Finding intersection of two algebraic curves

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Discussion Overview

The discussion revolves around finding the intersection points of two algebraic curves defined by polynomial equations in two variables. Participants explore both numerical approaches and theoretical limitations, particularly for higher-degree curves beyond quadratics and cubics.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant inquires about general numeric methods for finding intersections of algebraic curves of higher degrees, specifically mentioning degrees 10 or 12.
  • Another participant references the Abel-Ruffini theorem, indicating that there are no general solutions expressed with radicals for quintic equations and higher, suggesting inherent difficulties in the problem.
  • There is a suggestion to consider the resultant of the two polynomials as a potential method for finding intersections.
  • A participant provides a specific example of two polynomial equations and describes how they computed intersections for this simpler case, expressing uncertainty about applying the resultant method to higher degree curves.

Areas of Agreement / Disagreement

Participants express differing views on the feasibility of finding intersections for higher-degree curves. While some propose the resultant as a method, others highlight the limitations imposed by the Abel-Ruffini theorem, indicating that the discussion remains unresolved regarding effective approaches for complex cases.

Contextual Notes

Limitations include the complexity of higher-degree polynomial intersections and the potential inapplicability of certain methods, such as the resultant, for specific cases. The discussion does not resolve how to systematically find intersections for higher degree curves.

aheight
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TL;DR
Is there a standard numeric approach to finding the intersection of two algebraic curves?
Given two algebraic curves:

##f_1(z,w)=a_0(z)+a_1(z)w+\cdots+a_n(z)w^n=0##
##f_2(z,w)=b_0(z)+b_1(z)w+\cdots+b_k(z)w^k=0##

Is there a general, numeric approach to finding where the first curve ##w_1(z)## intersects the second curve ##w_2(z)##? I know for low degree like quadratic or cubics can find the intersection by brute force but was wondering if there is a more general approach for higher degrees say 10 or 12 each?
 
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Not expressed with radicals, at least.
If there would be then quintic equations would have such a solution. They do not in general: Abel-Ruffini theorem
 
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mfb said:
Not expressed with radicals, at least.
If there would be then quintic equations would have such a solution. They do not in general: Abel-Ruffini theorem

It appears to be a difficult problem even numerically. Was just wondering how others might approach it (numerically).
 
@aheight -- You should look at the resultant of the two polynomials. See the Resultant - Wikipedia article for more information, especially the section on Algebraic Geometry. HTH
 
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Petek said:
@aheight -- You should look at the resultant of the two polynomials. See the Resultant - Wikipedia article for more information, especially the section on Algebraic Geometry. HTH
Thanks for that. However, perhaps I should have stated above I'm interested in computing the intersections of the real and imaginary sheets of both algebraic curves. Consider a simple case I'm working on:

$$
\begin{align*}
f1(z,w)&=(-6z/5-2z^2+z^4)+(2/5-2 z^2/5)w+1/25 w^2=0\\
f2(z,w)&=(-1+z^2)-6/5 w=0
\end{align*}
$$
If I plot the real sheets of ##w_1## from f1 as the yellow and red surfaces and the real sheet of ##w_2## from f2 as the orange surface in the plot below, I obtain their intersections as the white curves. The white curves I computed by solving simple simultaneous equations for this simple case which I would not be able to do with higher degree curves and was wondering if there is a systematic way to find the intersections for the higher degree cases.

Guess I mean I don't see how the resultant can be used to find the white curves.
intersectionPlot.jpg
 
Last edited:

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