Rational interpolation to solutions of Riccati difference equations on elliptic lattices. Alphonse Magnus, Université Catholique de Louvain. http://www.math.ucl.ac.be/membres/magnus/ Elliptic Riccati, Luminy, July 2007. 1
y 3 (a) (b) y 3 y 1 y 2 y 1 y 2 x 3 x 1 x 2 x 3 x 1 x 2 (c) (d) y 3 y 4 y 3 y 2 y 1 y 2 y 1 x 1 x 1 x 2 x 3 x 2 x 3 x 4 Elliptic Riccati, Luminy, July 2007. 2
Marseille-Luminy July 2007 Rational interpolation to solutions of Riccati difference equations on elliptic lattices. The new new new difference calculus. Alphonse Magnus, Université Catholique de Louvain.... opresivo y lento y plural. J.L. Borges Elliptic Riccati, Luminy, July 2007. 3
Difference equations and lattices. (Df)(x) = f(ψ(x)) f(ϕ(x)), (1) ψ(x) ϕ(x) y 2 y 1 y 0 The simplest choice for ϕ and ψ is to take the two determinations of an algebraic function of degree 2, i.e., the two y roots of x 0 x 1 F (x, y) = X 0 (x) + X 1 (x)y + X 2 (x)y 2 = 0, (2a) where X 0, X 1, and X 2 are rational functions. Elliptic Riccati, Luminy, July 2007. 4
The difference equation at x = x 0 relates then f(y 0 ) to f(y 1 ), where y 1 is the second root of (2a) at x 0. We need x 1 such that y 1 is one of the two roots of (2a) at x 1, so for one of the roots of F (x, y 1 ) = 0 which is not x 0. Here again, the simplest case is when F is of degree 2 in x: F (x, y) = Y 0 (y) + Y 1 (y)x + Y 2 (y)x 2 = 0. (2b) Both forms (2a) and (2b) hold simultaneously when F is biquadratic: F (x, y) = 2 2 c i,j x i y j. (3) i=0 j=0 Elliptic Riccati, Luminy, July 2007. 5
Elliptic Riccati, Luminy, July 2007. 6
(x 2, y 3 )(x 3, y 3 ) (x 2, y 2 ) (x 0, y 0 ) (x 1, y 2 ) Elliptic Riccati, Luminy, July 2007. 7 (x 0, y 1 ) (x 1, y 1 )
Pons asinorum Find f such that Df(x) = x. ( Q(x) + ) ( P (x) Q(x) ) P (x) f f R(x) R(x) i.e., 2 = x, P (x)/r(x) where P, Q, R are given polynomials of degrees 4, 2, and 2! Elliptic Riccati, Luminy, July 2007. 8
Why elliptic? -Orthogonal polynomials on several intervals, -contined fraction continuation, -band structure in solid-state physics of Let S be a polynomial of degree 2 and we consider the root ζ 0 (z z 0 )(z x 0 )f 2 (z) 2S(z)f(z)+ζ 1 (z z 0 )(z x 1 ) = 0 (4) Elliptic Riccati, Luminy, July 2007. 9
which is regular at z 0. It is also f(z) = S(z) P (z) ζ 0 (z z 0 )(z x 0 ). f(z) = α 0 z + β 0 z z 0 (z z 0 ) 2 α 1 z + β 1 (z z 0) 2 α 2 z + β 2, (5) or f n (z) = with f 0 = f. z z 0 α n z + β n (z z 0 )f n+1 (z), n = 0, 1,... (6) Elliptic Riccati, Luminy, July 2007. 10
The form of f is kept in all the f n s (basically from Perron): ζ n (z z 0 )(z x n )f 2 n(z) 2S n (z)f n (x)+ζ n+1 (z z 0 )(z x n+1 ) = 0, (7) and we have the PropositionThe continued fraction expansion (5) of the quadratic function f defined by (4) involves a sequence of quadratic functions defined by (7). The related sequence {x n } is an elliptic sequence. Elliptic Riccati, Luminy, July 2007. 11
One has 2n z1 z 0 dt P (t) = x0 z 1 dt xn ± P (t) z 0 dt P (t) +2 j N j zj+1 z 1 (8) dt P (t), Elliptic functions, at last: θ(u; p) = (1 p j u)(1 p j+1 /u), j=0 x n = C θ(qn q 0 η 0 )θ(q n q 0 /η 0 ) θ(q n q 0 η )θ(q n q 0 /η ) Elliptic Riccati, Luminy, July 2007. 12
What s going on here. A special (bi-)orthogonal set of functions is known if one has 1. a formula for the scalar product, measure, weight, Stieltjes transform 2. a formula, or at least a recurrence relation formula, for the (bi-)orthogonal functions 3. difference relations and equations 4. hypergeometric expansions Elliptic Riccati, Luminy, July 2007. 13
Elliptic Pearson s equation. A famous theorem by Pearson relates the classical orthogonal polynomials to the differential equation w = rw satisfied by the weight function, where r is a rational function of degree 2. Theorem. Let {(x(s 0 + k), y(s 0 + k))} be an elliptic lattice built on the biquadratic curve (2a)-(2b)-(3). If there are polynomials a and c, with a(x(s 0 )) + (y(s 0 + 1) y(s 0 ))c(x(s 0 )) = 0, a(x(s 0 + N)) (y(s 0 + N + 1) y(s 0 + N))c(x(s 0 + N)) = 0, Elliptic Riccati, Luminy, July 2007. 14
such that w k+1 a(x Y 2 (y k+1 k) )(x k+1 x k ) w k Y 2 (y k )(x k x k 1 ) y k+1 y k [ ] = c(x w k+1 k) Y 2 (y k+1 )(x k+1 x k ) + w k Y 2 (y k )(x k x k 1 ), (9) k = 0, 1,..., N, where (x k, y k ) is a shorthand for (x(s 0 + Elliptic Riccati, Luminy, July 2007. 15
k), y(s 0 + k)), and w 0 = w N+1 = 0, then, f(x) = N k=1 w k x y k (10) satisfies f(ψ(x)) f(ϕ(x)) a(x)df(x) = a(x) ψ(x) ϕ(x) where d is a polynomial too. = c(x)[f(ϕ(x))+f(ψ(x))]+d(x) (11) Elliptic Riccati, Luminy, July 2007. 16
Elliptic logarithm We extend f(x) = log x a x b which satisfies f (x) = a b by looking for a function whose divided (x a)(x b) difference is a rational function of low degree. Answer: w k = (x k x k 1 )Y 2(y k ), Df(x) = (x N x 0)X 2 (x) (x x 0 )(x x N ). Elliptic Riccati, Luminy, July 2007. 17
Orthogonality and biorthogonality. Padé expansion to f about z 0 : denominators of q n (x) p n (x) = α 0+ α 0 x + β 0 x z 0 (x z 0 ) 2... α n 2 x + β n 2 + (x z 0) 2 α n 1 x + β n 1 satisfy p n+1 (x) = (α n x + β n )p n (x) (x z 0 ) 2 p n 1 (x), asks for f(z 0 ), f (z 0 ), f (z 0 ),... available if f is solution to a differential equation. Elliptic Riccati, Luminy, July 2007. 18
Recurrences of biorthogonal rational functions. q n (x) p n (x) = α 0+ α 0 x + β 0 x y 0 (x y 1 )(x y 2 )... which agree with a given set up to x = y 2n. The recurrence relations for p n and q n are α n 2 x + β n 2 + (x y 2n 3)(x y 2n α n 1 x + β n 1 Elliptic Riccati, Luminy, July 2007. 19
p n+1 (x) = (α n x + β n )p n (x) (x y 2n 1 )(x y 2n )p n 1 (x), q n+1 (x) = (α n x + β n )q n (x) (x y 2n 1 )(x y 2n )q n 1 (x), with q 0 = α 0, p 0 = 1, q 1 (x) = α 0(α 0 x + β 0 ) + x y 0, p 1 (x) = α 0 x + β 0. We could as well start with q 1 (x) = 1/(x y 1 ) and p 1 = 0. Consider now rational functions R n (x) = p n (x) (x y 2 )(x y 4 ) (x y 2n ) : (x y 2n+2 )R n+1 (x) = (α n x+β n )R n (x) (x y 2n 1 )R n 1 (x), Elliptic Riccati, Luminy, July 2007. 20
(x y 2n+1 )T n+1 (x) = (α n x + β n )T n (x) (x y 2n )T n 1 (x), which is of the same structure that the recurrence of the R n s, but with the odd x s interchanged with the even x s. Actually, T n (x) is a constant times the same p n (x) as before, divided by (x y 1 )(x y 3 )... (x y 2n 1 ). Let q n /p n interpolate a formal Stieltjes transform-like function f(x) = S dµ(t) x t, Elliptic Riccati, Luminy, July 2007. 21
then q n interpolates p n f at the 2n + 1 points y 0, y 1,..., y 2n. Also, for k < n, q(x) = q n (x)p k (x)(x y 2k+3 )(x y 2k+5 ) (x y 2n 1 ), still of degree < 2n, interpolates p n (x)p k (x))(x y 2k+3 )(x y 2k+5 ) (x y 2n 1 )f(x), still has a vanishing divided difference at these 2n + 1 points: [y 0,..., y 2n ] of p n (x)p k (x))(x y 2k+3 )(x y 2k+5 ) (x y 2n 1 )f(x) p n (t)p k (t))(t y 2k+3 )(t y 2k+5 ) (t y 2n 1 ) dµ(t) = = 0, (t y 0 )(t y 1 ) (t y 2n ) S as the divided difference of a rational function A(x)/(x t) Elliptic Riccati, Luminy, July 2007. 22
is A(t)/{(t y 0 )(t y 1 ) (t y 2n )} (Milne-Thomson 1.7). So, R n is orthogonal to T k with respect to the formal scalar product g 1, g 2 = g 1 (t)g 2 (t) dµ(t). S Elliptic Riccati, Luminy, July 2007. 23
Where are the orthogonal polynomials on elliptic lattices? Interpolation = Padé if z 0 = x is the limit of the x n s. point. A generic elliptic lattice has no convergence (However... ) Elliptic Riccati, Luminy, July 2007. 24
Definition Elliptic Riccati equations. An elliptic Riccati equation is f(ψ(x)) f(ϕ(x)) a(x) ψ(x) ϕ(x) = b(x)f(ϕ(x))f(ψ(x))+ c(x)(f(ϕ(x)) + f(ψ(x))) + d(x). (12) If x = x m, some point of our x lattice, then ϕ(x) = y m and ψ(x) = y m+1. Elliptic Riccati, Luminy, July 2007. 25
A first-order difference equation of the kind (13) relates f(y 0 ) to f(y 1 ) when x = x 0 ; f(y 1 ) to f(y 2 ) when x = x 1, etc. The direct relation is [ ] a ψ ϕ + c f(ϕ) + d f(ψ) = a ψ ϕ c bf(ϕ). It is sometimes easier to write (13) as e(x)f(ϕ(x))f(ψ(x))+g(x)f(ϕ(x))+h(x)f(ψ(x))+k(x) = 0, where e = b, g = a ψ ϕ c, h = a ψ ϕ c, and k = d. Elliptic Riccati, Luminy, July 2007. 26
However, if a, b, c, and d are rational functions, g and h are conjugate algebraic functions: h + g and hg are symmetric functions of ϕ and ψ, hence rational functions. This also happens with 2a = (h g)(ψ ϕ). Elliptic Riccati, Luminy, July 2007. 27
Now, if f n satisfies the Riccati equation a n (x) f n(ψ(x)) f n (ϕ(x)) ψ(x) ϕ(x) = b n (x)f(ϕ(x))f n (ψ(x)) + c n (x)(f(ϕ(x)) + f(ψ(x))) + d n (x), (13) Theorem. If f n satisfies the Riccati equation (14) with rational coefficients a n, b n, c n, and d n, and if f n (x) = x y 2n α n x + β n (x y 2n+1 )f n+1 (x), then f n+1 satisfies an equation with coefficients a n+1 etc. of same complexity (degree of the rational functions). Elliptic Riccati, Luminy, July 2007. 28
a n+1 = 2X 0 + (y 2n + y 2n+1 )X 1 + 2y 2n y 2n+1 X 2 a n 2F (x, y 2n ) +(X1 2 4X 0 X 2 ) [(y 2n+1 y 2n )c n + 2(α n y 2n+1 + β n )d n ]/X 2. 2F (x, y 2n ) (14) c n+1 = (h n+1 +g n+1 )/2 = 2X 0 + (y 2n + y 2n+1 )X 1 + 2y 2n y 2n+1 X 2F (x, y 2n ) + y 2n+1 y 2n 2F (x, y 2n ) X 2a n α n(2x 0 + y 2n+1 X 1 ) β n (X 1 + 2y 2n+1 X 2 ) 2F (x, y 2n ) d n Elliptic Riccati, Luminy, July 2007. 29
b n+1 = (ϕ y 2n+1)(ψ y 2n+1 ) d n = F (x, y 2n+1) d n, (ϕ y 2n )(ψ y 2n ) F (x, y 2n d n+1 = α ny 2n + β n F (x, y 2n ) X 2a n + b n + [α n(2x 0 + y 2n X 1 ) β n (X 1 + 2y 2n X 2 )]c n + (α 2 nx 0 α n β n X 1 + β F (x, y 2n ) c 2 n(x) b n (x)d n (x) X2 2(x) P (x) a 2 x x 2n 1 n(x) = C n x x 1 [ c 2 0 (x) b 0 (x)d 0 (x) X 2 2 (x) (15) Elliptic Riccati, Luminy, July 2007. 30
where C n = Y 2(y 2n 1 )Y 2 (y 2n 3 ) Y 2 (y 1 ) Y 2 (y 2n 2 )Y 2 (y 2n 4 ) Y 2 (y 0 ). Classical case. We keep the lowest possible degree, which is 3, considering that b n and d n must be X 2 (x) times a polynomial containing the factor x x 2n 1. Let d n (x) = ζ n (x x 2n 1 )X 2 (x), a n c n = X 2 times a polynomial of degree 1. of degree 3, and Elliptic Riccati, Luminy, July 2007. 31
b n+1 (x) = F (x, y 2n+1) F (x, y 2n ) ζ n(x x 2n 1 )X 2 (x) = Y 2(y 2n+1 ) Y 2 (y 2n ) ζ n(x x 2n+1 )X From the Riccati equation (14) at x = x 2n 1 and f n (y 2n ) = 0, we have a n (x 2n 1 ) y 2n y 2n 1 = c n (x 2n 1 ), allowing the divison of the left-hand side of (16), leaving c 2 n(x) b n (x)d n (x) X2 2(x) P (x) a 2 n(x) = C n (x x 2n 1 )Q(x), Elliptic Riccati, Luminy, July 2007. 32
where Q is a fixed polynomial of degree 5. At each of the four zeros z 1,..., z 4 of P, a n (z j ) = ± C n (z j x 2n 1 )Q(z j ), allowing to recover the third degree polynomial a n from four values... should the square roots be determined! Square rootfree relations come from (15) at z j, knowing that ϕ(z j ) = ψ(z j ), which we call ϕ j : a n+1 (z j ) = ϕ j y 2n+1 ϕ j y 2n a n (z j ) Elliptic Riccati, Luminy, July 2007. 33
Remark that, from (16), Q(z j ) = a 2 0(z j )/(z j x 1 ), so there is a subtle relation between the product of the (ϕ j y 2n+1 )/(ϕ j y 2n ) s and a square root of (z j x 2n 1 )/(z j x 1 ). Elliptic Riccati, Luminy, July 2007. 34
Linear difference relations and equations for the numerators and the denominators of the interpolants. g 0 (x m ) p n (y m+1 ) e n+1 (x m )(y m+1 y 0 )(y m+1 y 2 ) (y m+1 y 2n ) = h 0(x m 1 ) p n (y m 1 ) [ e n+1 (x m 1 )(y m 1 y 0 )(y m 1 y 2 ) (y m 1 y 2n ) hn+1 (x m ) e n+1 (x m ) g ] n+1(x m 1 ) p n (y m ) e n+1 (x m 1 ) (y m y 1 )(y m y 3 ) (y m y 2n 1 ), Elliptic Riccati, Luminy, July 2007. 35
Hypergeometric expansions. Building blocks: (Zhedanov) D (x y 0)(x y 1 ) (x y n 1 ) (x y 1 )(x y 2 ) (x y n) = C n X 2 (x) (x x 0)(x x 1 ) (x x n 2 ) (x x 0 )(x x 1 ) (x x n). Elliptic Riccati, Luminy, July 2007. 36
Recommended reading. V.P. Spiridonov, Elliptic hypergeometric functions, Abstract: This is a brief overview of the status of the theory of elliptic hypergeometric functions to the end of 2006 written as a complement to a Russian edition (to be published by the Independent University press, Moscow, 2007) of the book by G. E. Andrews, R. Askey, and R. Roy, Special Functions, Encycl. of Math. Appl. 71, Cambridge Univ. Press, 1999. Report number: RIMS- 1589 Cite as: http://arxiv.org/abs/0704.3099 : arxiv:0704.3099v1 [math.ca] Elliptic Riccati, Luminy, July 2007. 37
Spiridonov, V.P.; Zhedanov, A.S., Generalized eigenvalue problem and a new family of rational functions biorthogonal on elliptic grids, in Bustoz, Joaquin (ed.) et al., Special functions 2000: current perspective and future directions. Proceedings of the NATO Advanced Study Institute, Tempe, AZ, USA, May 29-June 9, 2000, Dordrecht: Kluwer Academic Publishers. NATO Sci. Ser. II, Math. Phys. Chem. 30, 365-388 (2001). V. P. Spiridonov and A. S. Zhedanov: Elliptic grids, rational functions, and the Padé interpolation The Ramanujan Journal 13, Numbers 1-3, June, 2007, p. 285 310. A.S. Zhedanov, Padé interpolation table and biorthogonal Elliptic Riccati, Luminy, July 2007. 38
rational functions. In: Proceedings of RIMS Workshop on Elliptic Integrable Systems. Kyoto, November 8-11 (2004) to be published http://www.math.kobe-u.ac.jp/publications/rlm18/20.pdf Elliptic Riccati, Luminy, July 2007. 39