4.1: Taylor Series

One series you have encountered before is Taylor’s series,

$$f(x) = \sum_{n=0}^{\infty} f^{(n)}(a)\frac{(x-a)^n}{n!}, \label{eq:IV:taylor}$$

where $f^{(n)}(x)$ is the $n$th derivative of $f$. An example is the Taylor series of the cosine around $x=0$ (i.e., $a=0$),

$$\begin{aligned} &&\qquad&\cos(0) &= 1,\nonumber\\ \cos'(x) &= -\sin(x),&&\cos'(0)&=0,\nonumber\\ \cos^{(2)}(x) &= -\cos(x),&&\cos^{(2)}(0)&=-1,\\ \cos^{(3)}(x) &= \sin(x),&&\cos^{(3)}(0)&=0,\nonumber\\ \cos^{(4)}(x) &= \cos(x),&&\cos^{(4)}(0)&=1.\end{aligned} \nonumber$$

Notice that after four steps we are back where we started. We have thus found (using $m=2n$ in ($\PageIndex{1}$)) )

$$\cos x = \sum_{m=0}^\infty \frac{(-1)^m}{(2m)!} x^{2m}, \nonumber$$

Exercise $\PageIndex{1}$

Show that $$\sin x = \sum_{m=0}^\infty \frac{(-1)^m}{(2m+1)!} x^{2m+1}. \nonumber$$

Answer

TBA