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Suppose that period of $f(x)=T$ and period of $g(x)=S$, I am interested what is a period of $f(x) g(x)$? period of $f(x)+g(x)$? What I have tried is to search in internet, and found following link for this.

Also I know that period of $\sin(x)$ is $2\pi$, but what about $\sin^2(x)$? Does it have period again $\pi n$, or? example is following function $y=\frac{\sin^2(x)}{\cos(x)}$ i can do following thing, namely we know that $\sin(x)/\cos(x)=\tan(x)$ and period of tangent function is $\pi$, so I can represent $y=\sin^2(x)/\cos(x)$ as $y=\tan(x)\times\sin(x)$,but how can calculate period of this?

Please help me.

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The period of $\sin(x)$ is not $\pi n$, but rather $2 \pi$. See this plot on W|A. –  Sasha Jun 28 '12 at 15:20
    
right thanks,thanks for correction –  dato datuashvili Jun 28 '12 at 15:22
    
Just use the definition of the period. –  Mercy Jun 28 '12 at 15:25
    
try $lcm$ of periods. –  Aang Jun 28 '12 at 15:28
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@dato: There is never any case where gcd is relevant here (except perhaps sometimes by accident). –  Henning Makholm Jun 28 '12 at 15:42

1 Answer 1

up vote 17 down vote accepted

We make a few comments only.

$1.$ Note that $2\pi$ is a period of $\sin x$, or, equivalently, $1$ is a period of $\sin(2\pi x)$.

But $\sin x$ has many other periods, such as $4\pi$, $6\pi$, and so on. However, $\sin x$ has no (positive) period shorter than $2\pi$.

$2.$ If $p$ is a period of $f(x)$, and $H$ is any function, then $p$ is a period of $H(f(x))$. So in particular, $2\pi$ is a period of $\sin^2 x$. However, $\sin^2 x$ has a period which is smaller than $2\pi$, namely $\pi$. Note that $\sin(x+\pi)=-\sin x$, so $\sin^2(x+\pi)=\sin^2 x$. It turns out that $\pi$ is the shortest period of $\sin^2 x$.

$3.$ For sums and products, the general situation is complicated. Let $p$ be a period of $f(x)$ and let $q$ be a period of $g(x)$. Suppose that there are positive integers $a$ and $b$ such that $ap=bq=r$. Then $r$ is a period of $f(x)+g(x)$, and also of $f(x)g(x)$.

So for example, if $f(x)$ has $5\pi$ as a period, and $g(x)$ has $7\pi$ as a period, then $f(x)+g(x)$ and $f(x)g(x)$ each have $35\pi$ as a period. However, even if $5\pi$ is the shortest period of $f(x)$ and $7\pi$ is the shortest period of $g(x)$, the number $35\pi$ need not be the shortest period of $f(x)+g(x)$ or $f(x)g(x)$.

We already had an example of this phenomenon: the shortest period of $\sin x$ is $2\pi$, while the shortest period of $(\sin x)(\sin x)$ is $\pi$. Here is a more dramatic example. Let $f(x)=\sin x$, and $g(x)=-\sin x$. Each function has smallest period $2\pi$. But their sum is the $0$-function, which has every positive number $p$ as a period!

$4.$ If $p$ and $q$ are periods of $f(x)$ and $g(x)$ respectively, then any common multiple of $p$ and $q$ is a period of $H(f(x), g(x))$ for any function $H(u,v)$, in particular when $H$ is addition and when $H$ is multiplication. So the least common multiple of $p$ and $q$, if it exists, is a period of $H(f(x),g(x))$. However, it need not be the smallest period.

$5.$ Periods can exhibit quite strange behaviour. For example, let $f(x)=1$ when $x$ is rational, and let $f(x)=0$ when $x$ is irrational. Then every positive rational $r$ is a period of $f(x)$. In particular, $f(x)$ is periodic but has no shortest period.

$6.$ Quite often, the sum of two periodic functions is not periodic. For example, let $f(x)=\sin x+\cos 2\pi x$. The first term has period $2\pi$, the second has period $1$. The sum is not a period. The problem is that $1$ and $2\pi$ are incommensurable. There do not exist positive integers $a$ and $b$ such that $(a)(1)=(b)(2\pi)$.

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Another potential problem can occur when we do have commensurable periods. Consider $f(x)=\sin(x)$ and $g(x)=x-\sin(x)$, for instance. –  Cameron Buie Jun 28 '12 at 17:04
    
@CameronBuie: I do not understand the $x-\sin x$, since it does not have a period. –  André Nicolas Jun 28 '12 at 17:18
    
/facepalm/ That was supposed to be $1-\sin(x)$. Of course the constant function $f+g$ is then periodic, but it is periodic of all periods. –  Cameron Buie Jun 28 '12 at 17:31
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@Dato: Division is the same, it is covered by item $4$, where $H(u,v)=u/v$. –  André Nicolas Jun 29 '12 at 6:48
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@Dato: It can be figured out. For if $f(x+p)=f(x)$ and $g(x+p)=g(x)$ then $H(f(x+p), g(x+p))=H(f(x),g(x))$. –  André Nicolas Jun 29 '12 at 6:57

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