Prove Euler Identity without using Euler Formula

[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))?

 

[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.

 

[maze]

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]

never mind

 

[maze]

never mind

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

 

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

 

[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.

 

[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.

 

[maze]

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 don't want to limit myself to that if other definitions are fruitful.

 

[arildno]

I tend to like defining it by the differential equation f' = f, f(0) = 1, though I don't 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]

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]

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

 

[gel]

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.

 

[maze]

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.

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]

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.

 

[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.

 

[morphism]

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

 

[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?

 

[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.

 

[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.

 

[morphism]

And why is that - i.e. why is g onto?

 

[gel]

And why is that - i.e. why is g onto?

intuitively, as |g'|=1, it must wrap the whole way around the circle.

For a mathematically rigorous proof you could first show that it is onto for the upper-right quadrant. The imaginary part, Im(g(x)) must be initially increasing up to 1 and its gradient decreasing to 0.

 

[morphism]

So, to sum it up, g(x)=cosx+isinx? :tongue2:

 

[gel]

yes, Euler's formula is never going to go away. It's just what you get by writing exp(ix) in real and imaginary parts. so exp(ipi)=-1 and cos(pi)=-1,sin(pi)=0 are equivalent statements. Just depends how you write it down.

 

[maze]

And why is that - i.e. why is g onto?

Because g is continuous and unit speed - consider what would happen if a point was missed by the parameterization. Then that point would split the circle into a curve topologically equivalent to a line segment. So you have a continuous function $\tilde{g}:[0,\infty)->(a,b)$ with slope 1, an impossibility. Here is a diagram:
http://img504.imageshack.us/img504/9121/circleunitspeedzi7.png

No need to invoke Euler's formula.

 

[morphism]

Can you prove that a circle with a point removed is homeomorphic to a line segment without using Euler's formula?

It wouldn't matter anyway, because

Euler's formula is still there: g=u+iv, u'=-v, v'=u, u(0)=1, u'(0)=0.

 

[maze]

Can you prove that a circle with a point removed is homeomorphic to a line segment without using Euler's formula?

It wouldn't matter anyway, because

Yes just use the arc length function on the punctured circle. I made a diagram above and edited it in just after your post apparently

 

[morphism]

But the point is, the behaviour of the arclength formula that you're taking for granted, and everything else, would be lost if we didn't have Euler's formula a priori.

All this proof does is try to conceal this fact. It's really a roundabout way of presenting the standard derivation of Euler's formula starting from the differential equation definitions of exp, cos and sin. In this sense it's not really "proving Euler's identity without using Euler's formula". To me, an acceptable proof would be one that, for instance, establishes the identities Phrak posted (in post #6) in a bare-handed manner (but this seems very unlikely).

 

[maze]

But the point is, the behaviour of the arclength formula that you're taking for granted, and everything else, would be lost if we didn't have Euler's formula a priori.

All we need is the existence of an arc length parameterization. This does not require Euler's formula.

So, we can either
(1) appeal to theorems from vector calc/geometry that show a punctured closed curve has an arc length parameterization (just as, say, a punctured closed surface in 3d has a 2d parameterization from the unit square), or
(2) simply construct (cos(x),sin(x)) and note it has all the necessary properties. (Note we don't actually need that (cos(x),sin(x)) equals e^i*x, we just need that it exists).

In fact, we can probably even show the punctured circle is equivalent to the line segment through purely topological arguments, though I am not an expert in such things.

None of these methods require any complex analysis, so we're all good

 

[morphism]

But Euler's formula is right there!

 

[maze]

But Euler's formula is right there!

Yeah you're right it would take only epsilon more effort, at any of several steps here, to get the full Euler's formula. Your post 19, for one, but you could also trivially get the full formula by noting uniqueness of unit speed parameterization up to sign and initial position.

 

[yasiru89]

Perhaps the following might help, due to John Bernoulli which we might adapt for this case;

Consider the area in the first quadrant of a unit circle centred about the origin.

$$A = \int_{0}^{1} (1 - x^{2})^{1/2} dx$$

With the change of variable [itex]u = ix[/tex] the integral is now,

$$A = -i \int_{0}^{i} (1 - u^{2})^{1/2} du$$

We already know that [itex]A = \frac{\pi}{4}[/tex] and by evaluating the integral [itex]A = -\frac{1}{2} i \log{i}[/tex]

Equating the expressions we have,

$$\frac{\pi}{2} = \frac{1}{i} \log{i} = \log{i^{\frac{1}{i}}}$$

Which in turn, on taking the exponent on both sides and raising to the [itex]i[/tex] th power yields;

$$e^{\frac{i \pi}{2}} = i$$

Take the square on both sides, et voila,
$$e^{i \pi} = -1$$ , as required!

 

[morphism]

There are a lot of things to be said about that proof. For one, how are you defining the complex logarithm and complex exponentiation?

 

[yasiru89]

Throughout the logarithm is applied only to real quantities (like [itex]i^{i}[/tex]), so simply treating it as the real logarithm whenever the argument can at least be 'made' real would be justified (of course, for this we may require an alternate proof that [itex]i^{i}[/tex] is real, finding one that doesn't rely on Euler's identity shouldn't be too hard).

Then the exponential is required only as the inverse of the said logarithm and we are done.

 

[morphism]

You're missing my point. i^i doesn't even make sense unless you define what complex exponentiation is. Same comment applies to "1/i logi = log(i^i)": this is meaningless unless you already have a complex logarithm which you know behaves like this (I'm not even going to mention branches). Remedying this will almost certainly require Euler's formula.