# 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).

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