Question

find the taylor polynomial of orders 0, 1, 2, and 3 generated by f at a
$f(x)=cos x, a=\frac{2 pi}{3}$
$p_{0}(x)=-\frac{1}{2}$
(type an exact answer, using $pi$ as needed)
$p_{1}(x)=-\frac{1}{2}-\frac{sqrt{3}}{2} x+\frac{sqrt{3} pi}{3}$
(type an exact answer, using $pi$ as needed)
$p_{2}(x)=-\frac{1}{2}-\frac{sqrt{3}}{2}left(x-\frac{2 pi}{3}
ight)+\frac{1}{4}left(x-\frac{2 pi}{3}
ight)^{2}$
(type an exact answer, using $pi$ as needed)
$p_{3}(x)=square$
(type an exact answer, using $pi$ as needed)

Explanation:

Step1: Recall the Taylor polynomial formula

The Taylor polynomial of order \(n\) for a function \(f(x)\) about \(x = a\) is given by \(P_{n}(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^{k}\), where \(f^{(k)}(x)\) is the \(k\) - th derivative of \(f(x)\).

For \(f(x)=\cos x\), we have:

• \(f^{(0)}(x)=\cos x\), so \(f^{(0)}(\frac{2\pi}{3})=\cos(\frac{2\pi}{3})=-\frac{1}{2}\)
• \(f^{(1)}(x)=-\sin x\), so \(f^{(1)}(\frac{2\pi}{3})=-\sin(\frac{2\pi}{3})=-\frac{\sqrt{3}}{2}\)
• \(f^{(2)}(x)=-\cos x\), so \(f^{(2)}(\frac{2\pi}{3})=-\cos(\frac{2\pi}{3})=\frac{1}{2}\)
• \(f^{(3)}(x)=\sin x\), so \(f^{(3)}(\frac{2\pi}{3})=\sin(\frac{2\pi}{3})=\frac{\sqrt{3}}{2}\)

Step2: Calculate \(P_{3}(x)\)

Using the Taylor polynomial formula \(P_{3}(x)=\frac{f( \frac{2\pi}{3})}{0!}(x-\frac{2\pi}{3})^{0}+\frac{f^{(1)}(\frac{2\pi}{3})}{1!}(x - \frac{2\pi}{3})^{1}+\frac{f^{(2)}(\frac{2\pi}{3})}{2!}(x-\frac{2\pi}{3})^{2}+\frac{f^{(3)}(\frac{2\pi}{3})}{3!}(x-\frac{2\pi}{3})^{3}\)

Substitute the values:

$$
LATEXBLOCK0
$$

Answer:

\(P_{3}(x)=-\frac{1}{2}-\frac{\sqrt{3}}{2}(x - \frac{2\pi}{3})+\frac{1}{4}(x-\frac{2\pi}{3})^{2}+\frac{\sqrt{3}}{12}(x-\frac{2\pi}{3})^{3}\)