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- Thread starter jsmith
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The Taylor expansions of the exponential function and the trigonometric functions are very similar. The Taylor series of sin(x) is very similar to odd terms of the Taylor series of exp(x). The only difference is that the sine series alternates in sign. The same is true for the Taylor series of cos(x) and the even terms of the Taylor series of exp(x). There's obviously a connection between the exponential function and those trigonometric functions.

Those sign alternations vanish when one looks at exp(ix) as opposed to exp(x). The real and imaginary parts of exp(ix) are the trigonometric functions.

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mathman

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phyzguy

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It's astonishing, isn't it? This is one of many examples of the deep underlying unity of mathematics. As you continue to study mathematics, you will find other examples of seemingly unrelated things which turn out to have an underlying connection. It always reminds me of the fable of the blind men and the elephant. Each branch of mathematics that we study is just a piece of a unified whole.

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Infrared

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A geometric way to think of the identity [itex] e^{i\pi}=-1 [/itex] is that magnitudes multiply and arguments add when multiplying complex numbers.

[tex] e^{i\pi}= \lim_{n \to \infty} (1+\frac{i\pi}{n})^n= \lim_{n \to \infty} (\sqrt{1+(\frac{\pi^2}{n^2})}z)^n [/tex]

where [itex] z [/itex] is a complex number with unit magnitude and argument [itex] \frac{\pi}{n} [/itex] (equal to ([itex] cos\frac{\pi}{n})+isin(\frac{\pi}{n})[/itex]).

So, [tex] \lim_{n \to \infty} (\sqrt{1+\frac{\pi^2}{n^2}}z)^n= \lim_{n \to \infty} (1+(\frac{\pi^2}{n^2}))^\frac{n}{2}z^n=\lim_{n \to \infty} z^n [/tex]. Since [itex] z [/itex] has magnitude 1 and argument [itex] \frac{\pi}{n} [/itex], the limit has magnitude 1 and argument [itex] \pi [/itex].

Therefore [itex] e^{i\pi}=-1 [/itex].

Note: I adapted this argument from The Princeton Companion to Mathematics. I thought it was neat and that I should share it. Choosing a general angle instead of [itex] \pi [/itex] gives the more general formula [itex] e^{i\theta}=cos\theta + isin\theta [/itex].

[tex] e^{i\pi}= \lim_{n \to \infty} (1+\frac{i\pi}{n})^n= \lim_{n \to \infty} (\sqrt{1+(\frac{\pi^2}{n^2})}z)^n [/tex]

where [itex] z [/itex] is a complex number with unit magnitude and argument [itex] \frac{\pi}{n} [/itex] (equal to ([itex] cos\frac{\pi}{n})+isin(\frac{\pi}{n})[/itex]).

So, [tex] \lim_{n \to \infty} (\sqrt{1+\frac{\pi^2}{n^2}}z)^n= \lim_{n \to \infty} (1+(\frac{\pi^2}{n^2}))^\frac{n}{2}z^n=\lim_{n \to \infty} z^n [/tex]. Since [itex] z [/itex] has magnitude 1 and argument [itex] \frac{\pi}{n} [/itex], the limit has magnitude 1 and argument [itex] \pi [/itex].

Therefore [itex] e^{i\pi}=-1 [/itex].

Note: I adapted this argument from The Princeton Companion to Mathematics. I thought it was neat and that I should share it. Choosing a general angle instead of [itex] \pi [/itex] gives the more general formula [itex] e^{i\theta}=cos\theta + isin\theta [/itex].

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There is a deep, underlying unity in a lot of math, that may be hard to see as long as we look at these things from a viewpoint of what problems they were invented to solve.

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So e^ix is a spiral that turns around the unit circle at a "natural" rate of e (whatever that means :P). And turning at this "natural" rate is complemented by pi which is half the circumference of the unit circle. Just my two cents.

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It's astonishing, isn't it? This is one of many examples of the deep underlying unity of mathematics. As you continue to study mathematics, you will find other examples of seemingly unrelated things which turn out to have an underlying connection. It always reminds me of the fable of the blind men and the elephant. Each branch of mathematics that we study is just a piece of a unified whole.

Considering math is one giant tautology, why do you find its underlying unity surprising?

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