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This is a very good point, but it took me a minute to get what you were saying beneath the snark. Translating without the snark:

There's a famous equation relating sin and cos to complex exponentiation. It also helps explain the Taylor expansions of sin and cos, which is one way to compute them and to find properties about them. It's a very important equation. It is:

     ix
    e    = cos x   +   i sin x

kazinator's point was that this equation relies on cos and sin taking radians as arguments. If they take turns instead, then you need to insert messy extra constants to state this equation!

jVinc's counter-point, made with lots of snark, is that there's an equation that's even nicer if you just instead measure angles in turns with ncos and nsin:

    (-1)^(2x) = ncos(x) + i nsin(x)

It's similar, but doesn't require the magic constant e.

A proof sketch that these are equivalent:

    (-1)^(2x) = e^ln((-1)^(2x)) = -e^(2x) = e^(i * (2 pi x)) using e^(pi i) = -1

Wow, this version makes intuitive sense when retroactively applying what I already know about the complex plane, and rotations about it. Thank you!

Perhaps an even nicer equation:

    1^x = ncos(x) + i nsin(x)

using a multi-valued definition of the exponentiation on the left hand side.

Isn’t this just changing units to suit your purpose?

Same way we might use electron volts rather than volts to make the equations nice.

Yes, it's exactly that. Changing the units from radians to turns, to make your equations nicer, because turns are evidently the more natural unit.

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.

>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.

That version of euler's formula might make a nice case for half turns. Then it's just

(-1)^x = ncos(x) + i nsin(x)

It's obvious how to handle it for integers (an even number of half turns is 1, an odd number is -1), and the extension to real numbers aids the intuition.

Or, depending on your focus, quarter turns are very clean too:

i^x = ncos(x) + i nsin(x)

Either way, turns > radians (it's what I think in when doing most fourier kinds of work anyways!).

>The actual equation you come to for ncos an nsin is:

>(-1)^(2x) = ncos(x) + i nsin(x)

Try to formally define this procedure, though. You end up going in circles.

Here's another version:

lim[N->infinity] (1 + ix/N)^N = cos(x) + i sin(x)

Now there are no "weird numbers", and both sides of the equation can be calculated directly, even by hand if you wanted.

If all you're teaching students is a bunch of formulas to be memorized, the (-1)^x notation is kind of cute. But usually when teaching math, we want to build some kind of understanding.

> Try to formally define this procedure, though. You end up going in circles.

The cos(x) + isin(x) formula gives us a way to find the point on the complex plane's unit circle corresponding to an angle x, given in radians. (Plus it does more, because the argument is complex valued.)

The new formula with ncos and nsin does the same thing for an angle given in turns. E.g 0.25 (90 degrees): -1^(0.5) = i. It's understandable in terms of roots of -1.

When you want to know the principal N-th root of number on the complex plane, you can simply divide its argument (i.e. angle) by N. The other roots are then equidistant points around the circle. So for instance, the square root of -1, which is sitting at 180 degrees, is found at 90 degrees, and is therefore i.

We can use -1 as the reference for measuring angles. The turns unit (one circle) is twice as far around the circle as as -1, so that's where we get the 2. Because 90 degrees in turns isn't 0.5, but 0.25.

We could use 1 directly, but then we need the first complex root of unity. For instance, here is the Wikimedia diagram of the fifth roots:

https://en.wikipedia.org/wiki/Root_of_unity#/media/File:One5...

That root which is close to i, has an angle which is exactly 1/5 turns. There is a relationship between turns and roots of unity, because N roots occupy N equidistanct points on the circle spaced by 1/N turns.

>It's understandable in terms of roots of -1.

You seem to have missed the point. You need the formula I gave to rigorously compute the roots of -1. Of course, you could notice that (cos(x) + i sin(x))^n = cos(nx) + i sin(nx), but that's what I meant by "going in circles". You end up defining (-1)^x in terms of sines and cosines, making the "formula" trivial. It's difficult, working this way, to understand why (-1)^(1/3) is (1 + isqrt(3))/2 and not just -1.

By contrast, the Bernoulli formula is actually computable. In fact, the CORDIC algorithm corresponds quite closely to computing the Bernoulli formula by repeated squaring. The use of arctan(2^(-n)) is just like taking (1 + i2^(-n))^(2^n).

There's a reason why math is structured the way it is.

> You end up defining (-1)^x in terms of sines and cosines

But we are explicitly doing that; we have "nsin" and "ncos" on the other side, and those are explicitly defined as just cos and sin with a scale factor applied to the argument.

The goal is simply, if there is a goal, can we have a nice correspondence between complex exponentiation of some base and the scaled sine and cosine that work with turns.

Hey look; if we change the angle coordinate so that a full circle is just 1 rather than an irrational number, then the transcendental e disappears from our version of this famous equation.

> if the formulas end up having weird numbers like 535.4916 or numbers like 2.718 or 6.28318 then obviously there's something suspicious about the equation.

Well, 2.718 is different than those numbers, because the derivative of 2.718^x is 2.178^x, which is a very interesting property of 2.718. The same cannot be said about 535.4. (6.283 is the ratio of a circle's, diameter to radius, which is just something intrinsic to the universe. I think it even transcends the universe, but that's hard for me to reason about. But basically, both 2*pi and e are fundamentally interesting.)

Circumference to radius.

But it really has nothing to do with the universe, except insofar as maths happen to (imperfectly) match it.

Presumably if the universe seemed to match some other maths, we would have invented that variety instead. The Greeks knew the Earth was round, yet made up plane geometry; and never touched on spherical geometry, as far as we know.

Astonishingly, the concept of the number line did not surface until 2000 years later. With the number line, school children can do on command what the best mathematicians of antiquity struggled with for centuries.