Hacker News Reader

Comment by kazinator | Hacker News Reader

That's a nice result. If we rearrange the products in the exponent we get

     2πix          πi2x          (  πi ) 2x
   e         ->   e         ->   (e    )

Where e^(πi) is -1. That shows there is something to the turns units; we can express the analog of the Euler identity using exponentiation using a base and factor which are integers.

Huge selling point for turns, IMHO.

View on HN

>Huge selling point for turns, IMHO.

Ok, then let’s measure angles in quarter-turns! Then the equation becomes even nicer:

i^x = cos(x) + isin(x)

Beautiful! :-0

Except not. Because you’re obscuring the connection of sin/cos with their hyperbolic counterparts. I.e. this is no longer true:

sinh(x) = -isin(ix)

cosh(x) = cos(ix)

Also, this new convention obscures the connection with the exponential map of Lie groups.

I.e. the exponential map of the complex unit circle as a Lie group is:

e^ix = cos(x) + isin(x)

Similarly, the exponential map of the unit hyperbola of the split-complex plane is:

e^jx = cosh(x) + jsinh(x)

Similarly, for the group of unit quaternions:

e^q = cos(|q|) + sin(|q|)(q/|q|)

These are deep connections, which would be obscured by using anything other than radians.

> Because you’re obscuring the connection of sin/cos with their hyperbolic counterparts.

Only because we forgot the name change: these are supposed to to be nsin and ncos.

Remember also that people use sin and cos with 360-degree degrees just fine; and don't worry about wrecking the connection to the hyperbolic counterparts --- and without changing the names, either.

Reply on news.ycombinator.com