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If $ f(x) $ has period $ \ a $ and $ g(x) $ has period $ \ b \ $ , then if $ \ \frac{a}{b} $ is rational number find the period of $ F(x) \equiv f(x)+g(x) $.

Answer:

I know that if $ f_1 , \ f_2 \ $ has periods $ T_1, \ T_2 $ respectively then the period of $ f_1(x)+f_2(x) $ is the $ l.c.m. $ of $ T_1 \ \ and \ \ T_2 \ $.

Following this principle , the period of $ F(x)=f(x)+g(x) $ is $ \ ab \ $.

But if $ \frac{a}{b} \ $ were not rational , would the logic hold ? Any help is there ?

MAS
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  • $f(x)=\sin x$ and $g(x)=\cos x$ both have period $a=b=2\pi$. Note that $a/b=1 \in \mathbb{Q}$. So by your logic $\sin x +\cos x$ should have period $(2\pi)(2\pi)=4\pi^2$. But that is definitely not true. – Anurag A Jul 18 '17 at 04:14
  • $ LCM ( 2 \pi , 2 \pi )=4 \pi $ , which is certainly a period. – MAS Jul 18 '17 at 04:18
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    $2\pi \not\in \mathbb{Z}$ so how are you defining lcm???? Moreover even if there was such a concept of lcm (which isn't) then why can't it be $2\pi$?? – Anurag A Jul 18 '17 at 04:19
  • @AnuragA Isn't that obvious? A multiple to $x$ would be a number that could be written $jx$ for some positive integer $j$ and a common is a multiple that is common to two number and the least common is the least of those. By this definition one can skip the requirement that $x$ is an integer also at least to me it sounds natural to define a multiple that way. – skyking Jul 18 '17 at 06:55

2 Answers2

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Yes the function has periods(*) being common multiples(**). A multiple of $a$ is a number of the form $ja$ where $j$ is a positive integer, similarily a multiple of $b$ is a number of the form $kb$ where $kb$ is a positive integer.

If $a/b$ is rational that is equal to $k/j$ for some positive integers $j$ and $k$ we have that since $a/b = k/j$ that $ja = kb$ that is we have a common multiple.

Note that it should not be required to use the least common multiplier since actually any multiplier is a period of the resulting function. It's not even sure that the least common multiplier results in the fundamental period since such is not required to exist if for example $f(x)+g(x)$ is a constant function. Also the fundamental period could be smaller than the least common multiplier, for example if $g(x) = f(x+a/2)$ the period would be $a/2$ yet $f$ and $g$ has a fundamental period of $a=b$.

(*) I use period to denote any positive number $T$ such that $f(x+T)=f(x)$ and not just the smallest such number.

(**) Both functions have the period of multiples of their period and especially if there's a common multiple they have a common period and the sum of two function with the same period is periodic (with the same period).

skyking
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Idea:

Since $a/b \in \mathbb{Q}$, therefore there exists integers $m,n \neq 0$ such that $\gcd(m,n)=1$ and $ma=nb$. So the sum function $F(x)=f(x)+g(x)$ will satisfy \begin{align*} F(x+ma)&=f(x+ma)+g(x+ma)\\ &=f(x+ma)+g(x+\color{red}{nb}) && (\because ma=nb)\\ &=f(x)+g(x)\\ &=F(x). \end{align*} So $F$ is periodic with $ma(\text{same as }nb)$. Assuming $a$ and $b$ are fundamental periods of $f$ and $g$ respectively, now you can also show that $F$ is periodic with fundamental period $ma$ (or $nb$).

Note: If $a/b \not\in \mathbb{Q}$, then the function will not be periodic. For example, if $f$ has period $2$ and $g$ has period $\pi$, then $f+g$ will not be periodic. If it were, then we would have $T=2j=\pi k$ for some integers $j,k$. But then this shows $\pi \in \mathbb{Q}$, which of course is not true.

Anurag A
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    Except that it doesn't say that $a$ and $b$ are fundamental periods of $f$ and $g$ respectively and even if it were there's no guarantee that $F$ has the fundamental period of $ma=nb$ - you're not even guaranteed that $F$ has a fundamental period. Also the lack of requirement that $a$ and $b$ are fundamental periods then you can't conclude that the sum is aperiodic just because $a/b\notin\mathbb Q$. – skyking Jul 18 '17 at 06:17
  • Also even if $a$ and $b$ are fundamental periods and $a/b\not\in\mathbb{Q}$, it is still possible that $f+g$ might be periodic. – Eric Wofsey Jul 18 '17 at 06:21
  • A counter example could be found here (also see the comment by @EricWofsey): https://math.stackexchange.com/a/1083/265767 – skyking Jul 18 '17 at 08:27