[scythe]

>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.

[kazinator]

> 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.

[scythe]

>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.

[kazinator]

> 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.