Euler's Formula Using Taylor Series Expansions

preview_player
Показать описание
Today, we prove Euler's formula using Taylor series.
Рекомендации по теме
Комментарии
Автор

how the hell can you explain that well. this channel is so underrated

joshuahector
Автор

Great explanation. I was always a fan of this derivation. I offer you another derivation that uses calculus.


Let y=cos(x)+i sin(x)


dy/dx = -sin(x)+i cos(x)
=i^2 sin(x) + i cos(x)
=i[ cos(x)+ i sin(x)] ---> looks a bit like what we started with
dy/dx = iy


Integrating this will lead to y=cos(x)+i sin(x).
So, e^(ix)=cos(x)+i sin(x)

jamesmott