
#1
Nov1408, 03:12 AM

P: 655

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))?




#2
Nov1408, 04:21 AM

P: 4,513

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. 



#3
Nov1408, 04:37 AM

P: 655





#4
Nov1408, 04:59 AM

P: 4,513

Prove Euler Identity without using Euler Formula
never mind




#5
Nov1408, 05:01 AM

P: 655





#6
Nov1408, 09:51 PM

P: 4,513

Thanks for being so gracious. I scibbled out a few things.
It's a blunt approach, but if you can prove these two series converge... [tex]1  \frac{\pi^2}{2!} + \frac{\pi^4}{4!}  ... \rightarrow 1[/tex] and [tex]\pi  \frac{\pi^3}{3!} + \frac{\pi^5}{5!}  ... \rightarrow 0[/tex] 



#7
Nov1508, 01:08 AM

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P: 9,428

this seems trivial: e^(x+y) = e^x e^y, so 1 = e^0 = e^ipi e^(ipi) = e^(ipi)/ e^ipi, oops, anything satisfies this.
well you could use the uniqueness theorem for diff eq's. or say that e^(inpi) = pooey. 



#8
Nov1508, 02:21 AM

P: 688

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. 



#9
Nov1508, 12:57 PM

P: 655





#10
Nov1508, 02:01 PM

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PF Gold
P: 12,016





#11
Nov1508, 02:46 PM

P: 532

Using f(0)=1,f'=f its not hard to see that g(x)=exp(ix) gives a periodic function mapping R to S^{1}={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. 



#12
Nov1508, 02:55 PM

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P: 2,020

How do you know that g(x)=1? And I'm not sure I see why g(L/2)=1 either.




#13
Nov1508, 03:08 PM

P: 532

[tex] \frac{d}{dx} g^2= \frac{d}{dx}(u^2+v^2)=2uu'+2vv'=2uv+2vu=0. [/tex] For, g(L/2)=1, it seems obvious to me that if you go half way round a unit circle centered at the origin and starting at (1,0) takes you to (1,0). With a bit of effort you can convert this to a rigorous argument. 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. 



#14
Nov1508, 03:16 PM

P: 655





#15
Nov1508, 03:50 PM

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And I actually don't buy that it's obvious that L/2=pi  at least not without knowing that e^ix=cosx+isinx. 



#16
Nov1508, 03:50 PM

P: 655

The result also shows g is unit speed. Take [itex]\frac{d}{dx}g'(x)^2 = \frac{d}{dx}g(x)^2 = 0[/itex] since g' = ig. Then g is a unit speed curve restricted to the unit circle.




#17
Nov1508, 03:57 PM

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P: 2,020

I'm probably being especially dense... but so what?




#18
Nov1508, 04:02 PM

P: 655

We have that g is a unit speed parameterization of a simple closed curve of finite length. Isnt this sufficient to show that g is periodic and the period is the arc length of the curve?



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