Q. 4 - Calculus

Question

What is a Taylor polynomial for a function f at a point $x_0$?

Step-by-Step Solution

Answer

${\mathrm{P}}_{\mathrm{n}}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{\mathrm{f}}^{\mathrm{k}}\left({\mathrm{x}}_0\right)}{\mathrm{k}!}{(\mathrm{x}-{\mathrm{x}}_0)}^{\mathrm{k}}$

Step 1. Given information is:

$\mathrm{We} \mathrm{have} \mathrm{a} \mathrm{function} \mathrm{f}\left(\mathrm{x}\right)$

Step 2. Finding Taylor Polynomial

$$\begin{gathered} \mathrm{Consider} \mathrm{that} \mathrm{the} \mathrm{function} \mathrm{f} \mathrm{is} \mathrm{a} \mathrm{function} \mathrm{with} \mathrm{a} \mathrm{derivative} \mathrm{of} \mathrm{order} \mathrm{n}, \\ \mathrm{then} \mathrm{the} \mathrm{taylor} \mathrm{polynomial} \mathrm{at} \mathrm{x}={\mathrm{x}}_0 \mathrm{is}, \\ {\mathrm{P}}_{\mathrm{n}}\left(\mathrm{x}\right)=\mathrm{f}\left({\mathrm{x}}_0\right) +\mathrm{f}\text{'}\left({\mathrm{x}}_0\right)(\mathrm{x}-{\mathrm{x}}_0)+\frac{\mathrm{f}\text{'}\text{'}\left({\mathrm{x}}_0\right)}{2!}{(\mathrm{x}-{\mathrm{x}}_0)}^2+....+\frac{{\mathrm{f}}^{\mathrm{n}}\left({\mathrm{x}}_0\right)}{\mathrm{n}!}{(\mathrm{x}-{\mathrm{x}}_0)}^{\mathrm{n}} \\ \mathrm{The} \mathrm{general} \mathrm{form} \mathrm{of} \mathrm{the} \mathrm{Taylor} \mathrm{polynomial} \mathrm{of} \mathrm{the} \mathrm{function} \mathrm{f} \mathrm{in} \mathrm{the} \mathrm{compact} \mathrm{form} \mathrm{is}: \\ {\mathrm{P}}_{\mathrm{n}}\left(\mathrm{x}\right)=\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{\mathrm{f}}^{\mathrm{k}}\left({\mathrm{x}}_0\right)}{\mathrm{k}!}{(\mathrm{x}-{\mathrm{x}}_0)}^{\mathrm{k}} \end{gathered}$$