I need some help here.
Find Maclaurin series representation for the function $f(x)=\cos(2x^3).$
I guess the easiest thing to do is using that $\displaystyle\cos x = \sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n)!}{x^{2n}}.$
Using that identity I end up with: $\displaystyle \sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n)!}{2x^{6n}},$
which is wrong. Could anyone help?
Best Answer
You were close, but you replaced $x$ in $\cos x$ with $(x^3)$ to get $(x^3)^{2n}$ and then multiplied this by one factor of $2$ to get $2(x^3)^{2n}$.
However, we need to replace $x$ with all of $(2x^3)$ to get $(2x^3)^{2n} = 2^{2n}\cdot (x^3)^{2n} = 4^nx^{6n}.$
Doing this gives us:
$$\cos(2x^3) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}(2x^3)^{2n} = \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} 4^nx^{6n}$$