Talk:Euler's formula
|
|
Can you show a proof of Euler's equation?
There is another way of demonstrating the formula
which I find to be more beautiful:
Let z = cos t + i sin t
then dz = (-sin t + i cos t) dt = i (cos t + i sin t) dt = i z dt.
Integrating:
int dz/z = int i dt
or
ln z = i t.
Exponentiating:
z = exp i t.
- let <math>\ln z = it + C_1<math>
- <math> z = e^{it +C_1} = e^{C_1}\cdot e^{it}<math>
- <math> C = e^{C_1} \rightarrow z = Ce^{it}<math>
The proof using Taylor series is silly! If one is allowed to assume the Taylor expansions of exp(x), sin(x) and cos(x), then just add the series for cos x + i sin x and note that it is the same as the series for exp(i x). --zero 09:38, 12 Oct 2003 (UTC)
You have an error anyway in your proof: i(-sin t + i cos t) = - (cos t + i sin t) = -z. I don't think you can differentiate like you're doing in any case since z is a complex variable (I could be wrong, I haven't done any complex analysis stuff for a while). Dysprosia 10:03, 12 Oct 2003 (UTC)
No, that part of the proof is fine. The only problematic step is the integration, since it really gives ln z = i t + C for a constant C. One then has to find an argument that C=0. --zero 12:46, 12 Oct 2003 (UTC)
The argument that C=0 can be easily found by substituting t=0 and evaluating. --Komp, 10th Sept 2004.
| Contents |
Taylor Series for e^x
I'm a little confused about one thing for the e^ix = cosx + (i)sinx derivation. It looks like the Taylor Series of e^ix is exanded around the point a = 0. Wouldn't that mean the proof is only valid near x = 0?
The series is valid for all x.
Charles Matthews 09:42, 18 Dec 2003 (UTC)
- Radius of convergence of exp x is infinite, btw. Dysprosia 09:48, 18 Dec 2003 (UTC)
That explains it, thanks a lot!
- You could expand it about any point, and as long as you took all (an infinite number of) the elements, it would still work. If you're only going to use a few terms you should expand it about whatever local operating point you're using. moink 05:12, 13 Jan 2004 (UTC)
I would like to suggest moving the complex analysis to the top, above the other one. In my experience it's much more common. moink 05:12, 13 Jan 2004 (UTC)
How about more -- split this into 2 articles; the two results have almost nothing to do with each other. They don't belong together.
I propose replacing the e^ix = cosx + (i)sinx derivation by the following simplified version.
It is known that exp(x), sin(x), and cos(x) have Talyor series which converge for all complex x:
- <math> \exp(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots <math>
- <math> \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots <math>
- <math> \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots <math>
Adding the series for cos(x) to i times the series for sin(x) gives the series for exp(ix).
- It misses some parts of the proof though; the periodicy where i^2 = -1 ; i^3 = -i ; i^4 = 1. ✏ Sverdrup 14:52, 6 May 2004 (UTC)
- Sverdup is right, but I think the notation in the current proof is more a hindrance than a help. It's much easier to visually see what's going on by writing "dot-dot-dot"'s and collecting terms than by using a jillion sigma notations.
- I suggest we use the proof on top of this talk page to motivate the formula, and keep the current taylor series proof as the proof. We need to be accurate, and we are also elegant if the math is done right with summation etc. ✏ Sverdrup 22:27, 6 May 2004 (UTC)
- I'm not sure what you mean by "the" proof -- most results have multiple proofs, and this is no exception. The proof using "dz/z = it dt" is good motivation, yes, but it's also a completely rigorous proof, as well, so by including it as a "real" proof, we would not lose any accuracy. I still maintain that the Taylor series proof is much easier to understand without sigma notation, without losing any rigor -- "dot-dot-dot's" are fully rigorous, as long as it's obvious what is intended, which is the case here if enough terms are spelled out. Having four different summations with "4n", "4n+1", "4n+2", and "4n+3" is only going to confuse people who aren't used to the notation for partitioning integers into congruence classes -- they will have to spell out what the sums say for themselves, so why not do it for them? (BTW, in case you wonder why the "dz/z = it dt" proof is rigorous, it comes down to this. We are basically dealing with the analytic continuation of the real exponential to the entire complex plane -- this is known to exist because the Taylor series at z = 0, say, has infinite radius of convergence. So, we can define the exponential as exp(z) = Taylor series. It's pretty trivial to show that d/dz(exp(z)) = exp(z) for all z, everything's abs/unif converg, etc. By the chain rule, d/dz(exp(iz)) = i*exp(iz), i.e. exp(iz) satisfies the diff eq w' = iw. Now, note that if w = cos(z) + isin(z), then this w also satisfies the equation; this means w = C*exp(iz) for some constant C; z = 0 gives 1 = w = C, so w = exp(iz) = cos(z) + isin(z). Now, just take z = x to be real. This is basically what is going on with the shorthand notation "dz/z = it dt". The shorthand notation proof is somewhat glossy over a couple of these details, but then again, a lot of proofs at wikipedia are really just "sketches of a proof".) Let me put a copy of what I would have as my Taylor series proof here, so people can see it and compare.
Here is my proposal to replace the current Taylor series proof:
Derivation
Here is a derivation of Euler's formula using Taylor series expansions as well as basic facts about the powers of i:
- <math>
i^0 = 1, \ i^1 = i, \ i^2 = -1, \ i^3 = -i, \ i^4 = 1, \ i^5 = i, \ i^6 = -1, \ i^7 = -i, \ i^8 = 1, \dots
<math>
The functions ex, cos(x) and sin(x) (assuming x is a real) can be written as:
- <math> e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots <math>
- <math> \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots
<math>
- <math> \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots
<math>
and for complex z we define each of these function by its series. This is possible because the radius of convergence of each series is infinite.
Now, take z = ix, where x is real, and note that
- <math>e^{ix} = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \frac{(ix)^6}{6!} + \frac{(ix)^7}{7!} + \frac{(ix)^8}{8!} + \cdots<math>
- <math>= 1 + ix - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \frac{ix^5}{5!} - \frac{x^6}{6!} - \frac{ix^7}{7!} + \frac{x^8}{8!} + \cdots<math>
- <math>= \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} + \cdots \right) + i\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \right) <math>
- <math>= \cos (x) + i\sin (x) <math>
The rearrangement of terms is justified because each series is absolutely convergent.
I think people will find it much easier to follow this proof.
- One problem is that you wrote "z = ix" but z is not defined and it does not appear elsewhere. Also, the proof works for all complex x but you limited it to real x. --Zero 09:27, 7 May 2004 (UTC)
- I see how might not be entirely clear -- actually, I did say what z is, when I said, "for complex z, we define each of these functions by these series". Also, yes, the proof works for all complex z, but "Euler's formula" is usually taken to mean when z is purely imaginary, primarily for historical reasons (Euler "derived" it for ix, not z); also, when most people say "Euler's formula", they're usually intimating at the periodic nature of exp around the unit circle. But, it's certainly true for any z, and I can add this.
- If z = a + bi; ez = eaebi, so there is no problem with complex x. ✏ Sverdrup 13:12, 7 May 2004 (UTC)
- I'm convinced, this looks very good. ✏ Sverdrup 13:12, 7 May 2004 (UTC)
- The "dz/z = i dt" argument can be made even more legit for most folks by taking the 2nd order linear diff eq, w'' = -w, gotten by iterating the 1st order one, then everything is real and you don't have to think about analytic continuation, etc.
- It would be interesting to note how Euler actually "discovered" this. The way he "proved" it is completely backwards from how it is usually presented in modern form -- he assumed DeMoivre's identity, and did some clever fooling around with i's (infinitely small numbers) and w's (infinitely large numbers), treating them as ordinary numbers. Of course, not rigorous at all, but historically very interesting, providing insight into Euler's brain circuitry.
Moved Euler characteristic material
I moved the material about the euler characteristic to the Euler_characteristic page. My reasons were
- elimination of duplicated material
- article should only focus on one topic and not on two totally unrelated topics.
For people who are looking for the euler characteristic on this page I have put a short note at the top.
I have tried to fix all broken links but I probably forgot some.MathMartin 22:18, 2 Aug 2004 (UTC)
The "by calculus" proof
...is wrong because it ignores the constant of integration. Please fix it! --Zero 03:23, 5 Dec 2004 (UTC)