Find upper and lower limit of $\cos(x^2)-\cos((x+1)^2)$.

Question

Find upper and lower limit of $\cos(x^2)-\cos((x+1)^2)$.

In previous topic was shown that 2 cannot be attain. but how to find upper bound?

Answer 1

Here's an argument that values arbitrarily close to $2$ are achieved:

As $x$ goes from $0$ towards $\infty$, $x^2$ will take on the value $2\pi k$ once for each $k\in\mathbb N$.

When $x^2$ is $2\pi k$, $(x+1)^2$ will be $2\pi k+2\sqrt{2\pi}\sqrt k + 1$, and in that case we have $$ \cos(x^2)-\cos((x+1)^2) = 1 - \cos(2\sqrt{2\pi}\sqrt k+1) $$

The values of $\sqrt k \bmod b$ are dense in $[0,b)$ for every $b>0$, thus particularly for $b=\frac12\sqrt{2\pi}$. This means that the values of $2\sqrt{2\pi}\sqrt k + 1 \bmod 2\pi$ are dense in $[0,2\pi)$. In particular, there will be $k$ such that this is arbitrarily close to $\pi$, in which case $\cos(2\sqrt{2\pi}\sqrt k + 1)$ is arbitrarily close to $-1$.

On the other hand, $2\sqrt{2\pi}\sqrt k + 1$ cannot be exactly $\pi$ modulo $2\pi$, because then $\pi$ would be algebraic, which is known not to be the case.

A similar argument shows that there are also values arbitrarily close to $-2$, but that exactly $-2$ is impossible.

Since it is also cleart the the function is continuous, its range must be the open interval $(-2,2)$.