Determine the $19$-th, $20$-th and $21$-st order Maclaurin polynomial for $f(x) = x\left(1-\cos{(2x^3)}\right)$

Question

Determine the $19$-th, $20$-th and $21$-st order Maclaurin polynomial for $f(x) = x\left(1-\cos{(2x^3)}\right)$.
We know that \begin{equation*} \cos{(x)} = \sum_{n=0}^{\infty} (-1)^n\frac{x^{2n}}{(2n)!} = 1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\ldots \end{equation*} Thus, \begin{equation*} \begin{split} x\left(1-\cos{(2x^3)}\right)&= x\left(1-\left(1-\frac{4x^6}{2}+\frac{16x^{12}}{24}-\frac{64x^{18}}{720}+O(x^{24})\right)\right) \\ &= x\left(2x^6-\frac{2x^{12}}{3}+\frac{4x^{18}}{45}-O(x^{24})\right) \\ &= 2x^7-\frac{2x^{13}}{3}+\frac{4x^{19}}{45}-O(x^{25}). \end{split} \end{equation*} Thus, the $19$-th, $20$-th and $21$-st order Maclaurin polynomial are all the same which are given by $2x^7-\frac{2x^{13}}{3}+\frac{4x^{19}}{45}$. Is this good?

Answer 1

Yes, your answer is correct and your calculations are valid. Since we have zero coefficients for terms $20, 21,22$ the polynomials of degree $19,20,21$ are all the same as the one with degree $19$