009C Sample Final 1, Problem 6 Detailed Solution

Find the Taylor polynomial of degree 4 of  $f(x)=\cos ^{2}x$  at  $a={\frac {\pi }{4}}$.

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Background Information:
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:
First, we make a table to find the coefficients of the Taylor polynomial.
$n$ $f^{(n)}(x)$ $f^{(n)}(a)$ ${\frac {f^{(n)}(a)}{n!}}$ $0$ $\cos ^{2}x$ ${\frac {1}{2}}$ ${\frac {1}{2}}$ $1$ $-2\cos x\sin x$ $-1$ $-1$ $2$ $2\sin ^{2}x-2\cos ^{2}x$ $0$ $0$ $3$ $8\sin x\cos x$ $4$ ${\frac {2}{3}}$ $4$ $8\cos ^{2}x-8\sin ^{2}x$ $0$ $0$

Step 2:
Since  $c_{n}={\frac {f^{(n)}(a)}{n!}},$  the Taylor polynomial of degree 4 of  $f(x)=\cos ^{2}x$  is
$T_{4}(x)={\frac {1}{2}}+-1{\bigg (}x-{\frac {\pi }{4}}{\bigg )}+{\frac {2}{3}}{\bigg (}x-{\frac {\pi }{4}}{\bigg )}^{3}.$

Final Answer:
${\frac {1}{2}}+-1{\bigg (}x-{\frac {\pi }{4}}{\bigg )}+{\frac {2}{3}}{\bigg (}x-{\frac {\pi }{4}}{\bigg )}^{3}$

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