A numerical curiosity

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

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

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

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

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

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

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!

Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

4 comments on “A numerical curiosity”

• Paul

Nearly rational with a smallish denominator, of course – all numbers are ‘nearly rational’.

• Colin

Quite so.

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I teach in my home in Abbotsbury Road, Weymouth.

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