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Prove Euler Identity without using Euler Formula

  1. Nov 14, 2008 #1
    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. jcsd
  3. Nov 14, 2008 #2
    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.
     
  4. Nov 14, 2008 #3
    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.
     
  5. Nov 14, 2008 #4
    never mind
     
  6. Nov 14, 2008 #5
    Hey no problem. I figured chances are someone would post that anyways.
     
  7. Nov 14, 2008 #6
    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]
     
  8. Nov 15, 2008 #7

    mathwonk

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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.
     
  9. Nov 15, 2008 #8
    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.
     
  10. Nov 15, 2008 #9
    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.
     
  11. Nov 15, 2008 #10

    arildno

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    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.
     
  12. Nov 15, 2008 #11

    gel

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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.
     
  13. Nov 15, 2008 #12

    morphism

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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.
     
  14. Nov 15, 2008 #13

    gel

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    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.
     
  15. Nov 15, 2008 #14
    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.

    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!
     
    Last edited: Nov 15, 2008
  16. Nov 15, 2008 #15

    morphism

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    Ahh.. of course. Clever!

    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.
     
  17. Nov 15, 2008 #16
    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.
     
    Last edited: Nov 15, 2008
  18. Nov 15, 2008 #17

    morphism

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    I'm probably being especially dense... but so what?
     
  19. Nov 15, 2008 #18
    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?
     
    Last edited: Nov 15, 2008
  20. Nov 15, 2008 #19

    morphism

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    It's still dubious - at least to me.

    In any case, Euler's formula is still there: g=u+iv, u'=-v, v'=u, u(0)=1, u'(0)=0.
     
  21. Nov 15, 2008 #20

    gel

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    yes, g is a "unit speed" parameterization of the unit circle. From that it follows that g(x+2pi)=g(x) and g(x+pi)=-g(x). Seems quite clear to me, and explains easily why g(pi)=-1.
     
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