Everybody knows the second derivative test for the convexity of Cartesian curves . What is the convexity test for polar curves ? Google search brought up Robert Israel’s answer on Math.SE: the relevant inequality is

But when using it, one should be aware of the singularity at the origin. For example, satisfies but the curve is not convex: it’s the cardioid.

The formula (1) was derived for ; the points with must be investigated directly. Actually, it is easy to see that when has a strict local minimum with value , 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 such that the polar curve is convex.

All values are ruled out by the cusp formed at . For we get a circle, obviously convex. When , some calculations are in order:

For this to be nonnegative for all , we need . Which is equivalent to two inequalities: and . Since , the first inequality says that , which is the opposite of the second.

*Answer*: and .

This exercise is relevant to the problem from the previous post, finding the “right” interpolation method in polar coordinates. Given a set of values, I interpolated with a trigonometric polynomial, and then raised that polynomial to power .

If the given points had Cartesian coordinates then this interpolation yields where depend on and satisfy . Using the exercise above, one can deduce that is the only nonzero power for which the interpolated curve is convex for any given points of the form .

In general, curves of the form , with a trigonometric polynomial, need not be convex. But even then they look more natural than their counterparts with powers . Perhaps this is not surprising: the equation has a decent chance of being algebraic when the degree of is low.

Random example at the end: