| | For all \(x\):
\[
\begin{aligned}
e^x &= \sum_{k=0}^\infty \frac{x^k}{k!} \\
\cos x &= \sum_{k=0}^\infty (-1)^k \frac{x^{2k}}{(2k)!} \\
\sin x &= \sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!} \\
\cosh x &= \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} \\
\sinh x &= \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!} \\
\end{aligned}
\] |
|
For \(\lvert x \rvert < 1\):
\[
\begin{aligned}
\frac{1}{1 - x} &= \sum_{k=0}^\infty x^k \\
\ln(1+x) &= \sum_{k=1}^\infty (-1)^{k+1} \frac{x^k}{k} \\
\arctan(x) &= \sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{2k+1} \\
(1 + x)^\alpha &= \sum_{k=0}^\infty {\alpha \choose k} x^k \\
\end{aligned}
\] |
Hyperbolic trigonometric functions
| |
\[
\begin{aligned}
\sinh(x) &= \frac{e^x - e^{-x}}{2} \\
\cosh(x) &= \frac{e^x + e^{-x}}{2} \\
\tanh(x) &= \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{\sinh(x)}{\cosh(x)} \\
\end{aligned}
\] |
|
|
The Taylor series about \(x = 0\) of any given function:
| |
\[
\begin{aligned}
\displaystyle f(x) = \sum_{k=0}^\infty f^{(k)}(0)\frac{x^k}{k!} \\
\end{aligned}
\] |
Some interesting observations:
- The long polynomial of \(\displaystyle e^x = \sum_{k=0}^\infty \frac{x^k}{k!}\) can easily be verified to be just the Taylor series about \(x = 0\) of \(e^x\)
- The Taylor series about \(x = 0\) of a polynomial is the polynomial per se
- Geometric series such as \(1 + x + x^2 + x^3 + \cdots\) can easily be verified to be the Taylor series about \(x = 0 \) of \(\displaystyle \frac{1}{1-x}\)
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 \(\cos x\).
