User:Reformbenediktiner

Head of my page

My name is Lion Emil Jann Fiedler and I live in Bamberg in Germany.

This is my German Wikipedia page:

https://de.wikipedia.org/wiki/Benutzer:Reformbenediktiner

My pseudonym on YouTube is Emil Jann Brahmeyer:

https://www.youtube.com/@emiljannbrahmeyer

From Principal quintics to Bring Jerrard quintics

Given Principal equation:

x^5 - u*x^2 + v*x - w = 0

Equation system for the initial clues on the A and B unknowns:

4*v*A - 3*u*B = 5*w

3*u^2*A^2 - 3*u*v*A + 25*w*A*B - 10*v*B^2 = 5*u*w - 2*v^2

Clue for the D unknown:

D = 4/5*v - 3/5*u*A

Clue for the C unknown:

u*C^3 + (5*w*A - 4*v*B + 3*u^2)*C^2 + (u*v*A^2 + 6*u*w*A + 5*w*B^2 - 8*u*v*B + 3*u^3 + v*w)*C + u^3*A^3 + v*w*A^3 - 2*u^2*B^3 + 2*u*v*A*B^2 - 2*u^2*A^2*D + 10*D^3 - 4*u^2*v*A^2 + v*w*A*B + 3*u*v*A*D + 2*u^2*w*A + 2*u^2*v*B - v^2*D + u^4 - 4*v^3 + 10*u*v*w = 0

Constructing the coefficients of the broken rational transformation key:

G = (-v*A^3 - A*B*D + w*A^2 + 2*v*A*B - A*C^2 - u*A*C + B^2*C - w*B + C*D - v*C + u*D - u*v)/(A^2*D - 2*A*B*C + B^3 + u*A^3 - 3*u*A*B - v*A^2 - B*D + C^2 + v*B + 2*u*C + u^2)

H = (A^3*w - 2*A*B*w - A*C*D - A*D*u + B^2*D + C*w + u*w)/(A^2*D - 2*A*B*C + B^3 + u*A^3 - 3*u*A*B - v*A^2 - B*D + C^2 + v*B + 2*u*C + u^2)

K = (-A*B+C+u)/(A^2-B)

L = (B^2-A*C-u*A)/(A^2-B)

Transformation key:

z = (x^2 + G*x + H)/(x^2 + K*x + L)

First equation:

((4*G*K^3*L + 12*G*K*L^2 + H*K^4 + 12*H*K^2*L + 6*H*L^2 + 6*K^2*L^2 + 4*L^3) + (6*G*K^2 + 4*G*L + 4*H*K + 4*K^3 + 12*K*L)*u - (4*G*K + H + 6*K^2 + 4*L)*v + (G + 4*K)*w + u^2)*M + + (-(5*K^4*L + 30*K^2*L^2 + 10*L^3) - (10*K^3 + 20*K*L)*u + (10*K^2 + 5*L)*v - 5*K*w - u^2)*N + + (5*G^4*H + 30*G^2*H^2 + 10*H^3) + (10*G^3 + 20*G*H)*u - (10*G^2 + 5*H)*v + 5*G*w + u^2 = 0

Second equation:

(H*L^4 + (G*K^4 + 12*G*K^2*L + 6*G*L^2 + 4*H*K^3 + 12*H*K*L + 4*K^3*L + 12*K*L^2)*w + (4*G*K + H + 6*K^2 + 4*L)*u*w - (G + 4*K)*v*w + w^2)*M + + (-L^5 - (K^5 + 20*K^3*L + 30*K*L^2)*w - (10*K^2 + 5*L)*u*w + 5*K*v*w - w^2)*N + + H^5 + (G^5 + 20*G^3*H + 30*G*H^2)*w + (10*G^2 + 5*H)*u*w - 5*G*v*w + w^2 = 0

Endform:

z^5 + M*z - N = 0

Other method with accurate examples

Verfahren der Teilschritte:

Grundlage:

x^5 - u*x^2 + v*x - w = 0

Gleichungssystem:

4*v*A - 3*u*B = 5*w

3*u^2*A^2 - 3*u*v*A + 25*w*A*B - 10*v*B^2 = 5*u*w - 2*v^2

Nachfolgende Gleichungen:

D = 4/5*v - 3/5*u*A

u*C^3 + (5*w*A - 4*v*B + 3*u^2)*C^2 + (u*v*A^2 + 6*u*w*A + 5*w*B^2 - 8*u*v*B + 3*u^3 + v*w)*C + + u^3*A^3 + v*w*A^3 - 2*u^2*B^3 + 2*u*v*A*B^2 - 2*u^2*A^2*D + 10*D^3 - 4*u^2*v*A^2 + v*w*A*B + 3*u*v*A*D + 2*u^2*w*A + 2*u^2*v*B - v^2*D + u^4 - 4*v^3 + 10*u*v*w = 0

Schlüssel für die Transformation:

y = x^4 + A*x^3 + B*x^2 + C*x + D

Verfahren der Teilschritte

(x^4 + A*x^3 + B*x^2 + C*x + D)^5/(x^5 - u*x^2 + v*x - w) = x^15 + 5*A*x^14 + E*x^13 + F*x^12 + G*x^11 + H*x^10 + I*x^9 + J*x^8 + K*x^7 + L*x^6 + M*x^5 + N*x^4 + O*x^3 + P*x^2 + Q*x + R + (-S*(x^4 + A*x^3 + B*x^2 + C*x + D) + T)/(x^5 - u*x^2 + v*x - w)

Koeffizienten des Polynoms fünfzehnten Grades:

E = 10*A^2 + 5*B

F = 5*C + 10*A^3 + 20*A*B + u

G = 20*A*C + 5*A^4 + 30*A^2*B + 10*B^2 + 2*u*A + 3*v

H = (30*A^2 + 20*B)*C + A^5 + 20*A^3*B + 30*A*B^2 + 20*A*D + u*E - 5*v*A + w

I = 10*C^2 + (20*A^3 + 60*A*B)*C + 5*A^4*B + 30*A^2*B^2 + 30*A^2*D + 10*B^3 + 20*B*D + u*F - v*E + 5*w*A

J = 30*A*C^2 + (5*A^4 + 60*A^2*B + 30*B^2 + 20*D)*C + 10*A^3*B^2 + 20*A*B^3 + 20*A^3*D + 60*A*B*D + u*G - v*F + w*E

K = (30*A^2 + 30*B)*C^2 + (20*A^3*B + 60*A*B^2 + 60*A*D)*C + 10*A^2*B^3 + 5*B^4 + 5*A^4*D + 60*A^2*B*D + 30*B^2*D + 10*D^2 + u*H - v*G + w*F

L = 10*C^3 + (10*A^3 + 60*A*B)*C^2 + (30*A^2*B^2 + 60*A^2*D + 60*B*D + 20*B^3)*C + 5*A*B^4 + 20*A^3*B*D + 60*A*B^2*D + 30*A*D^2 + u*I - v*H + w*G

M = 20*A*C^3 + (30*A^2*B + 30*B^2 + 30*D)*C^2 + (20*A*B^3 + 20*A^3*D + 120*A*B*D)*C + B^5 + 30*A^2*D^2 + 30*A^2*B^2*D + 30*B*D^2 + 20*B^3*D + u*J - v*I + w*H

N = (10*A^2 + 20*B)*C^3 + (30*A*B^2 + 60*A*D)*C^2 + (60*A^2*B*D + 5*B^4 + 60*B^2*D + 30*D^2)*C + 10*A^3*D^2 + 60*A*B*D^2 + 20*A*B^3*D + u*K - v*J + w*I

O = 5*C^4 + 20*A*B*C^3 + (10*B^3 + 30*A^2*D + 60*B*D)*C^2 + (60*A*D^2 + 60*A*B^2*D)*C + 5*B^4*D + 30*A^2*B*D^2 + 30*B^2*D^2 + 10*D^3 + u*L - v*K + w*J

P = 5*A*C^4 + (10*B^2 + 20*D)*C^3 + 60*A*B*D*C^2 + (30*A^2*D^2 + 20*B^3*D + 60*B*D^2)*C + 20*A*D^3 + 30*A*B^2*D^2 + u*M - v*L + w*K

Q = 5*B*C^4 + 20*A*D*C^3 + (30*B^2*D + 30*D^2)*C^2 + 60*A*B*D^2*C + 10*A^2*D^3 + 10*B^3*D^2 + 20*B*D^3 + u*N - v*M + w*L

R = C^5 + 20*B*D*C^3 + 30*A*D^2*C^2 + (30*B^2*D^2 + 20*D^3)*C + 20*A*B*D^3 + u*O - v*N + w*M

S = -5*D*C^4 - 30*B*D^2*C^2 - 20*A*D^3*C - 5*D^4 - 10*B^2*D^3 - u*P + v*O - w*N

T = D^5 + D*S + w*R

y^5 + S*y = T

Erstes Rechenbeispiel

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

u=5, v=5, w=5

Erster Lösungsweg vom ersten Rechenbeispiel

A=-1, B=-3, D=7

5*C^3 + 110*C^2 + 1100*C + 3080 = 0

C=-4*sqrt(3)*tanh(1/3*artanh(4/9*sqrt(3)))-2 = -4.268494999568071895959284940795497379614167933

E = -5

F = 5*C + 55 = 33.65752500215964052020357529602251310192916034

G = -20*C + 10 = 95.36989999136143791918569881590994759228335866

H = -30*C - 346 = -217.9451500129578431212214517761350786115749620

I = 10*C^2 + 185*C + 50 = -557.4710793067169597997475357093474157541519641

J = -30*C^2 + 110*C + 1320 = 303.864063207383068586318121499036489819034217

K = -60*C^2 - 925*C - 215 = 2640.154900920208458046017500208917479296290717

L = 10*C^3 + 220*C^2 - 135*C - 3205 = 601.93582399144787706493033675748638139908212

M = -20*C^3 + 190*C^2 + 2395*C - 453 = -5658.79229842233905302196993742276992758671759

N = -50*C^3 - 790*C^2 + 145*C + 4685 = -6439.16140617149471093190513669341071209135941

O = 5*C^4 + 110*C^3 - 70*C^2 - 2220*C + 6735 = 8040.5655109457491571159493206912429734422049

P = -5*C^4 + 80*C^3 + 810*C^2 - 3105*C - 7405 = 12725.2908526981776522296121660866288216580873

Q = -15*C^4 - 240*C^3 - 440*C^2 - 3105*C - 20715 = -1792.3722169251795378769955903014035538191141

R = C^5 + 25*C^4 + 280*C^3 + 3080*C^2 + 20240*C + 28565 = -16605.557480736173604702798226758969888697260

Koeffizienten der Bring Jerrard Endform:

S = 15*C^4 + 400*C^3 + 3960*C^2 + 10560*C + 4400 = 5346.765979941334291267356272672478351214203

S = -880*(C^2 + 16*C + 44)

T = 5*C^5 + 230*C^4 + 4200*C^3 + 43120*C^2 + 175120*C + 190432 = -28793.425544091527984642497225087500984986875

T = 704*(5*C^2 - 132)

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

C = -2-4*sqrt(3)*tanh(1/3*artanh(4/9*sqrt(3)))

y = x^4-x^3-3*x^2+C*x+7

y^5 - 880*(C^2 + 16*C + 44)*y = 704*(5*C^2 - 132)

Zweiter Lösungsweg vom ersten Rechenbeispiel

A=7/5, B=1/5, D=-1/5

5*C^3 + 106*C^2 + 620*C + 30488/25 = 0

C=-14/5-4/5*sqrt(15)*coth(1/3*arcoth(16/45*sqrt(15))) = -13.211735572933908537478732071425141899412474429210737

E = 103/5

F = 5*C + 951/25

G = 28*C + 7546/125

H = 314/5*C + 266982/3125

I = 10*C^2 + 2417/25*C + 72446/625

J = 42*C^2 + 19366/125*C + 126016/625

K = 324/5*C^2 + 24567/125*C + 192069/625

L = 10*C^3 + 2356/25*C^2 + 35749/125*C + 283163/625

M = 28*C^3 + 4174/25*C^2 + 73471/125*C + 2680691/3125

N = 118/5*C^3 + 3722/25*C^2 + 85933/125*C + 693573/625

O = 5*C^4 + 278/5*C^3 + 8578/25*C^2 + 153076/125*C + 1085823/625

P = 7*C^4 + 432/5*C^3 + 17106/25*C^2 + 62359/25*C + 2225123/625

Q = C^4 + 112/5*C^3 + 9544/25*C^2 + 241139/125*C + 2202873/625

R = C^5 + 25*C^4 + 1496/5*C^3 + 45192/25*C^2 + 703056/125*C + 4641913/625

Koeffizienten der Bring Jerrard Endform:

S = -9*C^4 - 272*C^3 - 61256/25*C^2 - 1223232/125*C - 9164368/625

S = 1936/125*(25*C^2 + 160*C + 332)

T = 5*C^5 + 634/5*C^4 + 7752/5*C^3 + 1191056/125*C^2 + 18799632/625*C + 125212192/3125

T = -23232/3125*(625*C^2 + 4800*C + 10668)

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

C = -14/5-4/5*sqrt(15)*coth(1/3*arcoth(16/45*sqrt(15)))

y = x^4+7/5*x^3+1/5*x^2+C*x-1/5

y^5 + 1936/125*(25*C^2 + 160*C + 332)*y = -23232/3125*(625*C^2 + 4800*C + 10668)

Quintic Principal Tschirnhaus Transformation Examples

Recipe for the Transformation Key

Given Principal Equation:

x^5 - u*x^2 + v*x - w = 0

Equation system for the unknowns A and B of the key:

4*v*A - 3*u*B = 5*w

3*u^2*A^2 - 3*u*v*A + 25*w*A*B - 10*v*B^2 = 5*u*w - 2*v^2

Equations for the unknowns D and C of the key:

D = 4/5*v - 3/5*u*A

u*C^3 + (5*w*A - 4*v*B + 3*u^2)*C^2 + (u*v*A^2 + 6*u*w*A + 5*w*B^2 - 8*u*v*B + 3*u^3 + v*w)*C + + u^3*A^3 + v*w*A^3 - 2*u^2*B^3 + 2*u*v*A*B^2 - 2*u^2*A^2*D + 10*D^3 - 4*u^2*v*A^2 + v*w*A*B + 3*u*v*A*D + 2*u^2*w*A + 2*u^2*v*B - v^2*D + u^4 - 4*v^3 + 10*u*v*w = 0

Resulting Transformation Key:

y = x^4 + A*x^3 + B*x^2 + C*x + D

Calculation Examples

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

C = -2-4*sqrt(3)*tanh(1/3*artanh(4/9*sqrt(3)))

y = x^4-x^3-3*x^2+C*x+7

y^5 - 880*(C^2+16*C+44)*y = 704*(5*C^2-132)

y^5 - 880*(C^2+16*C+44)*y = 4/5*(C+11)*880*(C^2+16*C+44)

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

C = -14/5-4/5*sqrt(15)*coth(1/3*arcoth(16/45*sqrt(15)))

y = x^4+7/5*x^3+1/5*x^2+C*x-1/5

y^5 + 1936/125*(25*C^2+160*C+332)*y = -23232/3125*(625*C^2+4800*C+10668)

y^5 + 1936/125*(25*C^2+160*C+332)*y = 12/125*(10*C+23)*1936/125*(25*C^2+160*C+332)

x^5 - 5*x^2 + 5*x - 8 = 0

C = -8/5-2/5*sqrt(546)*tanh(1/3*artanh(1/36*sqrt(546)))

y = x^4-x^3-4*x^2+C*x+7

y^5 - 455*(5*C^2+107*C+304)*y = 24843*(C^2+7*C+5)

y^5 - 455*(5*C^2+107*C+304)*y = 3/40*(9*C+160)*455*(5*C^2+107*C+304)

x^5 - 5*x^2 + 5*x - 8 = 0

C = -11/4-1/4*sqrt(273)*coth(1/3*arcoth(1/9*sqrt(273)))

y = x^4+2*x^3+C*x-2

y^5 + 455*(4*C^2+22*C+53)*y = -16562*(4*C^2+30*C+75)

y^5 + 455*(4*C^2+22*C+53)*y = 2/5*(4*C+11)*455*(4*C^2+22*C+53)

x^5 - 5*x^2 + 15*x - 12 = 0

C = -4-sqrt(10)*coth(1/3*arcoth(1/2*sqrt(10)))

y = x^4+x^3+C*x+9

y^5 + 675*(3*C^2 + 29*C + 78)*y = -2025*(7*C^2 + 69*C + 189)

y^5 + 675*(3*C^2 + 29*C + 78)*y = 3/10*(C-6)*675*(3*C^2 + 29*C + 78)

x^5 - 5*x^2 + 15*x - 12 = 0

C = 3-450^(1/3)

y = x^4+2*x^3+4*x^2+C*x+6

y^5 - 10125*(2*C+9)*y = 12150*(2*C+9)*(2*C-21)

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

C^3 + 1/433*(31*sqrt(8870)-2556)*C^2 - 1/187489*(31536*sqrt(8870)-125194)*C + 1/405913685*(44948782*sqrt(8870)-957483960) = 0

C = -1/1299*(31*sqrt(8870)-2556)+4/1299*sqrt(1774*(2069-9*sqrt(8870)))*cos(pi/6+1/6*arccos(324/34295*sqrt(8870)-15557/34295))

y = x^4 + 1/433*(401-3*sqrt(8870))*x^3 + 1/433*(187+4*sqrt(8870))*x^2 + C*x + 1/2165*(2935-9*sqrt(8870))

y^5 + (7096/187489*(2069-9*sqrt(8870))*C^2 - 113536/405913685*(704810+287*sqrt(8870))*C + 14192/175760625605*(1429371257-2976793*sqrt(8870)))*y = -14192/405913685*(212409*sqrt(8870)-18644740)*C^2 + 56768/175760625605*(34261965*sqrt(8870)-2500947946)*C - 28384/1902608772174125*(596854977381*sqrt(8870)-49163189125600)

x^5 + x^2 + 3*x - 7 = 0

C^3 + 1/33*(83*sqrt(895)-1123)*C^2 - 1/1089*(36602*sqrt(895)+946408)*C + 1/179685*(23002950+4579766*sqrt(895)) = 0

C = 1123/99 - (83 sqrt(895))/99 + 4/99 sqrt(19153 (134 - sqrt(895))) cos(pi/3-1/3*arccos(1/47 sqrt(214/235 (1799 - 22 sqrt(895)))))

y = x^4 + 1/33*(68-sqrt(895))*x^3 + 1/33*(4*sqrt(895)+113)*x^2 + C*x + 1/55*(200-sqrt(895))

y^5 + (76612/363*(134-sqrt(895))*C^2 - 153224/179685*(453480+13247*sqrt(895))*C + 153224/658845*(1126974+8183*sqrt(895)))*y = -16394968/16335*(173*sqrt(895)-3580)*C^2 + 32789936/5929605*(175774*sqrt(895)+3338171)*C - 32789936/1630641375*(60817981*sqrt(895)-750958700)

x^5 + 3*x^2 + 2*x - 1 = 0

C^3 - 5/114*(387-109*sqrt(5))*C^2 + 50/1083*(1661-643*sqrt(5))*C - 25/20577*(174995-76979*sqrt(5)) = 0

C = 5/342*(387-109*sqrt(5)) + 5/171 sqrt(94 (953 - 405 sqrt(5)))*cos(1/3*arccos(1/61 sqrt(47/610 (52673 - 23085 sqrt(5)))))

y = x^4-1/38*(45*sqrt(5)-89)*x^3+1/19*(20*sqrt(5)-29)*x^2+C*x+1/38*(221-81*sqrt(5))

y^5 + (1175/3249*(953-405*sqrt(5))*C^2 - 8225/123462*(53795-23363*sqrt(5))*C + 5875/2345778*(2295271-1003019*sqrt(5)))*y = 1380625/370386*(1821-811*sqrt(5))*C^2 - 276125/7037334*(1153697-514345*sqrt(5))*C + 55225/66854673*(215313100-96196009*sqrt(5))

Instructions

Principal Equation:

x^5 - u*x^2 + v*x - w = 0

y = x^4 + A*x^3 + B*x^2 + C*x + D

y^5 + S*y = T

First two Coefficients:

4*v*A - 3*u*B = 5*w

3*u^2*A^2 - 3*u*v*A + 25*w*A*B - 10*v*B^2 = 5*u*w - 2*v^2

Next two Coefficients:

D = 4/5*v - 3/5*u*A

u*C^3 + (5*w*A - 4*v*B + 3*u^2)*C^2 + (u*v*A^2 + 6*u*w*A + 5*w*B^2 - 8*u*v*B + 3*u^3 + v*w)*C + + u^3*A^3 + v*w*A^3 - 2*u^2*B^3 + 2*u*v*A*B^2 - 2*u^2*A^2*D + 10*D^3 - 4*u^2*v*A^2 + v*w*A*B + 3*u*v*A*D + 2*u^2*w*A + 2*u^2*v*B - v^2*D + u^4 - 4*v^3 + 10*u*v*w = 0

Schlüssel für die Transformation:

y = x^4 + A*x^3 + B*x^2 + C*x + D

Determination of the Bring Jerrard Endform:

Koeffizienten des Polynoms fünfzehnten Grades:

E = 10*A^2 + 5*B

F = 5*C + 10*A^3 + 20*A*B + u

G = 20*A*C + 5*A^4 + 30*A^2*B + 10*B^2 + 2*u*A + 3*v

H = (30*A^2 + 20*B)*C + A^5 + 20*A^3*B + 30*A*B^2 + 20*A*D + u*E - 5*v*A + w

I = 10*C^2 + (20*A^3 + 60*A*B)*C + 5*A^4*B + 30*A^2*B^2 + 30*A^2*D + 10*B^3 + 20*B*D + u*F - v*E + 5*w*A

J = 30*A*C^2 + (5*A^4 + 60*A^2*B + 30*B^2 + 20*D)*C + 10*A^3*B^2 + 20*A*B^3 + 20*A^3*D + 60*A*B*D + u*G - v*F + w*E

K = (30*A^2 + 30*B)*C^2 + (20*A^3*B + 60*A*B^2 + 60*A*D)*C + 10*A^2*B^3 + 5*B^4 + 5*A^4*D + 60*A^2*B*D + 30*B^2*D + 10*D^2 + u*H - v*G + w*F

L = 10*C^3 + (10*A^3 + 60*A*B)*C^2 + (30*A^2*B^2 + 60*A^2*D + 60*B*D + 20*B^3)*C + 5*A*B^4 + 20*A^3*B*D + 60*A*B^2*D + 30*A*D^2 + u*I - v*H + w*G

M = 20*A*C^3 + (30*A^2*B + 30*B^2 + 30*D)*C^2 + (20*A*B^3 + 20*A^3*D + 120*A*B*D)*C + B^5 + 30*A^2*D^2 + 30*A^2*B^2*D + 30*B*D^2 + 20*B^3*D + u*J - v*I + w*H

N = (10*A^2 + 20*B)*C^3 + (30*A*B^2 + 60*A*D)*C^2 + (60*A^2*B*D + 5*B^4 + 60*B^2*D + 30*D^2)*C + 10*A^3*D^2 + 60*A*B*D^2 + 20*A*B^3*D + u*K - v*J + w*I

O = 5*C^4 + 20*A*B*C^3 + (10*B^3 + 30*A^2*D + 60*B*D)*C^2 + (60*A*D^2 + 60*A*B^2*D)*C + 5*B^4*D + 30*A^2*B*D^2 + 30*B^2*D^2 + 10*D^3 + u*L - v*K + w*J

P = 5*A*C^4 + (10*B^2 + 20*D)*C^3 + 60*A*B*D*C^2 + (30*A^2*D^2 + 20*B^3*D + 60*B*D^2)*C + 20*A*D^3 + 30*A*B^2*D^2 + u*M - v*L + w*K

Q = 5*B*C^4 + 20*A*D*C^3 + (30*B^2*D + 30*D^2)*C^2 + 60*A*B*D^2*C + 10*A^2*D^3 + 10*B^3*D^2 + 20*B*D^3 + u*N - v*M + w*L

R = C^5 + 20*B*D*C^3 + 30*A*D^2*C^2 + (30*B^2*D^2 + 20*D^3)*C + 20*A*B*D^3 + u*O - v*N + w*M

S = -5*D*C^4 - 30*B*D^2*C^2 - 20*A*D^3*C - 5*D^4 - 10*B^2*D^3 - u*P + v*O - w*N

T = D^5 + D*S + w*R

Further concept of solving Principal Quintic Equations

Einfachste Rechenbeispiele

Erste Gleichung:

x^5 - 5*x^2 + 15*x - 7 = 0

(x^4+x^3+5/3*x^2-2/3*x+9)^5 - 338000/81*(x^4+x^3+5/3*x^2-2/3*x+9) - 8788000/243 = 0

Zweite Gleichung:

x^5 + 10*x^2 + 15*x - 4 = 0

(x^4-1/2*x^3+5/3*x^2+19/3*x+9)^5 - 10440125/1296*(x^4-1/2*x^3+5/3*x^2+19/3*x+9) - 177482125/3888 = 0

Erzeugungsalgorithmus

Prinzipielle Gleichung:

x^5 - u*x^2 + v*x - w = 0

Gleichungssystem mit zwei Unbekannten:

4*v*A - 3*u*B = 5*w 3*u^2*A^2 - 3*u*v*A + 25*w*A*B - 10*v*B^2 = 5*u*w - 2*v^2

Gleichungen für die weiteren Koeffizienten des Schlüssels:

D = 4/5*v - 3/5*u*A

u*C^3 + (5*w*A - 4*v*B + 3*u^2)*C^2 + (u*v*A^2 + 6*u*w*A + 5*w*B^2 - 8*u*v*B + 3*u^3 + v*w)*C + + u^3*A^3 + v*w*A^3 - 2*u^2*B^3 + 2*u*v*A*B^2 - 2*u^2*A^2*D + 10*D^3 - 4*u^2*v*A^2 + v*w*A*B + 3*u*v*A*D + 2*u^2*w*A + 2*u^2*v*B - v^2*D + u^4 - 4*v^3 + 10*u*v*w = 0

Schlüssel der Transformation:

y = x^4 + A*x^3 + B*x^2 + C*x + D

Koeffizienten der Polynomdivision:

H = (30*A^2 + 20*B)*C + A^5 + 20*A^3*B + 30*A*B^2 + 20*A*D + w - 5*v*A + 10*u*A^2 + 5*u*B

I = 10*C^2 + (20*A^3 + 60*A*B + 5*u)*C + 5*A^4*B - 8*u*A^3 + 30*A^2*B^2 + 18*v*A^2 + 11*v*B + 5*u*A*B + 10*B^3 + u^2

J = 30*A*C^2 + (5*A^4 + 60*A^2*B + 30*B^2 + 20*D + 20*u*A - 5*v)*C + 10*A^3*B^2 + 20*A*B^3 + 20*A^3*D + 40*A*B*D + 5*u*A^4 + 12*u*A^2*B + 7*u*B^2 + 2*u^2*A + 2*u*v - 2*v*A^3

K = (30*A^2 + 30*B)*C^2 + (20*A^3*B + 60*A*B^2 + 10*A*D + 35*u*B + 30*w)*C + 10*A^2*B^3 + 5*B^4 - 20*B^2*D + 20*D^2 - 2*u*A^5 - 22*u*A^3*B + u*w - u*v*A - 8*u^2*A^2 + 2*u^2*B + 7*v*A^4 + 18*v*A^2*B + 38*v*B^2 + 26*u*v*A - 11*v^2

L = 10*C^3 + (10*A^3 + 60*A*B + 10*u)*C^2 + (30*A^2*B^2 + 20*B^3 - 16*u*A^3 + 12*u*A*B + 5*u^2 + 34*v*A^2 + 28*v*B)*C + 5*A*B^4 + 10*A*B^2*D + 20*A*D^2 - 10*u*A^4*B - 8*u^2*A^3 + 11*u*v*B + 5*u^2*A*B + 4*u*B^3 + u^3 + 3*v*A^5 - 4*v*A^3*B + 30*v*A*B^2 + 2*v*w - 7*v^2*A + 14*u*v*A^2 - 5*u*v*B + 8*u*w*A

Nächste Stufe:

M = 20*A*C^3 + (30*A^2*B + 30*B^2 + 30*D)*C^2 + (20*A*B^3 + 20*A^3*D + 120*A*B*D)*C + B^5 + 30*A^2*D^2 + 30*A^2*B^2*D + 30*B*D^2 + 20*B^3*D + u*J - v*I + w*H

N = (10*A^2 + 20*B)*C^3 + (30*A*B^2 + 60*A*D)*C^2 + (60*A^2*B*D + 5*B^4 + 60*B^2*D + 30*D^2)*C + 10*A^3*D^2 + 60*A*B*D^2 + 20*A*B^3*D + u*K - v*J + w*I

O = 5*C^4 + 20*A*B*C^3 + (10*B^3 + 30*A^2*D + 60*B*D)*C^2 + (60*A*D^2 + 60*A*B^2*D)*C + 5*B^4*D + 30*A^2*B*D^2 + 30*B^2*D^2 + 10*D^3 + u*L - v*K + w*J

P = 5*A*C^4 + (10*B^2 + 20*D)*C^3 + 60*A*B*D*C^2 + (30*A^2*D^2 + 20*B^3*D + 60*B*D^2)*C + 20*A*D^3 + 30*A*B^2*D^2 + u*M - v*L + w*K

Koeffizienten der BJ Form:

S = -5*D*C^4 - 30*B*D^2*C^2 - 20*A*D^3*C - 5*D^4 - 10*B^2*D^3 - u*P + v*O - w*N T = D^5 + D*S + w*C^5 + 20*w*B*D*C^3 + 30*w*A*D^2*C^2 + (30*w*B^2*D^2 + 20*w*D^3)*C + 20*w*A*B*D^3 + u*w*O - v*w*N + w^2*M

Finale BJ Form der Transformation:

y^5 + S*y - T = 0

Akkurate Beispiele

Sehr bekanntes Beispiel:

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

Ausführung:

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

A = 401/433-3/433*sqrt(8870) = 0.273574983207144356854525060669

B = 187/433+4/433*sqrt(8870) = 1.301900022390474190860633252441

D = 2935/2165-9/2165*sqrt(8870) = 0.964144989924286614112715036401

C^3 + 1/433*(31*sqrt(8870)-2556)*C^2 - 1/187489*(31536*sqrt(8870)-125194)*C + 1/405913685*(44948782*sqrt(8870)-957483960) = 0

C = -1/1299*(31*sqrt(8870)-2556)+4/1299*sqrt(1774*(2069-9*sqrt(8870)))*cos(pi/6+1/6*arccos(324/34295*sqrt(8870)-15557/34295))

Berechnung der Koeffizienten:

H = 1/187489*(8838350-37540*sqrt(8870))*C + 1/15220870177393*(167857361941334+172425173618*sqrt(8870)) = 101.3838713311896890345642481862075568

I = 10*C^2 + 1/81182737*(1987271195-6636840*sqrt(8870))*C + 1/15220870177393*(889150922913648-2821275699115*sqrt(8870)) = 193.594880704790500968678478763099

J = 30/433*(401-3*sqrt(8870))*C^2 + 6/35152125121*(103116842455 + 2981785018*sqrt(8870))*C + 1/15220870177393*(608960088566976-4527197721838*sqrt(8870)) = 300.679179457897858384379535017224255

K = 60/187489*(160801-337*sqrt(8870))*C^2 + 1/35152125121*(1266401442975-7688909632*sqrt(8870))*C + 1/15220870177393*(341190567194247+4761774728111*sqrt(8870)) = 512.210447538549162263478951574907969

L = 10*C^3 + 50/81182737*(204603*sqrt(8870)-476188)*C^2 + 1/35152125121*(4382859911345-13695188068*sqrt(8870))*C + 3/15220870177393*(98464670363197-603965777854 sqrt(8870)) = 715.96498158419426770083994400225145

Nächste Stufe:

M = 20/433*(401-3*sqrt(8870))*C^3 + 1/81182737*(1321142870+41560704*sqrt(8870))*C^2 - 235/35152125121*(43119716*sqrt(8870)-4008069195)*C + 1/15220870177393*(941959891405511-2326092019423*sqrt(8870)) = 861.23875702962665069670109129185022

N = 10/187489*(402573+1058*sqrt(8870))*C^3 - 2/81182737*(18398502*sqrt(8870)-1336093445)*C^2 + 1/35152125121*(1495117291041+14640472168*sqrt(8870))*C + 1/15220870177393*(94696065336331+413515071290*sqrt(8870)) = 1012.27201768955591189993678152152316

O = 5*C^4 - 10/187489*(250395-2086*sqrt(8870))*C^3 + 1/81182737*(5545805010-11583382*sqrt(8870))*C^2 + 1/35152125121*(1349332968154-12759216412*sqrt(8870))*C + 1/76104350886965*(1537655855716955+3645674595942*sqrt(8870)) = 991.00371045848718460319041741076179

P = 5/433*(401-3*sqrt(8870))*C^4 + 8/187489*(188095+3169*sqrt(8870))*C^3 - 210/81182737*(-3682617+97213*sqrt(8870))*C^2 + 1/35152125121*(551425315187+10956051652*sqrt(8870))*C + 1/76104350886965*(3661529518488315-27543847813482*sqrt(8870)) = 804.18937869134555228873399337361968

Geschliffene Koeffizienten der Endform:

S = 7096/187489*(2069-9*sqrt(8870))*C^2 - 113536/405913685*(704810+287*sqrt(8870))*C + 14192/175760625605*(1429371257-2976793*sqrt(8870))

T = -14192/405913685*(212409*sqrt(8870)-18644740)*C^2 + 56768/175760625605*(34261965*sqrt(8870)-2500947946)*C - 28384/1902608772174125*(596854977381*sqrt(8870)-49163189125600)

Formale Plottung:

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

C^3 + 1/433*(31*sqrt(8870)-2556)*C^2 - 1/187489*(31536*sqrt(8870)-125194)*C + 1/405913685*(44948782*sqrt(8870)-957483960) = 0

C = -1/1299*(31*sqrt(8870)-2556)+4/1299*sqrt(1774*(2069-9*sqrt(8870)))*cos(pi/6+1/6*arccos(324/34295*sqrt(8870)-15557/34295))

y = x^4 + 1/433*(401-3*sqrt(8870))*x^3 + 1/433*(187+4*sqrt(8870))*x^2 + C*x + 1/2165*(2935-9*sqrt(8870))

y^5 + (7096/187489*(2069-9*sqrt(8870))*C^2 - 113536/405913685*(704810+287*sqrt(8870))*C + 14192/175760625605*(1429371257-2976793*sqrt(8870)))*y = -14192/405913685*(212409*sqrt(8870)-18644740)*C^2 + 56768/175760625605*(34261965*sqrt(8870)-2500947946)*C - 28384/1902608772174125*(596854977381*sqrt(8870)-49163189125600)

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

$y=x^{4}+{\frac {1}{433}}{\bigl (}401-3{\sqrt {8870}}{\bigr )}x^{3}+{\frac {1}{433}}{\bigl (}187+4{\sqrt {8870}}{\bigr )}x^{2}+$

$+{\biggl \{}{\frac {2}{41135}}{\sqrt {13012290{\sqrt {8870}}+334673970}}\cot {\bigl [}{\frac {\pi }{6}}+{\frac {1}{3}}\arctan {\bigl (}{\frac {1}{261}}{\sqrt {2610{\sqrt {8870}}-219066}}{\bigr )}{\bigr ]}+{\frac {1}{41135}}{\bigl (}27720+109{\sqrt {8870}}{\bigr )}{\biggr \}}x+$

$+{\frac {1}{2165}}{\bigl (}2935-9{\sqrt {8870}}{\bigr )}$

5*T/4/S = -(sqrt(8870)-82)/116*C - (196835+362*sqrt(8870))/125570

(5*T)^4/(4*S)^4*5/S =

= 1.19663220091271886926959406850172389677940858130431171845966625632465288719

Und aus den beiden anderen C_Lösungen:

= 8.65720164291023635432189644571896445631493417442280005262094668080634354873

= -0.00862520973605138819561746314908476390825838712155632909514232557414066630

1/S = (625(251909366410195+1924066033827sqrt(8870)))/161430602116240136192*((-514460 + 17856 sqrt(8870))/205675*C^2 + (31042913776 - 296153536 sqrt(8870))/445286375*C + (-105571339363640 + 247500682448 sqrt(8870))/964045001875)

Elliptischer Modul für diese Gleichung:

27571469241561972736 k^24 - 165428815449371836416 k^22 + 915305857698257235456 k^20 - 3060098480205377676800 k^18 - 692208622059345227885 k^16 + 19309708399526556771764 k^14 - 32642128148263000504974 k^12 + 19309708399526556771764 k^10 - 692208622059345227885 k^8 - 3060098480205377676800 k^6 + 915305857698257235456 k^4 - 165428815449371836416 k^2 + 27571469241561972736 = 0

k^24 - 6*k^22 + 29052602129/7876281216*k^20 - 291390542975/2625427072*k^18 - 33747920316865/1344218660864*k^16 + 235356249417409/336054665216*k^14 - 795716713584963/672109330432*k^12 + 235356249417409/336054665216*k^10 - 33747920316865/1344218660864*k^8 - 291390542975/2625427072*k^6 + 29052602129/7876281216*k^4 - 6*k^2 + 1 = 0

k^12+1-3*(k^10+k^2)-1/5250854144*(983008525*sqrt(8870)-63528962739)*(k^8+k^4)+1/5250854144*(1966017050*sqrt(8870)-100803654758)*k^6=0

k = sqrt(5/26912*sqrt(1/87*(39320341*sqrt(8870)-1280953515))*cos(1/3*arcsin(24 sqrt(3/145 (1280953515 + 39320341 sqrt(8870)))/318565)+pi/6) + 3/4) - sqrt(5/26912*sqrt(1/87*(39320341*sqrt(8870)-1280953515))*cos(1/3*arcsin(24 sqrt(3/145 (1280953515 + 39320341 sqrt(8870)))/318565)+pi/6) - 1/4)

k = sqrt(1/2-sqrt(3)/2*tan(1/6*arccos((135891098496*sqrt(8870)-10288155239785)/14715130587625)))

$k={\sqrt {{\tfrac {1}{2}}-{\tfrac {1}{2}}{\sqrt {3}}\tan {\bigl [}{\tfrac {1}{6}}\arccos {\bigl (}{\tfrac {135891098496}{14715130587625}}{\sqrt {8870}}-{\tfrac {10288155239785}{14715130587625}}{\bigr )}{\bigr ]}}}$

k = 0.54247018190725764574710293153684482480279268801231501035208038756875266423

s^3 + 25/336054665216*(39320341*sqrt(8870)-1280953515)*(-4*s+1)=0 for s=1/4+(1-k^2)^2/(4*k^2)

k^4*(-k^2+1)^2/(k^4-k^2+1)^3 = 256/14715130587625*(39320341*sqrt(8870)+1280953515)

k = tan(1/2*arccot(sqrt(sqrt(1/87 (-67467103347 + 983008525 sqrt(8870)))/53824*cot(1/6*arccos(324/34295*sqrt(8870)-15557/34295) + arctan(sqrt(174 (447 sqrt(8870) - 42019))/1131)+pi/6) + 1/8)))

k = tan(1/2*arccot(sqrt(sqrt(1/87 (-67467103347 + 983008525 sqrt(8870)))/53824*cot(arccot((-193366 - 5544 sqrt(8870) + 8227 sqrt(8870) C)/(3548 sqrt(37731 + 1467 sqrt(8870)))) + arctan(sqrt(174 (447 sqrt(8870) - 42019))/1131)) + 1/8)))

Complete elliptic solutions of principal quintic equations

Ermittlung des elliptischen Moduls

x^5 - u*x^2 + v*x - w = 0

Gleichungssystem für Transformation nach Adamchik und Jeffrey:

4*v*A - 3*u*B = 5*w

3*u^2*A^2 - 3*u*v*A + 25*w*A*B - 10*v*B^2 = 5*u*w - 2*v^2

Faktor der Linearkombination und zugleich Koeffizient der Brioschi Transformation:

P = ((27*u^4-160*v^3+300*u*v*w)*A-30*u^3*v+125*v^2*w)/(12*u^4-60*v^3+300*u*v*w)

Modulgleichung:

k^4*(-k^2+1)^2/(k^4-k^2+1)^3 = 256*u^2*(-4*v^2*P+20*u*w*P+5*v*w)/5/(15*w-16*u*P^2-12*v*P)^3

Erste durchgerechnete Gleichung

x^5 + 5*x^2 + 5*x - 6 = 0

27*k^4*(-k^2+1)^2/4/(k^4-k^2+1)^3 = (81/88)^2

k^6 - 1/54*(81+13*sqrt(21))*k^4 - 1/54*(81-13*sqrt(21))*k^2 + 1 = 0

k = sqrt(22/27*sin(1/3*arcsin(81/88))+3/4) - sqrt(22/27*sin(1/3*arcsin(81/88))-1/4)

k = cos(1/2*arcsin(sqrt(-729/239+351/239*sqrt(7)*tan(pi/3-1/3*arctan(622451/(255879 sqrt(7)))))))

k = sqrt(1/2+sqrt(3)/2*tan(1/3*arccos(81/88)))

k = 0.78524904412575068055871086982322788011467550335266512328053194380627812753

q = 0.05963619515075573277408024987905183643481517489470072405655924580865606967696

r = (2*theta(4,0,q^5)*theta(4,0,q^(1/5))-2*theta(4,0,q)^2)/(theta(4,0,q^(1/5))^2+5*theta(4,0,q^5)^2-4*theta(4,0,q)^2)

r = -0.7537978572717233955184462870828709259903045236185502763355728696109981879399296

z = r*(5*r^2-10*r+4)/(5*r^2-6*r+2)

z = -1.1575253093051875679824826927111080840076277720616441474343690050798733749551001822

x^4 + 2*x^2 + (3/2*sqrt(7)*cot(pi/6+1/3*arctan(sqrt(7)/9))+7/2)*x + 4 = (-5/4*sqrt(7)*cot(pi/6+1/3*arctan(1/9*sqrt(7)))-21/4)*z

x = 0.6910042313638778060039075030929807297194682102515707505178986679338019762

-72/91*(x^2-x+1)*(x+1)*(x-2) = z_1*z_2*z_3 = J

J = -72/91*x^4+144/91*x^3-72/91*x+144/91

J^2 = 88128/8281*x^4+186624/8281*x^3-57024/8281*x^2+767232/8281*x-311040/8281

J^3 = 234772992/753571*x^4-51881472/107653*x^3-36205056/57967*x^2+475891200/753571*x-95551488/753571

J^4 = -210637283328/68574961*x^4-126092132352/9796423*x^3+152804745216/68574961*x^2-823253704704/68574961*x+386983526400/68574961

Zweite durchgerechnete Gleichung

x^5 + 5*x^2 + 15*x - 6 = 0

27*k^4*(-k^2+1)^2/4/(k^4-k^2+1)^3 = (121*sqrt(77)/1176)^2

k = sqrt(126/121*sqrt(7/11)*sin(1/3*arcsin(121/168*sqrt(11/7))) + 3/4) - sqrt(126/121*sqrt(7/11)*sin(1/3*arcsin(121/168*sqrt(11/7))) - 1/4)

k = sqrt(1/2+sqrt(3)/2*tan(1/6*arccos(62267/98784)))

k = 0.79324374715564562455975157401671487681029101055142340366669515375968873721

q = 0.06169756600290766333368694743111437895094390597455230746707832430693247936189

r = (2*theta(4,0,q^5)*theta(4,0,q^(1/5))-2*theta(4,0,q)^2)/(theta(4,0,q^(1/5))^2+5*theta(4,0,q^5)^2-4*theta(4,0,q)^2)

r = -0.7379510313595900453447513829069863326367736667789303782693889087626802773440845

z = r*(5*r^2-10*r+4)/(5*r^2-6*r+2)

z = -1.1372912946877257755735674815125358327292192008579178352440233054521507354547503493

x^4 + 2*x^2 + (1/2*sqrt(143)*cot(pi/6+1/3*arctan(sqrt(13/11)))+7/2)*x + 12 = (-15/44*sqrt(143)*cot(1/6*pi+1/3*arctan(1/11*sqrt(143)))-39/4)*z

x = 0.35710487791136466208544001714283479964534181667612357460253736

-968/6201*(x^4-6*x^3-7*x-6) = z_1*z_2*z_3

Durchgerechnete Gleichung

x^5 + 10*x^2 + 15*x - 8 = 0

27*k^4*(-k^2+1)^2/4/(k^4-k^2+1)^3 = (3*sqrt(3)/14)^2

k^6 - 8 k^4 + 5 k^2 + 1 = 0

k = sqrt(9/8 - 13/24*sqrt(3)*tan(1/3*arctan(3/13*sqrt(3)))) - sqrt(1/8 - 13/24*sqrt(3)*tan(1/3*arctan(3/13*sqrt(3))))

k = sqrt(1/2+sqrt(3)/2*tan(1/6*arccos(-71/98)))

k = 0.92894384139046603714008382550030816096911667085237559011889968740168890560

q = 0.12176712871145015919097975166617402449922768147067164447123971571061305711146

r = (2*theta(4,0,q^5)*theta(4,0,q^(1/5))-2*theta(4,0,q)^2)/(theta(4,0,q^(1/5))^2+5*theta(4,0,q^5)^2-4*theta(4,0,q)^2)

r = -0.411594398344631625656328819791877387680121287087125351985841051334964768977295140

z = r*(5*r^2-10*r+4)/(5*r^2-6*r+2)

z = -0.693884590811116871540061610905190124132044403661222800482473272561073155056294093361

x^4 - 1/3*x^3 + 2*x^2 + (13/2+17/2/sqrt(3)*cot(1/3*arctan(1/3/sqrt(3))+pi/6))*x + 10 = (-17/2-17/sqrt(3)*cot(1/3*arctan(1/3/sqrt(3))+pi/6))*z

x = 0.416723308560700689327073547686522954734739750099217890859011012

-8/221 (x^4 - 6 x^3 + 2 x^2 - 2 x - 24) = z_1*z_2*z_3

Für die Bring Jerrard Form selbst

x^5 + 5*x - 4*f = 0

(4 k^12 - 12 k^10 - 512 k^8 f^8 - 414 k^8 f^4 - 512 k^8 f^6 sqrt(f^4 + 1) - 162 k^8 f^2 sqrt(f^4 + 1) - 3 k^8 + 1024 k^6 f^8 + 828 k^6 f^4 + 1024 k^6 f^6 sqrt(f^4 + 1) + 324 k^6 f^2 sqrt(f^4 + 1) + 26 k^6 - 512 k^4 f^8 - 414 k^4 f^4 - 512 k^4 f^6 sqrt(f^4 + 1) - 162 k^4 f^2 sqrt(f^4 + 1) - 3 k^4 - 12 k^2 + 4) = 0

((1-k^2)^2/(4*k^2)+1/4)^3 + (-4*((1-k^2)^2/(4*k^2)+1/4)+1)*1/256*(sqrt(f^4+1)+f^2)*(3*sqrt(f^4+1)+5*f^2)^3 = 0

k^2 = 1/2 + 1/2*f*sqrt(2*sqrt(f^4+1)-2*f^2)

k = (sqrt(sqrt(f^4+1)+1)+f)/sqrt(2*f^2+2+2*sqrt(f^4+1))

Elliptische Lösung einer Brioschi Gleichung

729 t^4 x^5 + 17280 t^2 x^3 + 184320 x - 262144 = 0

q = nome(1/2+1/2*sqrt(3)*tan(1/3*arctan(t)))

x = z_1*z_2*z_3

Von Bring Jerrard nach Brioschi

1 wird zu 13 - sqrt(97)

2 wird zu 208 - 8*sqrt(673)

1 wird zu 4/27*(11-sqrt(97))

2 wird zu 64/302589*(417-16*sqrt(673))

Complete calculation examples of Brioschi form

Principal core for the Brioschi pattern

5184*t^4*(t^2+1)*x^5 + (61440*F^3*t^2+69120*F^2*G*t^2-25920*F*G^2*t^4-3240*G^3*t^4)*x^2 + (184320*F^4-184320*F^3*g*t^2-51840*F^2*G^2*t^2-1215*G^4*t^4)*x -262144*F^5-184320*F^4*G-17280*F^2*G^3*t^2-729*G^5*t^4 = 0

729 t^4 B^5 + 17280 t^2 B^3 + 184320 B - 262144 = 0

x = (F*B+G)/(9/64*t^2*B^2+1)

Example:

x^5-u*x^2+v*x-w = 0

5/216*(512*F^3+576*F^2*G-216*F*G^2*t^2-27*G^3*t^2)/(t^2*(t^2+1)) = -u, 5/576*(4096*F^4-4096*F^3*G*t^2-1152*F^2*G^2*t^2-27*G^4*t^4)/(t^4*(t^2+1)) = v, (-262144*F^5-184320*F^4*G-17280*F^2*G^3*t^2-729*G^5*t^4)/(5184*t^4*(t^2+1))= -w

Example one

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

Elliptic nome:

$Q=q{\biggl \{}{\sqrt {{\tfrac {1}{2}}+{\tfrac {1}{2}}{\sqrt {3}}\tan {\bigl [}{\tfrac {1}{3}}\arctan {\bigl (}232^{-1}87^{-3/2}{\sqrt {983008525{\sqrt {8870}}-67467103347}}\,{\bigr )}{\bigr ]}}}\,{\biggr \}}$

Q = nome(1/2+1/2*sqrt(3)*tan(1/3*arctan(1/20184/sqrt(87)*sqrt(983008525*sqrt(8870)-67467103347))))

Q = 0.07586749185993868872658996634213628591559732724980749332173953859677738814

Jacobi Theta quotients:

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

y_{1} = (2*theta(3,0,Q^5)*theta(3,0,Q^(1/5))-2*theta(3,0,Q)^2)/(theta(3,0,Q^(1/5))^2+5*theta(3,0,Q^5)^2-4*theta(3,0,Q)^2)

y_{1} = 0.39463033878598007417484886271155487690287940131732677152451778251763441400428

$y_{2}={\frac {2\,\vartheta _{01}(Q^{5})\vartheta _{01}(Q^{1/5})-2\,\vartheta _{01}(Q)^{2}}{\vartheta _{01}(Q^{1/5})^{2}+5\,\vartheta _{01}(Q^{5})^{2}-4\,\vartheta _{01}(Q)^{2}}}$

y_{2} = (2*theta(4,0,Q^5)*theta(4,0,Q^(1/5))-2*theta(4,0,Q)^2)/(theta(4,0,Q^(1/5))^2+5*theta(4,0,Q^5)^2-4*theta(4,0,Q)^2)

y_{2} = -0.6390903532451355805215977950076780892310542613901676362469483683066183699121

$y_{3}={\frac {2\,\vartheta _{10}(Q^{5})\vartheta _{10}(Q^{1/5})-2\,\vartheta _{10}(Q)^{2}}{\vartheta _{10}(Q^{1/5})^{2}+5\,\vartheta _{10}(Q^{5})^{2}-4\,\vartheta _{10}(Q)^{2}}}$

y_{3} = (2*theta(2,0,Q^5)*theta(2,0,Q^(1/5))-2*theta(2,0,Q)^2)/(theta(2,0,Q^(1/5))^2+5*theta(2,0,Q^5)^2-4*theta(2,0,Q)^2)

y_{3} = -1.1026004408847745822765893660162869083332240051142820573778577815799612390581

Solutions of the Bring Jerrard Resolvent:

$T_{1R}={\frac {y_{1}(5y_{1}^{2}-10y_{1}+4)}{5y_{1}^{2}-6y_{1}+2}}$

T_{1R} = 0.7994367254391991483858581310261327714211754323089756226626663424430879436263159

$T_{2R}={\frac {y_{2}(5y_{2}^{2}-10y_{2}+4)}{5y_{2}^{2}-6y_{2}+2}}$

T_{2R} = -1.0087778570362330213391566905666618954518908612709341007600484641827379440753913

$T_{3R}={\frac {y_{3}(5y_{3}^{2}-10y_{3}+4)}{5y_{3}^{2}-6y_{3}+2}}$

T_{3R} = -1.58361280915163376523140885984387480890846942018749698776651264254507340384697656

$Z_{R}=T_{1R}T_{2R}T_{3R}$

Product of the Bring Jerrard Resolvent solutions:

Z_{R} = 1.27711099005753252926231503966574481871544686744469187876734972998463623021

Solution of the Principal Quintic equation:

x = (1/1560896*(2110353-1312*sqrt(8870))*Z_{R} + 1/29*(sqrt(8870)-82))/(1/84013666304*(983008525*sqrt(8870)-67467103347)*Z_{R}^2+3)

x = 0.5865437103233227180461879736270421554470797452310849347869301980618

Example two

$x^{5}+7x^{2}+11x-13=0$

Elliptic nome:

$Q=q{\biggl \{}{\sqrt {{\tfrac {1}{2}}+{\tfrac {1}{2}}{\sqrt {3}}\tan {\bigl [}{\tfrac {1}{3}}\arctan {\bigl (}12328^{-1}29279^{-3/2}{\sqrt {185557765540073175{\sqrt {622236795}}-4556700826849518431926}}\,{\bigr )}{\bigr ]}}}\,{\biggr \}}$

Q = nome(1/2+1/2*sqrt(3)*tan(1/3*arctan(sqrt(185557765540073175 *sqrt(622236795)-4556700826849518431926)/360951512/sqrt(29279))))

Q = 0.04830558400896237192730619096116805228935135359425303334539975988569219173

Jacobi Theta quotients:

y_{1} = 0.39982134311061550048817431831122685600386943005807090839755001529096294586548

y_{2} = -0.8480863243019138050936751180424934566610595905874916711381911814764775596252

y_{3} = -0.9393180638947115679319567725363316979060609323559222632020003192660940109859

Real solutions of the three Bring Jerrard Resolvents:

T_{1R} = 0.7999993621330883287607480767264685282684500890716315768013908703737283357403387

T_{2R} = -1.2760950036508141853618403878223302600923340546682774211745703822520252585007827

T_{3R} = -1.3882003714166771939210418500566398739199163418066618438152930274091344656371572

Product of the real Bring Jerrard Resolvent solutions:

Z_{R} = 1.41717931645917773115949067690545239349065678829215427222394878103412663990

Solution of the Principal Quintic equation:

x = (-3/84546394558784*(1466081212*sqrt(622236795)-27266853044217)*Z_{R}+(7*sqrt(622236795)-92511)/29279)/(27/244137191601133261451264*(185557765540073175*sqrt(622236795)-4556700826849518431926)*Z_{R}^2+3)

x = 0.77462127682345336120780799796447925310324039684142983979896307723798453065

Example three

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

$Q=q{\biggl \{}{\sqrt {{\tfrac {1}{2}}+{\tfrac {1}{2}}{\sqrt {3}}\tan {\bigl [}{\tfrac {1}{3}}\arctan {\bigl (}{\tfrac {13}{186484176}}{\sqrt {20738}}\,{\sqrt {41711089375{\sqrt {5}}+52939239767}}\,{\bigr )}{\bigr ]}}}\,{\biggr \}}$

Elliptic nome:

Q = nome(1/2+1/2*sqrt(3)*tan(1/3*arctan(13/186484176*sqrt(30738)*sqrt(41711089375*sqrt(5)+52939239767))))

Q = 0.15435129542272773902811068212919926997007289214226569340193009875470755959

Jacobi Theta quotients:

y_{1} = 0.361842180501371976283486477314205029180537580697474920950159730243598479901635

y_{2} = -0.306185315146311041738146518100120099450463871346185623596978182334716008419395

y_{3} = -1.08383645070215927354848849064207614055343952801479285274890696276088020868291

Real solutions of the three Bring Jerrard Resolvents:

T_{1R} = 0.77533901720713560285116712892096048204208984402078427955408614043471928542659643

T_{2R} = -0.535493425335378498680194816489266870099811857519215224207503727325997735978190226

T_{3R} = -1.5614538338090613666791327749741763843471196958380377148894694299473263485534732

Product of the real Bring Jerrard Resolvent solutions:

Z_{R} = 0.64829837167486583929139212349332049204392781931453115078423219986621652109915905

Solution of the Principal Quintic equation:

x = (-611/165763712*(641285*sqrt(5)+672931)*Z_{R}+1/218*(705*sqrt(5)+19))/(7943/126044074385408*(41711089375*sqrt(5)+52939239767)*Z_{R}^2+3)

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:

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

It shows the Tschirnhaus Transformation in this way:

${\biggl [}{\frac {129\,z-4}{1849\,z^{2}-3}}{\biggr ]}^{5}-5{\biggl [}{\frac {129\,z-4}{1849\,z^{2}-3}}{\biggr ]}^{2}+5{\biggl [}{\frac {129\,z-4}{1849\,z^{2}-3}}{\biggr ]}-5=0$

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:

${\bigl [}{\tfrac {1}{2}}(238528425{\sqrt {15085}}-29296305919){\bigr ]}^{2}z^{5}-10{\bigl [}{\tfrac {1}{2}}(238528425{\sqrt {15085}}-29296305919){\bigr ]}z^{3}+45z-1=0$

$z\approx 0.018787897470813223818839629483292225985487352060808662222311422852295054189328531$

Clues:

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

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

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

Solution:

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

$x\approx 0.808730600479392013738554526511400064951377351559313075548116401836543340748321$

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:

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

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

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

Corresponding Brioschi Quintic equation:

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

$z\approx 0.04103471241028970064518587406207759188088410335520608376935448159$

Real solution of the given Principal equation:

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

$x\approx 1.193859111321223012009020746298031124514524269486444509602081401596$

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:

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

$z\approx 0.01995485921964894576972367249477726279242885730382331060573983953100994109904667$

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

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

$x\approx 0.58654371032332271804618797362704215544707974523108493478693019806187056334481$

Hyperbolic 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:

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

Now a special substitution shall be made:

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

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

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:

$k=\cos {\bigl \{}{\frac {1}{2}}\operatorname {arccot} {\bigl [}{\sqrt {3}}\tanh(y){\bigr ]}{\bigr \}}$

$k^{*}=\sin {\bigl \{}{\frac {1}{2}}\operatorname {arccot} {\bigl [}{\sqrt {3}}\tanh(y){\bigr ]}{\bigr \}}$

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

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

$Q=\exp {\bigl [}-\pi \,K(k^{*})\div K(k){\bigr ]}$

$Q^{*}=\exp {\bigl [}-\pi \,K(k)\div K(k^{*}){\bigr ]}$

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:

$y={\frac {1}{3}}\operatorname {artanh} {\bigl (}{\frac {1}{8}}{\bigr )}$

$y\approx 0.0418857380468176796141896217336452799429827310106$

That inserted value belongs to this Brioschi equation:

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

The corresponding modulus k has the mentioned form:

$k=\cos {\bigl \{}{\frac {1}{2}}\operatorname {arccot} {\bigl [}{\sqrt {3}}\tanh(y){\bigr ]}{\bigr \}}$

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:

$y={\frac {1}{3}}\operatorname {artanh} {\bigl (}{\frac {1}{4}}{\bigr )}$

$y\approx 0.0851376039609984472009190160506103224796851327409613783629922596$

That inserted value belongs to this Brioschi equation:

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

The corresponding modulus k has the mentioned form:

$k=\cos {\bigl \{}{\frac {1}{2}}\operatorname {arccot} {\bigl [}{\sqrt {3}}\tanh(y){\bigr ]}{\bigr \}}$

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:

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

$x\approx 0.2024449345913542228580304306303447038721951474593556763277652278785$

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:

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

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:

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

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

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

$w=(1728r^{2}-r^{2}v)/5/(ru^{2}+su+5t)-(8ru^{3}+72su^{2}+360tu)/(ru^{2}+su+5t)$

This is the broken rational key for the Brioschi form:

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

The resulting Brioschi equation looks this way:

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

Der magische Saft

x^5 - u*x^2 + v*x - w = 0

40*(-s^3*t-3*r*s^2+3*r^2*s*t+r^3)/(27*(t^2+1))=u

20*(3*s^4-8*r*s^3*t-6*r^2*s^2-r^4)/(27*(t^2+1))=v

16*(8*s^5*t+15*r*s^4+10*r^3*s^2+3*r^5)/(81*(t^2+1))=w

729*t^4*z^5+17280*t^2*z^3+184320*z-262144 = 0

x = (1/2*s*t*z+4/3*r)/(9/64*t^2*z^2+1)

Quintische Gleichungen und ihre korrespondierenden elliptischen Module

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

$k={\sqrt {{\tfrac {1}{2}}+{\tfrac {1}{2}}{\sqrt {3}}\tan {\bigl [}{\tfrac {1}{2}}\arcsin {\bigl (}{\tfrac {5364}{465595}}{\sqrt {8870}}-{\tfrac {9917}{93119}}{\bigr )}+{\tfrac {1}{6}}\arcsin {\bigl (}{\tfrac {324}{34295}}{\sqrt {8870}}-{\tfrac {15557}{34295}}{\bigr )}-{\tfrac {1}{6}}\pi {\bigr ]}}}$

k = sqrt(1/2+1/2*sqrt(3)*tan(-pi/6+1/2*arcsin(5364/465595*sqrt(8870)-9917/93119)+1/6*arcsin(324/34295*sqrt(8870)-15557/34295)))

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

$k={\sqrt {{\tfrac {1}{2}}+{\tfrac {1}{2}}{\sqrt {3}}\tan {\bigl [}{\tfrac {1}{2}}\arcsin {\bigl (}{\tfrac {6929126}{7261745}}-{\tfrac {1593723}{7261745}}{\sqrt {5}}\,{\bigr )}+{\tfrac {1}{6}}\arcsin {\bigl (}{\tfrac {1084995}{1134905}}{\sqrt {5}}-{\tfrac {1340726}{1134905}}{\bigr )}{\bigr ]}}}$

k = sqrt(1/2+1/2*sqrt(3)*tan(1/2*arcsin(6929126/7261745-1593723/7261745*sqrt(5))+1/6*arcsin((1084995*sqrt(5)-1340726)/1134905)))

$x^{5}+7x^{2}+11x-13=0$

$k={\sqrt {{\tfrac {1}{2}}+{\tfrac {1}{2}}{\sqrt {3}}\tan {\bigl [}-{\tfrac {1}{2}}\arcsin {\bigl (}{\tfrac {13510624}{317372982505}}{\sqrt {622236795}}-{\tfrac {24325799463}{63474596501}}{\bigr )}-{\tfrac {1}{6}}\arcsin {\bigl (}{\tfrac {1524096}{31606815245}}{\sqrt {622236795}}-{\tfrac {19377815567}{31606815245}}{\bigr )}+{\tfrac {1}{6}}\pi {\bigr ]}}}$

k = sqrt(1/2+1/2*sqrt(3)*tan(-1/2*arcsin(13510624/317372982505*sqrt(622236795)-24325799463/63474596501)-1/6*arcsin(1524096/31606815245*sqrt(622236795) - 19377815567/31606815245)+pi/6))