find the taylor series for ( f(x) ) centered at the given value of ( a ). assume that ( f ) has a power…

Explanation:

Step1: Recall the Taylor series formula

The Taylor series of a function (f(x)) centered at (a) is given by (f(x)=\sum_{n = 0}^{\infty}\frac{f^{(n)}(a)}{n!}(x - a)^{n}). First, find the derivatives of (y=\sin(x)). The (n) - th derivative of (y = \sin(x)) is (y^{(n)}=\sin\left(x+\frac{n\pi}{2}\right)). Evaluate at (a=\pi): (y(\pi)=\sin(\pi) = 0) (y'(\pi)=\cos(\pi)=- 1) (y''(\pi)=-\sin(\pi)=0) (y'''(\pi)=-\cos(\pi)=1) (y^{(4)}(\pi)=\sin(\pi) = 0) The pattern for (y^{(n)}(\pi)) is: When (n = 2k) (even), (y^{(2k)}(\pi)=0); when (n = 2k + 1) (odd), (y^{(2k + 1)}(\pi)=(-1)^{k + 1})

Step2: Write out the Taylor series

(f(x)=\sum_{n = 0}^{\infty}\frac{f^{(n)}(\pi)}{n!}(x-\pi)^{n}) Since (f^{(2k)}(\pi)=0) for (k = 0,1,2,\cdots), let (n = 2k+1). (f(x)=\sum_{k = 0}^{\infty}\frac{(-1)^{k + 1}}{(2k+1)!}(x - \pi)^{2k+1})

Step3: Use the ratio test for radius of convergence

Let (a_{n}=\frac{(-1)^{n+1}}{(2n+1)!}(x - \pi)^{2n+1}) (\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_{n}}\right|=\lim_{n\rightarrow\infty}\left|\frac{\frac{(-1)^{n+2}(x-\pi)^{2n+3}}{(2n+3)!}}{\frac{(-1)^{n+1}(x-\pi)^{2n+1}}{(2n+1)!}}\right|) (=\lim_{n\rightarrow\infty}\left|\frac{(-1)(x - \pi)^{2}}{(2n+3)(2n+2)}\right|) (=\lim_{n\rightarrow\infty}\frac{(x - \pi)^{2}}{(2n+3)(2n+2)}=0) for all (x)

Answer:

The Taylor series is (f(x)=\sum_{n = 0}^{\infty}\frac{(-1)^{n+1}}{(2n+1)!}(x - \pi)^{2n+1}), and the radius of convergence (R=\infty)