Question

How do you find the Maclaurin Series for `F(x) = x*sin(x)`?

Answer 1

Maclaurin series for `xsinx` is

= `x^2-x^4/(3!)+x^6/(5!)-x^8/(7!)+x^10/(9!).......`

Explanation:

A Maclaurin series is a Taylor series expansion of a function about `0`. For example,

`f(x)=f(0)+f'(0)x+(f''(0))/(2!)x^2+(f^(3)(0))/(3!)x^3+(f^4(0))/(4!)x^4+.......`

Thus Maclaurin series for `sinx` would be

`sinx=sin0+cos0*x-sin0/(2!)x^2-cos0/(3!)x^3+sin0/(4!)x^4+.......`

= `0+1*x-0*x^2-1/(3!)x^3+1/(5!)x^5.......`

= `x-1/(3!)x^3+1/(5!)x^5-1/(7!)x^7+1/(9!)x^9.......`

Hence, Maclaurin series for `xsinx` is

= `x^2-x^4/(3!)+x^6/(5!)-x^8/(7!)+x^10/(9!).......`