# Prove Euler Identity without using Euler Formula

by maze
Tags: euler, formula, identity, prove
 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))?
 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.
P: 655
 Quote by Phrak 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.
That's all fine and good but its not what I asked. The purpose was to derive Euler's Identity without using Euler's formula.

P: 4,513

## Prove Euler Identity without using Euler Formula

never mind
P: 655
 Quote by Phrak never mind
Hey no problem. I figured chances are someone would post that anyways.
 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... $$1 - \frac{\pi^2}{2!} + \frac{\pi^4}{4!} - ... \rightarrow -1$$ and $$\pi - \frac{\pi^3}{3!} + \frac{\pi^5}{5!} - ... \rightarrow 0$$
 HW Helper Sci Advisor P: 9,371 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.
 P: 687 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.
P: 655
 Quote by Dodo 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.
I tend to like defining it by the differential equation f' = f, f(0) = 1, though I dont want to limit myself to that if other definitions are fruitful.
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Thanks
P: 11,935
 Quote by maze I tend to like defining it by the differential equation f' = f, f(0) = 1, though I dont want to limit myself to that if other definitions are fruitful.
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.
 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 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.
 HW Helper Sci Advisor P: 2,020 How do you know that |g(x)|=1? And I'm not sure I see why g(L/2)=-1 either.
P: 532
 Quote by morphism How do you know that |g(x)|=1? And I'm not sure I see why g(L/2)=-1 either.
You could use g'=ig, and, writing g(x)=u(x)+iv(x) gives u'=-v,v'=u. Then,
$$\frac{d}{dx} |g|^2= \frac{d}{dx}(u^2+v^2)=2uu'+2vv'=-2uv+2vu=0.$$

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.
P: 655
 Quote by arildno 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.
That's all fine and good but its not what I asked. The purpose was to derive Euler's Identity without using Euler's formula.

 Quote by gel 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.
Awesome, thanks gel. I was thinking along these same lines at first but then switched strategies and didn't take it to the logical conclusion. It seemed like showing |g(x)|=1 would require assuming the Euler formula, but thankfully it does not as you have shown!
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P: 2,020
 Quote by gel You could use g'=ig, and, writing g(x)=u(x)+iv(x) gives u'=-v,v'=u. Then, $$\frac{d}{dx} |g|^2= \frac{d}{dx}(u^2+v^2)=2uu'+2vv'=-2uv+2vu=0.$$
Ahh.. of course. Clever!

 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.
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?

And I actually don't buy that it's obvious that L/2=pi - at least not without knowing that e^ix=cosx+isinx.
 P: 655 The result also shows g is unit speed. Take $\frac{d}{dx}|g'(x)|^2 = \frac{d}{dx}|g(x)|^2 = 0$ since g' = ig. Then g is a unit speed curve restricted to the unit circle.
 HW Helper Sci Advisor P: 2,020 I'm probably being especially dense... but so what?
 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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