Why does e^{πi} + 1 = 0?This page is just a collection of a couple of answers on the LiveJournal Mathematics Community in a thread about e^{πi} + 1 = 0. Soon, I will whip them into a more coherent form. In collegiate calculus, you probably learned about something called Taylor series. You can use Taylor series to make polynomial approximations of infinitely-differentiable functions. If you take the Taylor series out infinitely, you actually have the function. The general form for a Taylor series is this: where the sum goes from n=0 to n=infinity and f^{(n)}(x_{0}) means the n-th derivative of f(x) evaluated at x_{0}. So, as long as there is some x_{0} at which we know infinitely many derivatives of f(x), we're set. Also, note that f^{(0)}(x) is just f(x). And, also note that 0! is defined to be one. Fortunately, we know all of the derivatives of e^{x}, sin x, and cos x. All of the derivatives of e^{x} are equal to e^{x}. And, e^{0} = 1. So, one convenient Taylor series expansion of e^{x} is: The derivatives of sin x are a bit more tricky. And, from there the pattern repeats... f^{(k+4)}(x) = f^{(k)}. Since sin 0 = 0 and cos 0 = 1, we can then make a Taylor series for sin x: And, in a similar manner, we can make a Taylor series for cos x: You can try taking these sums out a few places to verify to yourself that they really work for real numbers. Now, comes the fun part. What happens when you put in a complex number into these polynomials instead? The relevant interesting thing happens when you put ix into the series expansion for e^{ix}. The powers of i repeat every fourth power, just like the derivatives of sin and cos repeated every fourth time. So, when we plug ix into the expansion for e^{x}, we get: And, you can collect all the terms that contain an i to get: And that is just a series representation of cos x + i sin x. Another sort of way to go about it: That would have to go something like this, right: and since f(0) must be one, then A must be one. But, I'm uncomfortable with the absolute value signs there. |