Comment by drx | Hacker News Reader

Comment by drx | Hacker News Reader

Parent Full thread

It is actually far from just a convention. It has everything to do with the Taylor series of the functions e^x, sin x and cos x.

e^x = 1 + x/1 + x^2/(2!) + x^3/(3!) + ...

If you substitute ix for x, you get:

e^{ix} = 1 + ix/1 + (ix)^2/(2!) + (ix)^3/(3!) + ... = = 1 + ix + (-x^2)/(2!) + (-i)x^3/(3!) + ...

You can also derive the Taylor series for sin x and cos x, which are:

sin x = x - x^3/(3!) + x^5/(5!) + ...

cos x = 1 - x^2/(3!) + x^4/(4!) + ...

Therefore you can see that e^{ix} = cos x + i(sin x). Of course, I haven't shown how to derive the Taylor series (http://en.wikipedia.org/wiki/Taylor_series) for these functions, but I have to stop somewhere.

I hope that I have shown you that this isn't just an arbitrary convention, in fact far from it. The formula is so beautiful also because of the many intricate relationships between all these elementary mathematical concepts, including complex analysis, trigonometric functions, series, etc.

It should also be noted that this is just one of the many (in fact, aleph zero many) different ways to prove this equivalence.

Edit: Some formatting corrections, sorry, new here.

I can't resist not showing this clever derivation of the e Maclaurin series.

I referenced the fact that

e^x = 1 + x/1 + x^2/(2!) + x^3/(3!) + ...

Now, one of the many equivalent definitions of e is that it is the only real number for which this holds:

d/dt (e^t) = e^t

This means that e^t is a real function which is so smooth, that no matter how many times you differentiate it, you always get the same function (it turns out that it is the only such real function)

Now, we know that e^0 = 1 (since x^0 = 1, where x != 0). Therefore, in its Maclaurin series, the only term not depending on x should therefore be 1 (otherwise e^0 wouldn't be 1).

So know we know that e^x looks something like this:

e^x = 1 + (something)

Now we can ask ourselves which this question: since d/dx e^x = e^x, what must also be in the e^x series, if 1 belongs to it? Well, whatever differentiates to 1, so now we know that

e^x = 1 + x + (something)

(because d/dx (1 + x + something) = 1 + x + d/dx something)

Now we can again ask this question for our current form; what must we differentiate in order to obtain 1 + x? And thus we get

e^x = 1 + x + x^2/2 + something

This way, if we write the two (equivalent) series this way:

e^x = 1 + x + x^2/2 + d/dx e^x = 1 + x + x^2/2 + x^3/(2*3) +

And we can complete it with the infinite Maclaurin series.

Now this is less formal than it should be and it would probably make the formalists cringe, but I hope you get the idea. You can actually apply the same principle for sin x and cos x, except in their case, they're actually mutually derived from each other. I'll leave that as an exercise for the reader (oh how fun it is to say that after reading this phrase countless times)

Great. Now I'm going to spend the rest of the morning trying to figure out whether you can really do it that way.

[Edit: Ah yes. The formalism goes like this: Consider the above series. By the above argument, it differentiates to itself. QED. ]

Ah, but how do you know that e^x is the only thing that differentiates to itself? :)

You don't need to. The uniqueness part of the definition is never used in that argument. (In fact, a*e^x also differentiates to itself, for any a; but that's a trivial case.)

Uniqueness almost follows from that argument. It's now easy to see that exp is the only analytic function satisfying exp' = exp and exp(0) = 1: if you have another one, by the same argument, it has the same Maclaurin expansion, hence is the same function.

However, I don't know how to prove uniqueness over all functions, not just analytic ones.

Reply on news.ycombinator.com