Kinematic equation as infinite sum

Question

I'm not sure exactly how to phrase this question, but here it goes:

$v=\dfrac{dx}{dt}$ therefore $x=x_0+vt$

UNLESS there's an acceleration, in which case

$a=\dfrac{dv}{dt}$ therefore $x=x_0+v_0t+\dfrac{1}{2}at^2$

UNLESS there's a jerk, in which case

$j=\dfrac{da}{dt}$ therefore $x=x_0+v_0t+\dfrac{1}{2}a_0t^2+\dfrac{1}{6}jt^3$

Are you picking up on the pattern? Velocity is the first derivative of position with respect to time, acceleration is the second, jerk is the third, and the formula just gets longer and longer.

Let's say that, hypothetically, and object was moving such that $\dfrac{d^{500}x}{dt^{500}}$ was a constant greater than zero. Is there some formula for an object's position that implements an infinite summation of time derivatives of position? maybe following the form

$$x=\sum_{n=0}^{\infty}\dfrac{1}{n!}\dfrac{d^nx}{dt^n}t^n$$

Answer 1

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.

Answer 2

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 $$ \begin{aligned} v(t) =& \frac{{\rm d} x(t)}{{\rm d}t}\\a(t) =& \frac{{\rm d}^2 x(t)}{{\rm d}t^2} \\ j(t) =& \frac{{\rm d}^3 x(t)}{{\rm d}t^3} \\ \ldots \end{aligned} $$