It turns out $e^{x}$ can be defined as a long polynomial:

$$\begin{aligned} \displaystyle e^x = \sum_{k=0}^\infty \frac{x^k}{k!} \\ \end{aligned}$$

which has some nice properties.

For example, as a polynomial it's not hard to see that the derivative of $e^x$ is equal to itself, and the integral of $e^x$ is $e^x + C$.

Per Euler's formula,

$$\begin{aligned} \displaystyle e^{ix} = \cos(x) + i \sin(x) \\ \end{aligned}$$

but treating $e^{ix}$ as a long polynomial,

$$\begin{aligned} \displaystyle e^{ix} &= 1 + ix + \frac{i^2x^2}{2!} + \frac{i^3x^3}{3!} + \frac{i^4x^4}{4!} + \frac{i^5x^5}{5!} + \frac{i^6x^6}{6!} + \cdots \\ \displaystyle &= 1 + ix - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \frac{ix^5}{5!} - \frac{x^6}{6!} + \cdots \\ \displaystyle &= (1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots) + i(x - \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots) \\ \end{aligned}$$

Therefore,

$$\begin{aligned} \displaystyle \cos x &= \sum_{k=0}^\infty (-1)^k\frac{x^{2k}}{(2k)!} \\ \displaystyle \sin x &= \sum_{k=0}^\infty (-1)^k\frac{x^{2k+1}}{(2k+1)!} \\ \end{aligned}$$

Treating these as long polynomials, it's rather easy to verify that the derivative of $\cos x$ is $- \sin x$, and likewise the derivative of $\sin x$ is $\cos x$.

# posted by rot13(Unafba Pune) @ 8:50 AM