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I know that if $g(x)$ is periodic then $f(g(x))$ is periodic.

This is a sufficient condition but not necessary as $\sin(x)=f(g(x))$ is periodic where $g(x)=x$(non periodic) and $f(x)=\sin(x)$.

Can the above condition be made into a necessary and sufficient condition (assuming that $g(x)$ is not the identity function) or is there a better way to find the solution?

This is from a previous year college entrance exam. Any short trick rather to check for the periodicity of $f(g(x))$ when $g(x)$ is a complicated function would be helpful.

Robert Lee
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ThirstForMaths
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3 Answers3

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Suppose it is periodic with period $2\pi T$ for some $T>0$. This means that for all $x$, $\sin(x^2+3x+2)=\sin((x+2\pi T)^2+3(x+2\pi T)+2)$. Expanding the RHS gives $\sin(x^2+3x+2)=\sin(x^2+4\pi Tx+4\pi^2T^2+3x+6\pi T+2)$.

If $\sin(u)=\sin(v)$, then either $u=v-2\pi n$ or $u=\pi-v+2\pi n$ for some constant integer $n$. We will consider each of these cases:

  1. Suppose $x^2+3x+2=x^2+4\pi Tx+4\pi^2T^2+3x+6\pi T+2 - 2\pi n$. Then: $$2\pi n=4\pi Tx+4\pi^2T^2+6\pi T$$ $$n=Tx+\pi T+3T$$ $n$ is a constant, so $\frac\partial{\partial x}(Tx+\pi T+3T)=0$ (as is the case for all higher derivatives). Calculating the derivative shows that $T=0$, but we previously defined that $T>0$, a contradiction.

  2. Suppose $x^2+3x+2=\pi-x^2-4\pi Tx-4\pi^2T^2-3x-6\pi T-2 + 2\pi n$. Then: $$2x^2+6x+4-\pi+4\pi Tx+4\pi^2T^2+6\pi T = 2\pi n$$ $$n=\frac 1\pi x^2+\frac3\pi x+\frac 2\pi-\frac 12+2 Tx+2\pi T^2+3T$$ As in the previous case, $n$ is constant, so $\frac{\partial^2}{\partial x^2}(\frac 1\pi x^2+\frac3\pi x+\frac 2\pi-\frac 12+2 Tx+2\pi T^2+3T)=0$. Calculating this derivative gives the false statement $\frac 2\pi=0$, a contradiction.

lisyarus
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Moko19
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A continuous periodic function is uniformly continuous.

We let $f:x\mapsto x^2+3x+2$ and we note that for $n\geq 1$, there exists unique $x_n<y_n$ such that $f(x_n) = 2n\pi$ and $f(y_n) = \pi/4+2n\pi$. By the Mean Value Theorem, there is some $\xi_n\in (x_n,y_n)$ such that $$\frac{\sin(f(y_n))-\sin(f(x_n))}{y_n-x_n} = (2\xi_n+3)\cos(f(\xi_n))$$ Since $f$ is increasing over $[0,\infty)$, $f(x_n)\leq f(\xi_n)\leq f(y_n)$, hence $\cos(f(\xi_n))\geq 1/\sqrt 2$ and $$0<y_n-x_n\leq \frac{1}{2\xi_n+3}.$$

As a consequence, $y_n-x_n\to_n 0$, but $\sin(f(y_n))-\sin(f(x_n)) = 1/\sqrt 2$, hence $\sin\circ f$ is not uniformly continuous, hence not periodic.

Gabriel Romon
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We show the following lemma first:

Lemma. A continuously differentiable differentiable periodic function $f:\mathbb{R}\to\mathbb{R}$ has bounded derivative.

Proof. Suppose $f$ has period $T$. It is clear that $f'$ must also then have period $T$. By continuity, $\lvert f'\rvert$ attains a maximum on $[0,T]$, and by periodicity

$$\max_{x\in[0,T]}\lvert f'(x)\rvert =\max_{x\in\mathbb{R}}\lvert f'(x)\rvert.$$

Thus $f'$ is bounded. $\square$

We now move on to your function

$$f:\mathbb{R}\to\mathbb{R},\quad x\mapsto\sin(x^2+3x+2).$$

We can now prove that $f$ is not periodic if we can prove that $f'$ is unbounded. Now we have that

$$f'(x)=(2x+3)\cos(x^2+3x+2).$$

But notice that then, if we choose a sequence $\{x_n\}_{n\in\mathbb{N}}$ such that $x_n\to\infty$ and $x_n^2+3x_n+2=2\pi k_n$ for some sequence of integers $\{k_n\}_{n\in\mathbb{N}}$, then

$$\lvert f'(x_n)\rvert = \lvert 2x_n+3\rvert\to\infty$$

as $n\to\infty$, and so $f'$ is unbounded. It follows that $f$ is not periodic.

Lorago
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