Talk:Euler's formula
Can you show a proof of Euler's equation?
There is another way of demonstrating the formula
which I find to be more beautiful:
Let z = cos t + i sin t
then dz = (-sin t + i cos t) dt = i (cos t + i sin t) dt = i z dt.
Integrating:
int dz/z = int i dt
or
ln z = i t.
Exponentiating:
z = exp i t.
The proof using Taylor series is silly! If one is allowed to assume the Taylor expansions of exp(x), sin(x) and cos(x), then just add the series for cos x + i sin x and note that it is the same as the series for exp(i x). --zero 09:38, 12 Oct 2003 (UTC)
You have an error anyway in your proof: i(-sin t + i cos t) = - (cos t + i sin t) = -z. I don't think you can differentiate like you're doing in any case since z is a complex variable (I could be wrong, I haven't done any complex analysis stuff for a while). Dysprosia 10:03, 12 Oct 2003 (UTC)
No, that part of the proof is fine. The only problematic step is the integration, since it really gives ln z = i t + C for a constant C. One then has to find an argument that C=0. --zero 12:46, 12 Oct 2003 (UTC)
Taylor Series for e^x
I'm a little confused about one thing for the e^ix = cosx + (i)sinx derivation. It looks like the Taylor Series of e^ix is exanded around the point a = 0. Wouldn't that mean the proof is only valid near x = 0?
The series is valid for all x.
Charles Matthews 09:42, 18 Dec 2003 (UTC)
- Radius of convergence of exp x is infinite, btw. Dysprosia 09:48, 18 Dec 2003 (UTC)
That explains it, thanks a lot!
- You could expand it about any point, and as long as you took all (an infinite number of) the elements, it would still work. If you're only going to use a few terms you should expand it about whatever local operating point you're using. moink 05:12, 13 Jan 2004 (UTC)
I would like to suggest moving the complex analysis to the top, above the other one. In my experience it's much more common. moink 05:12, 13 Jan 2004 (UTC)