Draft:Brioschi quintic form

Template:Draft article The mathematical Brioschi quintic form is a special type of quintic polynome. This special polynome of fifth degree is used to solve quintic equations in an efficient way. All quintic equations can be transformed into the Brioschi form by Tschirnhaus transformations in such a way that the generated coefficients always have a quadratically radical relation to the original coefficients of the quintic initial equation. After the Italian mathematician Francesco Brioschi This special quintic equation form is named. Its efficiency in solving quintic equations was researched by that mathematician. He and Charles Hermite discovered efficient solution algorithms for quintic equations

Definition

This special Brioschi quintic polynome of fifth degree does contain a quintic, as well as a cubic, a linear and an absolute term, but it does neither contain a quartic term nor a quadratic term. The Brioschi quintic form has two saddle points in the corresponding function graph. That property distinguishes the Brioschi form from all other equation forms without quartic and quadratic terms. Therefore, the first derivative of the Brioschi form leads to the square of a quadratic equation. In the Brioschi form with alternating signs before the quintic and the cubic coefficient there is a third inflection point that in relation to the abscissa is the arithmetic mean of the two saddle points. On the whole the Brioschi form represents a point reflection symmetric function in which the mentioned third inflection point is the exact symmetry point. According to all those features, the Brioschi form generally looks like that:

$v^2{z}^5-10v{z}^3+45z-y=0$

The two saddle points appear as real values in the function graph if the value of v has a positive value in the portrayed Brioschi function equation. But if the value of v is negative, the quintic and the cubic coefficient have the same sign and then the complete elliptic function is a bijective function. The function of y is a point symmetric function to the origin of the function system.

Tschirnhaus transformations into the Brioschi form

Fundamental pattern of the transformation

In the following it is shown how some Principal Quintic equations are transformed into the Brioschi quintic forms. To fulfill this, only broken quadratic Tschirnhaus transformations are relevant. This is the initial Principal Quintic equation:

$x^5+r{x}^2+sx+t=0$

Now the coefficients of the broken rational key of the Brioschi form will be created by at first computing the value u for the absolute term in the numerator:

$(r^4-5s^3+25rst)u^2+(-11r^{3}s+125r{t}^2-50s^{2}t)u+64r^2{s}^2-135r^{3}t-125s{t}^2=0$

$v=1728-\frac{5(15t-r{u}^2+3su{)}^3}{r^2(s^{2}u-5rtu+5st)}$

$w=\frac{5(15t-r{u}^2+3su{)}^3-(s^{2}u-5rtu+5st)(40r{u}^3+360s{u}^2+1800tu)}{5(s^{2}u-5rtu+5st)(r{u}^2+su+5t)}$

This is the broken rational key for the Brioschi form:

$x=\frac{wz+u}{v{z}^2-3}$

The resulting Brioschi equation looks this way:

$v^2{z}^5-10v{z}^3+45z-1=0$

First Calculation example for the transformation

The Brioschi form shall be created in a certain calculation example of the Principal form. This is a Principal Quintic equation for which the Abel Ruffini theorem is valid. This equation can not be solved by elementary root expressions but can be solved by elliptic expressions:

$x^5-5x^2+5x-5=0$

The value for a was already computed in the mentioned calculation example:

$a=-1$

This solution brings those values:

$u=-4\text{and}v=1849\text{and}w=129$

This is the broken rational key:

$x=\frac{129z-4}{1849z^2-3}$

It shows the Tschirnhaus Transformation in this way:

$[\frac{129z-4}{1849z^2-3}{]}^5-5[\frac{129z-4}{1849z^2-3}{]}^2+5\left[\frac{129z-4}{1849z^2-3}\right]-5=0$

And this is the resulting Brioschi equation:

$3418801z^5-18490z^3+45z-1=0$

Second Calculation example for the transformation

Principal equation:

$x^5+x^2-1=0$

Brioschi equation:

$[\frac{1}{2}(238528425\sqrt{15085}-29296305919){]}^2{z}^5-[\frac{1}{2}(238528425\sqrt{15085}-29296305919)]z^3+45*z-1=0$

$z\approx 0.018787897470813223818839629483292225985487352060808662222311422852295054189328531$

Clues:

$u=-\frac{1}{2}(125-\sqrt{15085})$

$v=\frac{1}{2}(-29296305919+238528425\sqrt{15085})$

$w=\frac{1}{2}(382625-3117\sqrt{15085})$

Solution:

$x=\frac{\frac{1}{2}(382625-3117\sqrt{15085})z-\frac{1}{2}(125-\sqrt{15085})}{\frac{1}{2}(-29296305919+238528425\sqrt{15085})z^2-3}$

$x\approx 0.808730600479392013738554526511400064951377351559313075548116401836543340748321$

Third Calculation example for the transformation

This is a next Principal Quintic equation that can not be solved by elementary expressions either. But again an elliptic solution way does indeed exist. The Brioschi algorithm is shown:

$x^5-x^2-1=0$

These are the coefficients of the broken rational quadratic key:

$u=-\frac{1}{2}(\sqrt{16165}-125)$

$v=\frac{1}{2}(31757250331-249778425\sqrt{16165})$

$w=\frac{1}{2}(398625-3133\sqrt{16165})$

Corresponding Brioschi Quintic equation:

$[\frac{1}{2}(31757250331-249778425\sqrt{16165}){]}^2{z}^5-10[\frac{1}{2}(31757250331-249778425\sqrt{16165})]z^3+45z-1=0$

$z\approx 0.04103471241028970064518587406207759188088410335520608376935448159$

Real solution of the given Principal equation:

$x=\frac{\frac{1}{2}(398625-3133\sqrt{16165})z-\frac{1}{2}(\sqrt{16165}-125)}{\frac{1}{2}(31757250331-249778425\sqrt{16165})z^2-3}$

$x\approx 1.193859111321223012009020746298031124514524269486444509602081401596$

Fourth Calculation example for the transformation

Now this Principal Quintic equation is given:

$x^5+x^2+x-1=0$

The corresponding coefficients for the broken rational quadratic key are those:

$u=-\frac{1}{29}(\sqrt{8870}-82)$

$v=-\frac{1}{20511149}(983008525\sqrt{8870}-67467103347)$

$w=-\frac{1}{24389}(2110353-1312\sqrt{8870})$

Following expression represents the Brioschi Quintic for this calculation example:

$[-\frac{1}{20511149}(983008525\sqrt{8870}-67467103347){]}^2{z}^5-10[-\frac{1}{20511149}(983008525\sqrt{8870}-67467103347)]z^3+45z-1=0$

$z\approx 0.01995485921964894576972367249477726279242885730382331060573983953100994109904667$

The broken rational quadratic key gives the solution of the Principal Quintic equation:

$x=\frac{-\frac{1}{24389}(2110353-1312\sqrt{8870})z-\frac{1}{29}(\sqrt{8870}-82)}{-\frac{1}{20511149}(983008525\sqrt{8870}-67467103347)z^2-3}$

$x\approx 0.58654371032332271804618797362704215544707974523108493478693019806187056334481$

Solution of the Brioschi equation

Fundamental pattern of the solving

The Brioschi form with two real saddle points shall be solved in this section. Given is again the mentioned Brioschi Form:

$v^2{z}^5-10v{z}^3+45z-y=0$

Now a special substitution shall be made:

$v=1728\tanh (3y{)}^2$

$z=\frac{1}{9}x$

in the following by using the hyperbolic tangent function:

$4096\tanh (3y{)}^4{x}^5-1920\tanh (3y{)}^2{x}^3+405x-81=0$

According to a further Tschirnhaus transformation into the Bring Jerrard form and creating the elliptic modulus after the pattern of the Hermite essay Sur la résolution de l'Équation du cinquiéme degré Comptes rendus, the Legendre elliptic modulus k shall be generated after this pattern:

$k=\cos \{\frac{1}{2}\mathrm{arccot}[\sqrt{3}\tanh (y)\left]\right\}$

$k^{*}=\sin \{\frac{1}{2}\mathrm{arccot}[\sqrt{3}\tanh (y)\left]\right\}$

$k^{*}=\sqrt{1-k^2}$

In this way the Elliptic nome is generated after following formulas:

$Q=\exp [-\pi K(k^{*})÷K(k)]$

$Q^{*}=\exp [-\pi K(k)÷K(k^{*})]$

In the next step a fraction of Jacobi theta function expressions is created:

$f=\frac{2{\vartheta }_{00}(Q^5){\vartheta }_{00}(Q^{1/5})-2{\vartheta }_{00}(Q{)}^2}{{\vartheta }_{00}(Q^{1/5}{)}^2+5{\vartheta }_{00}(Q^5{)}^2-4{\vartheta }_{00}(Q{)}^2}$

Because of the Poisson summation formula the same value f appears via replacing Q by Q* indeed. For all real values y following expression solves the mentioned Brioschi quintic equation:

$\frac{12-32\tanh (3y)\coth (y)x}{64\tanh (3y{)}^2{x}^2-9}=\frac{f(5f^2-10f+4)}{5f^2-6f+2}$

One of the two real solutions of that quadratic equation is identical with the solution of the mentioned quintic equation in Brioschi form.

First calculation examples for solving a Brioschi equation

As an accurate calculation example we take following value:

$y=\frac{1}{3}\mathrm{artanh}\left(\frac{1}{8}\right)$

$y\approx 0.0418857380468176796141896217336452799429827310106$

That inserted value belongs to this Brioschi equation:

$x^5-30x^3+405x-81=0$

The corresponding modulus k has the mentioned form:

$k=\cos \{\frac{1}{2}\mathrm{arccot}[\sqrt{3}\tanh (y)\left]\right\}$

We get that value:

$k\approx 0.7322281040536863694792348185629455128841904674146670978796909090$

The nome is this value:

$Q\approx 0.047866673014485714950655122967535320490947054771013255128047166$

The Jacobi theta fraction is mentioned:

$f=\frac{2{\vartheta }_{00}(Q^5){\vartheta }_{00}(Q^{1/5})-2{\vartheta }_{00}(Q{)}^2}{{\vartheta }_{00}(Q^{1/5}{)}^2+5{\vartheta }_{00}(Q^5{)}^2-4{\vartheta }_{00}(Q{)}^2}$

In that example it has following value:

$f\approx 0.399849878028699235467562012341687999489926833880425671171043745076$

By entering the values of f and y we get this solution:

$\frac{12-32\tanh (3y)\coth (y)x}{64\tanh (3y{)}^2{x}^2-9}=\frac{f(5f^2-10f+4)}{5f^2-6f+2}$

$x\approx 0.2005971141471857121313807986552177730180286560025$

Second calculation examples for solving a Brioschi equation

As an accurate calculation example we take following value:

$y=\frac{1}{3}\mathrm{artanh}\left(\frac{1}{4}\right)$

$y\approx 0.0851376039609984472009190160506103224796851327409613783629922596$

That inserted value belongs to this Brioschi equation:

$16x^5-120x^3+405x-81=0$

The corresponding modulus k has the mentioned form:

$k=\cos \{\frac{1}{2}\mathrm{arccot}[\sqrt{3}\tanh (y)\left]\right\}$

We get that value:

$k\approx 0.7568160371064972605301165710544598383870177442240957097614220827$

The nome is this value:

$Q\approx 0.052953640059991015414100205940565208062046959019895010051080868$

The Jacobi theta fraction is mentioned:

$f=\frac{2{\vartheta }_{00}(Q^5){\vartheta }_{00}(Q^{1/5})-2{\vartheta }_{00}(Q{)}^2}{{\vartheta }_{00}(Q^{1/5}{)}^2+5{\vartheta }_{00}(Q^5{)}^2-4{\vartheta }_{00}(Q{)}^2}$

In that example it has following value:

$f\approx 0.399386396717083507551377733660924269800949276000648370570131646824$

By entering the values of f and y we get this solution:

$\frac{12-32\tanh (3y)\coth (y)x}{64\tanh (3y{)}^2{x}^2-9}=\frac{f(5f^2-10f+4)}{5f^2-6f+2}$

$x\approx 0.2024449345913542228580304306303447038721951474593556763277652278785$

Literature

• Francesco Brioschi: "Sul Metodo di Kronecker per la Risoluzione delle Equazioni di Quinto Grado", Atti Dell'i. R. Istituto Lombardo di Scienze, Lettere ed Arti. I, 1858: Pages 275–282.
• Francesco 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
• Raymond Garver: On the transformation which leads from the Brioschi quintic to a general principal quintic, Published 1 February 1930, Bulletin of the American Mathematical Society, DOI:10.1090/S0002-9904-1930-04902-2Corpus ID: 121786067