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I'm trying to get an intuitive sense of algebraic varieties from their equations. Some varieties are more easily imagined without a computer since either $y$ or $x$ can be isolated, producing functions, and these can be more easily graphed. But what about for varieties like $y^{4}x^{3}+x^{4}-y=0$? Is there any way to think of its shape without a computer? One approach would be to use the quartic formulas to isolate $y$, but this is time consuming. Are there better approaches, if not to obtain a perfect graph, at least to obtain an estimation?

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    $\begingroup$ I have had a good scientific calculator for some thirty years. It is programmable; once a function is defined, there is a "solve" function that finds a root between bounds I supply. The outcome is that I have always been able to draw curves in the plane. I claim that getting graph paper and pencil and plotting points on a curve leads to some understanding of it. You could try that. Well, the real part of the curve... $\endgroup$
    – Will Jagy
    Commented Apr 27, 2023 at 17:54
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    $\begingroup$ One can always generate a "slope field" by taking derivatives on both sides and then begin plotting slopes. Another approach that works well is decomposing the equation into parametric equations but that is an art and not an exact science. It takes a lot of intuition or trial and error to figure out what works. $\endgroup$ Commented Apr 27, 2023 at 17:57
  • $\begingroup$ Sometimes hyperbolic coordinates are useful for this. $\endgroup$
    – 2'5 9'2
    Commented Apr 27, 2023 at 18:39

3 Answers 3

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One of the most effective way I know is to get a parametric equation of the curve by intersecting it with a line having a variable slope $y-y_0=t(x-x_0)$ passing through a point $(x_0,y_0)$ of the curve. Here we can take plainly point $(x_0,y_0)=(0,0)$.

The abscissas of the points of intersection of the curve with the line having equation $y=tx$ verify the following equation :

$$t^4x^7+x^4-tx=0$$

factorizing $x$ and considering that $x \ne 0$ (the case $x=0$ corresponds to the "trivial" intersection situated at the origin), we get :

$$t^4x^6+x^3-t=0$$

which is a quadratic relationship in $X=x^3$, with parameter $t$ whose solution(s) will give a (double) expression $x=f(t)$.

(see Edit below)

Finaly, the two equations $x=f(t), y= t f(t)$ constitute a rather natural parametrization of the curve.

Edit :

  1. Details about function $f$. One finds, by solving the quadratic :

$$x=f(t)=\frac{1}{t}\sqrt[3]{\frac{-1 \pm \sqrt{1+4t^5}}{2 t}}\tag{1}$$

(for $t\ne 0$) (1) is valid under the condition that the square root makes sense, i.e., $t \ge t_{lim}:=-\tfrac{1}{\sqrt[5]{4}}.$

One finds, of course, similarities with the results obtained by Fabio Caiazzo ; however, using polar coordinates has a drawback : it doesn't give good results when the curve doesn't pass through the origin.

  1. Especialy interesting results can be obtained if the curve has a double point (i.e., is self-interesting) by taking this double point as point $(x_0,y_0)$. This is the case for example with the folium of Descartes (see here).

Have a look at the curve : the red part corresponds with the case of sign "+" in (1), the blue part to sign "-". The limit case as been placed into evidence by showing the associated secant $y=t_{lim} x$ (the grey straight line) which is in fact a tangent. This limit case separates the cases $t > t_{lim}$ whith two intersection points from the cases $t < t_{lim}$ where there are no intersection points.

enter image description here

Now, for infinite branches, as Somos has done, we can neglect one term after the other (this is authorized by the "technique" of Newton polygon) giving the following very satisfying approximations at the origin (maybe the most important) and when either $x$ or $y$ tend to $\infty$ :

enter image description here

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  • $\begingroup$ For asymptotes and so-called "parabolic branches", see the techniques connected to Newton polygons. $\endgroup$
    – Jean Marie
    Commented Apr 27, 2023 at 23:08
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You can convert it in polar equation: $$x=r\cos(\theta)\qquad y=r\sin(\theta)$$ So you have $$r^{4}\sin\left(\theta\right)^{4}r^{3}\cos\left(\theta\right)^{3}+r^{4}\cos\left(\theta\right)^{4}-r\sin\left(\theta\right)=0\\ r^{6}\sin\left(\theta\right)^{4}\cos\left(\theta\right)^{3}+r^{3}\cos\left(\theta\right)^{4}-\sin\left(\theta\right)=0\\ r^3:=R\\ R^2\sin\left(\theta\right)^{4}\cos\left(\theta\right)^{3}+R\cos\left(\theta\right)^{4}-\sin\left(\theta\right)=0 $$ Then you solve for $R$ (and then for $r=\sqrt[3]{R}$) and you have two functions in the form $$r_{1,2}\left(\theta\right)=\sqrt[3]{\frac{-\cos\left(\theta\right)^{3}\pm\sqrt{\cos\left(\theta\right)^{6}+4\sin\left(\theta\right)^{5}\cos\left(\theta\right)}}{2\sin\left(\theta\right)^{4}\cos\left(\theta\right)^{2}}}\\ r_{1,2}\left(\theta\right)=\frac{1}{\sin\left(\theta\right)}\sqrt[3]{\cot\left(\theta\right)\frac{-1\pm\sqrt{1+4\tan\left(\theta\right)^{5}}}{2}}$$ (You choose some values for $\theta$ and then you have $r$)

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The question is about the locus of all real points on the curve

$$ y^{4}x^{3}+x^{4}-y=0. \tag1 $$

In such a simple case, move the $y$ to the other side to get

$$ y = x^4 + x^3y^4. \tag2 $$

Define a function

$$ f(x, y) := x^4 + x^3y^4. \tag3 $$

Given an approximation of $\,y,\,$ then $\,f(x,y)\,$ is a better approximation.

For example, start with

$$ y_0 = O(x^0) $$

and iterate to get

$$ y_1 := f(x, y_0) = O(x^3), \\ y_2 := f(x, y_1) = x^4 + O(x^{15}), \\ y_3 := f(x, y_2) = x^4 + x^{19} + O(x^{30}), \\ y_4 := fx, (y_3) = x^4 + x^{19} + 4x^{34} + O(x^{45}). $$

The polynomial truncations can be easily graphed, however, the radius of convergence is small and so this will not give an accurate picture of the entire locus.


Another global method is to assume that one of the terms is small compared to the other two and can be removed without great distortion. Thus, this gives three modified equations whose solutions approximate the original locus on portions of the plane.

Consider $\, y^4 x^3+x^4=0.\,$ This simplifies to $\, y^4 + x = 0.\,$ The locus of solutions is a quartic parabola for $x$ in terms of $y$.

Consider $\, y^4 x^3 - y=0.\,$ This simplifies to $\, (xy)^3 - 1 = 0.\,$ The locus of solutions is a hyperbola $\, y = -1/x\,$ with two components.

Consider $\, x^4 - y = 0.\,$ This is exactly similar to the first case. The locus of solutions is a quartic parabola for $y$ in terms of $x$.

You can plot these three curves on the same plane and note where they transition to each other. That is, for $\,y>0\,$ the three curves transition into a single continuous curve. For $\,y<0\,$ only the first two curves transition into a single continuous curve confined to the third quadrant.


Define the function

$$ g(x, y) := y^4 x^3 + x^4 - y. \tag4 $$

The curve of interest is the locus of $\,g(x, y) = 0.\,$ Note that since

$$ g(x, y) = (xy)^4 g(-1/y, -1/x), \tag5 $$

that this is the same locus as $\,g(-1/y, -1/x) = 0.\,$ Thus, every $\,(x, y)\,$ solution has a corresponding $\,(-1/y, -1/x)\,$ solution. This involution is obvious for the hyperbola $\, y = -1/x\,$ while it swaps between the two quartic parabolas.

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  • $\begingroup$ In full agreement with your second part. $\endgroup$
    – Jean Marie
    Commented Apr 28, 2023 at 7:53
  • $\begingroup$ Very cool explanation on the second part, thanks a lot! $\endgroup$
    – shintuku
    Commented Apr 28, 2023 at 11:24

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