Question

How do you find the Maclaurin series for `sin(6x^2)` centered at 0?

Answer 1

`sin (6x^2) =sum_(n=0)^oo (-1)^n (6^(2n+1) x^(2(2n+1)))/((2n+1)!)`

Explanation:

We can start from the MacLaurin series of `sint`:

`sint = sum_(n=0)^oo (-1)^n t^(2n+1)/((2n+1)!)`

Now substitute `t=6x^2`:

`sin (6x^2) = sum_(n=0)^oo (-1)^n (6x^2)^(2n+1)/((2n+1)!)= sum_(n=0)^oo (-1)^n (6^(2n+1) x^(2(2n+1)))/((2n+1)!)`