3
$\begingroup$

If $f$ and $g$ are periodic functions, is $g \circ f$ periodic? If it is, what is the period?

So I know:

$f(x) = f(x + T), T \in R$

$g(x) = g(x + P), P \in R$

I have this question for my homework. I don't know how to start. Intuitively I would say that is some combination of periods of each function (T+P, T-P, or something else). Using some online graphing calculators and ploting $f(x)=tan(sin(x))$ and $f(x)=sin(tan(x))$ I came to conclusion that the period of the composition is the period of the "inner" function $f(x)$. But how to show/prove that?

$\endgroup$
1
  • 4
    $\begingroup$ Start by showing that $(g\circ f)(x+T) = (g\circ f)(x)$ for all $x$. So then you know that $T$ is a period of $g\circ f$. $\endgroup$ Commented Oct 5, 2014 at 11:42

1 Answer 1

7
$\begingroup$

Let $h(x)=g(f(x))$ then because $f(x)=f(x+T)$ one has $$h(x)=g(f(x))=g(f(x+T))=h(x+T)$$ Therefore $h(x)$ is periodic with the same period as $f(x)$.

$\endgroup$
2
  • 2
    $\begingroup$ I think part of the point of Daniel Fischer's comment is that $g \circ f$ could have a smaller period than $f$, e.g. compare the periods of $\sin x$ and $|\sin x|$. (Replace the absolute value function with a periodic function that agrees with it on $[-1,1]$ if you also want $g$ to be periodic.) $\endgroup$
    – Zoe H
    Commented Oct 5, 2014 at 12:06
  • $\begingroup$ @ Zoe: You are right that $T$ is the period of $h(x)$ as defined above but not necessarily the smallest one. $\endgroup$
    – Arian
    Commented Oct 5, 2014 at 13:37

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .