1

Obviously, the function $1/\cos(x)$ is unbounded. There are singularities at $x_k =k\pi + \pi/2$. Are there natural numbers that come arbitrarily close to $x_k$? Or is there any other way to prove that the assumption $|1/\cos(n)|< C$ leads to a contradiction?

  • You may want to have a look at the answer to [this question](https://math.stackexchange.com/questions/73262/proving-that-mn-sqrt2-is-dense-in-r). – Xander Henderson Nov 28 '17 at 17:03
  • This follows from [Weyl's Equidistribution Theorem](https://en.wikipedia.org/wiki/Equidistribution_theorem) and the irrationality of $\pi$. – lulu Nov 28 '17 at 17:04
  • Indeed that should solve the problem. But Dirichlet's approximation theorem (thanks to Xander) should also work. – Swanhild Bernstein Nov 28 '17 at 17:09

1 Answers1

0

Let us consider the continued fraction of $\frac{\pi}{2}$, which is infinite since $\pi\not\in\mathbb{Q}$: $$ \frac{\pi}{2}=\left[1; 1, 1, 3, 31, 1, 145, 1, 4, 2, 8, 1, 6, 1, 2, 3, 1, 4, 1, 5, 1, 41,\ldots\right] $$ There are infinite convergents of such continued fraction with an odd denominator: $$ \frac{2}{1},\frac{11}{7},\frac{344}{219},\frac{51819}{32989},\frac{52174}{33215},\frac{260515}{165849},\ldots$$ and their numerators give natural numbers $n$ such that $\cos(n)$ is closer and closer to zero.
Indeed, from $$ \left|q_n\frac{\pi}{2}-p_n\right|\leq \frac{1}{q_n} $$ and the Lipschitz-continuity of $\cos$ we get $$ \left|\cos(p_n)\right|=\left|\cos(p_n)-\cos(\pi q_n/2)\right|\leq\frac{1}{q_n}\approx\frac{\pi}{2p_n} $$ so $\left\{\frac{1}{\cos(n)}\right\}_{n\geq 1}$ is unbounded since $\{p_n\}_{n\geq 1}$ is unbounded.

Jack D'Aurizio
  • 338,356
  • 40
  • 353
  • 787