# Infinite series of derivatives of position when starting from rest

Suppose you have an object with zero for the value of all the derivatives of position. In order to get the object moving you would need to increase the value of the velocity from zero to some finite value. A change is velocity is acceleration, so the value of the acceleration would have to increase from zero to some value. You would also need to increase the jerk from zero to some value to have a change in acceleration. Is there an infinite series of higher derivatives of position for this to work? Or am I missing something?

• You have just proven why things can never move. It would take changing a inifinite series of derivatives from zero to a non-zero value. Another proof is that for a object to move a finite distance, it first has to move half that distance, then half the remaining distance, then half of that remaining distance. Obviously it would take forever to move the infinite number of segments. Realizing objects can't move also rather simplifies a lot of other physics. – Olin Lathrop May 5 '14 at 13:14
• Some higher order derivative is going to be non continuous, and a step function. – ja72 May 5 '14 at 16:42

I) Disclaimer: In this answer, we will only consider the mathematically idealized classical problem, which is interesting in itself, although admittedly academic and detached from actual physical systems.

II) It is in principle possible that all time derivatives of the position $x(t)$ vanishes at $t=0$, and yet the particle does move away (from where it was at $t=0$).

$\uparrow$ Fig. 1: A plot of position $x$ as a function of time $t$.

Example: Assume that the position is given by the following infinitely many times differentiable function

$$x(t)~:=~\left\{ \begin{array}{ccl} \exp\left(-\frac{1}{|t|}\right)&\text{for} & t~\neq~ 0, \\ \\ 0 &\text{for} & t~=~ 0. \end{array} \right.$$

Note that the Taylor series for the $C^{\infty}$-function $x:\mathbb{R}\to\mathbb{R}$ in the point $t=0$ is identically equal to zero.$^1$ So the function $x$ and its Taylor series are different for $t\neq 0$. In particular, we conclude that the smooth function $x$ is not a real analytic function.

III) If you like this Phys.SE question you may also enjoy reading about What situations in classical physics are non-deterministic? , Non-uniqueness of solutions in Newtonian mechanics and Norton's dome and its equation.

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$^1$ Sketched proof that $x\in C^{\infty}(\mathbb{R})$: Firstly, obviously, $x$ is $C^{\infty}$ for $t\neq 0$. By induction in $n\in\mathbb{N}_0$, for $t\neq 0$, it is straightforward to deduce that the $n$'th derivative is of the form $$x^{(n)}(t)~=~\frac{P_n(t,|t|)}{Q_n(t,|t|)}x(t), \qquad t\neq 0,$$ for some polynomials $P_n$ and $Q_n\neq 0$. Secondly, it follows that the $(n\!+\!1)$'th derivative at the origin $$x^{(n+1)}(0)~=~\lim_{t\to 0} \frac{x^{(n)}(t)}{t}~=~0$$ exists and is zero "because exponentials beat polynomials". $\Box$

• Is this because $\bf{e}$ is a transcendental number and thus the derivatives don't loose order. – ja72 Apr 24 '15 at 12:18
• This function has derivatives that look like a dirac delta function near zero. Essentially they provide an "impulse" to get things going as the value of the derivatives approaches infinity near $t=0$. – ja72 Apr 24 '15 at 13:00
• Here is a plot of the $C^{\infty}$-function in the example. – Qmechanic Apr 24 '15 at 15:32
• Yes, the trick is that the derivatives are almost like step functions. See plot from Wolfram Alpha. – ja72 Apr 24 '15 at 15:41
• The important point is that the derivatives in the example are still smooth. – Qmechanic Apr 24 '15 at 18:29

There are smooth functions that have all derivatives equal to zero at a point (in fact, outside a bounded interval) yet are not constant. You can for example find a smooth function $f$ such that $f(t)$ and all its derivatives are $0$ whenever $t \le 0$ and $f(t) = 1$ whenever $t \ge 1$ (and all the derivatives of $f$ are zero when $t \ge 1$). Then if you take the force in Newton's second law to be $$F(t) = mf''(t)$$ the solution to the IVP $$m\ddot{x} = F \qquad x(0) = \dot{x}(0) = 0$$ is clearly a smooth motion starting at $x = 0$ with all derivatives of displacement also zero, and ending at $x = 1$, with all derivatives of displacement also zero.

However for a time-independent force the IVP $$m\ddot{x}(t) = F(x(t)) \qquad x(0) = \dot{x}(0) = 0$$ when $F(x(0)) = 0$ the solution is a particle at rest. This follows from uniqueness of solutions to ODEs. Physically when $F = 0$ there is an equilibrium position, and a particle at rest at an equilibrium position remains at rest there. Of course this is really just Newton's first law.

• There are classical situations one can construct where the IVP of no inital force and no initial movement is not uniquely solved by the particle remaining at rest, see Norton's dome. By Picard-Lindelöf, you need $F(x)$ to be Lipschitz continuous to guarantee local uniqueness of the solution, which is a condition different from smooth as in "infinitely often differentiable", but it is not clear that all physical forces should be Lipschitz continuous. – ACuriousMind Mar 26 '15 at 14:00

Your question is "Is there an infinite series of higher derivatives of position for this to work?"

Acceleration can jump from zero to something. When it does, its derivative is not defined, so the series of position derivatives stops after the second one.

From the question: "A change is velocity is acceleration, so the value of the acceleration would have to increase from zero to some value."

The first part is true, we need non-zero acceleration. The second part is misleadingly phrased. It can be read as "acceleration has to increase continuously", which is false. In contrast to velocity, acceleration can jump from zero to something.

Philosophically speaking, there is no law that says "in physics, derivatives are always smooth". Some derivatives just jump.

• Why can the acceleration be discontinuous ? – Abc2000ro Apr 21 '15 at 9:35
• Think of why acceleration happens. It is caused by something. Can you turn that on or off? – Rainer Blome Apr 22 '15 at 20:21
• @Rainer Blome. In Physics you learn that a force is causing an acceleration. Ok, but then what is causing a jerk ? You cannot answer "an increasing force", because then what about the infinite derivative of position ? Is it caused by an infinite increasing force ? – Abc2000ro Apr 24 '15 at 9:20
• @Abc2000ro Ideed I do not answer "an increasing force", which might be interpreted as "continuously increasing". I answer "a force that changes from zero to something". With a suitable setup, you can turn some forces "on" (or "off"). For example in a trigger mechanism, or when a spring hits an end stop, or using electrical current. Since the force jumps, its derivative is not defined. – Rainer Blome Apr 25 '15 at 13:07

Mathematically, that's how it would work out. If you solved the initial value problem where all time derivatives of displacement were zero, then you would have zero displacement. Don't overthink it :)

Taking the Taylor's series approach at $t=0$ with respect to time, with all the derivatives zero the result would be zero motion.

If motion is present then some higher order derivatives must be an impulse at $t=0$ such that the derivative below that is a step function.

In valvetrain dynamics this is usually done by specifying linear jerk (step snap, impulse crackle, undefined pop).

So at $t=0$ you have the 4th derivative step from zero to $\ddddot{x}>0$ and making jerk linear $\dddot{x} = C_1 t$, acceleration a parabola $\ddot{x} = C_2 t^2$, velocity $\dot{x}=C_3 t^3$ and position $x=C_4 t^4$.

In some critical applications you might specify linear snap increasing the order of the curve by 1 to avoid certain undesirable dynamics of the valve spring.

NOTE: Snap Crackle and Pop are real terms.