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I ask for the equation of a quintic curve with 6 or at least 5 double points!
I have founded the equation with only 4 double points...
Help me, please!
Scifo from Italy
A more usual definition of a double point on a curve is that it is a point where the curve crosses itself. For a more precise definition, see here. On that same Wikipedia page, there is a graph of the folium of Descartes. That is a cubic curve, with the equationOriginally Posted by scifo
The double points of a curve F(x,y)=0 is the points where the prime
derivative respect to x e to y is zero (but not all the seconds derivative)
I think...it is correct?, which has a double point at the origin. The unit circle
cuts the folium of Descartes at four points. So if you multiply their equations together, you get a quintic curve
whose graph is the superposition of the graphs of the folium and the circle, and therefore has five double points.
I don't know whether a curve with a quintic equation can have six double points.
Hi Opalg!
I thank you very much for your intelligent solution
about the five double points.
At last I have a quintic five double points!
Perhaps you help me again for the following problems:
May you create a quintic five double points irreducible?
It is no hope for a quintic six double point?
I am not knowledgeable about algebraic geometry and I cannot answer those questions. The best information I could find on the web is a 1902 Cornell University PhD dissertation by Peter Field, which you can find here. Click on "Page 1" to get a pdf file of the first chapter, which seems to state that there are many irreducible quintic curves with six double points. The text is illustrated with hand-drawn diagrams of such curves (look at Plate I at the end of that chapter), but I could not find any explicit equations for them.
It seems that double points are known to those in the trade as "crunodes". If you have access to a university library with old books on algebraic geometry (and an Italian university ought to have lots of those) then look for "quintic" and "crunodes" in the index. You might get lucky.
Hello Opalg
I thank you for your useful information…
I wanted a your opinion on a method, than it has sprung to mind,
in order to find the a 6-double points quintic equation...
Practically, fixed in the cartesian plan 8 points, 6 double and 2 simple.
if F(x,y) is quintic polynomial and Fx(x,y) and Fy(x,y) its derivatives,
then for every Xi,Yi of the 6 double points I have the three equations
F(Xi,Yi)=0, Fx(Xi,Yi)=0, Fy(Xi,Yi)=0
and for every Xj,Yj of the 2 simple points I have the equation
F(Xj,Yj)=0.
I have so a linear omogeneous system of 20 equations in the 20 unknown quantities
that are the essential coefficients of the quintic, resolved which I have the equation…
That you think some?
That sounds fairly convincing to me: 20 equations for 20 unknowns should in general have a unique solution.
Sorry I don't have time to think about it more carefully now. I'm about to leave for a week's walking holiday in Scotland, away from the internet.
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