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Solving the cubic without complex numbers


Solving Cubic EquationProve or disprove this relation between one root of the quadratic and the cubic equation of a certain form, and linear recurrences.non-complex cubic roots formula?Solving a cubic with complex numbersSolving a complex cubic equationFinding parameters of an ellipse in terms of Semi-Latus Rectum and Directrix.Solving cubic polynomialsDepressed cubic






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margin-bottom:0;

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5















$begingroup$


I am trying to work out the resolution of the cubic equation without resorting to complex numbers at all.



The general equation can in all cases be reduced to the depressed form,



$$x^3+px+q=0.$$



By a suitable scaling of the variable, this can be further reduced to



$$x^3-frac34x-frac r4=0$$ where $r>0$, or



$$4x^3-3x=r.$$



Then depending on the magnitude of $r$, we write



$$4cos^3t-3cos t=cos3t=r,$$ $$x=cosfracarccos(r)+2kpi3$$



or
$$4cosh^3t-3cosh t=cosh3t=r,$$ $$x=coshfractextarcosh(r)3.$$



This correctly handles the cases of $1$ and $3$ real roots.



Unfortunately, the trick only works for $p<0$. How can I solve in a similar way when $p>0$ ?










share|cite|improve this question









$endgroup$














  • $begingroup$
    en.wikipedia.org/wiki/…
    $endgroup$
    – Claude Leibovici
    Sep 20 at 4:04










  • $begingroup$
    @ClaudeLeibovici: waw, great ! I am just lagging two centuries behind Chebyshev and his cube root.
    $endgroup$
    – Yves Daoust
    Sep 20 at 8:00


















5















$begingroup$


I am trying to work out the resolution of the cubic equation without resorting to complex numbers at all.



The general equation can in all cases be reduced to the depressed form,



$$x^3+px+q=0.$$



By a suitable scaling of the variable, this can be further reduced to



$$x^3-frac34x-frac r4=0$$ where $r>0$, or



$$4x^3-3x=r.$$



Then depending on the magnitude of $r$, we write



$$4cos^3t-3cos t=cos3t=r,$$ $$x=cosfracarccos(r)+2kpi3$$



or
$$4cosh^3t-3cosh t=cosh3t=r,$$ $$x=coshfractextarcosh(r)3.$$



This correctly handles the cases of $1$ and $3$ real roots.



Unfortunately, the trick only works for $p<0$. How can I solve in a similar way when $p>0$ ?










share|cite|improve this question









$endgroup$














  • $begingroup$
    en.wikipedia.org/wiki/…
    $endgroup$
    – Claude Leibovici
    Sep 20 at 4:04










  • $begingroup$
    @ClaudeLeibovici: waw, great ! I am just lagging two centuries behind Chebyshev and his cube root.
    $endgroup$
    – Yves Daoust
    Sep 20 at 8:00














5













5









5


3



$begingroup$


I am trying to work out the resolution of the cubic equation without resorting to complex numbers at all.



The general equation can in all cases be reduced to the depressed form,



$$x^3+px+q=0.$$



By a suitable scaling of the variable, this can be further reduced to



$$x^3-frac34x-frac r4=0$$ where $r>0$, or



$$4x^3-3x=r.$$



Then depending on the magnitude of $r$, we write



$$4cos^3t-3cos t=cos3t=r,$$ $$x=cosfracarccos(r)+2kpi3$$



or
$$4cosh^3t-3cosh t=cosh3t=r,$$ $$x=coshfractextarcosh(r)3.$$



This correctly handles the cases of $1$ and $3$ real roots.



Unfortunately, the trick only works for $p<0$. How can I solve in a similar way when $p>0$ ?










share|cite|improve this question









$endgroup$




I am trying to work out the resolution of the cubic equation without resorting to complex numbers at all.



The general equation can in all cases be reduced to the depressed form,



$$x^3+px+q=0.$$



By a suitable scaling of the variable, this can be further reduced to



$$x^3-frac34x-frac r4=0$$ where $r>0$, or



$$4x^3-3x=r.$$



Then depending on the magnitude of $r$, we write



$$4cos^3t-3cos t=cos3t=r,$$ $$x=cosfracarccos(r)+2kpi3$$



or
$$4cosh^3t-3cosh t=cosh3t=r,$$ $$x=coshfractextarcosh(r)3.$$



This correctly handles the cases of $1$ and $3$ real roots.



Unfortunately, the trick only works for $p<0$. How can I solve in a similar way when $p>0$ ?







roots real-numbers cubic-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 19 at 22:18









Yves DaoustYves Daoust

151k12 gold badges91 silver badges251 bronze badges




151k12 gold badges91 silver badges251 bronze badges














  • $begingroup$
    en.wikipedia.org/wiki/…
    $endgroup$
    – Claude Leibovici
    Sep 20 at 4:04










  • $begingroup$
    @ClaudeLeibovici: waw, great ! I am just lagging two centuries behind Chebyshev and his cube root.
    $endgroup$
    – Yves Daoust
    Sep 20 at 8:00

















  • $begingroup$
    en.wikipedia.org/wiki/…
    $endgroup$
    – Claude Leibovici
    Sep 20 at 4:04










  • $begingroup$
    @ClaudeLeibovici: waw, great ! I am just lagging two centuries behind Chebyshev and his cube root.
    $endgroup$
    – Yves Daoust
    Sep 20 at 8:00
















$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– Claude Leibovici
Sep 20 at 4:04




$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– Claude Leibovici
Sep 20 at 4:04












$begingroup$
@ClaudeLeibovici: waw, great ! I am just lagging two centuries behind Chebyshev and his cube root.
$endgroup$
– Yves Daoust
Sep 20 at 8:00





$begingroup$
@ClaudeLeibovici: waw, great ! I am just lagging two centuries behind Chebyshev and his cube root.
$endgroup$
– Yves Daoust
Sep 20 at 8:00











2 Answers
2






active

oldest

votes


















7

















$begingroup$

When $p>0$, you can apply Cardano's formula. It will give you the only real root of your cubic. And the formula will not have to deal with complex non-real numbers.






share|cite|improve this answer










$endgroup$













  • $begingroup$
    That's right, I forgot that. In fact the real case of Cardano also deals with my $r>1$, right ? I was hoping to find symmetric formulas.
    $endgroup$
    – Yves Daoust
    Sep 19 at 22:30


















3

















$begingroup$

Stupid me, I missed the relation



$$sinh 3t=4sinh^3t+3sinh t$$



which works without restrictions on the value of the LHS. It completely solves the case of $p>0$ as



$$sinhfractextarsinh(r)3^*.$$



Putting all together, one can solve all cases of the cubic by means of the following canonical "trisection" functions below, which solve



$$4y^3pm 3y=x.$$



The "hyperbolic" ones have expressions in terms of cubic roots, the trigonometric ones relate to the "casus irreductibilis".



enter image description here



Finally, it seems that it all boils down to the normalization of the equation to one of two forms, with or without an inflection, by an affine transformations of the argument, and getting rid of all coefficients.



This might look like a circular method (solving a cubic by solving a cubic), but we now have analytical expressions in terms of familiar transcendental, real functions.



Update:



These functions are directly related to the "Chebyshev cube roots".




$$^*sinhdfractextarsinh x3=dfrace^log(x+sqrtx^2+1)/3-e^-log(x+sqrtx^2+1)/32=frac12left(sqrt[3]x+sqrtx^2+1-dfrac1sqrt[3]x+sqrtx^2+1right)$$






share|cite|improve this answer












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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7

















    $begingroup$

    When $p>0$, you can apply Cardano's formula. It will give you the only real root of your cubic. And the formula will not have to deal with complex non-real numbers.






    share|cite|improve this answer










    $endgroup$













    • $begingroup$
      That's right, I forgot that. In fact the real case of Cardano also deals with my $r>1$, right ? I was hoping to find symmetric formulas.
      $endgroup$
      – Yves Daoust
      Sep 19 at 22:30















    7

















    $begingroup$

    When $p>0$, you can apply Cardano's formula. It will give you the only real root of your cubic. And the formula will not have to deal with complex non-real numbers.






    share|cite|improve this answer










    $endgroup$













    • $begingroup$
      That's right, I forgot that. In fact the real case of Cardano also deals with my $r>1$, right ? I was hoping to find symmetric formulas.
      $endgroup$
      – Yves Daoust
      Sep 19 at 22:30













    7















    7











    7







    $begingroup$

    When $p>0$, you can apply Cardano's formula. It will give you the only real root of your cubic. And the formula will not have to deal with complex non-real numbers.






    share|cite|improve this answer










    $endgroup$



    When $p>0$, you can apply Cardano's formula. It will give you the only real root of your cubic. And the formula will not have to deal with complex non-real numbers.







    share|cite|improve this answer













    share|cite|improve this answer




    share|cite|improve this answer










    answered Sep 19 at 22:22









    José Carlos SantosJosé Carlos Santos

    232k27 gold badges174 silver badges304 bronze badges




    232k27 gold badges174 silver badges304 bronze badges














    • $begingroup$
      That's right, I forgot that. In fact the real case of Cardano also deals with my $r>1$, right ? I was hoping to find symmetric formulas.
      $endgroup$
      – Yves Daoust
      Sep 19 at 22:30
















    • $begingroup$
      That's right, I forgot that. In fact the real case of Cardano also deals with my $r>1$, right ? I was hoping to find symmetric formulas.
      $endgroup$
      – Yves Daoust
      Sep 19 at 22:30















    $begingroup$
    That's right, I forgot that. In fact the real case of Cardano also deals with my $r>1$, right ? I was hoping to find symmetric formulas.
    $endgroup$
    – Yves Daoust
    Sep 19 at 22:30




    $begingroup$
    That's right, I forgot that. In fact the real case of Cardano also deals with my $r>1$, right ? I was hoping to find symmetric formulas.
    $endgroup$
    – Yves Daoust
    Sep 19 at 22:30













    3

















    $begingroup$

    Stupid me, I missed the relation



    $$sinh 3t=4sinh^3t+3sinh t$$



    which works without restrictions on the value of the LHS. It completely solves the case of $p>0$ as



    $$sinhfractextarsinh(r)3^*.$$



    Putting all together, one can solve all cases of the cubic by means of the following canonical "trisection" functions below, which solve



    $$4y^3pm 3y=x.$$



    The "hyperbolic" ones have expressions in terms of cubic roots, the trigonometric ones relate to the "casus irreductibilis".



    enter image description here



    Finally, it seems that it all boils down to the normalization of the equation to one of two forms, with or without an inflection, by an affine transformations of the argument, and getting rid of all coefficients.



    This might look like a circular method (solving a cubic by solving a cubic), but we now have analytical expressions in terms of familiar transcendental, real functions.



    Update:



    These functions are directly related to the "Chebyshev cube roots".




    $$^*sinhdfractextarsinh x3=dfrace^log(x+sqrtx^2+1)/3-e^-log(x+sqrtx^2+1)/32=frac12left(sqrt[3]x+sqrtx^2+1-dfrac1sqrt[3]x+sqrtx^2+1right)$$






    share|cite|improve this answer












    $endgroup$


















      3

















      $begingroup$

      Stupid me, I missed the relation



      $$sinh 3t=4sinh^3t+3sinh t$$



      which works without restrictions on the value of the LHS. It completely solves the case of $p>0$ as



      $$sinhfractextarsinh(r)3^*.$$



      Putting all together, one can solve all cases of the cubic by means of the following canonical "trisection" functions below, which solve



      $$4y^3pm 3y=x.$$



      The "hyperbolic" ones have expressions in terms of cubic roots, the trigonometric ones relate to the "casus irreductibilis".



      enter image description here



      Finally, it seems that it all boils down to the normalization of the equation to one of two forms, with or without an inflection, by an affine transformations of the argument, and getting rid of all coefficients.



      This might look like a circular method (solving a cubic by solving a cubic), but we now have analytical expressions in terms of familiar transcendental, real functions.



      Update:



      These functions are directly related to the "Chebyshev cube roots".




      $$^*sinhdfractextarsinh x3=dfrace^log(x+sqrtx^2+1)/3-e^-log(x+sqrtx^2+1)/32=frac12left(sqrt[3]x+sqrtx^2+1-dfrac1sqrt[3]x+sqrtx^2+1right)$$






      share|cite|improve this answer












      $endgroup$
















        3















        3











        3







        $begingroup$

        Stupid me, I missed the relation



        $$sinh 3t=4sinh^3t+3sinh t$$



        which works without restrictions on the value of the LHS. It completely solves the case of $p>0$ as



        $$sinhfractextarsinh(r)3^*.$$



        Putting all together, one can solve all cases of the cubic by means of the following canonical "trisection" functions below, which solve



        $$4y^3pm 3y=x.$$



        The "hyperbolic" ones have expressions in terms of cubic roots, the trigonometric ones relate to the "casus irreductibilis".



        enter image description here



        Finally, it seems that it all boils down to the normalization of the equation to one of two forms, with or without an inflection, by an affine transformations of the argument, and getting rid of all coefficients.



        This might look like a circular method (solving a cubic by solving a cubic), but we now have analytical expressions in terms of familiar transcendental, real functions.



        Update:



        These functions are directly related to the "Chebyshev cube roots".




        $$^*sinhdfractextarsinh x3=dfrace^log(x+sqrtx^2+1)/3-e^-log(x+sqrtx^2+1)/32=frac12left(sqrt[3]x+sqrtx^2+1-dfrac1sqrt[3]x+sqrtx^2+1right)$$






        share|cite|improve this answer












        $endgroup$



        Stupid me, I missed the relation



        $$sinh 3t=4sinh^3t+3sinh t$$



        which works without restrictions on the value of the LHS. It completely solves the case of $p>0$ as



        $$sinhfractextarsinh(r)3^*.$$



        Putting all together, one can solve all cases of the cubic by means of the following canonical "trisection" functions below, which solve



        $$4y^3pm 3y=x.$$



        The "hyperbolic" ones have expressions in terms of cubic roots, the trigonometric ones relate to the "casus irreductibilis".



        enter image description here



        Finally, it seems that it all boils down to the normalization of the equation to one of two forms, with or without an inflection, by an affine transformations of the argument, and getting rid of all coefficients.



        This might look like a circular method (solving a cubic by solving a cubic), but we now have analytical expressions in terms of familiar transcendental, real functions.



        Update:



        These functions are directly related to the "Chebyshev cube roots".




        $$^*sinhdfractextarsinh x3=dfrace^log(x+sqrtx^2+1)/3-e^-log(x+sqrtx^2+1)/32=frac12left(sqrt[3]x+sqrtx^2+1-dfrac1sqrt[3]x+sqrtx^2+1right)$$







        share|cite|improve this answer















        share|cite|improve this answer




        share|cite|improve this answer








        edited Sep 20 at 10:43

























        answered Sep 19 at 22:47









        Yves DaoustYves Daoust

        151k12 gold badges91 silver badges251 bronze badges




        151k12 gold badges91 silver badges251 bronze badges































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