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