Question

Question 18: In Problems, find a power series expansion for $f\left(X\right)$ , given the expansion for f(x).

role="math" localid="1664283848012" $f\left(x\right)=\sin x=\sum _{k=0}^{\infty }\frac{{(-1)}^k}{(2k+1)!}{x}^{2k+1}$

Short answer

The differentiation for the series

$f\left(x\right)=\sum _{n=0}^{\infty }{a}_n{x}^{n}f\left(x\right)=\sum _{n=0}^{\infty }\left(a_n{x}^{n=1}\right)$

The given series is

$f\left(x\right)=\sin x=\sum _{k=0}^{\infty }\frac{{(-1)}^k}{(2k+1)!}{x}^{2k+1}$

The differentiation of the above will be

$f\left(x\right)=\sum _{k=0}^{\infty }(2k+1)\left(\frac{{(-1)}^k}{(2k+1)!}{x}^{2k}\right)$

Step-by-step solution

differentiation of given power series

The differentiation for the series

$f\left(x\right)=\sum _{n=0}^{\infty }{a}_n{x}^{n}f\left(x\right)=\sum _{n=0}^{\infty }\left(a_n{x}^{n=1}\right)$

The given series is

$f\left(x\right)=\sin x=\sum _{k=0}^{\infty }\frac{{(-1)}^k}{(2k+1)!}{x}^{2k+1}$

The differentiation of the above will be

$f\left(x\right)=\sum _{k=0}^{\infty }(2k+1)\left(\frac{{(-1)}^k}{(2k+1)!}{x}^{2k}\right)$

Final proof

$f\left(x\right)=\sum _{k=0}^{\infty }(2k+1)\left(\frac{{(-1)}^k}{(2k+1)!}{x}^{2k}\right)$