have $f(x+L) = f(x)$, and it is accurate to say that $f(x)$ is $L$-periodic.

However, it is conceivable that $f(x)$ will have a shorter period than $L$. If you want to rule this out, then it is necessary and sufficient to require that $m$ and $n$ be coprime, i.e., that they do not share any prime factors. If they do, then they will have a nontrivial common divisor $k$, and $L/k$ will also preserve $f(x)$. The minimal period of $f(x)$ is then given by the greatest common divisor (gcd) of $m$ and $n$.

Correct answer by Emilio Pisanty on January 28, 2021

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