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Graphical solution of cubic with real roots

  1. Jul 18, 2007 #1
    Does anyone know whether the graphical solution of cubic equations with real roots by means of intersecting a circle and a parabola or hyperbola (or just a parabola and hyperbola) is known or not? That solution has to give the equations for the circles, parabolas and hyperbolas involved and not only for a special case. I have been surfing around for weeks now and got 1 hit: a 2-page document in the JSTOR library, but that is almost for sure not the one that I am looking for as it is more likely to be a 20 page document.

    A "yes" answer will do, a "no" answer will do, the "I don't know" answer is the one I already have. So, are there good detectives among all those good mathematicians here?
     
  2. jcsd
  3. Jul 19, 2007 #2

    Gib Z

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    Yes, in fact what you just said is a very famous achievement by Omar Khayyam. Best known worldwide for his poetry, he was also a mathematician: http://en.wikipedia.org/wiki/Omar_Khayyám#Mathematician

     
  4. Jul 19, 2007 #3
    graphical solution of the cubic with real roots

    Thanks Gib Z. I followed your link. I'm not sure whether Omar Khayyam solved that problem for all cubics or only for those with one real root. Wikipedia also leads to an article on Khayyam having found the generalization for all cubics with one real root, as it had been done before only for specific cubics. And believe it or not, within 1 hour after I posted my thread I tried for the 7.000.001st time on google and had a full hit in a minute. On the keyword as the title of this thread. You can't understand these things, maybe somebody had just triggered that document with the keyword quartic. Anyhow, Mr R.T. Running of the University of Michigan found the solution in 1943 (it's also a JSTOR document). He started with quartics and extended his solution to cubics by upgrading cubic polynomials to quartic polynomials. So in a way he was still solving quartics.

    I still am interested in solutions within the set of cubics. In a way I am not very interested in the answer that such solutions give as I found a solution years ago and I am convinced that they all are basically the same: intersecting conics. It's the background I am interested in as it might give "fresh blood" to my work, which is trying to come to some sort of general algebraic theory on the cubics.
    It was not my goal to find a graphical solution for the cubic with 3 real roots or,what is the same, finding an algebraic geometric solution for the trisection of an angle. My answer was just included in the basis of the theorie: the general solution of the cyclic system
    [itex]p{x_1}^2+qx_1+r=x_2[/itex]
    [itex]p{x_2}^2+qx_2+r=x_3[/itex]
    [itex]p{x_3}^2+qx_3+r=x_1[/itex]

    for any [itex]V = x^3+ax^2+bx+c=0[/itex] with no pair of equal roots.

    As you see, that is the equation for the parabola [itex]y=px^2+qx+r[/itex]
    Finding the circle that is passing through the points [itex]({x_j},{x_i})[/itex]
    was coming for free. The cyclic system is the 20 pages I spoke of.
    How did somebody else do it? That question is still open as far as I am concerned.
     
  5. Jul 19, 2007 #4

    Gib Z

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    Sorry It's 12:15pm here, the reason Im not at school is because I have a terrible cold and a headache, so please bear with me.

    .

    Unless you wish to find the solution of a cubic with complex coefficients, which I don't think you do, then that doesn't matter. All polynomials of odd degree and real coefficients will have at least 1 real root. As long as you find 1 root, you can do polynomial division, it becomes a quadratic and thats easy to solve.

    For the second part of your question, with the cyclic system's and stuff..I have no idea what exactly you are asking :(
     
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