Behavior of $x^2+y^2+\frac{y}{x}=1$.

Question

In my mathematical travels, I've stumbled upon the implicit formula $y^2+x^2+\frac{y}{x}=1$ and found that every graphing program I've plugged it in to seems to believe that there is large set of points which satisfy the equation $(y^2+x^2+\frac{y}{x})^{-1}=1$ which do not satisfy the original equation and this has me quite perplexed. I suspect that this is simply a glitch in the software and this question might therefore be better suited in the CS forum but I figured I would post it here first in the event that someone may have a mathematical explanation for this bizarre behavior. Any and all insights are welcome!

Answer 1

Here is a color plot (aka. scalar field) of $$x^2+y^2+{y\over x}$$ White is around zero, gray is positive and red is negative. The thick line is $0$ and the thin line is $1$, your original curve.

It's easy to see that by taking the inverse, the thin line will be preserved, but you will get a discontinuity at the thick line, because one side will go to $+\infty$ and the other to $-\infty$.

Now, from a CS point of view, plotting an implicit formula as a line drawing is not a trivial matter. It is usually done by sampling the implicit formula in many places, and then "closing in" around the desired value.

Many pieces of software are not sophisticated enough to deal with discontinuities, especially when plotting implicit formulas. Therefore when they see a negative value on one side and a positive on the other side, they will conclude it must pass from $1$ in between, where in fact it goes to $+\infty$ and comes back from $-\infty$.

Software is stupid like that :-)

Plots done with Grapher, a great plotting app that comes free with every Mac.

Answer 2

rotationally symmetric about the origin; your version does not allow $x=0$ but the curve becomes a smooth variety if $(0,0)$ is included

The smooth implicit function is $$ x^3 + x y^2 + y - x = 0 $$ with gradient $$ \left\langle 3 x^2 + y^2 - 1, 2xy + 1 \right\rangle $$

For large $|y|,$ solving the quadratic formula in $y$ shows $xy \approx -1.$ Indeed, large or small, for $x \neq 0$ we have $$ y = \frac{-1 \pm \sqrt {1 + 4 x^2 - 4 x^4}}{2x} $$ and we require $1 + 4 x^2 - 4 x^4 \geq 0.$