4

We know period of $\sin x$ is $2π.$ So period of $\sin cx$ will be $\frac{2π}{|c|}.$ Therefore period of $(\sin x+\sin cx)$ is:

$\text{LCM of }\left(2π, \frac{2π}{|c|}\right)=\frac{\text{LCM of }(2π, 2π)}{\text{HCF of }(1,|c|)}.$

Now if $c\in\mathbb R\smallsetminus\mathbb Q,$ then HCF of $1$(rational) and $|c|$(irrational) is not possible. But since $(\sin x+\sin cx)$ is periodic, so $c$ must be rational.

Conversely if $c\in\mathbb Q,$ then period of $(\sin x+\sin cx)$ is:

$\text{LCM of }\left(2π, \frac{2π}{|c|}\right)=\frac{\text{LCM of }(2π,2π)}{\text{HCF of }(1, |c|)}, \text{ which is possible as }|c|\in\mathbb Q.$

So the period of $f(x)=\sin x+\sin cx$ is $2π.$ Which is correct since $f(x+2π)=f(x).$

Hence the statement follows.

This was my approach. But I, personally, don't like this approach that much. So is there any other direct or obvious prove than this? Please suggest..

Dhrubajyoti Bhattacharjee
  • 1,265
  • 1
    While you can (with a bit of luck) define the LCM of two real numbers, your off-hand "period of $(sinx+sincx)$ is ..." (while true) is a non-trivial theorem you'd have to prove, first. Much simpler: if $f$ is periodic with period $L$, so is $g(x)=f(x)+f(x+\pi/c).$ –  May 31 '20 at 15:46
  • It is simply not true in general that the period of a sum of two functions is the LCM of their periods. See https://math.stackexchange.com/questions/1079/sum-of-two-periodic-functions for instance. – Eric Wofsey May 31 '20 at 16:42
  • @Eric Wofsey For continuous non-constant functions, it is true. –  May 31 '20 at 17:42
  • @ProfessorVector For $\sin x + \left(-\sin x\right)$ it isn't so I think you need the additional stipulation that the sum actually has a fundamental period. – Jam May 31 '20 at 18:19
  • Well now I'm confused. I don't know much about periodicity of functions and have very preliminary knowledge of it.. The formula, I used for calculating the period of $f(x),$ is the one that was taught in my school around 2 years ago. Although it was explicitly mentioned that the formula is not applicable always e.g., $|\sin x|+|\cos x|$ has period $\fracπ2$ instead of $π$. Now am in college and haven't read about periodicity of functions yet. I accidentally came across this statement and finding it interesting went to prove it but being unsatisfied with my own approach, posted it here. – Dhrubajyoti Bhattacharjee May 31 '20 at 18:57
  • @Jam, $\sin x+(-\sin x)$ is a constant function and constant functions don't possess a fundamental period, right? And $\sin x+\sin cx$ being a continuous and non-constant function must possess some fundamental period (for some rational $c$ as stated). Correct me if I'm wrong. – Dhrubajyoti Bhattacharjee May 31 '20 at 19:03
  • @ProfessorVector, in your first comment you said, "if $f$ is periodic with period $L$, so is $g(x)=f(x)+f(x+π/c).$" Would you please elaborate this considering $f(x)=\sin x$?! Thanks. – Dhrubajyoti Bhattacharjee May 31 '20 at 19:10
  • If $c > 0$ and $f(x) = \sin x + \sin(cx)$ is periodic with period $L$, then $f'(L) = f'(0) = 1 + c$ implies $\cos L = \cos(cL) = 1$. So, both $L$ and $cL$ would have to be integer multiples of $2\pi$. That proof breaks down for the case $c < 0$, though, so not posting as an answer. – Daniel Schepler May 31 '20 at 20:52

1 Answers1

0

Suppose $f(x)$ is periodic with period $L$. This means $f(x+L)=f(x)$ for all real $x$. Then, $g_a(x)=f(x)+f(x+a)$ (with some constant $a$) is periodic with period $L$, too: $g_a(x+L)=f(x+L)+f(x+a+L)=f(x)+f(x+a)=g_a(x).$ Now, consider $f(x)=\sin x+\sin cx.$ If it is periodic with period $L$, so is $$g_\pi(x)=\sin cx + \sin(cx+c\pi)=2\sin(cx+c\pi/2)\cos(c\pi/2)$$ (since $\sin x + \sin(x+\pi)=0. $) This can be $0$, if $\cos c\pi/2=0,$ but then, $c$ is rational, we know the zeroes of $\cos$. Otherwise, we must have $L=2k\pi/c,$ we know the period of $\sin.$ Also, $g_{\pi/c}(x)$ must have period $L,$ and that's $$\sin x + \sin(x+\pi/c)=2\sin(x+\pi/(2c))\cos(\pi/(2c)).$$ Same conclusion: $\cos(\pi/(2c))=0$ (i.e. $c$ rational), or $L=2l\pi$, i.e. $c=2l\pi/(2k\pi)=l/k.$

  • 1
    I think this answer relies on the inference of periodicity not just going from $f$ to $g$ but also back from $g$ to $f$, which I believe isn't necessarily true. I think we can construct a counterexample such that $f$ is aperiodic but $f(x)+f(x+a)$ is periodic using a method similar to (Overflow Q282756), where we make $f$ takes $3$ values for $x$ determined by equivalence relations dependent on $a$. – Jam Jun 01 '20 at 01:21
  • @Jam You may want to reread the answer: it relies on the special form of $g_a$ for special values of $a$. You'll find it hard to locate a counterexample for the periodicity of $\sin$. –  Jun 01 '20 at 05:25
  • 1
    I agree that $g$ is periodic and that $f$ being periodic implies that $g$ is also. And I agree with your constraints on $c$ for the periodicity of $g$. However, I don't agree that the relationship is necessarily biconditional and can be reversed; how do you justify that $f(x)+f(x+a)$ being periodic implies that $f$ is also periodic? – Jam Jun 01 '20 at 12:41
  • @Jam Please, reread it, again. And then, quote the part where I (allegedly) use "$f(x)+f(x+a)$ periodic, thus $f$ periodic," please. –  Jun 01 '20 at 14:16