taylor series 1 lecture participation: proble\n(2 points)\nwrite the taylor series for ( f(x)=sin (x) ) at (…

taylor series 1 lecture participation: proble\n(2 points)\nwrite 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}\right)^{n} ).\nfind the first five coefficients.\n( c_{0}= )\n( c_{1}= )\n( c_{2}= )\n( c_{3}= )\n( c_{4}= )\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have unlimited attempts remaining.

taylor series 1 lecture participation: proble\n(2 points)\nwrite 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}\right)^{n} ).\nfind the first five coefficients.\n( c_{0}= )\n( c_{1}= )\n( c_{2}= )\n( c_{3}= )\n( c_{4}= )\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have unlimited attempts remaining.

Answer

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})