# Kinematic equation as infinite sum

Any "reasonable" function $f$ (such functions are called *analytic*) has such an expansion, known as a *Taylor expansion* involving the derivatives of the function itself, which converges to it. Consider a position $x(t)$ of a particle:
\begin{aligned}
x(t) = x(t_0 + \Delta t) &= x(t_0) + x'(t_0)\Delta t + \tfrac{1}{2}x''(t_0)\Delta t^2 + \tfrac{1}{6}x'''(t_0)\Delta t^3 + \dotsb \\
&= \sum_{n=0}^\infty \tfrac{1}{n!}x^{(n)}(t_0)\Delta t^n.
\end{aligned}
As you mention, we expand the function to higher orders depending on the influence of the higher-order derivatives. However, note that in this expansion, the powers of $\Delta t$ increase with each additional term we add to the expansion. For small values of $\Delta t$, these terms have less and less influence. A small term cubed is smaller than a small term squared is less than a small term to the first power.

For that matter, this kind of series expansion is ubiquitous in physics, for example in the potential energy of a particle near a point of stable equilibrium $x_0$: \begin{aligned} U(x) = U(x_0 + \Delta x) &= U(x_0) + U'(x_0)\Delta x + \tfrac{1}{2}U''(x_0)\Delta x^2 + \dotsb \\ &= U(x_0) + \tfrac{1}{2}U''(x_0)\Delta x^2 + \dotsb \end{aligned} since the term involving the first derivative of $U$ is zero when $U$ is at a point of stable equilibrium (derivative zero at a minimum). If we rewrite this equation ignoring the higher order terms that contribute less and less, and define $U(x) - U(x_0) = \Delta U$ we have \begin{aligned} U(x) &= U(x_0) + \tfrac{1}{2}U''(x_0)\Delta x^2 \\ \Delta U &= \tfrac{1}{2}U''(x_0)\Delta x^2, \end{aligned} which you may see is suspiciously similar to the potential energy of a harmonic oscillator like a mass on the spring: $$ \Delta U = \tfrac{1}{2}k\Delta x^2. $$ It is the harmonic oscillator that is a special case of the more general expansion we are seeing above.

All you are doing is $$x = \int \int a(t)\, {\rm d}t$$ and when you have linearly varying acceleration (w/ jerk) you end up with a cubic function.

The key to all of this is the fact that $$ \frac{{\rm d}^i x^n}{{\rm d}x^i} = \frac{n!}{(n-i)!} x^{n-i} $$

and you are solving the equations $$ \begin{aligned} v(t) =& \frac{{\rm d} x(t)}{{\rm d}t}\\a(t) =& \frac{{\rm d}^2 x(t)}{{\rm d}t^2} \\ j(t) =& \frac{{\rm d}^3 x(t)}{{\rm d}t^3} \\ \ldots \end{aligned} $$