# 009C Sample Final 2, Problem 5

Find the Taylor Polynomials of order 0, 1, 2, 3 generated by  $f(x)=\cos(x)$ at  $x={\frac {\pi }{4}}.$ Foundations:
The Taylor polynomial of   $f(x)$ at   $a$ is

$\sum _{n=0}^{\infty }c_{n}(x-a)^{n}$ where $c_{n}={\frac {f^{(n)}(a)}{n!}}.$ Solution:

Step 1:
Let  $a={\frac {\pi }{4}}.$ First, we make a table to find the coefficients of the Taylor polynomial.
 $n$ $f^{(n)}(x)$ $f^{(n)}(a)$ ${\frac {f^{(n)}(a)}{n!}}$ $0$ $\cos x$ ${\frac {\sqrt {2}}{2}}$ ${\frac {\sqrt {2}}{2}}$ $1$ $-\sin x$ $-{\frac {\sqrt {2}}{2}}$ $-{\frac {\sqrt {2}}{2}}$ $2$ $-\cos x$ $-{\frac {\sqrt {2}}{2}}$ $-{\frac {\sqrt {2}}{4}}$ $3$ $\sin x$ ${\frac {\sqrt {2}}{2}}$ ${\frac {\sqrt {2}}{12}}$ Step 2:
Let  $T_{n}$ be the Taylor polynomial of order  $n.$ Since   $c_{n}={\frac {f^{(n)}(a)}{n!}},$ we have

$T_{0}={\frac {\sqrt {2}}{2}}$ $T_{1}={\frac {\sqrt {2}}{2}}-{\frac {\sqrt {2}}{2}}{\bigg (}x-{\frac {\pi }{4}}{\bigg )}$ $T_{2}={\frac {\sqrt {2}}{2}}-{\frac {\sqrt {2}}{2}}{\bigg (}x-{\frac {\pi }{4}}{\bigg )}-{\frac {\sqrt {2}}{4}}{\bigg (}x-{\frac {\pi }{4}}{\bigg )}^{2}$ $T_{3}={\frac {\sqrt {2}}{2}}-{\frac {\sqrt {2}}{2}}{\bigg (}x-{\frac {\pi }{4}}{\bigg )}-{\frac {\sqrt {2}}{4}}{\bigg (}x-{\frac {\pi }{4}}{\bigg )}^{2}+{\frac {\sqrt {2}}{12}}{\bigg (}x-{\frac {\pi }{4}}{\bigg )}^{3}.$ Let  $T_{n}$ be the Taylor polynomial of order  $n.$ $T_{0}={\frac {\sqrt {2}}{2}}$ $T_{1}={\frac {\sqrt {2}}{2}}-{\frac {\sqrt {2}}{2}}{\bigg (}x-{\frac {\pi }{4}}{\bigg )}$ $T_{2}={\frac {\sqrt {2}}{2}}-{\frac {\sqrt {2}}{2}}{\bigg (}x-{\frac {\pi }{4}}{\bigg )}-{\frac {\sqrt {2}}{4}}{\bigg (}x-{\frac {\pi }{4}}{\bigg )}^{2}$ $T_{3}={\frac {\sqrt {2}}{2}}-{\frac {\sqrt {2}}{2}}{\bigg (}x-{\frac {\pi }{4}}{\bigg )}-{\frac {\sqrt {2}}{4}}{\bigg (}x-{\frac {\pi }{4}}{\bigg )}^{2}+{\frac {\sqrt {2}}{12}}{\bigg (}x-{\frac {\pi }{4}}{\bigg )}^{3}$ 