How can I find the periods of difference of the sine waves that has irrational coefficients

Question

I'm working on a project right now. And now I need to find periods of difference of the sine waves and i'm stuck.

In few resources I found that I can find the periods of summed or differenced sine waves by using LCM(Least Common Multiple) and HCF(Highest Common Factor) of sine waves' periods. But the problem is: I'm dealing with sine waves that has irrational coefficients. And when we try to find the period of any sine wave we deal with the coefficient of "x". So when I jump into LCM and HCF it doesn't give any result I want.

After that I thought that I can make some approximations to the irrationals I deal with. Then I can use LCM and HCF. But I don't have a clue to how to do that either.

And I don't have a idea for how to apply LCM and HCF of rational approximations to irrational numbers either.

Is it possible to find the periods for such functions like this? If it's how can we?

I'm waiting for the answers thank you

Example for the sine functions I mentioned:

$$\sin(x)-\sin(\sqrt[7]{2}x).$$

Edit:

Couple graph images from the function type I want

$$\sin(x)-\sin(2^{12/12}x).$$

$$\sin(x)-\sin(2^{7/12}x).$$$$\sin(x)-\sin(2^{1/12}x).$$

As you can see these type of functions actually look like periodic functions. You can check out by yourself in any graph calculator.

Actually if you have knowledge about musical intervals,you can see that im dealing with sound waves of intervals between musical notes.

And notes that have "perfect" harmony such as octave$$(2^{12/12}).$$ or perfect fifth$$(2^{7/12}).$$ has much more smaller periods than the other intervals such as tritone etc.

*Of course when I say periodic its just a hypothesis.

Answer 1

At least to see that your example function is not periodic, suppose that it does have period $2L$. Then

$$\sin(x)-\sin(ax)=\sin(x+2L)-\sin(a(x+2L))$$

$$\Rightarrow \sin(x)-\sin(x+2L)=\sin(ax)-\sin(a(x+2L))$$

$$\sin(L)\cos(x+L)=\sin(aL)\cos(a(x+L))$$

for all $x$. Now, for irrational $a$ it is impossible that both $\sin(L)$ and $\sin(aL)$ are $0$. Note that $L>0$ (since $\sin(x)$ is non-constant) and if $\sin(L)=0$ then

$$\sin(L)=0$$

$$\Rightarrow L= k\pi;\ k\in\mathbb{N}$$

But if $\sin(aL)=0$ then we also have

$$\sin(L)=0$$

$$\Rightarrow aL= m\pi;\ m\in\mathbb{N}$$

Taken together, we get

$$L=k\pi=\frac{m}{a}\pi$$

$$\Rightarrow a=\frac{m}{k}$$

which is a contradiction. In fact, neither of $\sin(L)$ or $\sin(aL)$ can be zero since neither of $\cos(x+L)$ or $\cos(a(x+L))$ are constant functions. Thus we can rearrange the equation above to get

$$\frac{\cos(a(x+L))}{\cos(x+L)}=\frac{\sin(L)}{\sin(aL)}$$

The right hand side is constant so to arrive at a contradiction it suffices to find two values of $x$ which give different values on the left side. Note that at $x=-L$ we have

$$\left. \frac{\cos(a(x+L))}{\cos(x+L)}\right|_{x=-L}=\frac{\cos(0)}{\cos(0)}=1$$

However, at $x=\frac{\pi}{2a}-L$ we have

$$\left. \frac{\cos(a(x+L))}{\cos(x+L)}\right|_{x=\frac{\pi}{2a}-L}=\frac{\cos\left(\frac{\pi}{2}\right)}{\cos\left(\frac{\pi}{2a}\right)}=\frac{0}{\cos\left(\frac{\pi}{2a}\right)}=0$$

And we are done... as long as the denominator is not $0$. Of course, if it was then

$$\cos\left(\frac{\pi}{2a}\right)=0$$

$$\Rightarrow \frac{\pi}{2a}=\pi\left(k+\frac{1}{2}\right)$$

$$a=\frac{2}{k+\frac{1}{2}}$$

which is a contradiction. So yes, we are done.

Answer 2

Under a musical context, it makes sense to use rational approximation, as $2^{1/12},2^{7/12}$, etc. represent frequency ratios (given by certain musical intervals) in perfectly tuned equal temperament ($12$-TET). In fact, if you use other temperaments such as meantone temperament or Pythagorean tuning, the ratios will automatically be rational, and then the difference function will be periodic. Indeed, for $q\in\Bbb Q$, write $q=\dfrac rs $ with $r,s\in\Bbb Z$, then

$$\sin(x)-\sin(qx)=2\sin(\dfrac{1-q}{2}x)\cos(\dfrac{1+q}{2}x)=2\sin(\dfrac{s-r}{2s}x)\cos(\dfrac{s+r}{2s}x)$$

has $T=4s\pi$ as a period. Moreover, the above function represents the waveform from the inteferrence of a sine wave with period $2\pi$ and another with period $\dfrac{2\pi}{|q|}$. Therefore, it should suffice to approximate the irrational ratios either by some other temperament, in which case you aren't really doing any approximation but simply copying the estimates, or use any of the common/standard approximation methods, for instance Newton's method: to approximate $2^{5/12}$, for example, you will iteratively approximate the positive real root of $x^{12}=2^5=32$. The later has the additional benefit that it's possible to get arbitrarily accurate, which means for the interference waveforms with irrational frequency ratios, you can find "almost-periods" as good as you like. Formally,

$$\forall\varepsilon\gt 0, \exists T\in\Bbb R\text{ such that }|\sin(x+T)-\sin(q(x+T))-\sin(x)+\sin(qx)|\lt\varepsilon$$

And for musically related purposes, that's way much more precision than one should ever need.