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as a function of a real variable, apparently. Part of the freedom in choosing a proof is that you get to choose what definition of $e^{ix}$ to start from -- do you use a differential equation? a power series? a definition in terms of trig functions? Another bit of freedom is that you get to choose what definition of $\pi$ to start from.

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closed as primarily opinion-based by Daniel Fischer Mar 7 at 21:31

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise. If this question can be reworded to fit the rules in the help center, please edit the question.

    
Even if x is complex, $\exp(ix)$ is still periodic. –  J. M. Sep 2 '10 at 9:05
    
No takers for using the ODE?! –  Mariano Suárez-Alvarez Sep 2 '10 at 12:48
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Maybe I didn't really frame the question clearly enough, but these are all kind of boring proofs. Perhaps I'd get some more interesting responses on MathOvervlow? –  user1678 Sep 3 '10 at 6:55
    
I suppose the recognition that e^ix = cosx + isinx from Taylor series is really surprising when you first see it. But I find it inelegant because you're reasoning about an infinite number of terms (granted, they're really simple terms) in order to understand something about a finite number of functions. I kind of assumed there'd be another proof out there that doesn't rely so heavily on Taylor series... –  user1678 Sep 3 '10 at 7:02

3 Answers 3

My favorite has always been Walter Rudin's proof in the prologue to his "Real and Complex Analysis" (2nd Ed.). Here's a sketch:

  • Define $\exp$ in terms of the power series.

  • By manipulating the series, deduce that $\exp$ is a homomorphism from the additive group to the group of complex units.

  • Show it satisfies the usual first order ODE.

  • Define $\cos$ and $\sin$ as the real and imaginary parts of $\exp$, respectively.

  • Define $\pi$ as twice the smallest positive real root of $\cos$.

  • Deduce that $\exp( i \pi / 2) = i$.

  • By multiplying, conclude that $2 \pi i$ is a period of $\exp$.

  • Show, by means of the preceding properties, that no smaller period exists.

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The first time I ever read that prologue, it gave me goosebumps. –  heropup Mar 7 at 20:26

$$ e^{ix} = \cos x + i \sin x \ . $$

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That pushes the question back: how do you define cosine and sine, and then--based on that definition--how do prove that their common period is 2 pi? One would like a definition that makes it relatively easy to prove the important properties of exp, especially that it is an entire analytic function of the complex plane. –  whuber Sep 2 '10 at 16:07
    
You're right, but assuming the knowledge of cosine, sine and Euler identity it's an streamlined proof, isn't it? :-) –  a.r. Sep 2 '10 at 17:32
    
Well, yes, but then it's a tautology. Now you have to prove that the lcm of the periods of sine and cosine equals $2 \pi$! –  whuber Sep 2 '10 at 22:10
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Define (cos x, sin x) to be the rotation image of (1, 0) by x about the origin. They are thus obviously periodic with period 2π. The Maclaurin series for sine and cosine follow from this definition. Define exp(z) by its power series and it follows that $e^{ix}=\cos x+i\sin x$. –  Isaac Sep 2 '10 at 22:14

$$e^{ix} = e^{i(x+T)} = e^{ix}e^{iT}$$

We have to find $T$ for which $e^{iT} = 1$

$$\rightarrow cos(T) + isin(T) = 1$$

$$\rightarrow sin(T) = 0$$ for all $$T = 2n\pi , n = 0,1,2,3...$$

So, period is $2\pi$.

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Can you explain the last line? You go from $T = 2n\pi$ to $2\pi$ –  Tyler Hilton Sep 10 '10 at 18:51