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This question relates only to $x \in \mathbb{R}^+$. The function $f(x) = \cos (\log (x))$ is clearly defined on the positive reals, with a monotonic decreasing period $p(x)$ which is defined at the limits of this range by

$$\underset{x\to 0^+}{\text{lim}}p(f(x)) = \infty$$ $$\underset{x\to \infty}{\text{lim}}p(f(x)) = 0$$

$p$ still needs clear definition for any given $0 < x_0 < \infty$. So, let us define $p(x_ 0)$ as the smallest positive real $a$ for which this pair of equations hold:

$$\cos (\log (x_0)) = \cos (a \log (x_0))$$ $$\text{sgn}\left(\frac{\partial \cos \left(\log \left(x_0\right)\right)}{\partial x_0}\right) = \text{sgn}\left(\frac{\partial \cos \left(a \log \left(x_0\right)\right)}{\partial x_0}\right)$$

Is the right approach? And how do I proceed?

Richard Burke
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  • Are those limits reversed? As $x\to0^{+}$, it would seem the "period" goes to zero and the frequency goes to $\infty$. – 2'5 9'2 Mar 02 '21 at 17:30
  • Why did you choose a multiplicative period? It seems $\cos\log x_0 = \cos \log(a+ x_0)$ would be a more natural definition. – Greg Martin Mar 02 '21 at 17:35

2 Answers2

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I would approach this as follows.

The period of $\cos(x)$ is $2\pi$, everywhere. At a particular $x_0$, what $a$ must you add to $x_0$ so that $\ln(x_0+a)$ is $2\pi$ greater than $\ln(x_0)$? In other words, solve for $a$ in

$$\ln(x_0)+2\pi=\ln(x_0+a)$$

The solution is

$$a=\left(e^{2\pi}-1\right)x_0$$

It would also make sense to me to go backwards and solve $$\ln(x_0)-2\pi=\ln(x_0-a)$$ where the solution is $$a=\left(1-e^{-2\pi}\right)x_0$$

You could take the two solutions and average them somehow. It might make sense to take their geometric mean:

$$a=\left(e^{\pi}-e^{-\pi}\right)x_0$$

2'5 9'2
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$\cos x = \cos y\iff y = \pm x + 2k \pi$ for some integer $k$

$\log w = \log u \iff w = u$ so

$\cos (\log x) = \cos (\log y) \iff \log x = \pm \log y + 2k \pi $

$\iff \log x \mp \log y = 2k\pi\iff \log (\begin{cases}xy\\\frac xy\end{cases}) = 2k\pi$

And if there were a period, $p > 0$, so that $\cos (\log (x + p))=\cos (\log x)$ for all $x$ we would have that for every $x$ that either $\log (x(x+p))$ or $\log \frac x{x+p} = 2k\pi$.

But $\log(x(x+p))$ and $\log\frac x{x+p}$ are both continuous and not constant they can't be restricted to only multiples of $2\pi$, $\cos (\log x)$ is not periodic.

fleablood
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  • It's not entirely clear form the post, but I interpreted the question as looking for some local notion of period that comes from transforming the regular $\cos(x)$ function. – 2'5 9'2 Mar 02 '21 at 19:25