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
$T_{0}={\frac {\sqrt {2}}{2}}$
$T_{1}={\frac {\sqrt {2}}{2}}-{\frac {\sqrt {2}}{2}}{\bigg (}x-{\frac {\pi }{4}}{\bigg )}$
$T_{2}={\frac {\sqrt {2}}{2}}-{\frac {\sqrt {2}}{2}}{\bigg (}x-{\frac {\pi }{4}}{\bigg )}-{\frac {\sqrt {2}}{4}}{\bigg (}x-{\frac {\pi }{4}}{\bigg )}^{2}$
$T_{3}={\frac {\sqrt {2}}{2}}-{\frac {\sqrt {2}}{2}}{\bigg (}x-{\frac {\pi }{4}}{\bigg )}-{\frac {\sqrt {2}}{4}}{\bigg (}x-{\frac {\pi }{4}}{\bigg )}^{2}+{\frac {\sqrt {2}}{12}}{\bigg (}x-{\frac {\pi }{4}}{\bigg )}^{3}.$

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

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