Question

How do you find the power series representation for the function `f(x)=sin(x^2)` ?

Answer 1

Since

`sinx=sum_{n=0}^infty(-1)^n{x^{2n+1}}/{(2n+1)!}`,

by replacing `x` by `x^2`,

`Rightarrow f(x)=sum_{n=0}^infty(-1)^n{(x^2)^{2n+1}}/{(2n+1)!}`

`=sum_{n=0}^infty(-1)^n{x^{4n+2}}/{(2n+1)!}`