Question

taylor series 1 lecture participation: proble
(2 points)
write the taylor series for ( f(x)=sin (x) ) at ( x=\frac{pi}{3} ) as ( sum_{n = 0}^{infty} c_{n}left(x-\frac{pi}{3}
ight)^{n} ).
find the first five coefficients.
( c_{0}= )
( c_{1}= )
( c_{2}= )
( c_{3}= )
( c_{4}= )
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Explanation:

Step1: Recall the Taylor series formula

The Taylor series of a function \(f(x)\) about \(x = a\) is given by \(\sum_{n = 0}^{\infty}c_{n}(x - a)^{n}\), where \(c_{n}=\frac{f^{(n)}(a)}{n!}\). Here \(a=\frac{\pi}{3}\) and \(f(x)=\sin(x)\).

Step2: Calculate \(f(\frac{\pi}{3})\)

\(f(x)=\sin(x)\), so \(f(\frac{\pi}{3})=\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}\), and \(c_{0}=\frac{f(\frac{\pi}{3})}{0!}=\frac{\sqrt{3}}{2}\) (since \(0!=1\)).

Step3: Calculate \(f^{\prime}(x)\) and \(f^{\prime}(\frac{\pi}{3})\)

\(f^{\prime}(x)=\cos(x)\), then \(f^{\prime}(\frac{\pi}{3})=\cos(\frac{\pi}{3})=\frac{1}{2}\), and \(c_{1}=\frac{f^{\prime}(\frac{\pi}{3})}{1!}=\frac{1}{2}\).

Step4: Calculate \(f^{\prime\prime}(x)\) and \(f^{\prime\prime}(\frac{\pi}{3})\)

\(f^{\prime\prime}(x)=-\sin(x)\), so \(f^{\prime\prime}(\frac{\pi}{3})=-\sin(\frac{\pi}{3})=-\frac{\sqrt{3}}{2}\), and \(c_{2}=\frac{f^{\prime\prime}(\frac{\pi}{3})}{2!}=\frac{-\frac{\sqrt{3}}{2}}{2}=-\frac{\sqrt{3}}{4}\).

Step5: Calculate \(f^{(3)}(x)\) and \(f^{(3)}(\frac{\pi}{3})\)

\(f^{(3)}(x)=-\cos(x)\), then \(f^{(3)}(\frac{\pi}{3})=-\cos(\frac{\pi}{3})=-\frac{1}{2}\), and \(c_{3}=\frac{f^{(3)}(\frac{\pi}{3})}{3!}=\frac{-\frac{1}{2}}{6}=-\frac{1}{12}\).

Step6: Calculate \(f^{(4)}(x)\) and \(f^{(4)}(\frac{\pi}{3})\)

\(f^{(4)}(x)=\sin(x)\), so \(f^{(4)}(\frac{\pi}{3})=\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}\), and \(c_{4}=\frac{f^{(4)}(\frac{\pi}{3})}{4!}=\frac{\frac{\sqrt{3}}{2}}{24}=\frac{\sqrt{3}}{48}\).

Answer:

\(c_{0}=\frac{\sqrt{3}}{2}\), \(c_{1}=\frac{1}{2}\), \(c_{2}=-\frac{\sqrt{3}}{4}\), \(c_{3}=-\frac{1}{12}\), \(c_{4}=\frac{\sqrt{3}}{48}\)