[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.