Principal equation form

In mathematics and, more specifically, in theory of equations, the principal form of an irreducible polynomial of degree at least three is a polynomial of the same degree n without terms of degrees n−1 and n−2, such that each root of either polynomial is a rational function of a root of the other polynomial.

The principal form of a polynomial can be found by applying a suitable Tschirnhaus transformation to the given polynomial.

Definition

Let

$f(x)=x^{n}+a_{1}x^{n-1}+\cdots +a_{n-1}x+a_{n}$

be an irreducible polynomial of degree at least three.

Its principal form is a polynomial

$g(y)=y^{n}+b_{3}y^{n-3}+\cdots +b_{n-1}y+b_{n},$

together with a Tschirnhaus transformation of degree two

$\varphi (x)=x^{2}+\alpha x+\beta$

such that, if r is a root of f, $\phi (r)$ is a root of ⁠$g$⁠.

Expressing that ⁠$g$⁠ does not has terms in ⁠$y^{n-1}$⁠ and ⁠$y^{n-2}$⁠ leads to a system of two equations in ⁠$\alpha$⁠ and ⁠$\beta$⁠, one of degree one and one of degree two. In general, this system has two solutions, giving two principal forms involving a square root. One passes from one principal form to the secong by changing the sign of the square root.

Cubic case

Tschirnhaus transformation with three clues

This is the given cubic equation:

$x^{3}-ax^{2}+bx-c=0$

Following quadratic equation system shall be solved:

$\mathrm {First\,clue}$	$au+3v+a^{2}-2b=0$
$\mathrm {Second\,clue}$	$bu^{2}-3v^{2}+abu-3cu-2ac+b^{2}=0$
$\mathrm {Third\,clue}$	$w=cu^{3}+acu^{2}+bcu+v^{3}+c^{2}$

And so exactly this Tschirnhaus transformation appears:

$(x^{2}+ux+v)^{3}-w=0$

Cubic calculation examples

Plastic constant:

$x^{3}-x-1=0$

${\bigl [}x^{2}+({\tfrac {1}{6}}{\sqrt {69}}-{\tfrac {3}{2}})x-{\tfrac {2}{3}}{\bigr ]}^{3}-{\tfrac {23}{54}}(3{\sqrt {69}}-23)=0$

Supergolden constant:

$x^{3}-x^{2}-1=0$

${\bigl [}x^{2}+({\tfrac {1}{2}}{\sqrt {93}}-{\tfrac {11}{2}})x-{\tfrac {1}{6}}{\sqrt {93}}+{\tfrac {3}{2}}{\bigr ]}^{3}-{\tfrac {31}{18}}(29{\sqrt {93}}-279)=0$

Tribonacci constant:

$x^{3}-x^{2}-x-1=0$

${\bigl [}x^{2}+({\tfrac {1}{4}}{\sqrt {33}}-{\tfrac {9}{4}})x-{\tfrac {1}{12}}{\sqrt {33}}-{\tfrac {1}{4}}{\bigr ]}^{3}-{\tfrac {11}{72}}(19{\sqrt {33}}-99)=0$

Quartic case

Tschirnhaus transformation with four clues

This is the given quartic equation:

$x^{4}-ax^{3}+bx^{2}-cx+d=0$

Now this quadratic equation system shall be solved:

$\mathrm {First\,clue}$	$at+4u+a^{2}-2b=0$
$\mathrm {Second\,clue}$	$bt^{2}-6u^{2}+abt-3ct-2ac+b^{2}+2d=0$
$\mathrm {Third\,clue}$	$v=ct^{3}+act^{2}-4dt^{2}-3adt+bct+4u^{3}-2bd+c^{2}$
$\mathrm {Fourth\,clue}$	$w=dt^{4}-u^{4}+uv+adt^{3}+bdt^{2}+cdt+d^{2}$

And so accurately that Tschirnhaus transformation appears:

$(x^{2}+tx+u)^{4}-v(x^{2}+tx+u)+w=0$

Quartic calculation examples

The Tschirnhaus transformation of the equation for the Tetranacci constant contains only rational coefficients:

$x^{4}-x^{3}-x^{2}-x-1=0$

$y=x^{2}-3x$

$y^{4}-11y-41=0$

In this way following expression can be made about the Tetranacci constant:

$x^{2}-3x=({\tfrac {41}{3}})^{1/4}{\sqrt {\sinh {\bigl [}{\tfrac {1}{3}}\operatorname {arsinh} ({\tfrac {363}{26896}}{\sqrt {123}}){\bigr ]}}}-$

$-({\tfrac {41}{3}})^{1/4}{\bigl \{}{\tfrac {11}{4}}({\tfrac {3}{41}})^{3/4}{\sqrt {\operatorname {csch} {\bigl [}{\tfrac {1}{3}}\operatorname {arsinh} ({\tfrac {363}{26896}}{\sqrt {123}}){\bigr ]}}}-\sinh {\bigl [}{\tfrac {1}{3}}\operatorname {arsinh} ({\tfrac {363}{26896}}{\sqrt {123}}){\bigr ]}{\bigr \}}^{1/2}$

That calculation example however does contain the element of the square root in the Tschirnhaus transformation:

$x^{4}+x^{3}+x^{2}-x-1=0$

$y=x^{2}+{\tfrac {1}{5}}(19+4{\sqrt {21}})x+{\tfrac {1}{5}}(6+{\sqrt {21}})$

$y^{4}-{\tfrac {1}{125}}(38267+8272{\sqrt {21}})y-{\tfrac {1}{625}}(101277{\sqrt {21}}+463072)=0$

Special form of the quartic

In the following we solve a special equation pattern that is easily solvable by using elliptic functions:

$x^{4}-6x^{2}-8{\sqrt {S^{2}+1}}\,x-3=0$

$Q=q{\bigl \{}\tanh {\bigl [}{\tfrac {1}{2}}\operatorname {arsinh} (S){\bigr ]}{\bigr \}}=q{\bigl [}S\div ({\sqrt {S^{2}+1}}+1){\bigr ]}$

$x={\frac {3\,\vartheta _{01}(Q^{3})^{2}}{\vartheta _{01}(Q)^{2}}}$

These are important additional informations about the elliptic nome and the mentioned Jacobi theta function:

$q(\varepsilon )=\exp {\bigl [}-\pi K({\sqrt {1-\varepsilon ^{2}}})\div K(\varepsilon ){\bigr ]}$

$\vartheta _{01}(r)=\sum _{k=-\infty }^{\infty }(-1)^{k}r^{k^{2}}=\prod _{n=1}^{\infty }(1-r^{2n})(1-r^{2n-1})^{2}$

Computation rule for the mentioned theta quotient:

${\frac {3\,\vartheta _{01}\{q[\kappa ^{3}\div ({\sqrt {\kappa ^{6}+1}}+1)]^{3}\}^{2}}{\vartheta _{01}\{q[\kappa ^{3}\div ({\sqrt {\kappa ^{6}+1}}+1)]\}^{2}}}={\sqrt {2{\sqrt {\kappa ^{4}-\kappa ^{2}+1}}-\kappa ^{2}+2}}+{\sqrt {\kappa ^{2}+1}}$

Accurately the Jacobi theta function is used for solving that equation.

Now we create a Tschirnhaus transformation on that:

$x^{4}-6x^{2}-8{\sqrt {S^{2}+1}}\,x-3=0$

$y=x^{2}-2({\sqrt {S^{2}+1}}-S)x-3$

$y^{4}+64\,S^{2}(4S^{2}+1-4S{\sqrt {S^{2}+1}})y-384\,S^{3}({\sqrt {S^{2}+1}}-S)=0$

Elliptic solving of principal quartics

Now this solving pattern shall be used for solving some principal quartic equations:

First calculation example:

$x^{4}+x-1=0$

$Q=q{\bigl \{}\tanh {\bigl [}{\tfrac {1}{2}}\operatorname {artanh} ({\tfrac {31}{100}}+{\tfrac {1}{300}}{\sqrt {849}}-{\tfrac {1}{300}}{\sqrt {386{\sqrt {849}}-1902}}){\bigr ]}{\bigr \}}$

$x={\frac {4}{{\sqrt {2{\sqrt {849}}+18}}-6}}{\biggl [}{\frac {9\,\vartheta _{01}(Q^{3})^{4}}{\vartheta _{01}(Q)^{4}}}-{\frac {1}{4}}{\sqrt {32+2{\sqrt {6{\sqrt {849}}-54}}}}\,{\frac {3\,\vartheta _{01}(Q^{3})^{2}}{\vartheta _{01}(Q)^{2}}}-3{\biggr ]}$

Second calculation example:

$x^{4}+2x-1=0$

$Q=q{\bigl \{}\tanh {\bigl [}{\tfrac {1}{2}}\operatorname {artanh} ({\tfrac {1}{10}}+{\tfrac {1}{30}}{\sqrt {129}}-{\tfrac {1}{30}}{\sqrt {26{\sqrt {129}}-102}}){\bigr ]}{\bigr \}}$

$x={\frac {2}{{\sqrt {2{\sqrt {129}}+18}}-6}}{\biggl [}{\frac {9\,\vartheta _{01}(Q^{3})^{4}}{\vartheta _{01}(Q)^{4}}}-{\frac {1}{2}}{\sqrt {8+2{\sqrt {6{\sqrt {129}}-54}}}}\,{\frac {3\,\vartheta _{01}(Q^{3})^{2}}{\vartheta _{01}(Q)^{2}}}-3{\biggr ]}$

Third calculation example:

$x^{4}+5x-3=0$

$Q=q{\bigl \{}\tanh {\bigl [}{\tfrac {1}{2}}\operatorname {artanh} ({\tfrac {239}{5092}}+{\tfrac {75}{5092}}{\sqrt {881}}-{\tfrac {5}{5092}}{\sqrt {11618{\sqrt {881}}-112750}}){\bigr ]}{\bigr \}}$

$x={\frac {4}{{\sqrt {2{\sqrt {881}}+50}}-10}}{\biggl [}{\frac {9\,\vartheta _{01}(Q^{3})^{4}}{\vartheta _{01}(Q)^{4}}}-{\frac {1}{4}}{\sqrt {32+10{\sqrt {2{\sqrt {881}}-50}}}}\,{\frac {3\,\vartheta _{01}(Q^{3})^{2}}{\vartheta _{01}(Q)^{2}}}-3{\biggr ]}$

Clue for the computation of the S value for a given principal quartic:

$x^{4}+\psi x-\omega =0$

If this equation pattern is given, the modulus tangent duplication value S can be determined in this way:

$\psi ^{4}{\bigl [}384\,S^{3}({\sqrt {S^{2}+1}}-S){\bigr ]}^{3}=\omega ^{3}{\bigl [}64\,S^{2}(4S^{2}+1-4S{\sqrt {S^{2}+1}}){\bigr ]}^{4}$

The solution of the now mentioned formula always is in pure biquadratic radical relation to psi and omega and therefore it is a useful tool to solve principal quartic equations.

$Q=\exp {\bigl \langle }-\pi K{\bigl \{}\operatorname {sech} {\bigl [}{\tfrac {1}{2}}\operatorname {arsinh} (S){\bigr ]}{\bigr \}}\div K{\bigl \{}\tanh {\bigl [}{\tfrac {1}{2}}\operatorname {arsinh} (S){\bigr ]}{\bigr \}}{\bigr \rangle }$

And this can be solved in that way:

$x={\frac {\omega [64\,S^{2}(4S^{2}+1-4S{\sqrt {S^{2}+1}})]}{\psi [384\,S^{3}({\sqrt {S^{2}+1}}-S)]}}{\biggl [}{\frac {9\,\vartheta _{01}(Q^{3})^{4}}{\vartheta _{01}(Q)^{4}}}-2({\sqrt {S^{2}+1}}-S){\frac {3\,\vartheta _{01}(Q^{3})^{2}}{\vartheta _{01}(Q)^{2}}}-3{\biggr ]}$

Quintic case

Synthesis advice for the quadratic Tschirnhaus key

This is the given quintic equation:

$x^{5}-ax^{4}+bx^{3}-cx^{2}+dx-e=0$

That quadratic equation system leads to the coefficients of the quadratic Tschirnhaus key:

$\mathrm {First\,clue}$	$as+5t+a^{2}-2b=0$
$\mathrm {Second\,clue}$	$bs^{2}-10t^{2}+abs-3cs-2ac+b^{2}+2d=0$

By polynomial division that Tschirnhaus transformation can be made:

$(x^{2}+sx+t)^{5}-u(x^{2}+sx+t)^{2}+v(x^{2}+sx+t)-w=0$

Calculation examples

This is the first example:

$x^{5}-x^{4}-x^{2}-1=0$

$y=x^{2}-{\tfrac {1}{4}}(19-{\sqrt {265}})x-{\tfrac {1}{20}}({\sqrt {265}}-15)$

$y^{5}+{\tfrac {1}{80}}(24455-1501{\sqrt {265}})y^{2}-{\tfrac {1}{160}}(5789{\sqrt {265}}-93879)y-{\tfrac {1}{4000}}(5393003{\sqrt {265}}-87785025)=0$

And this is the second example:

$x^{5}+x^{4}+x^{3}+x^{2}-1=0$

$y=x^{2}+{\tfrac {1}{3}}({\sqrt {30}}-3)x+{\tfrac {1}{15}}{\sqrt {30}}$

$y^{5}-{\tfrac {1}{45}}(465-61{\sqrt {30}})y^{2}+{\tfrac {2}{45}}(1616-289{\sqrt {30}})y-{\tfrac {1}{1125}}(33758{\sqrt {30}}-183825)=0$

Solving the principal quintic via Adamchik transformation

The mathematicians Victor Adamchik and David Jeffrey found out how to solve every principal quintic equation. In their essay Polynomial Transformations of Tschirnhaus, Bring and Jerrard they wrote this way down. These two mathematicians solved this principal form by transforming it into the Bring Jerrard form. Their method contains the construction of a quartic Tschirnhaus transformation key. For the construction of that key they executed a disjunction of the linear term coefficient of the key in order to get a system that solves all other terms in a quadratic radical way and to only solve a further cubic equation to get the coefficient of the linear term of the Tschirnhaus key.

In their essay they constructed the quartic Tschirnhaus key in this way:

$y^{5}-uy^{2}+vy-w=0$

In order to do the transformation Adamchik and Jeffrey constructed equation system that generates the coefficients of the cubic, quadratic and absolute term of the Tschirnhaus key:

$4v{\color {crimson}\alpha }-3u{\color {green}\beta }=5w$

$3u{\color {crimson}\alpha }+5{\color {blue}\delta }=4v$

$15w{\color {crimson}\alpha }{\color {green}\beta }-3u{\color {crimson}\alpha }{\color {blue}\delta }-6v{\color {green}\beta }^{2}-v{\color {blue}\delta }=3uw-2v^{2}$

And for receiving the coefficient of the linear term this cubic equation shall be solved successively:

$u{\color {orange}\gamma }^{3}+(5w{\color {crimson}\alpha }-4v{\color {green}\beta }+3u^{2}){\color {orange}\gamma }^{2}+(uv{\color {crimson}\alpha }^{2}-8v{\color {crimson}\alpha }{\color {blue}\delta }+5w{\color {green}\beta }^{2}+6u{\color {green}\beta }{\color {blue}\delta }-8uv{\color {green}\beta }+3u^{3}+9vw){\color {orange}\gamma }+$

$+u^{3}{\color {crimson}\alpha }^{3}+vw{\color {crimson}\alpha }^{3}+u^{2}{\color {green}\beta }^{3}-10uw{\color {crimson}\alpha }^{2}{\color {green}\beta }+4v^{2}{\color {crimson}\alpha }^{2}{\color {green}\beta }+9u^{2}{\color {crimson}\alpha }^{2}{\color {blue}\delta }-uv{\color {crimson}\alpha }{\color {green}\beta }^{2}+15u{\color {crimson}\alpha }{\color {blue}\delta }^{2}+10{\color {blue}\delta }^{3}-$

$-u^{2}v{\color {crimson}\alpha }^{2}-3u^{3}{\color {crimson}\alpha }{\color {green}\beta }+6vw{\color {crimson}\alpha }{\color {green}\beta }-12uv{\color {crimson}\alpha }{\color {blue}\delta }+5uw{\color {green}\beta }^{2}-4v^{2}{\color {green}\beta }^{2}+uv^{2}{\color {crimson}\alpha }-u^{2}w{\color {crimson}\alpha }+2u^{2}v{\color {green}\beta }+u^{4}-4v^{3}+8uvw=0$

The solution of that system then has to be entered in that mold here:

$z=y^{4}+{\color {crimson}\alpha }y^{3}+{\color {green}\beta }y^{2}+{\color {orange}\gamma }y+{\color {blue}\delta }$

$z^{5}+\lambda z-\mu =0$

The coefficients Lambda and My can be found out by doing a polynomial division of z^5 divided by the initial principal polynome and reading the resulting remainder rest. So a Bring Jerrard equation appears that contains only the quintic, the linear and the absolute term. That equation can be solved by an elliptic Jacobi theta quotient that contains the fifth powers and the fifth roots of the corresponding elliptic nome in the theta function terms.

Examples of solving the principal form

Along with the Abel Ruffini theorem the following equation is an example that can not be solved in an elementary way, but can be reduced to the Bring Jerrard form by only using cubic radical elements. This shall be demonstrated here:

$y^{5}-5y^{2}+5y-5=0$

For this given equation the quartic Tschirnhaus key now shall be synthesized. For that example the values u = v = w = 5 are the combination. Therefore, this is the mentioned equation system for that example:

$20{\color {crimson}\alpha }-15{\color {green}\beta }=25$

$15{\color {crimson}\alpha }+5{\color {blue}\delta }=20$

$75{\color {crimson}\alpha }{\color {green}\beta }-15{\color {crimson}\alpha }{\color {blue}\delta }-30{\color {green}\beta }^{2}-5{\color {blue}\delta }=25$

So these are the coefficients of the cubic, quadratic and absolute term of the Tschirnhaus transformation key:

${\color {crimson}\alpha }=-1,\,{\color {green}\beta }=-3,\,{\color {blue}\delta }=7$

That cubic equation leads to the coefficient of the linear term of the key:

$5{\color {orange}\gamma }^{3}+110{\color {orange}\gamma }^{2}+1100{\color {orange}\gamma }+3080=0$

${\color {orange}\gamma }={\frac {8}{3}}{\sqrt {11}}\sinh {\bigl [}{\frac {1}{3}}\operatorname {arsinh} {\bigl (}{\frac {4}{11}}{\sqrt {11}}{\bigr )}{\bigr ]}-{\frac {22}{3}}$

By doing a polynomial division on the fifth power of the quartic Tschirnhaus transformation key and analyzing the remainder rest the coefficients of the mold can be determined. This is the result:

$y^{5}-5y^{2}+5y-5=0$

$z=y^{4}-y^{3}-3y^{2}+{\bigl \{}{\frac {8}{3}}{\sqrt {11}}\sinh {\bigl [}{\frac {1}{3}}\operatorname {arsinh} {\bigl (}{\frac {4}{11}}{\sqrt {11}}{\bigr )}{\bigr ]}-{\frac {22}{3}}{\bigr \}}y+7$

$z^{5}+{\frac {14080}{3}}{\bigl \{}11-2{\sqrt {22}}\cosh {\bigl [}{\frac {1}{3}}\operatorname {arcosh} {\bigl (}{\frac {7}{22}}{\sqrt {22}}{\bigr )}{\bigr ]}{\bigr \}}z+{\frac {11264}{3}}{\bigl \{}11-110{\sqrt {2}}\sinh {\bigl [}{\frac {1}{3}}\operatorname {arsinh} {\bigl (}{\frac {1}{22}}{\sqrt {2}}{\bigr )}{\bigr ]}{\bigr \}}=0$

These are the approximations of the solution:

$z\approx -4.87187987090341241739191116705958390845844658170795795268900739402026742$

$y\approx 1.56670895425072582758152133323240667646412076995364965189840377191745$

Literature

• "Polynomial Transformations of Tschirnhaus", Bring and Jerrard, ACM Sigsam Bulletin, Vol 37, No. 3, September 2003
• F. Brioschi, Sulla risoluzione delle equazioni del quinto grado: Hermite — Sur la résolution de l'Équation du cinquiéme degré Comptes rendus —. N. 11. Mars. 1858. 1. Dezember 1858, doi:10.1007/bf03197334
• Bruce and King, Beyond the Quartic Equation, Birkhäuser, 1996.