A numerical curiosity

A numerical curiosity today, all to do with $\mathrm{i}$th powers.

Euler noticed, some centuries ago, that $13(2^{\mathrm{i}}+2^{-\mathrm{i}})$ is almost exactly $20$. As you would, of course. But why? And more to the point, how do you work out an $\mathrm{i}$th power?

It’s all to do with the exponential form, of course

You can write $2^{\mathrm{i}}$ as $e^{\mathrm{i}\ln (2)}$, which is the same as $\cos (\ln (2))+\mathrm{i}\sin (\ln (2))$.

Similarly, $2^{-\mathrm{i}}=e^{-\mathrm{i}\ln (2)}=\cos (\ln (2))-\mathrm{i}\sin (\ln (2))$. Add ‘em up and you get:

$13(2^{\mathrm{i}}+2^{-\mathrm{i}})=26\cos (\ln (2))$.

So… that’s only 20 if $\cos (\ln (2))\simeq \frac{10}{13}$. Why should that be so?

Coincidence?

I think so. I’ve looked at expansions for $\cos (x)$ and for $\ln (x)$ and I’m completely perplexed as to why the cosine of this angle - about $0.693$ radians, or 39.7 degrees, should be nearly rational.

In the absence of any better ideas, I’m going to chalk it up as a coincidence - after all, there’s no reason the cosine of one number should be closer-to-rational than any other.

If you have a better explanation than ‘it just is’, I’d love to hear it… drop me a comment!