3 ms·
As I understand, "complex exponential" function f(x) = e^ix must satisfy only two equalities: f(0) = 1 f'(x) = i f(x) So any function that satisfies t
by nlitened 3y ago
As I understand, "complex exponential" function f(x) = e^ix must satisfy only two equalities:
f(0) = 1
f'(x) = i f(x)
So any function that satisfies these equalities can work as a "complex exponential" function which we denote as e^ix.
So we can define a function with period of 1, and use it everywhere — then "2π" vanishes from most equations, and the complex math still works and all equalities hold.*
- messe 3y agoThe only function that satisfies those two equalities is e^ix which has period 2π.
- nlitened 3y agoHey, you're actually right! My bad
- jojobas 3y agoConsider simultaneous functional equations: f(x) = dg(x)/dx g(x) = df(x)/dx Its only linearly independent solutions are sin(x+c) and cos(x+c) with, x being in radians, periodic by 2pi.