Question

How do you find the maclaurin series expansion of `f(x)=cos(5x^2)`?

Answer 1

`1-25/2 x^4+ 625/24 x^8 - 3125/144 x^12+cdots`

Explanation:

Take the Maclaurin expansion for `cos(x)` and replace `x` with `5x^2`:

`cos(x)=1-x^2/(2!)+x^4/(4!)-x^6/(6!)+cdots`

so

`cos(5x^2)=1-(5x^2)^2/2+(5x^2)^4/24-(5x^2)^6/720+cdots`

`=1-25/2 x^4+ 625/24 x^8 - 3125/144 x^12+cdots`

This converges for all `x`.