# A equation of a quintic curve with 6 or at least 5 double points

• Jun 1st 2010, 01:12 AM
scifo
A equation of a quintic curve with 6 or at least 5 double points
Hi(Hi)

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... (Worried)

Scifo from Italy
• Jun 1st 2010, 01:22 AM
Prove It
Quote:

Originally Posted by scifo
Hi(Hi)

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... (Worried)

Scifo from Italy

What do you mean by double points?
• Jun 1st 2010, 03:29 AM
scifo
double points of F(x,y)=0 if Fx=Fy=0
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?(Thinking)
• Jun 1st 2010, 05:22 AM
Opalg
Quote:

Originally 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?(Thinking)

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 equation \$\displaystyle x^3+y^3 - 3xy = 0\$, which has a double point at the origin. The unit circle \$\displaystyle x^2+y^2-1 = 0\$ cuts the folium of Descartes at four points. So if you multiply their equations together, you get a quintic curve \$\displaystyle (x^3+y^3 - 3xy)(x^2+y^2 - 1) = 0\$ 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.
• Jun 1st 2010, 07:40 AM
scifo
A quintic five double points irreducible
Hi Opalg!(Hi)

I thank you very much for your intelligent solution
about the five double points.
At last I have a quintic five double points!(Clapping)

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?
• Jun 1st 2010, 11:37 AM
Opalg
Quote:

Originally Posted by scifo
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 (Happy) ) then look for "quintic" and "crunodes" in the index. You might get lucky.
• Jun 4th 2010, 07:36 AM
scifo
Perhaps a method for to find the 6-double points quintic equation?
Hello Opalg(Hi)

I thank you for your useful information…(Clapping)

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?(Wondering)
• Jun 4th 2010, 08:28 AM
Opalg
Quote:

Originally Posted by scifo
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 homogeneous system of 20 equations in the 20 unknown quantities
that are the essential coefficients of the quintic, resolved which I have the equation…

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.