b8 - taylor and maclaurin series: prol(2 points)\nfind the taylor series for ( f(x)=sin (x) ) centered at (…

b8 - taylor and maclaurin series: prol(2 points)\nfind the taylor series for ( f(x)=sin (x) ) centered at ( c=\frac{pi}{2} ).\n( sin (x)=sum_{n = 0}^{infty} )\non what interval is the expansion valid? give your answer using interval notation. if yo is the only point in the interval of convergence, you would answer with 0.\nthe expansion is valid on\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have 15 attempts remaining.

b8 - taylor and maclaurin series: prol(2 points)\nfind the taylor series for ( f(x)=sin (x) ) centered at ( c=\frac{pi}{2} ).\n( sin (x)=sum_{n = 0}^{infty} )\non what interval is the expansion valid? give your answer using interval notation. if yo is the only point in the interval of convergence, you would answer with 0.\nthe expansion is valid on\nnote: you can earn partial credit on this problem.\npreview my answers submit answers\nyou have attempted this problem 0 times.\nyou have 15 attempts remaining.

Answer

Answer:

$$\sin(x)=\sum_{n = 0}^{\infty}\frac{(- 1)^{n}}{(2n)!}\left(x-\frac{\pi}{2}\right)^{2n}$$ The expansion is valid on $(-\infty,\infty)$

Explanation:

Step1: Recall the Taylor series formula

The Taylor series of a function (f(x)) centered at (x = c) is given by (f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(c)}{n!}(x - c)^{n})

Step2: Find the derivatives of (y=\sin(x)) and evaluate at (c=\frac{\pi}{2})

  • (f(x)=\sin(x)), (f\left(\frac{\pi}{2}\right)=1)
  • (f^{\prime}(x)=\cos(x)), (f^{\prime}\left(\frac{\pi}{2}\right)=0)
  • (f^{\prime\prime}(x)=-\sin(x)), (f^{\prime\prime}\left(\frac{\pi}{2}\right)=- 1)
  • (f^{(3)}(x)=-\cos(x)), (f^{(3)}\left(\frac{\pi}{2}\right)=0)
  • (f^{(4)}(x)=\sin(x)), (f^{(4)}\left(\frac{\pi}{2}\right)=1)

We can observe the pattern: (f^{(2n)}\left(\frac{\pi}{2}\right)=(-1)^{n}) and (f^{(2n + 1)}\left(\frac{\pi}{2}\right)=0) for (n = 0,1,2,\cdots)

Step3: Substitute into the Taylor series formula

[ \begin{align*} \sin(x)&=\sum_{n = 0}^{\infty}\frac{f^{(n)}\left(\frac{\pi}{2}\right)}{n!}\left(x-\frac{\pi}{2}\right)^{n}\ &=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}\left(x-\frac{\pi}{2}\right)^{2n}+\sum_{n = 0}^{\infty}\frac{0}{(2n+1)!}\left(x-\frac{\pi}{2}\right)^{2n + 1}\ &=\sum_{n = 0}^{\infty}\frac{(-1)^{n}}{(2n)!}\left(x-\frac{\pi}{2}\right)^{2n} \end{align*} ]

Step4: Determine the interval of convergence

We use the ratio test. Let (a_{n}=\frac{(-1)^{n}}{(2n)!}\left(x-\frac{\pi}{2}\right)^{2n})

[ \begin{align*} \lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_{n}}\right|&=\lim_{n\rightarrow\infty}\left|\frac{\frac{(-1)^{n+1}(x-\frac{\pi}{2})^{2(n + 1)}}{(2(n+1))!}}{\frac{(-1)^{n}(x-\frac{\pi}{2})^{2n}}{(2n)!}}\right|\ &=\lim_{n\rightarrow\infty}\left|\frac{(-1)(x-\frac{\pi}{2})^{2}}{(2n + 2)(2n+1)}\right|\ &=0 \end{align*} ]

Since (\lim_{n\rightarrow\infty}\left|\frac{a_{n + 1}}{a_{n}}\right| = 0<1) for all (x\in(-\infty,\infty)), the interval of convergence is ((-\infty,\infty))