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I have a function which is written as $$ \left\|r_1+\frac{(1-r_1^2)r_2e^{-i\delta}}{1-r_2^2e^{-i\delta}}\right\|^2 $$ where $r_1, r_2$ are constants, and $\delta$ is variable. This is originally from physics, which is related with interference of thin film.

I want to determine the period of this function.

Simplifying above equation, $$ a+\frac{be^{-ix}}{1-ce^{-ix}} =\frac{a+(b-ac)e^{-ix}}{1-ce^{-ix}} =\frac{a+de^{-ix}}{1-ce^{-ix}} =\frac{d}{c}\frac{a/d+e^{-ix}}{1/c-e^{-ix}} \approx\frac{p+e^{-ix}}{q-e^{-ix}} $$ Then whole squared term is $$ \|\frac{p+e^{-ix}}{q-e^{-ix}}\|^2 =\frac{p^2+2p\cos(x)+1}{(q-\cos(x))^2+\sin^2(x)} =\frac{p^2+2p\cos(x)+1}{q^2-2q\cos(x)+1} $$ which I can further simplify approximately as $$ \frac{r+\cos(x)}{s-\cos(x)} $$ Both denominator and numerator are periodic function with $2\pi$, but does division of these two still give $2\pi$ period?

Ѕᴀᴀᴅ
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user65452
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2 Answers2

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If $f(x)$ and $g(x)$ have period $T$, then their quotient $f / g$ satisfies $$\boxed{\left(\tfrac{f}{g}\right)(x + T) = \frac{f(x + T)}{g(x + T)} = \frac{f(x)}{g(x)} = \left(\tfrac{f}{g}\right)(x)}$$ and so also has period $T$.

Remark Even if $T$ is the minimal period for both $f, g$, however, it need not be the minimal period for the quotient. For example, $\sin$ and $\cos$ both have minimal period $2 \pi$, but their quotient, $\tan$, has minimal period $\pi$.

Travis Willse
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Hint:

$$f(0)=\left\|r_1+\frac{(1-r_1^2)r_2}{1-r_2^2}\right\|^2$$

and $$f(\pi)=\left\|r_1+\frac{(1-r_1^2)r_2}{1+r_2^2}\right\|^2$$

and the period is not $\pi$. (And of course $f(0)=f(2\pi)$.)

  • Well, it could work out if I had known that $2\pi$ is the period at the beginning, but I was wondering about more general way of finding periodicity... – user65452 Oct 01 '18 at 19:10
  • @user65452: you draw $2\pi$ from the period of the constituting functions, which is obvious. If you don't state the real question, don't expect a real answer. –  Oct 01 '18 at 19:12
  • Oh. sorry. I meant, general way to determine that $2\pi$ is period. No offense. – user65452 Oct 01 '18 at 19:17