\[f(x)\]
can be represented by a - possibly infinite - polynomial, called a Taylor series. The the function is repeatedly differentiated at the derivatives are evaluated at a point.\[\begin{equation} \begin{aligned} f(x) &= f(x_0) + \frac{\frac{df}{dx}}{1!}(x-x_0)+ \frac{\frac{d^2f}{dx^2}}{2!}(x-x_0)^2 \\ &+ \frac{\frac{d^3f}{dx^3}}{3!}(x-x_0)^3+...+ \frac{\frac{d^nf}{dx^n}}{n!}(x-x_0)^n+... \\ &=\sum_{n=1}^{\infty} \frac{\frac{d^nf}{dx^n}}{n!}(x-x_0)^n \end{aligned} \end{equation}\]
Example:
\[f(x)=sinx, \; \frac{df}{dx}=cosx, \; \frac{d^2f}{dx^2}=-sinx, \; \frac{d^3f}{dx^3}=-cosx, \; \frac{d^4f}{dx^4}=sinx,...\]
Evaluated at
\[x= \frac{\pi}{2}\]
the respective values are \[1, \; 0, \; -1, \; 0, \; 1,...\]
and the Taylor series is\[\begin{equation} \begin{aligned} sinx &= 1+\frac{0}{1!}(x- \frac{\pi}{2})+\frac{-1}{2!}(x- \frac{\pi}{2})^2 \\ &+ \frac{0}{3!}(x- \frac{\pi}{2})^3+\frac{1}{4!}(x- \frac{\pi}{2})^4 \\ &=1- \frac{1}{2}(x- \frac{\pi}{2})^2+ \frac{1}{24}(x- \frac{\pi}{2})^4 \end{aligned} \end{equation} \]
.A Mclaurin series is a Taylor series at
\[x_0=0\]
. Every power series - so called because they contain powers of \[x\]
- has an interval of convergence, within which the service converges. The general form of the series above is \[a_n=\frac{(-1)^{n}}{(2n)!}(x- \frac{\pi}{2})^(2n)\]
and the series converges if \[-1 \lt \frac{a_{n+1}}{a_n} \lt 1 \rightarrow -1 \lt \frac{\frac{(-1)^{n+1}}{(2(n+1))!}(x- \frac{\pi}{2})^{2(n+1)}}{\frac{(-1)^{n}}{(2n)!}(x- \frac{\pi}{2})^(2n)} = \frac{-(x- \frac{\pi}{2})^2}{2n(2n-1)} \lt 1\]
.For a fixed
\[x, \; lim_{n \rightarrow \infty} \frac{-(x- \frac{\pi}{2})^2}{(2n(2n-1)}=0\]
so the series converges for all \[x\]
.