Question

find the taylor polynomial of orders 0, 1, 2, and 3 generated by f at a

f(x)=cosx, 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)=\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 \(f^{(2)}(\frac{2\pi}{3})=-\cos(\frac{2\pi}{3})=\frac{1}{2}\)

Step2: Calculate \(P_{2}(x)\)

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

We know \(P_{1}(x)=-\frac{1}{2}-\frac{\sqrt{3}}{2}(x-\frac{2\pi}{3})\) (since \(P_{1}(x)=\frac{f( \frac{2\pi}{3})}{0!}+\frac{f^{\prime}(\frac{2\pi}{3})}{1!}(x - \frac{2\pi}{3})\))

Now, \(\frac{f^{(2)}(\frac{2\pi}{3})}{2!}(x - \frac{2\pi}{3})^{2}=\frac{\frac{1}{2}}{2}(x-\frac{2\pi}{3})^{2}=\frac{1}{4}(x - \frac{2\pi}{3})^{2}\)

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Answer:

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