Trigonometric Polynomials and Quartic Hyperbolic Ternary Forms

Transcription

1 International Journal of Algebra, Vol. 5, 2011, no. 4, Trigonometric Polynomials and Quartic Hyperbolic Ternary Forms Seiji Kitamura Graduate School of Science and Technology Hirosaki University, Hirosaki , Japan Course of Safety Science and Technology, Safety System Engineering Hirosaki University, Hirosaki , Japan Hiroshi Nakazato Graduate School of Science and Technology Hirosaki University, Hirosaki , Japan Department of Mathematical Sciences Faculty of Science and Technology Hirosaki University, Hirosaki , Japan Abstract The 4 4 matrix associated with a hyperbolic ternary form of degree 4 is provided constructively. Mathematics Subject Classification: 15A60 Keywords: Numerical range; algebraic curves 1. Introduction In the paper [1], a 4 4 matrix a A = A(a) = 2 1 a ( a>1) is associated with the form F (t, x, y) = det(ti n + x/2(a + A ) yi/2(a A )),

2 160 S. Kitamura and H. Nakazato and the affine curve F (1,x,y) = 0 is parametrized as 2 2 x = R( {a exp(2iθ)+exp( iθ)}),y = I( {a exp(2iθ)+exp( iθ)}), a 2 1 a 2 1 (0 θ 2π). A real ternary form G(t, x, y) defined as det(ti n + x/2(b + B ) yi/2(b B )) by an n n matrix B and its adjoint (conjugate transpose) B is hyperbolic with respect to (1, 0, 0), that is, the univariate polynomial t G(t + t 0,x 0,y 0 ) has all real roots, for all vectors (t 0,x 0,x 0 ) R 3 independent of (1, 0, 0). The converse of this statement was conjectured by Fiedler in 1981 and it was proved to be true (cf. [2], [3],[4]). However the proof is not so elementary. We consider a trigonometric polynomial φ(exp(iθ)) = exp(2iθ) s 1 exp(iθ)+s 2 s 3 exp( iθ), (1.1) with 4 complex coefficients 1, s 1, s 2, s 3 for which the polynomial exp(iθ)φ(exp(iθ)) = exp(3iθ) s 1 exp(2iθ)+s 2 exp(iθ) s 3, (1.2) is factorized as (exp(iθ) a 1 )(exp(iθ) a 2 )(exp(iθ) a 3 ), (1.3) by some complex numbers a 1,a 2,a 3 with max{ a 1, a 2, a 3 } < 1. (1.4) Then the following relations hold: s 1 = a 1 + a 2 + a 3, s 2 = a 1 a 2 + a 1 a 3 + a 2 a 3, s 3 = a 1 a 2 a 3. We consider a curve {φ(exp(iθ)) : θ R, 0 θ 2π}, (1.5) on the Gaussian plane C and a quartic form F (t, x, y) satisfying F (1, R(φ(exp(iθ)))), I(φ(exp(iθ)) )) = 0, (1.6) for every 0 θ 2π and F (1, 0, 0) = 1. We shall show that the function f(θ) = Arg(φ(exp(iθ))) = I(Logφ(exp(iθ)))

3 Quartic hyperbolic forms 161 is a strictly increasing function in θ. Since the equation 4 f(θ) = Arg((exp(iθ) a j ) exp( iθ/2)), j=1 holds, it is sufficient to prove that the function g(θ) = Arg((exp(iθ) a) exp( iθ/2)), is a strictly increasing function for any a C with a < 1. By the equation exp(iθ/2) a exp(iψ) exp( iθ/2) = {exp(i (θ/2 ψ/2)) a exp( i (θ/2 ψ/2))} exp(iψ/2), we may assume that 0 a<1. In this case we have g(θ) = θ + I(Log(1 a exp( iθ))), 2 g (θ) = 1 ai exp( iθ) + I( 2 1 a exp( iθ) ) 1 a 2 = 2(1 2a cos θ + a 2 ) > 0. Hence we proved that f is strictly increasing. Since the function f(θ) satisfies 2π 0 dg(θ) =2π π = π, 2π 0 df (θ) =4π, that is, the winding number of φ(θ) around 0 for 0 θ 2π is 2. The curve (1.5) does not pass through the origin 0 of the Gaussian plane. So the equation F (t, x 0,y 0 )=0inthave 4 real solutions for every (x 0,y 0 ) R 2. Thus the form F (t, x, y) ishyperbolic with respect to (1, 0, 0). In this paper we present 4 4 matrix X satisfying det(ti 4 +(x/2)(x + X ) yi/2(x X )) = F (t, x, y) for some class of the quartic forms F. For the simplicity we restrict our attention to the case the coefficients s j s are real. In this case the polynomial F (t, 1, 0) satisfies the equation F (t, 1, 0)

4 162 S. Kitamura and H. Nakazato 1 = {t } (t )(t ) 1 s 2 + s 1 s 3 s s 1 + s 2 + s 3 1 s 1 + s 2 s 3 1 = {t 1 j<k 3(1 a j a k ) }2 1 {t (1 + a 1 )(1 + a 2 )(1 + a 3 ) }{t 1 (1 a 1 )(1 a 2 )(1 a 3 ) }, In the above the denominators 1 s 2 + s 1 s 3 s 2 3 =(1 a 1 a 2 )(1 a 1 a 3 )(1 a 2 a 3 ), (1.7) 1+s 1 + s 2 + s 3 =(1+a 1 )(1 + a 2 )(1 + a 3 ), (1.8) 1 s 1 + s 2 s 3 =(1 a 1 )(1 a 2 )(1 a 3 ), (1.9) are positive real numbers under the assumption (1.4). In fact, each one of these expressions is the product of real factors of the form 1 t j ( 1 <t j < 1) or (1 t j )(1 t j )(I(t j ) 0, t j < 1). If all the roots a j s are real, then the polynomial F (t, 1, i) is given by F (t, 1, i) =t 3 2a 2 k (t 3j=1 k=1 (1 a j a k ) ). We shall give the explicit form of the polynomial F by using the 3 real coefficients s 1, s 2, s 3. The polynomial F (t, x, y) =c0,0 F (t, x, y) = F (t, x, y : a 1,a 2,a 3 ) satisfying F (t, 0, 0) = c 00 t 4, c 00 =(1+s 1 + s 2 + s 3 )(1 s 1 + s 2 s 3 )(1 s 2 + s 1 s 3 s 2 3) 2 is given by the following equation F (t, x, y) =c 40 x 4 + c 22 x 2 y 2 + c 04 y 4 + c 30 tx 3 + c 12 txy 2 +c 20 t 2 x 2 + c 02 t 2 y 2 + c 10 t 3 x + c 00 t 4, where the 9 coefficients c ij are given by the following : c 40 = c 04 =1, c 22 =2, c 30 = 2 {2s 2 s 1 s 3 + s 2 3 },c 12 = 2{2s 2 s 1 s 3 3s 2 3 }, c 20 = 2 s s2 2 4s 1 s 3 6s 1 s 2 s 3 + s s2 1 s2 3 +6s 2 s 2 3 2s 1 s s 4 3, c 02 = 2 s s s 1 s 3 2s 1 s 2 s 3 + s s 2 1 s 2 3 6s 2 s 2 3

5 Quartic hyperbolic forms s 1 s s4 3, c 10 = 2(1 s 2 + s 1 s 3 s 2 3)( s s 2 +2s 2 2 3s 1 s 3 s 1 s 2 s 3 + s 2 s 2 3), c 00 =(1+s 1 + s 2 + s 3 )(1 s 1 + s 2 s 3 )(1 s 2 + s 1 s 3 s 2 3) 2. Theorem 2.1 Suppose that φ(exp(iθ)) = exp(2iθ) s 1 exp(iθ)+s 2 s 3 exp( iθ), (2.1) is a trigonometric polynomial with real coefficients s 1, s 2, s 3 satisfying the condition t 3 s 1 t 2 + s 2 t s 3 =(t a 1 )(t a 2 )(t a 3 ), (2.2) for some complex numbers a j s with a j < 1(j =1, 2, 3, 4). If the coefficient s 3 vanishes, or equivalently, a 1 a 2 a 3 =0, (2.3) then there exists a 4 4 real matrix in the form for which the quartic form satisfies the condition x 1 b 12 b 13 0 b A = 12 x 2 b 23 b 24 b 13 b 23 x 3 0, (2.4) 0 b 24 0 x 4 F (t, x, y) = det(ti 4 + x/2(a + A ) yi/2(a A )), (2.5) for 0 θ 2π. F (1, R(φ(exp(iθ))), I(φ(exp(iθ)))) = 0, (2.6) Proof. We define the matrix A in the form (2.4) by the equations x 1 = x 2 = 1, 1 s 2 + s 1 s 3 s 2 3 b 12 = x 3 = 1 1, x 4 =, 1+s 1 + s 2 + s 3 1 s 1 + s 2 s 3 B 13, b 23 = B 23, b 24 = B 12, b 13 = B 24.

6 164 S. Kitamura and H. Nakazato Then the above x j s satisfy the equation c 00 det(ti 4 + x/2(a + A )) = F (t, x, 0). Moreover, the equation c 00 det(ti 4 + x/2(a + A ) yi/2(a A )) = F (t, x, y). in b 12,b 13,b 23,b 24 is expressed as the following c 00 det(ti 4 + x/2(a + A ) yi/2(a A )) F (t, x, y) = y 2 {L 1 (B 12,B 13,B 23,B 24 ; s 1,s 2,s 3 ) t 2 + L 2 (B 12,B 13,B 23,B 24 ; s 1,s 2,s 3 ) x 2 +L 3 (B 12,B 13,B 23,B 24 ; s 1,s 2,s 3 ) tx+ L 4 (B 13 B 24 ; s 1,s 2,s 3 ) y 2 } =0, (2.8) where B 12 = b 2 12,B 13 = b 2 13,B 23 = b 2 23,B 24 = b 2 24, and L j s for j =1, 2, 3 are expressed as L j (B 12,B 13,B 23,B 24 ; s 1,s 2,s 3 ) = φ j,1 (s 1,s 2,s 3 ) B 12 + φ j,2 (s 1,s 2,s 3 ) B 13 +φ j,3 (s 1,s 2,s 3 ) B 23 + φ j,4 (s 1,s 2,s 3 ) B 24 + φ j,0 (s 1,s 2,s 3 ). for some polynomials φ j,k s with integral coefficients. The coefficient function L 4 is expressed as L 4 (B 13 B 24 ; s 1,s 2,s 3 ) =(1 s 1 + s 2 s 3 )(1 + s 1 + s 2 + s 3 )(1 s 2 + s 1 s 3 s 2 3) 2 B 13 B We solve the simultaneous equations L 1 = L 2 = L 3 = L 4 =0inB 12,B 13,B 23,B 24. We take the resultant R 1 (B 12,B 24 )ofl 1, L 2 with respect to B 23. We take the resultant R 2 (B 12,B 24 )ofl 1, L 3 with respect to B 23. We eliminate B 24 from the equations R 1 = R 2 = 0 and obtain an equation in B 12. We solve this equation and obtain the solution : B 12 =(s 1 3s 3 ) 2 /[(2 + s 1 s 3 )(2 s 1 + s 3 )(1 s 2 + s 1 s 3 s 2 3 )2 ]. (2.9) We substitute this expression of B 12 into the equation R 2 = 0 and obtain an equation in B 24. We solve this equation and obtain the solution : B 24 =(4 s 1 s 3 )(1 s 3 )/[2(1 s 1 + s 2 s 3 )(2 s 1 + s 3 )(1 s 2 + s 1 s 3 s 2 3)]. (2.10) We substitute this expression of B 24 into the equation L 4 = 0 and obtain an equation in B 13. We solve this equation and obtain the solution :

7 Quartic hyperbolic forms 165 B 13 = 2(2 s 1 + s 3 )/[(4 s 1 s 3 )(1 + s 1 + s 2 + s 3 )(1 s 3 )(1 s 2 + s 1 s 3 s 2 3 )]. (2.11) We substitute the expressions of B 12,B 24,B 13 in to the equation L 1 = 0 and obtain an equation in B 23. We solve this equation and obtain the equation B 23 =(s 1 + s 3 )(3s 1 13s 3 + s 1 s s3 3 )/[2(4 s 1 s 3 )(1 + s 1 + s 2 + s 3 ), (2 + s 1 s 3 )(1 s 3 )(1 s 2 + s 1 s 3 s 2 3 )]. (2.12) Conversely the rational functions B 12,B 13,B 23,B 24 in s 1,s 2,s 3 given by (2.9), (2.10), (2.11), (2.12) satisfy the simultaneous equations L 1 = L 2 = L 3 = L 4 =0. The equation (2.8) in b 12,b 13,b 23,b 24 has a real solution if and only if the denominators of (2.9), (2.10), (2.11), (2.12) are non-zero and the functions B 12,B 13,B 23,B 24 take non-negative values under the assumption s 1 s 3 =0. Under the assumption s 3 = 0, we have B 12 = s 2 1 /[(2 + s 1)(2 s 1 )(1 s 2 ) 2 ]. B 24 =(4 s 1 )/[2(1 s 1 + s 2 )(2 s 1 )(1 s 2 )]. B 13 = 2(2 s 1 )/[(4 s 1 )(1 + s 1 + s 2 )(1 s 2 )]. B 23 =3s 2 1 /[2(4 s 1)(1 + s 1 + s 2 )(2 + s 1 )(1 s 2 )]. We may assume that a 3 = 0 and s 1 = a 1 + a 2, s 2 = a 1 a 2. Hence 1+s 1 +s 2 =1+a 1 +a 2 +a 1 a 2 = (1+a 1 )(1+a 2 ) > 0, 1 s 1 +s 2 =(1 a 1 )(1 a 2 ) > 0, 1 s 2 =1 a 1 a 2 > 0, 1+s 2 =1+a 1 a 2 > 0 by the assumption a j < 1(j =1, 2). Thus the equation (2.8) has a nonnegative real solutions (b 12,b 13,b 23,b 24 ). References [1] M. T. Chien and H. Nakazato, Numerical range for orbits under a central force, to appear in Mathematical Physics, Analysis and Geometry. [2] M. Fiedler, Geometry of the numerical range of matrices, Linear Algebra Appl. 37(1981), [3] P. D. Lax, Differential equations, difference equations and matrix theory. Comm. Pure Appl. Math., 6(1958),

8 166 S. Kitamura and H. Nakazato [4] A. S. Lewis, P. A. Parrilo and M. V. Ramana, The Lax conjecture is true, Proc. Amer. Math. Soc. 133(2005), Received: September, 2010