The problem is:
f(x) = [cos(3x^2)-1]/(x^2) at x=0
I know that the Mclaurin series for cos(x) = 1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + ....
That is where I'm stuck, I can't remember how to use the Mclaurin series with this function
The problem is:
f(x) = [cos(3x^2)-1]/(x^2) at x=0
I know that the Mclaurin series for cos(x) = 1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + ....
That is where I'm stuck, I can't remember how to use the Mclaurin series with this function
Learn and watch.
$\displaystyle \cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-...$
Thus, (composition of two analyic functions at a point is analytic),
$\displaystyle \cos (3x^2)=1-\frac{(3x^2)^2}{2!}+\frac{(3x^2)^4}{4!}-...$
Thus,
$\displaystyle \cos (3x^2)-1=-\frac{9x^4}{2!}+\frac{81x^8}{4!}-...$
When you divide by $\displaystyle x^2$, is the same as dividing term by term, thus,
$\displaystyle -\frac{9x^2}{2!}+\frac{81x^6}{4!}-...$