Talk:Euler's Formula

I vote we remove the boxes around the formulas on this page. Who's with me? --Cynic 14:17, 27 April 2008 (UTC)

I'm for that --Joe 14:20, 27 April 2008 (UTC)

Looks better, I think. Also, I think the derivative of the quotient expression ought to be illustrated as well.--MathMonkeyMan 20:51, 27 April 2008 (UTC)

Was the last box left in intentionally? It actually might be a good idea to box the final step in all the proofs so it's clear when you have reached the end and you can see what was being proved. Or add a "Theorem:" line to the beginning of each proof.--Cynic 21:28, 27 April 2008 (UTC)

I think the boxes are a good idea, I vote we leave them.

Euler's formula can be taken as a definition rather than something one can prove: the exponential is defined for real numbers, and one wants to extend it to complex numbers in some way. You can prove that the expression must be that, but one should clearly state the assumptions; for example, that the only holomorphic extension of the exponential to the complex plane must be given by that expression, or that the only continuous extension is given by that expression. These are different results, and here it is not clear what the assumptions are. When you want to prove that $e^{i\theta}$ has a certain expression, what is your definition of $e^{i\theta}$?--Cañizo 12:34, 19 February 2009 (UTC)

Don't take me wrong, I just say this to make the page better... For the moment I'm enthusiastic about it, and find it real fun to edit things, then find out that other people edited them, try to discuss about this or that... I can disagree on how to best write things, but that's the point! A lot of disagreement may converge to something nice. Of course this is just starting, and the questions you mention have to be taken one by one.--Cañizo 00:16, 21 February 2009 (UTC)

Added a "proof" which works more as a definition of the (complex) logarithm function than a proof itself. It's wordy and long because it appeals to intuition and common insights about complex numbers.--Misael.G.Mx 00:01, 6 December 2011 (CST)

It certainly opens up a few avenues which we haven't got round to doing yet (I was going to get round to doing complex analysis once I'd covered the appropriate topological background but never got back to it). Mind, it needs a lot of tidying up and serious restructuring. "Wordy and long" is not the usual style that ProofWiki's philosophy is based on. It's got definitions, lemmas, more lemmas, subproofs, who knows what-all, and all of those really belong on their own pages. Lemmas are often useful somewhere else not just in the proof they are raised. And there's some of colloquial language used grammatically inaccurately, which makes it look unprofessional. I'm also not too sure about circularity. However, I'm not in the mood for going over it at the moment, I'm doing other stuff and (my usual complaint) I'm seriously mentally exhausted by the day job at the moment. If anyone else wants to take this one on, feel free. --prime mover 15:19, 6 December 2011 (CST)

I too was worried about circularity, so I've done some tidying up of some sections.

1. the definition of the argument

2. the definition of the polar form of complex numbers

3. the proof that the argument of the product is the sum of the arguments.

If I didn't mess up, Euler's formula now is equivalent to the definition of the logarithm of complex numbers, which is now based on 3. which is based on 2. which is based on 1. (i.e. the definition of argument is the most fundamental concept). I guess this all could go to the section for the definition of logarithm of complex numbers, which (as now) states the definition without any justification whatsoever (or any mention of Euler's formula). BTW, this "proof" is useless if we accept that definitions can be given without any intuitive or operational justification; which I guess is reasonable, since proofwiki is not a textbook. I'd still argue that the proof based on properties of the argument is intuitive and in a sense elementary, since it doesn't use taylor series and maybe, if more thought is given to it, it can be done on purely arithmetic-geometric arguments.

Also, English is not my first language, so the unprofessional writing will probably have to be fixed by someone else; sorry about that. Good luck, I'm eager to see the evolution of all this.--Misael.G.Mx 16:52, 6 December 2011 (CST)

I don't understand what you mean "if we accept that definitions can be given without any intuitive or operational justification" - when a definition is given on this site, it is alwys to be defined in terms of previously defined entities, and the existence of the entity is always to be backed up by some justification. In intent it is to be more rigorous than a textbook, which (unless about the foundations of logic and mathematics) always has to start from some basic assumptions. On ProofWiki everything should be traceable right back to the source axioms of Zermelo-Fraenkel set theory and the axioms of propositional logic. (There are some gaps which still need filling, but we have reason to be pleased with what we have done so far.) --prime mover 17:05, 6 December 2011 (CST)

Ok, sorry for the broad generalization. What I mean is that the logarithm of $z$ can be defined straightforwardly as $\log{z}=i \arg{z} + \left\vert {z} \right\vert$. Notice this is defined on terms of the previously defined entities $i$, $\arg$, and $\left\vert {z} \right\vert$, and it's clear that it exists; yet it has a major problem: it makes Euler's formula trivial. We can all agree that Euler's formula is not trivial: it summarizes deep and rich mathematical insight. Where should that insight be placed on proofwiki? That's what I mean by "textbook" and "intuitive justification": in a textbook, the quest for a $\log$ function precedes it's definition; it's not about rigor, it's about clarity. So, erase this last proof from Euler's Formula page and write up the justification on the definition of the logarithm?

Basically: yes. That's what I mean about circularity: the plan would be to derive the definition of the logarithm from the complex exponential.--prime mover 00:27, 7 December 2011 (CST)

First Proof

Hey, the first proof is bothering me, as it doesn't address the constant of integration or the domain of $\ln$. Wikipedia has a version of the proof, though it skips steps, that shows that both sides of Euler's Formula satisfy:

$D_z f\left({z}\right) = i \cdot f\left({z}\right)$, $f\left({0}\right) = 1$

and that the solution to the above is unique. This is similar to the second proof. This has the advantage of not needing to address $\ln$ at all. Should I replace it? I don't think it's worth adding as another proof because of its similarity to the other ones. I also added half of the same proof commented out trying to address the problems, but I got stuck. --GFauxPas 08:24, 7 December 2011 (CST)

A proof based on De Moivres formula

Hey I'm new here. I want to add this simple, but maybe overlooked, proof. Whats the etiquette here? I don't want to tread on anybody's toes. May I just go ahead and add it to the article, or is there discussion first?

Eulers formula, for real x, may be obtained from De Moivres formula, for integer n,

$ (\cos(\theta) + i \sin(\theta))^n = \cos(n \theta) + i \sin(n \theta) $

Let $\theta = \frac{x}{n}$, and take the limit as n tends to infinity;

$\ds \lim_{n \to \infty}(\cos(\frac{x}{n}) + i \sin(\frac{x}{n}))^n = \cos(\frac{n x}{n}) + i \sin(\frac{n x}{n}) $

Using the power series expansions,

$\cos(\frac{x}{n}) = 1 + \frac{0}{n} + \frac{?}{n^2} + ...$

$\sin(\frac{x}{n}) = 0 + \frac{x}{n} + \frac{?}{n^2} + ...$

gives,

$\cos(\frac{x}{n}) + i \sin(\frac{x}{n}) = 1 + \frac{i x}{n} + \frac{?}{n^2} + ...$

In the limit of the binomial expansion of <math>(\cos(\frac{x}{n}) + i \sin(\frac{x}{n}))^n</math>, it can be shown that the sum of all the terms arising from <math>\frac{?}{n^2}</math> and higher power terms will go to zero as n goes to infinity. So,

$\ds \lim_{n \to \infty}((\cos(\frac{x}{n}) + i \sin(\frac{x}{n})^n) = \lim_{n \to \infty}((1 + \frac{i x}{n})^n) = \cos(x) + i \sin(x)$

From the definition of exponentiation for complex numbers,

$\ds \lim_{n \to \infty}(1+\frac{z}{n})^n = \sum_{k=0}^\infty \frac{z^k}{k!} = e^z $

so where <math>z = i x</math>

$\ds \lim_{n \to \infty}(1 + \frac{i x}{n})^n = e^{i x} = \cos(x) + i \sin(x)$

Thpigdog (talk) 06:29, 7 March 2014 (UTC)

It appears not to add to what we already have. We have already got a power series expansion proof (Proof 3), so just putting n=1 into your proof instantly gets you where you want to go without needing to go near nth powers. It seems to go off on a limb and come back down that same limb for no reason.

Maybe I'm not getting its subtleties, but then I'm not a subtle man. --prime mover (talk) 13:43, 7 March 2014 (UTC)

Certainly, if you just want to prove the result, the power series proof is the simplest. But it gives you the answer with no explanation as to why. The power series on the two sides of the equation just happen to be equal, like a miracle.

What I claim (rightly or wrongly) is that the above proof gives some intuition as to why Euler's formula is correct. It says that, in the complex number domain, a small rotation, repeated many times, with multiplication implementing rotation looks like $(1 + \frac{i x}{n})^n = ((1 + \frac{1}{n})^n)^{i x} = e^{i x}$. This is my personal answer to why is Euler's formula correct. Is this important? I don't know.

Thepigdog (talk) (talk) 04:21, 8 March 2014 (UTC)

My point is that you get to here:

$\cos \left({\dfrac x n}\right) + i \sin \left({\dfrac x n}\right) = 1 + \dfrac {i x} n + \dfrac{\left({i x}\right)} {2! n^2} + ...$

(No need to be coy about the constants, by the way, we have already established what they are in the construction of the power series which we are invoking in the expansions of $\cos$ and $\sin$.)

... and then it is noticed that the power expansion on the RHS is the power expansion of $\exp \left({\dfrac {i x} n }\right)$ from whichever page establishes this fact. And suddenly you have:

$\cos \left({\dfrac x n}\right) + i \sin \left({\dfrac x n}\right) = \exp \left({\dfrac {i x} n }\right)$

whatever $\dfrac x n$ happens to be. Put $z = \dfrac x n$ -- hey presto!

What you seem to be doing is deliberately to ignore that obvious step and instead go via a completely different technique for defining $e^z$, skipping over the important part with the handwavery of "it can be shown that" ... which is at best frowned on in ProofWiki.

The next step in your argument is already covered here: Equivalence of Definitions of Exponential Function (although be aware that this page is up for a refactoring so as to extract the proof into its own page).

I understand that your aim is to demonstrate a geometric, intuitive perception on why it works, but until you have established that using some sort of diagram or something, that intuition does not come through in your exposition. This needs to be shown explicitly -- and that we have not got round to doing yet. --prime mover (talk) 09:25, 8 March 2014 (UTC)

The hand wavery bit can be proved, but I don't know of a neat proof. It's an obvious thing which is messy to. prove.

I just want to get to,

$\ds \lim_{n \to \infty}((\cos(\frac{x}{n}) + i \sin(\frac{x}{n}))^n) = \lim_{n \to \infty}((1 + \frac{i x}{n})^n)$

This step is often glossed over so I introduced the power series and the consideration of the second power to make this accurate. But as you say, why not just consider the whole power series, as this takes you straight there without needing to deal with any messy bits. Because I dont want to. Aaaggggh. Let me think about it.

Thepigdog (talk) (talk) 12:04, 8 March 2014 (UTC)

Why is the Proof Wiki different from Wikipedia? Is it Mediawiki? I see there is no Lucerne Search. There also seems no way of editing sections. I used to use MathJax with Wikipedia and my own Mediawiki installation but have noticed recently it has started centering my formulas. Also the limits above look ugly. I guess there is some history to this.

Thepigdog (talk) (talk) 12:18, 8 March 2014 (UTC)

$\mathsf{Pr} \infty \mathsf{fWiki}$ is different from Wikipedia because certain of the "editorial team" on $\mathsf{Pr} \infty \mathsf{fWiki}$ have fundamental philosophical differences of opinion with those of Wikipedia. We are not Wikipedia and we have absolutely no desire to be considered associated with it.

Don't know what Lucerne search is, presume you might mean Lucene. Never heard of it before.

No, we cannot edit sections, a) because it compromises our house style; when you edit a section it automatically removes the deliberately-inserted extra line between it and the following section, which detracts from the page's neatness, and b) we have an extension in place on this website to enhance the management of transclusions which proved to be incompatible with the ability to edit sections.

MathJax only centres your formulae if you use the < math > delimiters (which I explained are not properly supported by MathJax). If you stick to the programming source-code style that is espoused on this site (as noted above) then you will not go far wrong.

If you want your limits to look prettier, then I encourage you to put \ds at the start of any $\LaTeX$ string that includes such limits. See the house style guide.

Your limits are the least part of your $\LaTeX$ style issues: I contend that your $\cos_{\frac{x}{n}}$ looks utterly revolting -- not to mention hard to read. What's with the underscores? Is this a preferred convention for denoting arguments of functions that is taught somewhere? If so we do not endorse it.

The presumption is that someone who is fluent in mathematics should have no trouble learning our ways. We encourage and nurture the ability to learn for oneself from the available documentation. It can be done. A recent new editor was creating house-style-perfect contributions from the very start.

Sorry typo I meant Lucene search. I have updated the style, but the proof is till unsatisfactory.

I could use Taylors theorem but I still need this identity.

$ \ds \lim_{n \to \infty}(1 + \frac{a}{n} + \frac{b}{n^2})^n = \lim_{n \to \infty}(1 + \frac{a}{n})^n $

Any suggestions other than "scrap the whole approach".

Thepigdog (talk) (talk) 23:21, 8 March 2014 (UTC)

Beyond proving $\ds \lim_{n \to \infty}\left({1 + \frac a n + \frac b {n^2}}\right)^n = \lim_{n \to \infty}\left({1 + \frac a n}\right)^n$, not really.

The aim of $\mathsf{Pr} \infty \mathsf{fWiki}$ is not really to use it as a whiteboard for trying out ingenious proofs -- particularly in this area of well-established mathematics it's more of a dictionary of results. As a general rule, if a published work demonstrates a proof, then we make an effort to include it. If a published work merely states a result, then we report the result and generate the proof ourselves (or crib it from somewhere if we can find the thing). It is rarely the case that non-professional mathematicians actually find a proof which is genuinely new and elegant; frequently such attempts at a new approach are merely taking the long way round and are embarrassingly laughable (see P-Product Metrics on Real Vector Space are Topologically Equivalent for a good example of a duffer trying to prove something obvious and producing something egregiously ridiculous).

In short (tl;dr) if it were me, I would not bother pursuing the above, unless I were *really* attached to it, because (as I say) it takes a long route round to a fairly well-established result. However, the concept of a multiplication by $e^{i\theta}$ being a rotation around the angle $\theta$ has *not* been explicitly stated, although it falls out directly from the demonstration of the behaviour of multiplication in polar form. --prime mover (talk) 00:54, 9 March 2014 (UTC)

Here is the result of going through with the proof,

• User:Thpigdog/Eulers formula proof based on De Moivres formula

I derived an OK way to proving,

$\ds \lim_{n \to \infty}\left({1 + \frac a n + \frac b {n^2}}\right)^n = \lim_{n \to \infty}\left({1 + \frac a n}\right)^n$

• User:Thpigdog/Limit power identity

As you say it is a long way round to prove a result so easily proven using a power series. I guess that is why it is not generally done that way.

I am not trying to do anything new or noteworthy here, it's more of an emotional attachment to this old derivation. I read a book on complex numbers when I was about 16, and it had a lasting effect on me. That's some time ago now.

Instead of directly deriving the Sine of Sum and Cosine of Sum using Angle sum and difference identitiesthe book showed how complex numbers and polar co-ordinates could give the result trivially (proof 1 with $e^{i \theta}$ replaced with $[1, \theta]$ representing polar co-odinates). It built up the idea of rotations and translations being operators, which give an interpretation of complex numbers, as vectors in 2D, that followed the axioms of numbers, under the operations of translation (as addition) and rotation (as multiplication).

From distant memory the book had the style of proof I gave earlier, but it glossed over some key steps, which I am trying to clean up.

For me the part that is intuitive is its relationship to the derivative of a power,

$\ds \dfrac{\mathrm d a^x}{\mathrm d x} = \lim_{\delta \to 0} \frac{a^{x+\delta} - a^x}{\delta} = a^x \lim_{\delta \to 0} \frac{a^{\delta} - 1}{\delta}$

and then defining k as,

$\ds \lim_{\delta \to 0} (k = \frac{a^{\delta} - 1}{\delta})$

Gives (with some loose interpretation),

$\ds a = \lim_{\delta \to 0} (1 + k \delta)^{\frac{1}{\delta}}$

Or for comparison set $\delta = \frac{1}{n}$

$\ds a = \lim_{n \to \infty} (1 + \frac{k}{n})^n = \lim_{m \to \infty} (1 + \frac{1}{m})^{m k} = e^k$

Here 1 + small amount, raised to an equally large power, equals e.

I appreciate that some liberties have been taken here but you can see the comparison to,

$\ds \lim_{n \to \infty}(1 + \frac{i x}{n})^n = e^{i x} = \cos(x) + i \sin(x)$

Here, a small rotation, represented as 1 + i * a small amount, applied an equally large number of times, equals $e^i$.

The Analytic representation of a function allows us to define a complex function by its mapping in the real domain. This is because the mapping of an analytic function in the real domain completely determines coefficients of the power series, and so determines the mapping in the complex domain. So results not immediately provable in the complex domain may be generalized from the real domain.

Anyway, thanks for your help.

Thepigdog (talk) (talk) 05:30, 9 March 2014 (UTC)