Prove that $\cos\sqrt x$ is not a periodic function.

Question

Prove that $\cos\sqrt x$ is not a periodic function.

My solution goes like this:

If $\cos\sqrt x$ is a periodic function , then let $T$ be its period . Now, $\cos\sqrt x=\cos\sqrt {x+T}$ . So, we get, $\sqrt {x+T}\pm\sqrt {x}=2k\pi$. Now, for a particular $k\in\mathbb {Z}$ , this identity is impossible because the left member is a variable continous argument in $x$ , while its right member is a constant. Hence, $\cos\sqrt x$ is not a periodic function.

Is the above solution correct? Is it valid? If not, where is it going wrong?...

Answer 1

The statement is equivalent to:

Find the constant $T≠0$, such that $\cos {\sqrt x}=\cos \sqrt{x+T}$ holds, $\forall x≥0$.

This also implies that, $T>0$.

We claim that, such constant $T≠0$ doesn't exist.

Proof: Let $x=0$. Then you have:

$$\begin{align}&\cos \sqrt T=1\\ \implies &\sqrt T=2πn,\; n\in\mathbb Z^{+}\\ \implies &T=4π^2n^2,\;n\in\mathbb Z^{+}.\end{align}$$

Now, let $x=4π^2$. We obtain:

$$ \begin{align}&\cos 2π=\cos \sqrt {4π^2+T}=1\\ \implies &\cos 2π\sqrt {n^2+1}=1\\ \implies &\sqrt {n^2+1}=k,\;k\in\mathbb Z^{+}\\ \implies &n^2+1=k^2\\ \implies &(k-n)(k+n)=1\\ \implies &n=0 \;\text{or}\; T=0\\ &\text {A contradiction .}\end{align} $$

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Explanation:

Just because you missed a small detail, your proof couldn't work.

You derived the following relationship:

$$\sqrt {x+T}\pm\sqrt {x}=2kπ,\,k\in\mathbb Z$$

Then, that's correct and you are right.

But observe that, this doesn't imply us, you can consider $2πk$ as a constant for a particular $k\in\mathbb Z$.

This implies that,

For all $x≥0$, does there always exist $k\in\mathbb Z^{+}$ and an independent non-zero constant $T$, such that: $$ \begin{align}\sqrt {x+T}\pm\sqrt {x}&=2k\pi,k\in\mathbb Z\end{align} $$ holds $?$

In other words, you can also understand this statement as follows:

If $\cos \sqrt x=\cos \sqrt{x+T}$, then

$$\sqrt {x+T}\pm\sqrt {x}=2\pi k,\,k\in\mathbb Z$$

holds, for some $k\in\mathbb Z$.

Therefore, we cannot say that the right-hand side should be a constant even for a particular $k\in\mathbb Z$.

For instance, you can take

$$\cos 2π=\cos 4π=1$$

This implies,

$$4π-2π=2π×\color {red}{1}$$

or

$$4π+2π=2π×\color {red}{3}$$

Now take,

$$\cos 2π=\cos 6π=1$$

This yields,

$$6π-2π=2π×\color {red}{2}$$

or

$$6π+2π=2π×\color {red}{4}$$

We see that, since $k$ is not a constant, we cannot consider the right-hand side as a constant for any particular $k\in\mathbb Z$.

Answer 2

The idea is essentially correct but you should make it more precise (also, it is false that $\cos(a)=\cos(b)$ iff the variables differ by a multiple of $2\pi$).

Suppose that $T>0$ is (one of) the periods of $f(x):=\cos(\sqrt{x})$. Then $$ \forall x \in \mathbf{R}, \quad f(x)=f(x+T). \quad \quad (\star) $$ Now, $\cos(a)=\cos(b)$ if and only if there exists $k \in \mathbf{Z}$ such that $a+b=2k\pi$ or $a-b=2k\pi$.

Pick $x_0 \in \mathbf{R}$ such that $0<|\sqrt{x+T}-\sqrt{x}|<2\pi$ for all $x\to x_0$, which is possible since $\sqrt{x+T}-\sqrt{x}\to 0$ as $x\to \infty$.

Pick some $x_\star\ge x_0$ such that $\sqrt{x_\star}+\sqrt{x_\star+T}$ is not a multiple of $2\pi$ (which is possible simply because it attains uncountably many values, while the multiple of $2\pi$ are countably many).

For such choice of $x_\star$, we have by construction that $\sqrt{x_\star+T}+\sqrt{x_\star}$ is not a multiple of $2\pi$ and, in addition, $0<\sqrt{x_\star+T}-\sqrt{x_\star}<2\pi$ (hence also $\sqrt{x_\star+T}-\sqrt{x_\star}$ cannot be a multiple of $2\pi$).

It follows that $f(x_\star) \neq f(x_\star+T)$, which contradicts $(\star)$.