- #1
maze
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Is it possible to prove Euler's identity (e^i*pi = -1) without simply taking it as a special case of Euler's formula (e^i*x = cos(x) + i sin(x))?
Phrak said:You can derive Euler's equation by Taylor expanding e^ix.
Collecting the odd powers of x will give you the Taylor expansion of i sin(x).
Collecting the even powers of x will give you the Taylor expansion of cos(x).
Setting x=pi gets you what you want.
Phrak said:never mind
Dodo said:One question: how do you define the complex exponential, in the first place? (Without using Euler's formula).
If it is defined as a Taylor series, then there's no way but to start with one, I think.
Well, that leads you directly to the series representation of the function, which then can be rearranged into Euler's formula, whereby your result is proven.maze said:I tend to like defining it by the differential equation f' = f, f(0) = 1, though I don't want to limit myself to that if other definitions are fruitful.
morphism said:How do you know that |g(x)|=1? And I'm not sure I see why g(L/2)=-1 either.
arildno said:Well, that leads you directly to the series representation of the function, which then can be rearranged into Euler's formula, whereby your result is proven.
gel said:Using f(0)=1,f'=f its not hard to see that g(x)=exp(ix) gives a periodic function mapping R to S1={x in C: |x|=1}. If its period is L then g(L/2)=-1.
As pi is defined to be half the circumference of a unit circle in the Euclidean plane, then pi=L/2 follows almost by definition.
Ahh.. of course. Clever!gel said:You could use g'=ig, and, writing g(x)=u(x)+iv(x) gives u'=-v,v'=u. Then,
[tex]
\frac{d}{dx} |g|^2= \frac{d}{dx}(u^2+v^2)=2uu'+2vv'=-2uv+2vu=0.
[/tex]
I'm still not fully convinced. Is it easy to prove that g isn't multiply periodic? And is it actually immediate that g(L/2)=1 => g has period L/2 or smaller?Alternatively, to make things easier you can use g(x+y)=g(x)g(y). If L is the period then g(L/2)^2=g(L)=g(0)=1, so g(L/2) is either 1 or -1. You can rule out g(L/2)=1, because that would imply that g has period L/2 or smaller. So, g(L/2)=-1.
morphism said:And why is that - i.e. why is g onto?
morphism said:And why is that - i.e. why is g onto?
Euler's formula is still there: g=u+iv, u'=-v, v'=u, u(0)=1, u'(0)=0.
morphism said:Can you prove that a circle with a point removed is homeomorphic to a line segment without using Euler's formula?
It wouldn't matter anyway, because
morphism said:But the point is, the behaviour of the arclength formula that you're taking for granted, and everything else, would be lost if we didn't have Euler's formula a priori.
morphism said:But Euler's formula is right there!