Prove Euler Identity without using Euler Formula


by maze
Tags: euler, formula, identity, prove
maze
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#1
Nov14-08, 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))?
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Phrak
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#2
Nov14-08, 04:21 AM
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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.
maze
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#3
Nov14-08, 04:37 AM
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Quote Quote by Phrak View Post
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.

Phrak
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#4
Nov14-08, 04:59 AM
P: 4,513

Prove Euler Identity without using Euler Formula


never mind
maze
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#5
Nov14-08, 05:01 AM
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Quote Quote by Phrak View Post
never mind
Hey no problem. I figured chances are someone would post that anyways.
Phrak
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#6
Nov14-08, 09:51 PM
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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]
mathwonk
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#7
Nov15-08, 01:08 AM
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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.
dodo
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#8
Nov15-08, 02:21 AM
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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.
maze
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#9
Nov15-08, 12:57 PM
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Quote Quote by Dodo View Post
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.
arildno
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#10
Nov15-08, 02:01 PM
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Quote Quote by maze View Post
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.
gel
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#11
Nov15-08, 02:46 PM
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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.
morphism
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#12
Nov15-08, 02:55 PM
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How do you know that |g(x)|=1? And I'm not sure I see why g(L/2)=-1 either.
gel
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#13
Nov15-08, 03:08 PM
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Quote Quote by morphism View Post
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,
[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.
maze
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#14
Nov15-08, 03:16 PM
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Quote Quote by arildno View Post
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 Quote by gel View Post
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!
morphism
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#15
Nov15-08, 03:50 PM
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Quote Quote by gel View Post
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]
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.
maze
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#16
Nov15-08, 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.
morphism
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#17
Nov15-08, 03:57 PM
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I'm probably being especially dense... but so what?
maze
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#18
Nov15-08, 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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