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Prove that $$ \limsup_{x\to\infty}\left(\cos x + \sin\left(\sqrt2 x\right)\right) = 2 $$

Pretty much always when I ask a question here I do provide some trials of mine to give some background. Unfortunately, this one is such a tough one for me that I don't even see a starting point.

Here are some observations though. Let's denote the function under the limsup as $f(x)$: $$ f(x) = \cos x + \sin\left(\sqrt2 x\right) $$

Since sin's argument contains an irrational multiplier the function itself is not periodic, perhaps this may be used somehow. I've tried assuming that there exists $x$ such that the equality holds, namely: $$ \cos x + \sin\left(\sqrt2 x\right) =2 $$

Unfortunately, I was not able to solve it for $x$. I've then tried to use Mathematica for a numeric solution, but NSolve didn't output anything in Reals.

The problem becomes even harder since there are some constraints on the tools to be used. It is given at the end of the chapter on "Limit of a function". Before the definition of derivatives, so the author assumes the statement might be proven using more or less elementary methods.

Also, I was thinking that it could be possible to consider $f(n),\ n\in\Bbb N$ rather than $x\in\Bbb R$ and use the fact that $\sin(n)$ and $\cos(n)$ are dense in $[-1, 1]$. But not sure how that may help.

What would the argument be to prove the statement in the problem section?

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  • $\begingroup$ As far as I know, this is an almost periodic function: en.wikipedia.org/wiki/Almost_periodic_function. You might start from there. $\endgroup$ – pluton Apr 25 at 17:32
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    $\begingroup$ Take $x=2n\pi$ such that $\cos x=1$. Then, the question amounts to proving that $\limsup\sin(2\sqrt{2}n\pi)=1$. This is probably possible using a Diophantine type of approximation? $\endgroup$ – Stan Tendijck Apr 25 at 17:33
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    $\begingroup$ Yes, $\exp(2\sqrt{2}n\pi i)$ for $n\in\mathbb{Z}$ is a subgroup of $S^1$. Therefore, it is either finite or dense. If it were finite, then $\sqrt{2}$ would be rational. Therefore, it is dense. So, you can find a sequence $n_1,n_2,...$ such that $\exp(2\sqrt{2}n\pi i)\to i$. $\endgroup$ – user647486 Apr 25 at 17:38
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In $1901$, Minkowski has used his geometry of numbers and proved following theorem${}^{\color{blue}{[1],[2]}}$.

Given any irrational $\theta$ and non-integer real number $\alpha$ such that $x - \theta y - \alpha = 0$ has no solutions in integers. Then for any $\epsilon > 0$, there are infinitely many pairs of integers $p,q$ such that $$|q(p - \theta q - \alpha)| < \frac14 \quad\text{ and }\quad |p - \theta q - \alpha| < \epsilon$$

Take $(\theta,\alpha) = (\sqrt{2},-\frac14)$, this choice satisfies the condition in above theorem. This means for any $\epsilon > 0$, there are infinitely many pairs of integers $p,q$ such that

$$\left|p - \sqrt{2} q + \frac14\right| < \frac{\epsilon}{6\pi}$$

For such a pair $p,q$ with $q \ne 0$, define

$$(P,Q) = \begin{cases}(p,q), & q > 0\\(-3p-1,-3q) & q < 0\end{cases}$$

If we set $x$ to $2\pi Q$, we will have

$$\sqrt{2}x = 2\pi P + \frac{\pi}{2} + \eta\quad\text{ for some } |\eta| < \epsilon$$

This leads to

$$\cos x = 1 \land \sin(\sqrt{2}x) \ge 1 - |\eta| > 1 - \epsilon \quad\implies\quad \cos x + \sin(\sqrt{2}x) > 2 - \epsilon$$

Since there are infinitely many such $x$ and they can be as large as one wish, we obtain

$$\limsup_{x\to \infty}\, ( \cos x + \sin(\sqrt{2}x) ) \ge 2 - \epsilon$$

Since $\epsilon$ is arbitrary and $\cos x + \sin(\sqrt{2}x)$ is bounded from above by $2$, we can conclude $$\limsup_{x\to \infty}\, ( \cos x + \sin(\sqrt{2}x) ) = 2$$

Update

Thinking more about this, since we only use the part $| p - \theta q - \alpha| < \epsilon$ in Minkowski's theorem. This theorem is an overkill. The fact "the fractional part of $\sqrt{2} q$ is dense in $[0,1]$" is enough to derive above limit. Since Minkowski's theorem is a useful theorem to know for such problems, I will leave the answer as is.

References

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