A curious identity | Colin

There’s something neat about an identity or result that seems completely unexpected, and this one is an especially nice one:

$$ e^{2\pi \sin( i \ln(\phi) }= -1$$

(where $\phi$ is the golden ratio.)

It’s one of those that just begs, “prove me!” So, here goes!

I’d start with the identity $2\sin(x) \equiv \frac{e^{ix} – e^{-ix}}{i}$. In particular, the power on the left hand side is $2\pi \sin( i \ln(\phi) ) = \frac{\pi}{i} \left( e^{-\ln(\phi)} – e^{\ln(\phi)} \right)$.

We’ve got $e$s and $\ln$s mixed up in the brackets, which is a Clue if ever I saw one: since $e^{\ln(x)} = x$, the…

Continue reading at:
http://ift.tt/25vHwvt

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s