I understand that the equation for kinematic displacement is:
$x = v_{0x}t+\frac{1}{2}a_xt^2$
Perhaps my understanding is naive, but it seems like this leaves out higher order rates of change. Why wouldn't the equation be like:
$x = v_{0x}t+\frac{1}{2}a_xt^2+\frac{1}{6}j_xt^3+\frac{1}{24}s_xt^4+\frac{1}{120}c_xt^5+. . . $
where $j_x$ represents jerk, $s_x$ represents snap, $c_x$ represents crackle, and so on for $n$ number of higher-order terms, perhaps as an infinite series?
Let's first number the equations for convenience:
\begin{align} x&=x_0+vt+\frac 1 2 at^2\tag{1}\\ &\qquad\text{and}\\ x&=x_0+vt+\frac 1 2 at^2+\frac 1 6 j t^3+\frac 1 {24} st^4 +\frac 1 {120} ct^5\dots\tag{2}\\ &=\frac 1 {0!} x_0+\frac 1 {1!} vt +\frac 1 {2!}at^2+\frac 1 {3!} jt^3+\frac 1 {4!} st^4 +\frac 1 {5!} ct^5\dots \tag{3} \end{align}
Now the equations $(2)$ and $(3)$ are the most general equations for any particle's motion, whereas the equation $(1)$ is the equation which holds true only in the special case where the acceleration is constant. This implies that the higher derivatives of position are zero (i.e. $a=\text{constant}\implies j=s=c=\dots=0$). Thus the equations $(2)$ and $(3)$ reduce to equation $(1)$.
In essence, the equation $(2)$ is just the Taylor expansion of displacement $x$ and therefore it's the most general form of representing the displacement $x$ and it holds true in all cases.
You're just mixing displacement Taylor expansion around t=0 for any acceleration and kinematics of the constant acceleration motion.
The first relation you wrote $$ x= v_{0 \mathrm{x}} t + \frac{1}{2} a_\mathrm{x} t^2 $$ holds true for constant acceleration only but it is exact.
The latter is more generic, but it's an approximate expression which holds true for any acceleration.
By the way it is consistent with the first one given naught higher derivatives like jerk and beyond.
