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Solving the cubic without complex numbers

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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 $p>0$ ?

roots real-numbers cubic-equations

share|cite|improve this question

asked Sep 19 at 22:18

Yves DaoustYves Daoust

151k1212 gold badges9191 silver badges251251 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  
  

  
  
  
  

add a comment
 | 

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 $p>0$ ?

roots real-numbers cubic-equations

share|cite|improve this question

asked Sep 19 at 22:18

Yves DaoustYves Daoust

151k1212 gold badges9191 silver badges251251 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  
  

  
  
  
  

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 Sep 19 at 22:18

Yves DaoustYves Daoust

151k1212 gold badges9191 silver badges251251 bronze badges

$endgroup$

share|cite|improve this question

asked Sep 19 at 22:18

Yves DaoustYves Daoust

151k1212 gold badges9191 silver badges251251 bronze badges

share|cite|improve this question

asked Sep 19 at 22:18

Yves DaoustYves Daoust

151k1212 gold badges9191 silver badges251251 bronze badges

asked Sep 19 at 22:18

Yves DaoustYves Daoust

151k1212 gold badges9191 silver badges251251 bronze badges

asked Sep 19 at 22:18

Yves DaoustYves Daoust

151k1212 gold badges9191 silver badges251251 bronze badges

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

answered Sep 19 at 22:22

José Carlos SantosJosé Carlos Santos

232k2727 gold badges174174 silver badges304304 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
  Sep 19 at 22:30
  

  
  
  
  

add a comment
 | 

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".

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

edited Sep 20 at 10:43

answered Sep 19 at 22:47

Yves DaoustYves Daoust

151k1212 gold badges9191 silver badges251251 bronze badges

$endgroup$

add a comment
 | 

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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

answered Sep 19 at 22:22

José Carlos SantosJosé Carlos Santos

232k2727 gold badges174174 silver badges304304 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
  Sep 19 at 22:30
  

  
  
  
  

add a comment
 | 

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

answered Sep 19 at 22:22

José Carlos SantosJosé Carlos Santos

232k2727 gold badges174174 silver badges304304 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
  Sep 19 at 22:30
  

  
  
  
  

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 Sep 19 at 22:22

José Carlos SantosJosé Carlos Santos

232k2727 gold badges174174 silver badges304304 bronze badges

$endgroup$

share|cite|improve this answer

answered Sep 19 at 22:22

José Carlos SantosJosé Carlos Santos

232k2727 gold badges174174 silver badges304304 bronze badges

answered Sep 19 at 22:22

José Carlos SantosJosé Carlos Santos

232k2727 gold badges174174 silver badges304304 bronze badges

answered Sep 19 at 22:22

José Carlos SantosJosé Carlos Santos

232k2727 gold badges174174 silver badges304304 bronze badges

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".

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

edited Sep 20 at 10:43

answered Sep 19 at 22:47

Yves DaoustYves Daoust

151k1212 gold badges9191 silver badges251251 bronze badges

$endgroup$

add a comment
 | 

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".

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

edited Sep 20 at 10:43

answered Sep 19 at 22:47

Yves DaoustYves Daoust

151k1212 gold badges9191 silver badges251251 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.

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

edited Sep 20 at 10:43

answered Sep 19 at 22:47

Yves DaoustYves Daoust

151k1212 gold badges9191 silver badges251251 bronze badges

$endgroup$

share|cite|improve this answer

edited Sep 20 at 10:43

answered Sep 19 at 22:47

Yves DaoustYves Daoust

151k1212 gold badges9191 silver badges251251 bronze badges

answered Sep 19 at 22:47

Yves DaoustYves Daoust

151k1212 gold badges9191 silver badges251251 bronze badges

answered Sep 19 at 22:47

Yves DaoustYves Daoust

151k1212 gold badges9191 silver badges251251 bronze badges