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.

FooPlot graphics today: fooplot.com
FooPlot graphics today: fooplot.com

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, finding 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}}

Celebration of power -1/2
Celebration of power -1/2

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