Question

How do you find the Maclaurin series for `f(x)= x cos (1/2x^2)`?

Answer 1

`xcos(1/2x^2) = sum_(n=0)^oo ((-1)^n)/2^(2n)x^(4n+1)/((2n)!)`

Explanation:

We can start from the MacLaurin series for `cost`:

`cost = sum_(n=0)^oo (-1)^n ^(2n)/((2n)!)`

substitute `t=1/2x^2`

`cos(1/2x^2) = sum_(n=0)^oo (-1)^n (1/2x^2)^(2n)/((2n)!)=sum_(n=0)^oo (-1)^n 1/2^(2n)x^(4n)/((2n)!)`

and finally multiply by `x`:

`xcos(1/2x^2) = sum_(n=0)^oo ((-1)^n)/2^(2n)x^(4n+1)/((2n)!)`