Fundamental period of $e^{ik_1x}+e^{ik_2x}$?

Question

Is the function $f(x)=e^{ik_1x}+e^{ik_2x}$ periodic? Clearly, the first term is $T_1=2\pi/k_1$-periodic and the second is $T_2=2\pi/k_2$-periodic. Can I say something about $f$?. I read that if $T_1/T_2\in \mathbb{Q}$ then $f$ is, but what is its fundamental period?

Answer 1

Note that $$e^{ik_{1}x}+e^{ik_{2}x} = \cos(k_{1}x)+\cos(k_{2}x) + i[\sin(k_{1}x)+\sin(k_{2}x)] = 2\cos\bigg{[}\frac{(k_{1}+k_{2})x}{2}\bigg{]}\cos\bigg{[}\frac{(k_{1}-k_{2})x}{2}\bigg{]}+2i \sin\bigg{[}\frac{(k_{1}+k_{2})x}{2}\bigg{]}\cos\bigg{[}\frac{(k_{1}-k_{2})x}{2}\bigg{]} = 2\cos\bigg{[}\frac{(k_{1}-k_{2})x}{2}\bigg{]}e^{i\frac{(k_{1}+k_{2})x}{2}}$$ Now, let $T_{1} = \frac{4\pi}{k_{1}-k_{2}}$, the period of $\cos\frac{(k_{1}-k_{2})x}{2}$, (assuming $k_{1} \neq k_{2}$) and $T_{2} = \frac{4\pi}{k_{1}+k_{2}}$ the period of $e^{i\frac{(k_{1}+k_{2})x}{2}}$. Suppose $\frac{T_{1}}{T_{2}}$ is rational, that is: $$\frac{k_{1}+k_{2}}{k_{1}-k_{2}} = \frac{n}{n}$$ for some $n,m \in \mathbb{Z}$. Now, take $T = T_{1}m$. We have: $$\frac{k_{1}-k_{2}}{2}T_{1}m = \frac{k_{1}-k_{2}}{2}\frac{4\pi}{k_{1}-k_{2}}m = 2\pi m$$ and $$\frac{k_{1}+k_{2}}{2}T_{1}m = \frac{k_{1}+k_{2}}{2}\frac{4\pi}{k_{1}-k_{2}}m = 2\pi m\frac{k_{1}+k_{2}}{k_{1}-k_{2}} = 2\pi m \frac{n}{m} = 2\pi n $$ Thus, $T$ is a period for $e^{i k_{1}x}+e^{ik_{2}x}$.