Find the values of $a$ for which $f(x)$ is periodic

Question

Let a function $f(x)$ be defined as

$$f(x)=x^a+\sin{x}-ax$$

Then find the number of values of $a$ for which $f(x)$ is periodic.

My Attempt:

I was only able to find two values i.e. $a=0, a=1$.

But how can I prove that these are the only solutions.

Answer 1

Well, if $f(x)$ is periodic, then $f(x)$ cannot have a limit as $x \to \infty$ unless it is a constant function. This is because if $h>0$ is the period of $f$, then starting from any $f(y) \neq f(z)$ we can produce two constant subsequences $f(y + nh), f(z+nh)$ as $n \to \infty$, and these two will have different finite limits, hence $f(x)$ won't have a limit as $x \to \infty$. Note that the limit cannot even be infinite, since in that case every such sequence must go to (plus or minus) infinity, but we have created a sequence that does not do that.

Clearly, $f(x)$ is periodic if $x = 0$ and $x = 1$.

Let $a < 0$ now.Then $x^a \to 0$ as $a \to \infty$, and $ax \to - \infty$. Since $|\sin x| \leq 1$, the function has limit $-\infty$ by the squeeze theorem.

Let $a \in (0,1)$. Then, again you see that the limit as $x \to \infty$ must be $-\infty$.

Finally, if $a > 1$, then the limit would be $+\infty$. Consequently, we complete the proof.

---

Also, note that if $f(x)$ is periodic, then $f'$ if it exists, must also be periodic, and therefore the function $f'(x)= ax^{a-1} + \cos x - a$ must be periodic. This is periodic if and only if the function $ax^{a-1} + \cos x$ is periodic. Now, use the limit argument for this function when $a \neq 0,1$ to conclude similarly.

Answer 2

Hint If $a>1$ then $\lim_{x \to \infty} f(x)=\infty$.

If $a \in (0,1)$ then $\lim_{x \to \infty} f(x)=-\infty$.

If $a<0$ then $\lim_{x \to 0^+} f(x)=\infty$.

Answer 3

If $f$ has period $p$ then $f(np)$ is bounded. This implies $(np)^{a}-anp$ is bounded. It is easy to see from this that $a$ must be $0$ or $1$. For $a>1$ write $(np)^{a}-anp$ as $n^{a} (p^{a}-apn^{1-a})$ and let $n \to \infty$. For $0<a<1$ write $(np)^{a}-anp$ as $n(-ap-n^{a-1} p^{a})$ and let $n \to \infty$.

Answer 4

Apparently, $a=0,1$ are two proper values of $a$ satisfying the requirement.

Case 1

Assume that $a>1$ and $f(x)$ has a positive period $T$. Then $f(0)=f(0+nT),(n=1,2,\cdots)$, namely $$\frac{\sin (nT)}{anT}=-\frac{(nT)^a}{anT}+1=-\frac{(nT)^{a-1}}{a}+1,$$which holds for all $n=1,2,\cdots.$ Hence, let $n \to \infty$ and take the limits of both sides. We obtain $$0=-\infty,$$ which is absurd.

Case 2

Assume that $0<a<1$ and $f(x)$ has a positive period $T'$. Then $f(0)=f(0+nT'),(n=1,2,\cdots)$, namely $$\frac{\sin (nT')}{anT'}=-\frac{(nT')^a}{anT'}+1=-\frac{(nT')^{a-1}}{a}+1,$$which holds for all $n=1,2,\cdots.$ Hence, let $n \to \infty$ and take the limits of both sides. We obtain $$0=1,$$ which is absurd.

Case 3

Assume that $a<0$ and $f(x)$ has a positive period $T''$. Notice that $f(x)$ has no definition at $x=0.$ Then we should have $f(x_0)=f(x_0+nT''),(x_0 \neq 0,~~~n=1,2,\cdots)$, namely $$x_0^a+\sin x_0=(x_0+nT)^a+\sin(x_0+nT)-anT\geq (x_0+nT)^a-1-anT$$which holds for all $n=1,2,\cdots.$ Hence, let $n \to \infty$ and take the limits of both sides. We obtain $$x_0^a+\sin x_0 \geq +\infty,$$ which is absurd.

Summing up the three cases, we may know that $a=0,1$ are the all values of $a$ such that $f(x)$ is periodic.