I have a question about the represantation of a function.
$$\ddot {\vec r}= -g\vec e_z +2\vec v \times \vec \omega - \vec \omega \times (\vec \omega \times \vec v)$$ with $\vec v = \dot {\vec r}$. In the next step it is stated that you can write $\ddot {\vec r}$ the following way:
$$\ddot {\vec r}=\ddot {\vec r}^{(0)}+\omega\ddot {\vec r}^{(1)}+\omega^2\ddot {\vec r}^{(2)}+... $$ The number in the round brackets is called the order of the function. Why can you represent the function like this? I think it looks a bit like a Taylor-Series.
Edit as a responds to the first comment:
I only did a Taylor series for one-dimensional functions until now, but $\ddot {\vec r}$ is a vector function in 3 dimensions. The funtion $\ddot {\vec r}$ only depends on the time $t$, so it would be $\ddot {\vec r}: \mathbb{R} \rightarrow \mathbb{R}^3$
$\vec \omega = \omega \begin{pmatrix} \omega_x \\ 0 \\ \omega_z \end{pmatrix} $, $\vec r(t) = \begin{pmatrix} r_x(t) \\ r_y(t) \\ r_z(t) \end{pmatrix} $ and $\vec v =\dot {\vec r(t)} = \begin{pmatrix} \dot r_x(t) \\ \dot r_y(t) \\ \dot r_z(t) \end{pmatrix} $. Initial conditions for $t=0$: $\vec r(0) = \begin{pmatrix} 0 \\ 0 \\ h \end{pmatrix} $ ($h$ is the height of the particle) and $\vec v(0) =\dot {\vec r(0)} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} $
I would compare the components of $\ddot {\vec r}$ separately $\ddot r_z(t)=\sum \limits_{n=0}^{\infty}\omega^n\cdot {\ddot r_z}^{(n)}$ to the general form of a Taylor-Series of a function $f(t)$ developed at point $a$: $$f(t)=\sum \limits_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(t-a)^n$$ So I would identify $f(t)=\ddot r_z(t)$ and $a=0$. The zero order with $n=0$ would then be $\ddot r_z^{(0)}=\frac{\ddot r_z(0)}{0!}(t-0)^0=-g$
The first order would be: $\ddot r_z^{(1)}=\ddot r_z^{(0)}+\frac{\dddot r_z(0)}{1!}t^1=-g+\dddot r_z(0)\cdot t$