International Journal of Pure and Applied Mathematics ————————————————————————– Volume 18 No. 2 2005, 213-222 NEW WAY FOR A TWO-PARAMETER CANONICAL FORM OF SEXTIC EQUATIONS AND ITS SOLVABLE CASES Yoshihiro Mochimaru Department of International Development Engineering Graduate School of Science and Engineering Tokyo Institute of Technology 2-12-1, O-okayama, Meguru-ku, Tokyo 152-8550, JAPAN e-mail: [email protected] Abstract: A new way for a two-parameter canonical form of sextic equations is given, through a Tschirnhausian transform by multiple steps, supplemented with some solvable cases at most in terms of hypergeometric functions. AMS Subject Classification: 11D41 Key Words: sextic equation, Tschirnhausian transform 1. Introduction 6 6 6−n Actually the general sextic equation x + anx = 0 can be solved in nX=1 terms of Kamp´ede F´eriet functions, and it can be reduced to a three-parameter Received: October 21, 2004 c 2005, Academic Publications Ltd. 214 Y. Mochimaru 6 4 2 resolvent of the form y + ay + by + cy + c = 0 by Joubert [3]. Such a type of the sextic equation was treated by Felix Klein and a part was given in [4]. Other contributions to treatment on sextic equations were given e.g. by Cole [2] and Coble [1]. In this paper, a new way leading to a two-parameter canonical form of sextic equations is given. 2. Analysis 2.1. General Let the original given sextic equation be expressed as 6 ∗ 5 ∗ 4 ∗ 3 ∗ 2 ∗ ∗ x + a x + b x + c x + d x + e x + f = 0. (1) 3 Here the quantity 5a∗ − 18a∗b∗ + 27c∗ corresponding to the polynomial of the left hand side of equation (1) is invariant with respect to the parallel shift x → x+constant. 3 2.2. In Case of 5a∗ − 18a∗b∗ + 27c∗ = 0 Let x = y − a∗/6, then equation (1) becomes 6 5 ∗2 ∗ 4 25 ∗4 1 ∗2 ∗ ∗ 2 y + − a + b y + a − a b + d y 12 432 6 1 ∗5 1 ∗3 ∗ 1 ∗ ∗ ∗ + − a + a b − a d + e y 81 27 3 35 ∗6 1 ∗4 ∗ 1 ∗2 ∗ 1 ∗ ∗ ∗ + a − a b + a d − a e + f = 0. (2) 46656 432 36 6 3 2.3. In Case of 5a∗ − 18a∗b∗ + 27c∗ = 0 Now apply a Tschirnhausian transform of the following type: 2 −y = p + qx + x . (3) 2 Eliminating x from equation (3) and equation (1) with p = −(a∗ −2b∗ −a∗q)/6, using Sylvester’s resultant, gives 6 4 3 2 y + b0y + c0y + d0y + e0y + f0 = 0, (4) NEW WAY FOR A TWO-PARAMETER CANONICAL... 215 2 b0 ≡ 15p + 5py5 + y4, (5) 3 2 5 ∗3 2 ∗ ∗ ∗ 3 c0 ≡ 20p + 10p y5 + 4py4 + y3 = − a + a b − c q 27 3 5 ∗4 22 ∗2 ∗ 4 ∗2 ∗ ∗ ∗ 2 + a − a b + b + 3a c − 4d q 9 9 3 5 ∗5 26 ∗3 ∗ 26 ∗ ∗2 10 ∗2 ∗ ∗ ∗ 13 ∗ ∗ ∗ + − a + a b − a b − a c + 3b c + a d − 5e q 9 9 9 3 3 5 ∗6 10 ∗4 ∗ 14 ∗2 ∗2 4 ∗3 4 ∗3 ∗ 8 ∗ ∗ ∗ + a − a b + a b − b + a c − a b c 27 9 9 27 3 3 ∗2 4 ∗2 ∗ 2 ∗ ∗ ∗ ∗ ∗ + c − a d + b d + 2a e − 2f , (6) 3 3 4 3 2 d0 ≡ 15p + 10p y5 + 6p y4 + 3py3 + y2, (7) 5 4 3 2 e0 ≡ 6p + 5p y5 + 4p y4 + 3p y3 + 2py2 + y1, (8) 6 5 4 3 2 f0 ≡ p + p y5 + p y4 + p y3 + p y2 + py1 + y0, (9) ∗ ∗2 ∗ y5 ≡−a q + a − 2b , (10) ∗ 2 ∗ ∗ ∗ ∗ ∗ ∗2 ∗ y4 ≡b q + (−a b + 3c ) q − 2a c + b + 2d , (11) ∗ 3 ∗ ∗ ∗ 2 ∗ ∗ ∗ ∗ ∗ y3 ≡−c q + (a c − 4d ) q + (3a d − b c − 5e ) q ∗ ∗ ∗ ∗ ∗2 ∗ +2a e − 2b d + c − 2f , (12) ∗ 4 ∗ ∗ ∗ 3 ∗ ∗ ∗ ∗ ∗ 2 y2 ≡d q + (−a d + 5e ) q + (−4a e + b d + 9f ) q ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗2 + (−5a f + 3b e − c d ) q + 2b f − 2c e + d , (13) ∗ 5 ∗ ∗ ∗ 4 ∗ ∗ ∗ ∗ 3 y1 ≡−e q + (a e − 6f ) q + (5a f − b e ) q ∗ ∗ ∗ ∗ 2 ∗ ∗ ∗ ∗ ∗ ∗ ∗2 + (−4b f + c e ) q + (3c f − d e ) q − 2d f + e , (14) ∗ 6 ∗ ∗ 5 ∗ ∗ 4 ∗ ∗ 3 ∗ ∗ 2 ∗ ∗ ∗2 y0 ≡f q − a f q + b f q − c f q + d f q − e f q + f . (15) Thus if q is to be one of the roots of the cubic equation c0 = 0, then equation (1) becomes 6 4 2 y + b0y + d0y + e0y + f0 = 0. (16) Therefore, now consider a sextic equation of the form: 6 ∗ 4 ∗ 2 ∗ ∗ y + B y + D y + E y + F = 0. (17) 216 Y. Mochimaru 2.4. In Case of B∗ = 0 Apply a Tschirnhausian transform of the type: 2 3 4 −z = p + qy + ry + sy + y . (18) Eliminating y from equation (18) and equation (17), using Sylvester’s resultant, gives 6 n (p + z) Gn (q,r,s)=0 with G6 = 1, (19) Xn=0 ∗ ∗2 ∗ G5 ≡−2B r + 2B − 4D , (20) ∗ 2 ∗2 ∗ ∗2 ∗ 2 ∗ G4 ≡ B q + −2B + 4D qs + B + 2D r + 5E rs ∗3 ∗ ∗ ∗ 2 ∗ ∗3 ∗ ∗ ∗ + B − 3B D + 3F s + 5E q + −2B + 2B D + 6F r ∗ ∗ ∗4 ∗2 ∗ ∗2 ∗ ∗ − 7B E s + B − 4B D + 6D − 4B F , (21) ∗ 2 ∗ 2 ∗ 2 ∗ ∗ ∗ G3 ≡−4D q r − 5E q s − 5E qr + (4B D − 12F ) qrs ∗ ∗ 2 ∗ ∗ ∗ 3 ∗ ∗ 2 + 7B E qs + (−2B D − 2F ) r − 3B E r s ∗2 ∗ ∗ ∗ ∗2 2 ∗2 ∗ ∗ ∗ 3 + −2B D + 2B F + 4D rs + −3B E + 3D E s ∗ ∗ ∗ 2 ∗ ∗ ∗2 ∗ ∗ + (2B D − 6F ) q + 4B E qr + −8D + 16B F qs ∗2 ∗ ∗2 ∗ ∗ 2 ∗2 ∗ ∗ ∗ + 4B D − 4D − 4B F r + 6B E − 2D E rs ∗2 ∗ ∗2 ∗2 ∗ ∗ ∗ 2 ∗2 ∗ ∗ ∗ + 5E + 2B D − 4B F − 2D F s + B E − 11D E q ∗3 ∗ ∗ ∗2 ∗2 ∗ ∗ ∗ ∗2 + −2B D + 2B D + 10B F − 14D F + 5E r ∗3 ∗ ∗ ∗ ∗ ∗ ∗ + −3B E + 6B D E + 11E F s ∗3 ∗ ∗2 ∗2 ∗ ∗ ∗ ∗ ∗2 ∗3 ∗2 − 4B F + 2B D + 8B D F − 4B E − 4D + 2F . (22) G2, G1, G0 are polynomials of degree 4, 5, and 6 with respect to the combination of q, r, and s respectively, and are not shown here. Equation (19) can be rearranged as a polynomial of z: 5 6 n z + Hnz = 0, (23) nX=0 NEW WAY FOR A TWO-PARAMETER CANONICAL... 217 where H5 = 6p + G5, (24) 2 H4 = 15p + 5pG5 + G4, (25) 3 2 H3 = 20p + 10p G5 + 4pG4 + G3, (26) 4 3 2 H2 = 15p + 10p G5 + 6p G4 + 3pG3 + G2, (27) 5 4 3 2 H1 = 6p + 5p G5 + 4p G4 + 3p G3 + 2pG2 + G1, (28) 6 5 4 3 2 H0 = p + p G5 + p G4 + p G3 + p G2 + pG1 + G0. (29) Since H5 is linear in p,q,r,s, and H4 is of second degree, and H3 is of third degree, one way of determing p,q,r,s to satisfy H3 = H4 = H5 = 0 is as follows: at first p,q,r, and s are selected with at least one free parameter so as to satisfy H5 = H4 = 0, then the parameter is determined from the restriction H3 = 0. That is, let q0 be one of the roots of the quadratic equation: ∗ 2 ∗ 2 ∗4 8 ∗2 ∗ ∗ ∗ 2 ∗2 B q0 + 5E q0 − B + B D − 4B F − D = 0. (30) 3 3 3 Using q0, the quantities q1, r1,s1 are defined as ∗ ∗ q1 ≡ 2B q0 + 5E , (31) 4 ∗3 14 ∗ ∗ ∗ r1 ≡ B − B D + 6F , (32) 3 3 ∗2 ∗ ∗ ∗ s1 ≡ −2B + 4D q0 − 7B E . (33) In case of q1 = 0, i.e. ∗2 ∗ 2 ∗4 8 ∗2 ∗ ∗ ∗ 2 ∗2 25E + 4B B − B D + 4B F + D = 0, 3 3 3 q ≡ q0 − (r1r + s1s)/ q1, (34) r ≡ λϕ0, (35) s ≡ µϕ0, (36) where λ, µ are eigenvalues corresponding to the homogeneous part (second de- gree) of H4 = 0. It is necessary only to get the ratio λ/µ or µ/λ through the homogeneous quadratic equation. That is, 2 ∗2 ∗ 16 ∗7 112 ∗5 ∗ 196 ∗3 ∗2 ∗4 ∗ − B + 2D + B − B D + B D + 16B F 3 9 9 9 218 Y. Mochimaru ∗2 ∗ ∗ ∗ ∗2 2 2 − 56B D F + 36B F q1 λ . ∗ 16 ∗6 88 ∗4 ∗ 112 ∗2 ∗2 ∗3 ∗ ∗ ∗ ∗ + 5E + − B + B D − B D −24B F +48B D F 3 3 3 2 56 ∗5 ∗ 196 ∗3 ∗ ∗ ∗2 ∗ ∗ 2 ×q0/ q1 + − B E + B D E − 84B E F q1 3 3 8 ∗5 44 ∗3 ∗ 56 ∗ ∗2 ∗2 ∗ ∗ ∗ + B − B D + B D + 12B F − 24D F q1 λµ 3 3 3 ∗3 ∗ ∗ ∗ ∗5 ∗3 ∗ ∗ ∗2 2 2 + B − 3B D + 3F + 4B − 16B D + 16B D q0 /q1 ∗4 ∗ ∗2 ∗ ∗ 2 ∗3 ∗2 2 + 28B E − 56B D E q0/q1 + 49B E /q1 ∗4 ∗2 ∗ ∗2 + −4B + 16B D − 16D q0/q1 ∗3 ∗ ∗ ∗ ∗ 2 + −14B E + 28B D E /q1 µ = 0 . (37) In case of q1 = 0, i.e. ∗2 ∗ 2 ∗4 8 ∗2 ∗ ∗ ∗ 2 ∗2 25E + 4B B − B D + 4B F + D = 0, 3 3 3 q ≡ q0 + µϕ0, (38) r ≡−s1λϕ0, (39) s ≡ r1λϕ0, (40) ∗ 2 ∗2 ∗ ∗3 ∗ ∗ ∗ 2 B µ + −2B + 4D r1µλ + B − 3B D + 3F r1 2 ∗2 ∗ 2 ∗ 2 + − B + 2D s1 − 5E r1s1 λ = 0. (41) 3 ϕ0 can be determined by H3 = 0 (at most a cubic equation), since all the coefficients appearing in H3,H4, and H5 are rational and each component as a polynomial is irreducible. Finally equation (23) is expressed in the form 6 2 z + Dz + Ez + F = 0.