Taylor series, derivation of

Let $f(x)$ be given by the following power series:

$\displaystyle f(x)=c_{0}+c_{1}(x-a)+c_{2}(x-a)^{2}+\cdots+c_{n}(x-a)^{n}+% \cdots=\sum_{k=0}^{\infty}c_{k}(x-a)^{k}$

Now let’s compute a few derivatives at $x=a.$

$\displaystyle f(a)=c_{0};\quad f^{\prime}(a)=c_{1};\quad f^{\prime\prime}(a)=2% c_{2};\quad f^{(3)}(a)=6c_{3};\quad f^{(n)}(a)=n!c_{n}$

From this, it is clear that $\displaystyle c_{n}=\frac{f^{(n)}(a)}{n!}$ , thus the series can be written as:

$\displaystyle T_{n}=\sum_{k=0}^{n}c_{k}(x-a)^{k}=\sum_{k=0}^{n}\frac{f^{(k)}(a% )}{k!}(x-a)^{k}$

where $f(x)=T_{\infty}$ .