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Playing around with circles led me to the following observations:

For the equation $x^n+y^n=r^n$ and any real $r$:

• For any positive real $n$, there seem to be three possible lines that can be drawn: rounded squares, the top-right (bottom-right for a negative $r$) corner of a rounded square, and a negatively sloped line with a bulge in it around the origin that resembles the top-right (bottom-right for negative $r$) corner of a rounded circle. $r$ can change without affecting the shape of the curve; all it affects is the x- and y-intercepts (always equivalent to $r$).


• When $n$ is a negative real, the options are inverted: imagine drawing a square and extending the lines outward to ±∞ on both axes; these are the asymptotes for the curves drawn, and while sometimes there are four curves, other times there are two, and other times just one.

Pictures:

$n=3$

$n=4$

$n=4.3$

$n=-3$

$n=-4$

$n=-4.3$

For integers the pattern is pretty straightforward: the first type for an odd $n$, while the second for an even one. The third type never appears for an integer $n$.

What I can’t figure out is the pattern when $n$ is not an integer; there doesn’t seem to be a pattern. The odd/even thing fails.

Is this a known phenomenon? Is there a way to predict what the line will look like given...something about $n$?

(As an aside, posting a polar formula equivalent to these would be very helpful, possibly.)

Using spherical coordinates, you have $cos^nphi + sin^nphi = 1$. For negative $n$, write $m = -n$ to give $cos^mphi + sin^mphi = cos^mphi cdot sin^mphi$ which explains the obvious difference between positive and negative $n$.

The reason why for non-integer values there are problems in other than the first quadrant can be explained by trigonometric features of the angular functions.

For example, take the negative angle. Since $cos (- phi) = cos phi $ and $sin (- phi) = - sin phi $ you get

$cos^n(-phi) + sin^n(-phi) = cos^nphi + (-1)^n sin^nphi = 1$. For $n = pm 4$, as in your examples, this means that the solution in the first quadrant will be mirrored in the third. However, for non-integer $n$, you are facing $(-1)^n = exp (i pi n)$ which in general is complex, so there will be no real solution at all. The same arguments hold for an angular shift of $pm pi/2$.

This explains why you have only a solution in the first quadrant for non-integer $n$.

Other than that, the shape of the curve smoothly varies in the first quadrant as you change $n$. For example, your curves for $n=4.3$ will lie in between those for $n=4$ and $n=5$ in the first quadrant.