# Prove Euler Identity without using Euler Formula

1. Nov 14, 2008

### maze

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. Nov 14, 2008

### 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.

3. Nov 14, 2008

### maze

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.

4. Nov 14, 2008

### Phrak

never mind

5. Nov 14, 2008

### maze

Hey no problem. I figured chances are someone would post that anyways.

6. Nov 14, 2008

### Phrak

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
$$\pi - \frac{\pi^3}{3!} + \frac{\pi^5}{5!} - ... \rightarrow 0$$

7. Nov 15, 2008

### mathwonk

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. Nov 15, 2008

### 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.

9. Nov 15, 2008

### 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.

10. Nov 15, 2008

### 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.

11. Nov 15, 2008

### 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.

12. Nov 15, 2008

### morphism

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

13. Nov 15, 2008

### gel

You could use g'=ig, and, writing g(x)=u(x)+iv(x) gives u'=-v,v'=u. Then,
$$\frac{d}{dx} |g|^2= \frac{d}{dx}(u^2+v^2)=2uu'+2vv'=-2uv+2vu=0.$$

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. Nov 15, 2008

### maze

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
15. Nov 15, 2008

### morphism

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.

16. Nov 15, 2008

### maze

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.

Last edited: Nov 15, 2008
17. Nov 15, 2008

### morphism

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

18. Nov 15, 2008

### maze

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
19. Nov 15, 2008

### morphism

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.

20. Nov 15, 2008

### gel

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.