# Convexity of polar curves

Everybody knows the second derivative test for the convexity of Cartesian curves ${y=y(x)}$. What is the convexity test for polar curves ${r=r(\theta)}$? Google search brought up Robert Israel’s answer on Math.SE: the relevant inequality is

$\displaystyle \mathcal{C}[r]:=r^2+2(r')^2-r\,r''\ge 0 \ \ \ \ \ (1)$

But when using it, one should be aware of the singularity at the origin. For example, ${r=1+\cos \theta}$ satisfies $\mathcal{C}[r] = 3(1+\cos \theta)\ge 0$ but the curve is not convex: it’s the cardioid.

The formula (1) was derived for ${r> 0}$; the points with ${r=0}$ must be investigated directly. Actually, it is easy to see that when ${r }$ has a strict local minimum with value ${0}$, the polar curve has an inward cusp and therefore is not convex.

As usual, theoretical material is followed by an exercise.

Exercise: find all real numbers ${p}$ such that the polar curve ${r=(1+\cos 2\theta)^p}$ is convex.

All values ${p>0}$ are ruled out by the cusp formed at ${\theta=\pi/2}$. For ${p=0}$ we get a circle, obviously convex. When ${p<0}$, some calculations are in order:

$\displaystyle \mathcal{C}[r] = (1+4p+4p^2+(1-4p^2)\cos 2\theta)(1+\cos 2\theta)^{2p-1}$

For this to be nonnegative for all ${\theta}$, we need ${1+4p+4p^2\ge |1-4p^2|}$. Which is equivalent to two inequalities: ${p(4+8p) \ge 0}$ and ${2+4p\ge 0}$. Since ${p<0}$, the first inequality says that ${p\le -1/2}$, which is the opposite of the second.

Answer: ${p=0}$ and ${p=-1/2}$.

This exercise is relevant to the problem from the previous post Peanut allergy and Polar interpolation about the search for the “right” interpolation method in polar coordinates. Given a set of ${(r,\theta)}$ values, I interpolated ${(r^{1/p},\theta)}$ with a trigonometric polynomial, and then raised that polynomial to power ${p}$.

If the given points had Cartesian coordinates ${(\pm a,0), (0,\pm b)}$ then this interpolation yields ${r=(\alpha+\beta \cos \theta)^p}$ where ${\alpha,\beta}$ depend on ${a,b}$ and satisfy ${\alpha>|\beta|}$. Using the exercise above, one can deduce that ${p=-1/2}$ is the only nonzero power for which the interpolated curve is convex for any given points of the form ${(\pm a,0), (0,\pm b)}$.

In general, curves of the form ${r=P(\theta)^{-1/2}}$, with ${P}$ a trigonometric polynomial, need not be convex. But even then they look more natural than their counterparts with powers ${p\ne -1/2}$. Perhaps this is not surprising: the equation ${r^2P(\theta)=1}$ has a decent chance of being algebraic when the degree of ${P}$ is low.

Random example at the end: ${r=(3-\sin \theta +2\cos \theta+\cos 2\theta-2\sin 2\theta)^{-1/2}}$

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