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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 general cubic without complicated substitutionsSolving a cubic with complex numbersFinding parameters of an ellipse in terms of Semi-Latus Rectum and Directrix.Solving cubic polynomialsDepressed cubic

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4

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 $p>0$ ?

roots real-numbers cubic-equations

share|cite|improve this question

asked 8 hours ago

Yves DaoustYves Daoust

146k1010 gold badges8989 silver badges248248 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  en.wikipedia.org/wiki/…
  $endgroup$
  – Claude Leibovici
  2 hours ago
  

  
  
  
  

add a comment
 | 

4

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 $p>0$ ?

roots real-numbers cubic-equations

share|cite|improve this question

asked 8 hours ago

Yves DaoustYves Daoust

146k1010 gold badges8989 silver badges248248 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  en.wikipedia.org/wiki/…
  $endgroup$
  – Claude Leibovici
  2 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

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 $p>0$ ?

roots real-numbers cubic-equations

share|cite|improve this question

asked 8 hours ago

Yves DaoustYves Daoust

146k1010 gold badges8989 silver badges248248 bronze badges

$endgroup$

share|cite|improve this question

asked 8 hours ago

Yves DaoustYves Daoust

146k1010 gold badges8989 silver badges248248 bronze badges

share|cite|improve this question

asked 8 hours ago

Yves DaoustYves Daoust

146k1010 gold badges8989 silver badges248248 bronze badges

asked 8 hours ago

Yves DaoustYves Daoust

146k1010 gold badges8989 silver badges248248 bronze badges

asked 8 hours ago

Yves DaoustYves Daoust

146k1010 gold badges8989 silver badges248248 bronze badges

2 Answers
2

active

oldest

votes

5

$begingroup$

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

answered 8 hours ago

José Carlos SantosJosé Carlos Santos

215k2626 gold badges167167 silver badges291291 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  8 hours ago
  

  
  
  
  

add a comment
 | 

2

$begingroup$

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".

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.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Yves DaoustYves Daoust

146k1010 gold badges8989 silver badges248248 bronze badges

$endgroup$

add a comment
 | 

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5

$begingroup$

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

answered 8 hours ago

José Carlos SantosJosé Carlos Santos

215k2626 gold badges167167 silver badges291291 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  8 hours ago
  

  
  
  
  

add a comment
 | 

5

$begingroup$

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

answered 8 hours ago

José Carlos SantosJosé Carlos Santos

215k2626 gold badges167167 silver badges291291 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  8 hours ago
  

  
  
  
  

add a comment
 | 

$begingroup$

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

answered 8 hours ago

José Carlos SantosJosé Carlos Santos

215k2626 gold badges167167 silver badges291291 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

José Carlos SantosJosé Carlos Santos

215k2626 gold badges167167 silver badges291291 bronze badges

answered 8 hours ago

José Carlos SantosJosé Carlos Santos

215k2626 gold badges167167 silver badges291291 bronze badges

answered 8 hours ago

José Carlos SantosJosé Carlos Santos

215k2626 gold badges167167 silver badges291291 bronze badges

2

$begingroup$

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".

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.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Yves DaoustYves Daoust

146k1010 gold badges8989 silver badges248248 bronze badges

$endgroup$

add a comment
 | 

2

$begingroup$

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".

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.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Yves DaoustYves Daoust

146k1010 gold badges8989 silver badges248248 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

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".

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.

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Yves DaoustYves Daoust

146k1010 gold badges8989 silver badges248248 bronze badges

$endgroup$

share|cite|improve this answer

edited 7 hours ago

answered 8 hours ago

Yves DaoustYves Daoust

146k1010 gold badges8989 silver badges248248 bronze badges

answered 8 hours ago

Yves DaoustYves Daoust

146k1010 gold badges8989 silver badges248248 bronze badges

answered 8 hours ago

Yves DaoustYves Daoust

146k1010 gold badges8989 silver badges248248 bronze badges