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

# posted by rot13(Unafba Pune) @ 8:49 PM 0 comments

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 $\displaystyle \frac{1}{1-x}$

# posted by rot13(Unafba Pune) @ 4:37 PM 0 comments

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 0 comments