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)=\square
(type an exact answer, using \pi as needed.)

Explanation:

Step1: Recall the formula for Taylor polynomial

The Taylor polynomial of order \(n\) for a function \(f(x)\) about \(x = a\) is \(P_n(x)=\sum_{k = 0}^{n}\frac{f^{(k)}(a)}{k!}(x - a)^k\), where \(f^{(k)}(x)\) is the \(k\) - th derivative of \(f(x)\).
First, find the derivatives of \(f(x)=\cos x\).
\(f(x)=\cos x\), \(f^{\prime}(x)=-\sin x\), \(f^{\prime\prime}(x)=-\cos x\), \(f^{\prime\prime\prime}(x)=\sin x\)

Step2: Evaluate the function and its first - order derivative at \(a=\frac{2\pi}{3}\)

• Evaluate \(f(x)\) at \(x = a=\frac{2\pi}{3}\): \(f(\frac{2\pi}{3})=\cos(\frac{2\pi}{3})=-\frac{1}{2}\)
• Evaluate \(f^{\prime}(x)\) at \(x = a=\frac{2\pi}{3}\): \(f^{\prime}(\frac{2\pi}{3})=-\sin(\frac{2\pi}{3})=-\frac{\sqrt{3}}{2}\)

Step3: Find \(P_1(x)\)

For \(n = 1\), \(P_1(x)=f(a)+f^{\prime}(a)(x - a)\)
Substitute \(a=\frac{2\pi}{3}\), \(f(a)=-\frac{1}{2}\), and \(f^{\prime}(a)=-\frac{\sqrt{3}}{2}\) into the formula:
\(P_1(x)=-\frac{1}{2}-\frac{\sqrt{3}}{2}(x-\frac{2\pi}{3})\)

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Answer:

\(P_1(x)=-\frac{1}{2}-\frac{\sqrt{3}}{2}x+\frac{\sqrt{3}\pi}{3}\)