Thread: Computer 10th derivative using Mclaurin series

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

Originally Posted by pakman

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.


Thus, (composition of two analyic functions at a point is analytic),

Thus,

When you divide by , is the same as dividing term by term, thus,