Least positive integer such that $\cos^k \left(\frac{\pi}{2k}\right)\geq \frac{99}{100}$

Question

I wrote a program to figure this out and found $k=123$. Writing $f(k)$ as the function, I showed that $f$ is strictly increasing on the positive integers, and Wolfram agrees that $f(122)<99/100<f(123)$, so this is the integer I was looking for. I'm wondering whether it's possible to find this value without a computer?

Answer 1

You can get a good approximation by writing that $\cos(x)\approx 1-\frac{x^2}{2}$ and thus $\cos^k(\alpha/k)\approx \left(1-\frac{\alpha^2}{2k^2}\right)^{k}$ and then use that $\left(1-\frac{x}{n}\right)^n \approx e^{-x}$, so you get that, for $k$ reasonably large:

$$\cos^k\left(\frac{\pi}{2k}\right)\approx e^{-\pi^2/(8k)}$$

Setting this to $99/100$, we get:

$$\frac{\pi^2}{8k}<-\log(99/100)\\ k> \frac{\pi^2}{-8\log(99/100)}$$

You still need a calculator to compute right side, but it is faster than trial and error, and it gives you $k>122.75217148152386$, which is close enough for seeking an integer.

You'd have to do more work to know when you'd get this value exactly via this technique. You can get pretty easily that $\cos^k(\alpha/k)=\left(1-\frac{\alpha^2}{2k^2}\right)^k + O\left(\frac{1}{k^3}\right)$. This in turn lets you see that:

$$k\log \cos(\alpha/k)=\frac{-\alpha^2}{2k} + O\left(\frac{1}{k^2}\right)$$

You'd have to get a reasonable value for the constant for the $O()$ here, which is a fair amount of work. That constant depend on $\alpha,$ of course.

That means if you want to find $k$ so that $\cos^k(\alpha/k)>1-\beta$, you'd approximately want:

$$\frac{\alpha^2}{2k}<-\log(1-\beta)\\ k>\frac{\alpha^2}{-2\log(1-\beta)}$$

You can use $\beta < -\log(1-\beta)<\frac{\beta}{1-\beta}$ to eliminate the logarithm.