Why does this graph behave the way it does?

Question

A little while ago, I learned how to superimpose the graphs of multiple equations using one equation. Suppose I want to superimpose: $$x-y=0$$ $$x^2-y=0$$ $$x^3-y=0$$ . The equation* I need is: $$\prod_{n=1}^{3}\left(x^{n}-y\right)=0$$ Recently I have discovered that some very interesting graphs can be obtained by changing the value of $C$ in equations of the form: $$\prod_{n=a}^{b}\left(f\left(x\right)-y\right)=C$$ . Starting with $C=1$, for example, and having it approach $0$ can reveal a lot about the nature of the left-hand expression. As I was doing this with the starred equation, I observed that an oval-like shape appeared and disappeared in a certain space on the graph. It takes its elliptical form beginning at $C\approx0.00001$, and grows smaller and smaller until it disappears at $C\approx0.0041596$. I can see how the individual curves come together to form the superimposition at $C=0$, but what is so special about $C\approx0.0041596$ and why does the oval disappear/reappear at that specific point? Thanks in advance.

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Answer 1

What you're graphing is known as a "level curve" for the function. (You can read up on them in more detail, for example, here.) Your equation is essentially $f(x,y)=C$ for a function $f$ (in this case, a product). This graphs to show all of the points where $f$ is equal to that constant $C$.

This can be best visualized if you use three dimensions in your case. Let $x,y$ remain as they are, and let $z$ be $f(x,y)$, i.e. your product. Then this produces a three-dimensional graph. Seeing $f(x,y)=C$ - that is, $z=C$ - as you have, thus produces a cross-section of the whole graph of $f$, represented in two dimensions.

You can graph such a function in software like GeoGraba. If you do so, you can get a blue surface and can examine it. Let's look at what I think is the bit you're referring to in your post, which is roughly around $x=1/2$:

The area in the black area is the one of concern. Though it's exceptionally subtle (though it shouldn't be surprising considering you're dealing with values of $C \in [0,0.005]$), the function has a local maximum in that region. It can be seen as a result of the gray bleeding through, since that's where $f$ dips below $0$.

If you were to take a cross-section parallel to the $z$ plane (the gray square) containing that line, you'd find a single line or something of the sort. If you move the cross-section above that (i.e. raise $C$ sufficiently high), then you're no longer cutting into that shallow hill, and the oval that makes its cross section disappears. Below that point, to an extent, you cut into the hill, and reveal the oval structure again.

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In summary, setting $(x-y)(x^2 - y)(x^3 - y) = C$ is finding the level curves of the function $f(x,y) = (x-y)(x^2-y)(x^3 - y)$, and varying $C$ is like taking repeat cross-sections of the surface $f$ produces, almost like mapping out the topography of it without trying to view the three-dimensional structure $f$ produces directly. The oval structure you see likely disappears at the specified $C$ value due to that $z=f(x,y)$ value being a local extremum for the function, or at least very close to one.