## Points of maximal curvature

In Calculus I students are taught how to find the points at which the graph of a function has zero curvature (that is, the points of inflection). The points of maximal curvature are usually not discussed. This post attempts to rectify this omission.

The (signed) curvature of ${y=f(x)}$ is

${\displaystyle \kappa = \frac{y''}{(1+(y')^2)^{3/2}}}$

We want to maximize the absolute value of ${\kappa}$, whether the function is positive or negative there (so both maxima and minima of ${\kappa}$ can be of interest). The critical points of ${\kappa}$ are the zeros of

${\displaystyle \kappa' = \frac{y''' (1+(y')^2) - 3 y' (y'')^2 }{(1+(y')^2)^{5/2}}}$

So we are lead to consider a polynomial of the first three derivatives of ${y}$, namely ${p := y''' (1+(y')^2) - 3 y' (y'')^2 }$.

Begin with some simple examples:

${y = x^2}$ has ${p = -24x}$ so the curvature of a parabola is maximal at its vertex. No surprise there.

${y = x^3}$ has ${p = 6(1 - 45x^4)}$, indicating two symmetric points of maximal curvature, ${x\approx \pm 0.386}$, pretty close to the point of inflection.

${y=x^4}$ has ${p = 24 x (1 - 56 x^6)}$. This has three real roots, but ${x=0}$ actually minimizes curvature (it vanishes there).

More generally, ${y=x^n}$ with positive integer ${n}$ yields ${\displaystyle p = n(n-1)x^{n-3} (n - 2 - (2n^3-n^2) x^{2n-2})}$ indicating two points ${\displaystyle x = \pm \left(\frac{n-2}{2n^3-n^2} \right)^{1/(2n-2)}}$ which tend to ${\pm 1}$ as ${n}$ grows.

The graph of a polynomial of degree ${n}$ can have at most ${n-2}$ points of zero curvature, because the second derivative vanishes at those. How many points of maximal curvature can it have? The degree of expression ${p}$ above is ${3n-5}$ but it is not obvious whether all of its roots can be real and distinct, and also be the maxima of ${|\kappa|}$ (unlike ${x=0}$ for ${y=x^4}$). For ${n=2}$ we do get ${3n-5 = 1}$ point of maximal curvature. But for ${n=3}$, there can be at most ${2}$ such points, not ${3n-5 = 4}$. Edwards and Gordon (Extreme curvature of polynomials, 2004) conjectured that the graph of a polynomial of degree ${n}$ has at most ${n-1}$ points of maximal curvature. This remains open despite several partial results: see the recent paper Extreme curvature of polynomials and level sets (2017).

A few more elementary functions:

${y = e^x}$ has ${p = e^x(1-2e^{2x})}$, so the curvature is maximal at ${x=-\log(2)/2}$. Did I expect the maximum of curvature to occur for a negative ${x}$? Not really.

${y=\sin x}$ has ${p = (\cos 2x - 3)\cos x}$. The first factor is irrelevant: the points of maximum curvature of a sine wave are at its extrema, as one would guess.

${y = \tan x}$ has ${p = -6\tan^8(x) - 16\tan^6(x) - 6\tan^4(x) + 8\tan(x)^2 + 4}$ which is zero at… ahem. The expression factors as

${\displaystyle p = -2(\tan^2 x+1)(3\tan^6(x) + 5\tan^4(x) - 2\tan^2(x) - 2)}$

Writing ${u = \tan^2(x)}$ we can get a cubic equation in ${u}$, but it is not a nice one. Or we could do some trigonometry and reduce ${p=0}$ to the equation ${8 - 41\cos 2x + \cos 6x =0}$. Either way, a numerical solution is called for: ${x \approx \pm 0.6937}$ (and ${+\pi n}$ for other periods).

## Extreme values of a reproducing kernel for polynomials

For every nonnegative integer ${n}$ there exists a (unique) polynomial ${K_n(x, y)}$ of degree ${n}$ in ${x}$ and ${y}$ separately with the following reproducing property:

${\displaystyle p(x) = \int_{-1}^1 K_n(x, y)p(y)\,dy}$

for every polynomial ${p}$ of degree at most ${n}$, and for every ${x}$. For example, ${K_1(x, y)= (3xy+1)/2}$; other examples are found in the post Polynomial delta function.

This fact gives an explicit pointwise bound on a polynomial in terms of its integral on an interval:

${\displaystyle |p(x)| \le M_n(x) \int_{-1}^1 |p(y)|\,dy}$

where ${M_n(x) = \sup\{|K(x, y)| \colon y\in [-1, 1]\}}$. For example, ${M_1(x) = (3|x|+1)/2}$.

Although in principle ${x}$ could be any real or complex number, it makes sense to restrict attention to ${x\in [-1, 1]}$, where integration takes place. This leads to the search for extreme values of ${K}$ on the square ${Q=[-1, 1]\times [-1, 1]}$. Here is how this function looks for ${n=1, 2, 3}$:

The symmetries ${K(x, y)=K(-x, -y) = K(y, x)}$ are evident here.

Explicitly,

${\displaystyle K_n(x, y) = \sum_{k=0}^n \frac{2k+1}{2} P_k(x)P_k(y)}$

where ${P_k}$ is the Legendre polynomial of degree ${k}$ and the factor ${(2k+1)/2}$ is included to make the polynomials an orthonormal set in ${L^2(-1, 1)}$. Since ${P_k}$ oscillates between ${-1}$ and ${1}$, it follows that

${\displaystyle |K_n(x, y)|\le \sum_{k=0}^n \frac{2k+1}{2} = \frac{(n+1)^2}{2}}$

and this bound is attained at ${K(1, 1)=K(-1,-1)=(n+1)^2/2}$ because ${P_k(1)=1}$ and ${P_k(-1)=(-1)^k}$.

Is

${\displaystyle K_n(-1, 1) =\sum_{k=0}^n (-1)^k\frac{2k+1}{2} = (-1)^n \frac{n+1}{2}}$

the minimum value of ${K}$ on the square ${Q}$? Certainly not for even ${n}$. Indeed, differentiating the sum

${\displaystyle S_n(x) = K_n(x, 1) = \sum_{k=0}^n \frac{2k+1}{2} P_k(x)}$

with respect to ${x}$ and using ${P_k'(-1) =(-1)^{k-1}k(k+1)/2}$, we arrive at

${\displaystyle S_n'(-1) = (-1)^{n-1} \frac{n(n^2+3n+2)}{4}}$

which is negative if ${n}$ is even, ruling out this point as a minimum.

What about odd ${n}$, then: is it true that ${K_n \ge -(n+1)/2}$ on the square ${Q}$?

${n=1}$: yes, ${K_1(x, y) = (3xy+1)/2 \ge (-3+1)/2 = -1}$ is clear enough.

${n=3}$: the inequality ${K_3\ge -2}$ is also true… at least numerically. It can be simplified to ${35(xy)^3 + 9(xy)^2 + 15xy \ge (21x+21y+3)(x^2+y^2)}$ but I do not see a way forward from there. At least on the boundary of ${Q}$ it can be shown without much work:

${\displaystyle K_3(x, 1) + 2 = \frac{5}{4}(x+1)(7x^2-4x+1)}$

The quadratic term has no real roots, which is easy to check.

${n=5}$: similar story, the inequality ${K_5\ge -3}$ appears to be true but I can only prove it on the boundary, using

${\displaystyle K_5(x, 1)+3 = \frac{21}{16}(x + 1)(33 x^4 - 18x^3 - 12x^2 + 2x + 3)}$

The quartic term has no real roots, which is not so easy to check.

${n=7}$: surprisingly, ${K_7(4/5, 1) = -2229959/500000}$ which is about ${-4.46}$, disproving the conjectural bound ${K_7\ge -4}$. This fact is not at all obvious from the plot.

Questions:

• Is ${K_n \ge -Cn}$ on the square ${Q = [-1, 1]\times [-1, 1]}$ with some universal constant ${C}$?
• Is the minimum of ${K_n}$ on ${Q}$ always attained on the boundary of ${Q}$?
• Can ${M_n(x) = \sup\{|K(x, y)| \colon y\in [-1, 1]\}}$ be expressed in closed form, at least for small degrees ${n}$?

## Need for speed vs bounded position and acceleration

You are driving a car with maximal acceleration (and deceleration) A on a road that’s been blocked off in both directions (or, if you prefer, on the landing strip of an aircraft carrier). Let L be the length of the road available to you.
What is the maximal speed you can reach?

Besides A and L, the answer also depends on your mood: do you want to live, or are you willing to go out in a blaze of glory? In the latter case the answer is obvious: position the car at one end of the interval, and put the pedal to the metal. The car will cover the distance L within the time ${\sqrt{2L/A}}$, reaching the speed ${v=\sqrt{2AL}}$ at the end. In the former scenario one has to switch to the brake pedal midway through the distance, so the maximal speed will be attained at half the length, ${\sqrt{AL}}$.

Rephrased in mathematical terms: if ${f}$ is a twice differentiable function and ${M_k = \sup|f^{(k)}|}$ for ${k=0,1,2}$, then ${M_1^2 \le 4M_0M_2}$ if ${f}$ is defined on a half-infinite interval, and ${M_1^2 \le 2M_0M_2}$ if the domain of ${f}$ is the entire line. To connect the notation, just put ${L=2M_0}$ and ${A=M_1}$ in the previous paragraph… and I guess some proof other than “this is obvious” is called for, but it’s not hard to find one: this is problem 5.15 in Rudin’s Principles of Mathematical Analysis.

Perhaps more interesting is to study the problem in higher dimensions: one could be driving in a parking lot of some shape, etc. Let’s normalize the maximal acceleration as 1, keeping in mind it’s a vector. Given a set E, let S(E) be the square of maximal speed attainable by a unit-acceleration vehicle which stays in E indefinitely. Also let U(E) be the square of maximal speed one can attain while crashing out of bounds after the record is set. Squaring makes these quantities scale linearly with the size of the set. Both are monotone with respect to set inclusion. And we know what they are for an interval of length L: namely, ${S = L}$ and ${U=2L}$, so that gives some lower bounds for sets that contain a line interval.

When E is a circle of radius 1, the best we can do is to drive along it with constant speed 1; then the centripetal acceleration is also 1. Any higher speed will exceed the allowable acceleration in the normal direction, never mind the tangential one. So, for a circle both S and U are equal to its radius.

On the other hand, if E is a disk of radius R, then driving along its diameter is better: it gives ${S\ge 2R}$ and ${U\ge 4R}$.

Some questions:

1. If E is a convex set of diameter D, is it true that ${S(E) = D}$ and ${U(E) = 2D}$?
2. Is it true that ${U\le 2S}$ in general?
3. How to express S and U for a smooth closed curve in terms of its curvature? They are not necessarily equal (like they are for a circle): consider thin ellipses converging to a line segment, for which S and U approach the corresponding values for that segment.

The answer to Question 1 is yes. Consider the orthogonal projection of E, and of a trajectory it contains, onto some line L. This does not increase the diameter or the acceleration; thus, the one-dimensional result implies that the projection of velocity vector onto L does not exceed ${\sqrt{D}}$ (or ${\sqrt{2D}}$ for the crashing-out version). Since L was arbitrary, it follows that ${S(E) \le D}$ and ${U(E) \le 2D}$. These upper bounds hold for general sets, not only convex ones. But when E is convex, we get matching lower bounds by considering the longest segment contained in E.

I don’t have answers to questions 2 and 3.