Marseille-Luminy July Rational interpolation to solutions of Riccati difference equations on elliptic lattices.

Transcription

1 Marseille-Luminy July 2007 Rational interpolation to solutions of Riccati difference equations on elliptic lattices. Alphonse Magnus, Institut de Mathématique Pure et Appliquée, Université Catholique de Louvain, Chemin du Cyclotron,2, B-1348 Louvain-la-Neuve (Belgium) (0)(10)473157, opresivo y lento y plural. J.L. Borges This version: June 28, 2007 (incomplete and unfinished) The present file is Abstract. An elliptic lattice, or grid, {x 0, x 1,...} is built with the help of a biquadratic curve F (x, y) = 2 i=0 2 j=0 c i,jx i y j by the following rules: (1) let y n and y n+1 be the two y roots of F (x n, y) = 0, (2) then, x n+1 is found as the remaining x root of F (x, y n+1 ) = 0. There is also a direct symmetric biquadratic relation E(x n, x n+1 ) = 0, see 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 Numerators and denominators of rational interpolants on such lattices satisfy interesting difference equations when the interpolated function f itself satisfies a Riccati difference equation on the same lattice: a(x n ) f(y n+1) f(y n ) y n+1 y n = b(x n )f(y n )f(y n+1 ) + c(x n )(f(y n ) + f(y n+1 )) + d(x n ), where a, b, c, and d are polynomials. Contents 1. Difference equations and lattices Elliptic grid, or lattice Definition of elliptic grid Periodicity, theta functions A special product Elliptic Pearson s equation Theorem Elliptic logarithm Recurrences of biorthogonal rational functions Padé and interpolatory continued fractions

2 Marseille-Luminy July 2007 Elliptic lattices. 1 Diff. eq. & lattices Biorthogonality and orthogonality Example: the exponential function (Iserles [18]) Elliptic Riccati equations Definition Rational interpolation Theorem Classical case Linear difference relations and equations for the numerators and the denominators of the interpola Difference equation for the denominator p n Hypergeometric expansions References Difference equations and lattices. Simplest difference equations relate two values of the unknown fuction f: say, f(ϕ(x)) and f(ψ(x)). Most instances [29] are (ϕ(x), ψ(x)) = (x, x + h), or the more symmetric (x h/2, x + h/2), or also (x, qx) in q difference equations [14]. Recently, more complicated forms (r(x) s(x), r(x) + s(x)) have appeared [5,8,22,23,30,31], where r and s are rational functions. This latter trend will be examined here: we need, for each x, two values f(ϕ(x)) and f(ψ(x)) for f. A first-order difference equation is F (x, f(ϕ(x)), f(ψ(x))) = 0, or f(ϕ(x)) f(ψ(x)) = G (x, f(ϕ(x)), f(ψ(x))) if we want to emphasize the difference of f. There is of course some freedom in this latter writing. Only symmetric forms in ϕ and ψ will be considered here: where D is the divided difference operator (Df)(x) = F (x, f(ϕ(x)), f(ψ(x))), (1) (Df)(x) = f(ψ(x) f(ϕ(x), (2) ψ(x) ϕ(x) and where F is a symmetric function of its two last arguments. For instance, a linear difference equation of first order may be written as as well as a(x)f(ϕ(x)) + b(x)f(ψ(x)) + c(x) = 0, α(x)(df)(x) = β(x)[f(ϕ(x)) + f(ψ(x))] + γ(x), with α(x) = [b(x) a(x)][ψ(x) ϕ(x)]/2, β(x) = [a(x) + b(x)]/2, and γ(x) = c(x).

3 Marseille-Luminy July 2007 Elliptic lattices. 2 Diff. eq. & lattices 3 y 2 y 1 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 y 0 x 0 x 1 F (x, y) = X 0 (x) + X 1 (x)y + X 2 (x)y 2 = 0, (3a) where X 0, X 1, and X 2 are rational functions. But difference equations must allow the recovery of f on a whole set of points! An initialvalue problem for a first order difference equation starts with a value for f(y 0 ) at x = x 0, where y 0 is one root of (3a) at x = x 0. The difference equation at x = x 0 relates then f(y 0 ) to f(y 1 ), where y 1 is the second root of (3a) at x 0. We need x 1 such that y 1 is one of the two roots of (3a) 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. (3b) Both forms (3a) and (3b) hold simultaneously when F is biquadratic: F (x, y) = 2 2 c i,j x i y j. (4) i=0 j=0 2. Elliptic grid, or lattice.

4 Marseille-Luminy July 2007 Elliptic lattices. 2 Diff. eq. & lattices Definition of elliptic grid. 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 Various forms of the curve F (x, y) = 0 are degenerate parabolas (two parallel lines) or hyperbolas (two lines), or generic conics, or a full biquadratic curve y = X 1(x) ± X1(x) 2 4X 0 (x)x 2 (x). 2X 2 (x) To one x = x n correspond the two ordinates y n and y n+1. One has y n + y n+1 = X 1(x n ) X 2 (x n ), and y ny n+1 = X 0(x n ) X 2 (x n ).

5 Marseille-Luminy July 2007 Elliptic lattices. 2 Diff. eq. & lattices 5 (x 2, y 3 ) (x 3, y 3 ) (x 2, y 2 ) (x 0, y 0 ) (x 1, y 2 ) (x 0, y 1 ) (x 1, y 1 ) Also, to one ordinate y = y n correspond the two abscissae x n and x n 1, and we now have x n + x n 1 = Y 1(y n ) Y 2 (y n ), x nx n 1 = Y 0(y n ) Y 2 (y n ). (5) A relation involving only x n and x n 1 is obtained by the elimination of y n through the resultant of the two polynomials in y n P 1 (y n ) = (x n + x n 1 )Y 2 (y n ) + Y 1 (y n ) and P 2 (y n ) = x n x n 1 Y 2 (y n ) Y 0 (y n ). The form of this resultant is most easily found through interpolation at the two zeros u and v of Y 2 : let Y 2 (t) = α(t u)(t v), Y 0 (t) = β(t u)(t v)+β (t u)+β, Y 1 (t) = γ(t (x n + x n 1 )α + γ γ γ 0 u)(t v)+γ (t u)+γ, then the resultant is R = 0 (x n + x n 1 )α + γ γ γ x n x n 1 α β β β 0, 0 x n x n 1 α β β β which is clearly a polynomial of degree 2 in x n +x n 1 and x n x n 1, so a symmetric biquadratic relation: Definition. An elliptic lattice, or grid, is a sequence satisfying. E(x n, x n 1 ) = d 0,0 +d 0,1 (x n +x n 1 )+d 0,2 (x 2 n+x 2 n 1)+d 1,1 x n x n 1 +d 1,2 x n x n 1 (x n +x n 1 )+d 2,2 x 2 nx 2 n 1 = 0. (6) We also get a linear recurrence relation between three x s by adding (??) for s and s + 1: x(s + 1) + 2x(s) + x(s 1) = x(s + 1) + α(y(s + 1) + y(s)) 2δ γ = α2 x(s) + αβ 2δ, so γ 2γ α2 αβ 2δ x(s) + x(s 1) = γ γ (7)

6 Marseille-Luminy July 2007 Elliptic lattices. 2 Diff. eq. & lattices 6 x(s + 1) Why elliptic? x(s) x(s 1) The learned answer is that a biquadratic curve (4) or (6) is normally of genus one, and receives therefore a parametric representation involving elliptic functions [42]. It is also known that formulas like (6) appear in Euler s pionneering work on what is now called addition x(s) x(s + 1) formulas for elliptic functions, that s why the authors of [42] use the letter E in (6). Here is an explanation based on special Padé approximations, periodic continued fractions, and orthogonal polynomials on several intervals: Let S be a polynomial of degree 2 S(z) = γ + δ(z z 0 ) + ξ(z z 0 ) 2, and we consider the root of ζ 0 (z z 0 )(z x 0 )f 2 (z) 2S(z)f(z) + ζ 1 (z z 0 )(z x 1 ) = 0 (8) which is regular at z 0. It is also f(z) = S(z) P (z) ζ 0 (z z 0 )(z x 0 ), where the choice of the square root of P (z) = S 2 (z) ζ 0 ζ 1 (z z 0 ) 2 (z x 0 )(z x 1 ) = c(z z 1 )(z z 2 )(z z 3 )(z z 4 ) in a neighbourhood of z 0 is such that the value of this square root is S(z 0 ) = γ. Actually, this regular root even vanishes at z 0, and can be represented by the continued fraction 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, (9) or f n (z) = z z 0, n = 0, 1,... (10) α n z + β n (z z 0 )f n+1 (z) with f 0 = f. Remark that α n z+β n is the Taylor approximation of degree 1 to (z z 0 )/f n (z). We can therefore recover α n and β n from the behaviour of f n near z 0. The form of f is kept in all the f n s (basically from Perron [35, 60, eq. (5)-(14)]): and we have the 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, (11) Proposition. The continued fraction expansion (9) of the quadratic function f defined by (8) involves a sequence of quadratic functions defined by (11). The related sequence {x n } is an elliptic sequence. We first show that the quadratic equation (11) holds for all n. Indeed, if f n is the root of (11) such that f n (z) = S n(z) P (z) ζ n (z z 0 )(z x n ), with S n(z) 2 ζ n ζ n+1 (z z 0 ) 2 (z x n )(z x n+1 ) =

7 Marseille-Luminy July 2007 Elliptic lattices. 2 Diff. eq. & lattices 7 P (z), we have f n (z) = S2 n(z) P (z) = ζ n ζ n+1 (z z 0 ) 2 (z x n )(z x n+1 ) ζ n (z z 0 )(z x n )[S n (z) + P (z)] = z z 0 α n z + β n ζ n+1(α n z + β n )(z x n+1 ) S n (z) P (z) ζ n+1 (z x n+1 ) = (z z 0 ) S n (z) + P (z) ζ n+1 (z x n+1 ) showing that f n+1 (z) = S n+1(z) P (z) ζ n+1 (z z 0 )(z x n+1 ), as expected, where S n+1(z) = ζ n+1 (α n z + β n )(z x n+1 ) S n (z), α n z + β n being the Taylor approximant of degree 1 to [S n (z) + P (z)]/[ζn+1 (z x n+1 )] about z = z 0. This way to build S n+1 ensures that the fourthdegree polynomial Sn+1 2 P has factors (z z 0 ) 2 and ζ n+1 (z x n+1 ). Let us call the remaining factor ζ n+2 (z x n+2 ), and this completes the definition of f n+1. And here is how the present x n and x n+1 actually satisfy the elliptic lattice equation (6): We expand Sn(z) P 2 (z) = ζ n ζ n+1 (z z 0 ) 2 (z x n )(z x n+1 ) with S n (z) = γ +δ(z z 0 )+ ξ n (z z 0 ) 2 and P (z) = γ 2 +2γδ(z z 0 )+P (z 0 )(z z 0 ) 2 /2+P (z 0 )(z z 0 ) 3 /6+P (z 0 )(z z 0 ) 4 /24 (so that the expansion of the square root of P starts indeed with γ + δ(z z 0 )). 2γξ n + δ 2 P (z 0 )/2 = ζ n ζ n+1 (z 0 x n )(z 0 x n+1 ), (12a) 2δξ n P (z 0 )/6 = ζ n ζ n+1 (2z 0 x n x n+1 ), (12b) ξn 2 P (z 0 )/24 = ζ n ζ n+1. (12c) The last equation yields ζ n ζ n+1 as a polynomial of degree 2 in ξ n, therefore the two first equations give the sum and the product of x n and x n+1 as rational functions of degree 2 of an intermediate parameter, and this is the structure of (3a)-(3b)-(4): z 0 x n and z 0 x n+1 are the two roots of (z 0 x) 2 2δξ n P (z 0 )/6 ξn 2 P (z 0 )/24 (z 0 x) + 2γξ n + δ 2 P (z 0 )/2 = 0, ξn 2 P (z 0 )/24 so F (x, y) = [y 2 P (z 0 )/24](z 0 x) 2 [2δy P (z 0 )/6](z 0 x) + 2γy + δ 2 P (z 0 )/2 = 0. So, the continued fraction expansion (9)-(11) of a quadratic algebraic function leads to an elliptic lattice. Conversely, can we find P, z 0,etc. from a given elliptic lattice (6)? From a solution x n = x n+1 = z 0 of (12a)-(12c) (when ξ n = ), z 0 is one of the four roots of E(z 0, z 0 ) = d 0,0 + 2d 0,1 z 0 + (2d 0,2 + d 1,1 )z d 1,2z d 2,2z 4 0 = 0. Moreover, (6) yields y = x n±1 as a quadratic function of x = x n as y = d 1,2x 2 d 1,1 x d 0,1 ± (d 1,2 x 2 + d 1,1 x + d 0,1 ) 2 4(d 2,2 x 2 + d 1,2 x + d 0,2 )(d 0,2 x 2 + d 0,1 x + d 0,0 ). 2(d 2,2 x 2 + d 1,2 x + d 0,2 ) Let us look now at the (S n, P, ζ n (z x n )) construction as a way to find x n+1 from x n : if x n is known, Sn 2 P must vanish at x = x n S n (x n ) = γ +δ(x n z 0 )+ξ n (x n z 0 ) 2 = ± P (x n ), giving two possible values for ξ n. Then, as already seen, we factor Sn 2 P as a constant

8 Marseille-Luminy July 2007 Elliptic lattices. 2 Diff. eq. & lattices 8 times (z z 0 ) 2 (z x n ) times a last factor which must be a constant times z x n+1, yielding for x n+1 an expression containing P (x n ), so that we obtain P from (d 1,2 x 2 + d 1,1 x + d 0,1 ) 2 4(d 2,2 x 2 + d 1,2 x + d 0,2 )(d 0,2 x 2 + d 0,1 x + d 0,0 ) = const. P (x). This allows to interpret any elliptic lattice in terms of the continued fraction expansion of a quadratic algebraic function The elliptic functions, at last. Let A n (z) B n (z) = z z 0 (z z 0 ) 2, (13) α 0 z + β 0 (z z 0 ) 2 α 1 z + β 1... α 2 z + β 2 α n 1 z + β n 1 with A 0 = 0, A 1 (z) = z z 0, A k+1 (z) = (α k z + β k )A k (z) (z z 0 ) 2 A k 1 (z), B 0 = 1, B 1 (z) = α 0 z + β 0, B k+1 (z) = (α k z + β k )B k (z) (z z 0 ) 2 B k 1 (z), or also B 1 = 0, A 1 (z) = 1/(z z 0 ). Ratios g k+1 = λb k+1 A k+1 satisfy g k+1 = α k z + β k (z z 0 ) 2 g k /g k, or = λb k A k z z 0 z z 0, i.e., the recurrence (10) of the f k s, which correspond to λ = f (as α k z + β k g k+1 g 0 = λ/a 1 = (z z 0 )λ, so g 0 (z)/(z z 0 ) = f 0 (z) = f(z) if λ = f). The quadratic function B n f A n = (z z 0 ) n f 0 f 1 f n has an extremely peculiar set of zeros and poles: let f conj be the conjugate function to f, i.e., the quadratic function S + P where the sign of the square root of P has been changed, an elementary ζ 0 (z z 0 )(z x 0 ) way to deal with two-sheeted Riemann surfaces. Then, the set of zeros and poles of B n f A n is a part of the set for (B n f A n )(B n f conj A n ) (B n f A n )(B n f conj A n ) = (z z 0 ) 2n f 0 f n f conj 0 fn conj = (z z 0 ) 2 S 0 P ζ 0 (z x 0 ) Sn P S 0 + P ζ n (z x n ) ζ 0 (z x 0 ) S 0 + P ζ n (z x n ) = (z z 0) 2n (z x n+1 ), ζ 0 ζ n (z x 0 ) a very limited set! Actually, most zeros and poles are concentrated at z 0 and : B n f A n has a zero of order 2n + 1 at z 0 (Padé property), B n f conj A n a simple pole; both functions have a pole of order n at What is (are) the period(s)?

9 Marseille-Luminy July 2007 Elliptic lattices. 2 Diff. eq. & lattices 9 x 0 S z 0 z 1 z 4 z 2 Consider the [ integral of ] dt πi P (t) log Bn 1 (t)f(t) A n 1 (t) on a B n 1 (t)f conj (t) A n 1 (t) big contour, result is 0, as everything is regular when t is large; and we shrink the contour about z 0 and the zero and pole x 0 and x n : z1 0 = 2n z 0 dt x0 + P (t) z z 3 1 dt ± P (t) xn z 1 dt P (t) + 2 j N j zj+1 z j dt P (t) } {{ } periods It happens that, knowing n, P, and x 0, (14) allows to find the remaining unknowns, including the ± signs (Jacobi problem, [1, 3, 26, 32, 33,?]). There is absolutely no need for n to be an integer in the description (14) of the Jacobi problem. To see how this x n is a function of n, we... take the derivative of (14) with respect to n (!!): 1 dx n h = ± dn, z1 where h = 2 z 0 (14) P (xn ) dt. We have a differential equation for x n. An initial condition P (t) consists of x 0 and a sign (= a place on the Riemann surface of P ). Generalization. Hyperelliptic case, generalization of Padé approximation and continued fraction (recurrence relations) constructions: see [?,4,33,44]. We then have a vector of length g (genus),..., x n (g) ] of unknowns which is a well defined function (Jacobi-Abel function) of the left-hand side [nh 0,..., nh g 1 ]. [x (1) n 2.2. Periodicity, theta functions. x n is kept unchanged if nh + h 0 in (14) is augmented by integers times the integrals zj+1 t k 2ω j = 2 dt (periods). P (t) z j So, x n is some periodic function (elliptic function) of nh + h 0. There are two zeros and two poles in a fundamental parallelogram of periods (2ω 1, 2ω 2 ), they are given by 0 dt dt nh + h 0 = ± and ±, say, ±ζ 0 and ±ζ. P (t) P (t) z 1 z 1 We expect x n to involve standard functions of nh + h 0 ± ζ 0 and nh + h 0 ± ζ. With the p theta function θ(u; p) = (1 p j u)(1 p j+1 /u), j=0 which vanishes at log u = all the integer multiples of log p plus all the integer multiples of 2πi, we consider [37, 2]

10 Marseille-Luminy July 2007 Elliptic lattices. 2 Diff. eq. & lattices 10 x n = C θ ( θ exp ( exp ( iπ nh + h 0 ζ 0 ω 1 ( iπ nh + h 0 ζ ω 1 )) )) θ ( θ exp ( exp ( iπ nh + h )) 0 + ζ 0 ω ( 1 iπ nh + h )) 0 + ζ see also [42, 4], there is a simple relation between the p theta function and the Jacobi theta functions. Let q = exp(iπh/ω 1 ), q 0 = exp(iπh 0 /ω 1 ), η 0 = exp(iπζ 0 /ω 1 ), η = exp(iπζ /ω 1 ), so 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 ) x n x m, where n and m need not be integers, is another (much more interesting) elliptic function of n with the same poles, but with zeros such that n = m is one of them. This leads to replace η 0 by 1/(q 0 q m ): ω 1 For the y s, one has x n x m = C θ(qn m )θ(q n+m q 2 0 ) θ(q n q 0 η )θ(q n q 0 /η ) y n y m = C θ(q n m )θ(q n+m q 0 2 ) θ(q n q 0η )θ(q n q 0/η ) with the same θ function and the same q (15) (16) 2.3. A special product. We will have to considerate the special algebraic function and its conjugate X n (x) = (ψ(x) y 1)(ψ(x) y 3 ) (ψ(x) y 2n 1 ) (ϕ(x) y 0 )(ϕ(x) y 2 ) (ϕ(x) y 2n 2 ) X conj. n (x) = (ϕ(x) y 1)(ϕ(x) y 3 ) (ϕ(x) y 2n 1 ) (ψ(x) y 0 )(ψ(x) y 2 ) (ψ(x) y 2n 2 ) the product is F (x, y 1)F (x, y 3 ) F (x, y 2n 1 ) F (x, y 0 )F (x, y 2 ) F (x, y 2n 2 ) = Y 2(y 1 )Y 2 (y 3 ) Y 2 (y 2n 1 ) Y 2 (y 0 )Y 2 (y 2 ) Y 2 (y 2n 2 ) The value of (17) is well defined when x is some x m : x x 2n 1 x x 1. X n (x m ) = (y m+1 y 1 )(y m+1 y 3 ) (y m+1 y 2n 1 ) (y m y 0 )(y m y 2 ) (y m y 2n 2 ) Now, following (15), y r y s is a ratio of products with numerator θ(q r s )θ(q r+s q 2 0 ), so (17) X n (x m ) = θ(qm )θ(q m+2 q 0 2 )θ(q m 2 )θ(q m+4 q 0 2 ) θ(q m 2n+2 )θ(q m+2n q 2 θ(q m )θ(q m q 0 2 )θ(q m 2 )θ(q m+2 q 0 2 ) θ(q m 2n+2 )θ(q m+2n 2 q 2 and what remains is X n (x m ) = θ(qm+2n q 0 2 ) θ(q m q 2 0 ) 0 ) 0 ) (18)

11 Marseille-Luminy July 2007 Elliptic lattices. 3 Pearson 11 and Xn conj (x m ) = θ(qm 2n+1 ) θ(q m+1 ) from the identities θ(pu) = θ(1/u) = θ(u)/u. = θ(pq2n 1 m ) θ(pq m 1 ) (19) 3. 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 Theorem. Let {(x(s 0 + k), y(s 0 + k))} be an elliptic lattice built on the biquadratic curve (3a)-(3b)-(4), with s 0 / Z. If there are polynomials a and c, with a(x(s 0 )) (y(s 0 + 1) y(s 0 ))c(x(s 0 )) = a(x(s 0 + N)) (y(s 0 + N + 1) y(s 0 + N))c(x(s 0 + N)) = 0, such that w k+1 a(x k ) Y 2 (y k+1 )(x k+1 x k ) w k Y 2 (y k )(x k x k 1 ) [ ] y k+1 = c(x y k ) 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 ), (20) k = 0, 1,..., N, where (x k, y k ) is a shorthand for (x(s 0 +k), y(s 0 +k)), and w 0 = w N+1 = 0, then, N w k f(x) = (21) x y k satisfies k=1 f(ψ(x)) f(ϕ(x)) a(x)df(x) = a(x) = c(x)[f(ϕ(x)) + f(ψ(x))] + d(x), (22) ψ(x) ϕ(x) where d is a polynomial too. Indeed, = N 1 f(ψ(x)) f(ϕ(x)) ψ(x) ϕ(x) = N w k X 2 (x) Y 2 (y k )(x x k 1 )(x x k ) = X 2(x) with w 0 = w N+1 = 0, f(ψ(x)) + f(ϕ(x)) = = N 1 N w k [X 1 (x) + 2y k X N 2(x)] Y 2 (y k )(x x k 1 )(x x k ) = 1 1 N w k (ϕ(x) y k )(ψ(x) y k ) = N 0 1 w k X 2 (x) F (x, y k ) w k+1 Y 2 (y k+1 )(x k+1 x k ) w k Y 2 (y k )(x k x k 1 ) x x k w k [X 1 (x) + 2y k X N 2(x)] X 2 (x)(ϕ(x) y k )(ψ(x) y k ) = 0 w k+1 [X 1 (x) + 2y k+1 X 2(x)] Y 2 (y k+1 )(x k+1 x k ) 1 x x k w k [X 1 (x) + 2y k X 2(x)] F (x, y k ) w k[x 1 (x) + 2y k X 2(x)] Y 2 (y k )(x k x k 1 ) therefore the rational functions adf and c(f(ϕ) + f(ψ)) differ by a polynomial if all the residues are equal:,

12 Marseille-Luminy July 2007 Elliptic lattices. 4 Biorthogonal rationa functions 12 [ ] a(x k )X 2(x k ) w k+1 Y 2 (y k+1 )(x k+1 x k ) w k Y 2 (y k )(x k x k 1 [ ) = c(x wk+1 [X 1 (x k k) ) + 2y k+1 X 2(x k )] Y 2 (y k+1 )(x k+1 x k ) w k[x 1 (x k ) + 2y k X 2(x k )] ] Y 2 (y k )(x k x k 1 ) for k = 0, 1,..., N. Or, as X 1 (x k ) = (y k + y k+1 )X 2(x k ), [ ] a(x k ) w k+1 Y 2 (y k+1 )(x k+1 x k ) w k Y 2 (y k )(x k x k 1 [ ) = c(x k)(y k+1 y k) 3.2. Elliptic logarithm. w k+1 Y 2 (y k+1 )(x k+1 x k ) + ] w k Y 2 (y k )(x k x k 1 ) We extend f(x) = log x a x b which satisfies f a b (x) = by looking for a (x a)(x b) function whose divided 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 (23) x N ). 4. Recurrences of biorthogonal rational functions. From excerpts of Spiridonov & Zhedanov [40], also A. Zhedanov, Biorthogonal rational functions and generalized eigenvalue problem, J. Approx. Theory 101 (1999), no. 2, , and [51]. Also Brezinski, Iserles, Ismail, Masson, Norsett Padé and interpolatory continued fractions. α Padé. α 1 matches a given Laurent expansion c 0 /x+ x β x β α n 1 x β n 1 c 1 /x 2 + at up to the c 2n+1 /x 2n+2 term. Numerators and denominators satisfy the recurrence relation P n+1 (x) = (x β n )P n (x)+α n P n 1 (x), suggesting some kind of (formal?) orthogonality. This is even more obvious in the matrix-vector setting β 0 α1 α1 β 1 α P 0 (x) P 0 (x) P 1 (x). = x P 1 (x)

13 Marseille-Luminy July 2007 Elliptic lattices. 4 Biorthogonal rationa functions 13 If one wants to approximate a Taylor expansion about the origin, just take z = 1/x and rewrite the continued fraction as α 0 z α 1 z 2 which matches a given Taylor-Maclaurin expansion up to the z 2n 1 zβ β 1 z + + α n 1z 2 1 β n 1 z term Interpolation. Rational interpolations to a given set of values at x = y 0, y 1,... (yes, the relevant set will be a y lattice) are achieved by q n (x) p n (x) = α 0 + x y 0 (x y 1 )(x y 2 ) α 0 x + β 0... α n 2 x + β n 2 + (x y 2n 3)(x y 2n 2 ) α n 1 x + β n 1 which agree with a given set up to x = y 2n. The recurrence relations for p n and q n are 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. We also have the Casorati relation (24) p n (x)q n 1 (x) p n 1 (x)q n (x) = (x y 0 )(x y 1 ) (x y 2n 3 )(x y 2n 2 ). (25) 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), so that the matrix-vector setting is now β 0 y 2 R 0 (x) α 0 1 R 0 (x) y 1 β 1 y 4 R 1 (x) = x 1 α 1 1 R 1 (x) So, {R 0, R 1,... } is a right eigenvector which is in some way biorthogonal to the set of left eigenvectors {T 0, T 1,... } satisfying the recurrence (x y 2n+1 )T n+1 (x) = (α n x + β n )T n (x) (x y 2n )T n 1 (x),

14 Marseille-Luminy July 2007 Elliptic lattices. 4 Biorthogonal rationa functions 14 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 ). That p n is invariant under matrix transposition is clear: p n (x) is the determinant α 0 x + β 0 (x y 2 ) (x y 1 ) α 1 x + β 1 (x y 4 ) (x y 2n 3 ) α n 1 x + β n Biorthogonality and orthogonality. But what is the bilinear form exhibiting the biorthogonality condition? Let q n /p n interpolate a formal Stieltjes transform-like function f(x) = S dµ(t) x t, 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 2n 1 ), still of degree < 2n, interpolates p n (x)p k (x))(x y 2k+3 ) (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 2n 1 )f(x) p n (t)p k (t))(t y 2k+3 ) (t y 2n 1 ) dµ(t) = (t y 0 ) (t y 2n ) S = 0, as the divided difference of a rational function A(x)/(x t) is A(t)/{(t y 0 ) (t y 2n )} (Milne-Thomson [29,...]). 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). Even simpler: if f(x) = N 1 0 ρ k, x x k S N 1 g 1, g 2 = ρ k g 1 (x k)g 2 (x k) Example: the exponential function (Iserles [18]). Rational interpolations to e ax at x = 0, h, 2h,...

15 Marseille-Luminy July 2007 Elliptic lattices. 5 Riccati 15 The hypergeometric expansions are p n (x) = C n p n (x)e ax = C n n ( 1) k (2n k)(2n k 1) (n k + 1)t k x(x h) (x (k 1)h), h k (1 + t) k k! k=0 k=0 (2n k)(2n k 1) (n k + 1)t k x(x h) (x (k 1)h), h k k! = q n (x) + O(x(x h) (x 2nh)), where t = e ah 1. The constant C n is needed to allow the form of the recurrence relation (24). (26) 5.1. Definition. 5. Elliptic Riccati equations. An elliptic Riccati equation is f(ψ(x)) f(ϕ(x)) a(x) = b(x)f(ϕ(x))f(ψ(x)) + c(x)(f(ϕ(x)) + f(ψ(x))) + d(x). (27) ψ(x) ϕ(x) If x = x m, some point of our x lattice, then ϕ(x) = y m and ψ(x) = y m+1. A first-order difference equation of the kind (27) 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 (27) as e(x)f(ϕ(x))f(ψ(x)) + g(x)f(ϕ(x)) + h(x)f(ψ(x)) + k(x) = 0, (28) where e = b, g = a ψ ϕ c, h = a c, and k = d. ψ ϕ 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) Rational interpolation. We consider now rational interpolation according to the setting of above. Why is this relevant? x y 2n From f n (x) = α n x + β n (x y 2n+1 )f n+1 (x), α nx + β n is the polynomial interpolant of degree 1 to (x y 2n )/f n (x) at y 2n+1 and y 2n+2, so we need f n (y 2n+1 ) and f n (y 2n+2 ) in order to find α n and β n.

16 Marseille-Luminy July 2007 Elliptic lattices. 5 Riccati 16 Now, if f n satisfies the Riccati equation a n (x) f n(ψ(x)) f n (ϕ(x)) ψ(x) ϕ(x) or equivalently = b n (x)f(ϕ(x))f n (ψ(x)) + c n (x)(f(ϕ(x)) + f(ψ(x))) + d n (x), e n (x)f n (ϕ(x))f n (ψ(x)) + g n (x)f n (ϕ(x)) + h n (x)f n (ψ(x)) + k n (x) = 0, (30) where e n = b n, g n = a n ψ ϕ c n, h n = a n ψ ϕ c n, and k n = d n, one finds at x = x 2n, ϕ(x) = y 2n, ψ(x) = y 2n+1, and f n (y 2n ) = 0, so f n (y 2n+1 ) = y 2n+1 y 2n α n y 2n+1 + β n = and at x = x 2n+1, keeping the e g h k form, (29) (y 2n+1 y 2n )d n (x 2n ) a n (x 2n ) (y 2n+1 y 2n )c n (x 2n ) = k n(x 2n ) h n (x 2n ), (31) f n (y 2n+2 ) = y 2n+2 y 2n = g n(x 2n+1 )k n (x 2n ) k n (x 2n+1 )h n (x 2n )., (32) α n y 2n+2 + β n e n (x 2n+1 )k n (x 2n ) + h(x 2n+1 )h(x 2n ) which shows how to extract α n and β n from a n,... at x 2n and x 2n+1. Another form of (32) is (y 2n+2 y 2n )[(y 2n+1 y 2n )e n (x 2n+1 ) + (α n y 2n+1 + β n )h n (x 2n+1 )] + (α n y 2n+2 + β n )[(y 2n+1 y 2n )g n (x 2n+1 ) + (α n y 2n+1 + β n )k n (x 2n+1 )] = 0. (33) Furthermore, the Riccati form is well suited to continued fraction progression: 5.3. Theorem. If f n satisfies the Riccati equation (29) with rational coefficients a n, b n, c n, and d n, and if x y 2n f n (x) = α n x + β n (x y 2n+1 )f n+1 (x), then f n+1 satisfies an equation of same complexity (degree of the rational functions) of its coefficients. x y 2n Indeed, enter f n (x) = α n x + β n (x y 2n+1 )f n+1 (x) in the Riccati equation (27) for f n, actually using the (30) form: ϕ(x) y 2n ψ(x) y 2n e n (x) α n ϕ(x) + β n (ϕ(x) y 2n+1 )f n+1 (ϕ(x)) α n ψ(x) + β n (ψ(x) y 2n+1 )f n+1 (ψ(x)) ϕ(x) y 2n +g n (x) α n ϕ(x) + β n (ϕ(x) y 2n+1 )f n+1 (ϕ(x)) +h ψ(x) y 2n n(x) α n ψ(x) + β n (ψ(x) y 2n+1 )f n+1 (ψ(x)) +k n(x) = 0. or e n (x)(ϕ(x) y 2n )(ψ(x) y 2n ) + g n (x)(ϕ(x) y 2n )[α n ψ(x) + β n (ψ(x) y 2n+1 )f n+1 (ψ(x))] +h n (x)(ψ(x) y 2n )[α n ϕ(x) + β n (ϕ(x) y 2n+1 )f n+1 (ϕ(x))] +k n (x)[α n ψ(x)+β n (ψ(x) y 2n+1 )f n+1 (ψ(x))][α n ϕ(x)+β n (ϕ(x) y 2n+1 )f n+1 (ϕ(x))] = 0 We therefore have a relation between f n+1 (ϕ(x)) and f n+1 (ψ(x)) of the form ẽ n+1 (x)f n+1 (ϕ(x))f n+1 (ψ(x))+ g n+1 (x)f n+1 (ϕ(x))+ h n+1 (x)f n+1 (ψ(x))+ k n+1 (x) = 0, (34)

17 Marseille-Luminy July 2007 Elliptic lattices. 5 Riccati 17 where ẽ n+1 (x) = (ψ(x) y 2n+1 )(ϕ(x) y 2n+1 )k n (x), g n+1 (x) = (ϕ(x) y 2n+1 ) [(ψ(x) y 2n )h n (x) + (α n ψ(x) + β n )k n (x)], h n+1 (x) = (ψ(x) y 2n+1 ) [(ϕ(x) y 2n )g n (x) + (α n ϕ(x) + β n )k n (x)], k n+1 (x) = (ϕ(x) y 2n )(ψ(x) y 2n )e n (x) + (ϕ(x) y 2n )(α n ψ(x) + β n )g n (x) + (ψ(x) y 2n )(α n ϕ(x) + β n )h n (x) + (α n ψ(x) + β n )(α n ϕ(x) + β n )k n (x). The tilde ẽ,etc. notation is needed because (34) is not yet the equation (30) with n replaced by n + 1: the coefficients of (34) will have to be divided by common factors first. The equation (34) is already a Riccati equation, introducing the difference and the sum of f n+1 (ϕ(x)) and f n+1 (ψ(x)) as f n+1 (ψ(x)), f n+1 (ϕ(x)) = f n+1(ϕ(x)) + f n+1 (ψ(x)) ±(ψ(x) ϕ(x)) f n+1(ψ(x)) f n+1 (ϕ(x)), 2 2(ψ(x) ϕ(x)) then (34) takes the form of (27) ( h n+1 (x) g n+1 (x))(ψ(x) ϕ(x)) f n+1(ψ(x)) f n+1 (ϕ(x)) 2(ψ(x) ϕ(x)) (35) = ẽ n+1 (x)f n+1 (ϕ(x))f n+1 (ψ(x)) (h n+1 (x) + g n+1 (x)) f n+1(ϕ(x)) + f n+1 (ψ(x)) k n+1 (x). 2 The coefficients are now symmetric functions of ϕ and ψ, therefore rational functions, as ϕ + ψ = X 1 /X 2 and ϕψ = X 0 /X 2. 2ã n+1 (x) = (h n+1 (x) g n+1 (x))(ψ(x) ϕ(x)) = [(ψ(x) y 2n+1 )(ϕ(x) y 2n )+(ϕ(x) y 2n+1 )(ψ(x) y 2n )]a n (x) +(ψ(x) ϕ(x)) 2 [(y 2n+1 y 2n )c n (x) + (α n y 2n+1 + β n )d n (x)], 2 c n+1 (x) = h n+1 (x)+g n+1 (x) = (y 2n+1 y 2n )a n (x)+[(ψ(x) y 2n+1 )(ϕ(x) y 2n )+(ϕ(x) y 2n+1 )(ψ(x) y 2n )]c +[(ϕ(x) y 2n+1 )(α n ψ(x) + β n ) + (ψ(x) y 2n+1 )(α n ϕ(x) + β n )]d n (x) We must now be sure that no increase of complexity occurs in the new Riccati equation! From ϕ + ψ = X 1 /X 2 and ϕψ = X 0 /X 2, where the X s are second degree polynomials, we see that the new coefficients (h n+1 (x) g n+1 (x))(ψ(x) ϕ(x)), e n+1 (x), h n+1 (x)+g n+1 (x), and k n+1 (x) are rational functions of denominator X 2, and sometimes X 2 2. This problem is settled by multiplying the four coefficients by X 2, assuming that b n, c n, and d n already have the factor X 2. The new coefficients are now polynomials, but of higher degree than before! Fortunately, they have convenient common factors, which can be removed: (1) The four coefficients of (34) vanish at x = x 2n. Indeed, at x = x 2n, ψ(x) = y 2n+1, so that ẽ n+1 and h n+1 do already vanish. Moreover, with ϕ(x) = y 2n, one sees that g n+1 and k n+1 are products containing the factor (y 2n+1 y 2n )h n (x 2n ) + (α n y 2n+1 + β n )k n (x 2n ), which must vanish, according to (31). Remark that x x 2n+1 is an obvious factor of ẽ n+1 and g n+1 ; also a (much less obvious) factor of k n+1, as k n+1 at x = x 2n+1 from (35) gives (33).

18 Marseille-Luminy July 2007 Elliptic lattices. 5 Riccati 18 (2) When n > 0, the four coefficients of (34) vanish at x = x 2n 1. According to the remark just above, if n 1, e n, g n, and k n already vanish at x 2n 1, so ẽ n+1 and h n+1 vanish too. The values of g n+1 and k n+1 at x 2n 1 from (35) contain the same remaining term (ψ(x) y 2n )h n (x) which vanishes at x = x 2n 1 too. If a n, b n, c n, and d n are polynomials, we recover polynomials without increasing the degrees by multiplying ( h n+1 (x) g n+1 (x))(ψ(x) ϕ(x))/2, ẽ n+1 (x), [ h n+1 (x)+ g n+1 (x)], X 2 (x) and k n+1 (x) by, or, considering that (x x 2n 1 )(x x 2n ) (ϕ(x) y 2n )(ψ(x) y 2n ) = X 0(x) + X 1 (x)y 2n + X 2 (x)y 2 2n X 2 (x) we may as well divide by (ϕ(x) y 2n )(ψ(x) y 2n ): = F (x, y 2n) X 2 (x) = Y 2 (y 2n ) (x x 2n)(x x 2n 1 ), X 2 (x) 1 [e n+1, g n+1, h n+1, k n+1 ] = (ϕ y 2n )(ψ y 2n ) [ẽ n+1, g n+1, h n+1, k n+1 ], and the division of (35) by (ϕ(x) y 2n )(ψ(x) y 2n ) yields at last the Riccati coefficients at the (n + 1) th step: g n+1 h n+1 e n+1 k n+1 or g n+1 = M n g n. 0, ϕ y 2n+1, 0, (ϕ y 2n+1)(α n ψ + β n ) ϕ y 2n (ϕ y 2n )(ψ y 2n ) g n ψ y 2n+1, 0, 0, (ψ y 2n+1)(α n ϕ + β n ) = ψ y 2n (ϕ y 2n )(ψ y 2n ) h n (ϕ y 2n+1 )(ψ y 2n+1 ) 0, 0, 0, (ϕ y 2n )(ψ y 2n ) e n α n ψ + β n α n ϕ + β n (α n ψ + β n )(α n ϕ + β n ),, 1, k n, ψ y 2n ϕ y 2n (ϕ y 2n )(ψ y 2n ) A very interesting identity is g n+1 h n+1 e n+1 k n+1 = (ϕ y 2n+1)(ψ y 2n+1 ) [g n h n e n k n ], (ϕ y 2n )(ψ y 2n ) which, by (ϕ(x) y 2n+1)(ψ(x) y 2n+1 ) (ϕ(x) y 2n )(ψ(x) y 2n ) becomes = F (x, y 2n+1) F (x, y 2n ) g n (x)h n (x) e n (x)k n (x) = 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 ) = Y 2(y 2n+1 ) Y 2 (y 2n ) (36) (x x 2n )(x x 2n+1 ) (x x 2n 1 )(x x 2n ), x x 2n 1 [g 0 (x)h 0 (x) e 0 (x)g 0 (x)]. x x 1 (37) The polynomial coefficients are then recovered by a n+1 = (h n+1 g n+1 )(ψ ϕ)/2, b n+1 = e n+1, c n+1 = (h n+1 + g n+1 )/2, and d n+1 = k n+1, using the rational functions ϕ + ψ = X 1 /X 2, ϕψ = X 0 /X 2, (ψ ϕ) 2 = (X 2 1 4X 0 X 2 )/X 2 2, (ϕ(x) y m )(ψ(x) y m ) = (X 0 (x) + y m X 1 (x) + y 2 mx 2 (x))/x 2 (x) = F (x, y m )/X 2 (x) = Y 2 (y m )(x x m )(x x m 1 )/X 2 (x) :

19 Marseille-Luminy July 2007 Elliptic lattices. 6 Classical. 19 g n+1 ±h n+1 = ϕ y 2n+1 h n (ϕ y 2n+1)(α n ψ + β n ) k n ψ y 2n+1 g n (ψ y 2n+1)(α n ϕ + β n ) ϕ y 2n (ϕ y 2n )(ψ y 2n ) ψ y 2n (ϕ y 2n )(ψ y 2n ) = 1 [ ϕ y2n+1 + ψ y ] 2n+1 (h n ± g n ) 1 [ ϕ y2n+1 ψ y ] 2n+1 (h n g n ) 2 ϕ y 2n ψ y 2n 2 ϕ y 2n ψ y 2n (ϕ y 2n+1)(α n ψ + β n ) ± (ψ y 2n+1 )(α n ϕ + β n ) k n (ϕ y 2n )(ψ y 2n ) = 1 2 so, or [ ] 2X0 + (y 2n + y 2n+1 )X 1 + 2y 2n y 2n+1 X 2 X 2 F (x, y 2n ) (h n ± g n ) (y 2n+1 y 2n )(ϕ ψ) (g n h n ) 2F (x, y 2n ) α n[(1 ± 1)ϕψ y 2n+1 (ψ ± ϕ)] + β n [ϕ ± ψ y 2n+1 (1 ± 1)] k n F (x, y 2n ) a n+1 = (h n+1 g n+1 )(ψ ϕ)/2 = 2X 0 + (y 2n + y 2n+1 )X 1 + 2y 2n y 2n+1 X 2 2F (x, y 2n ) [ +(ψ ϕ) 2 y2n+1 y 2n c n + α ] ny 2n+1 + β n d n 2F (x, y 2n )/X 2 F (x, y 2n )/X 2 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. (38) 2F (x, y 2n ) 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 2 c n 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 ) d n 2F (x, y 2n ) 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 + (αnx 2 0 α n β n X 1 + βnx 2 2 )d n F (x, y 2n ) From the theorem of 5.3, a n+1,..., d n+1 must remain polynomials of fixed degree. This can be rechecked, knowing that X 2 is a factor of b n, c n, and d n, that b n (x 2n 1 ) = d n (x 2n 1 ) = 0. And (37) becomes, using g n = a n /(ψ ϕ) c n, h n = a n /(ψ ϕ) c n, and (ψ ϕ) 2 = P/X2 2 = (X1 2 4X 0 X 2 )/X2 2 [ c 2 0 (x) b 0 (x)d 0 (x) c 2 n(x) b n (x)d n (x) P (x) a 2 x x 2n 1 X n(x) = C 2(x) 2 n x x 1 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 ). X 2 2(x) a n ] P (x) a 2 0(x), (39) k n

20 Marseille-Luminy July 2007 Elliptic lattices. 7 Linear difference equations 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 of degree 3, and c n = X 2 times a polynomial of degree 1. 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 2 (x) From the Riccati equation (29) 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 (39), leaving c 2 n(x) b n (x)d n (x) P (x) a 2 X n(x) = C 2(x) 2 n (x x 2n 1 )Q(x), 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 root-free relations come from (38) 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 ) Remark that, from (39), 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 ). Indeed, as explained in 2.3, eq. (17), the product is the value of the algebraic function X n at z j, where X = its conjugate X conj, so that the value is a square root of X (z j )X conj (z j ) = C n (z j y 2n+1 )/(z j x 1 ). Well, from (14), x m = z j means mh + h 0 = a sum of half-periods, so q m q 0 = 1 or ±p 1/2, and we have from (18)-(19) where σ j = ±1 or ±p 1/2. a n (z j ) = θ(q2n q 0σ j ) θ(q 0 σ j) a 0(z j ), j = 1,..., 4 7. Linear difference relations and equations for the numerators and the denominators of the interpolants., the recurrence relations for p n and q n being now ( 4.1.2, p. 13) p n+1 (x) = (α n x + β n )p n (x) (x y 2n 1 )(x y 2n )p n 1 (x), or p n 1 (x) = (α nx + β n )p n (x) p n+1 (x). (x y 2n 1 )(x y 2n )

21 Marseille-Luminy July 2007 Elliptic lattices. 7 Linear difference equations. 21 We now consider the linear recurrence satisfied by combinations of such products, i.e., by combinations of p n (ϕ)p n (ψ), p n (ϕ)q n 1 (ψ), q n 1 (ϕ)p n (ψ), and q n 1 (ϕ)q n 1 (ψ). We just have to consider a product r n s n, kowing that r n 1 = (α nϕ + β n )r n r n+1 (ϕ y 2n+1 )(ϕ y 2n+2 ), and s n 1 = (α nψ + β n )s n s n+1 (ψ y 2n 1 )(ψ y 2n ). Simplest way is to write it as matrix-vector recurrence. r n 1 s n 1 r n s n 1 r n 1 s n = r n s n A(ϕ) A(ψ) B(ψ) B(ϕ) A(ϕ)B(ψ) A(ψ)B(ϕ) A(ϕ)A(ψ) B(ϕ)B(ψ) 1 where A(t) = (t y 2n 1 )(t y 2n ) and B(t) = α n t + β n (t y 2n 1 )(t y 2n ). The matrix is D 1 n M T ne 1 n, where ψ y 2n 1 D n = ϕ y 2n 1 ϕ y 2n+1 E n = ψ y 2n r n+1 s n r n s n+1 r n+1 s n+1 r n s n, (ϕ y 2n 1 )(ψ y 2n 1 ) (ϕ y 2n+1 )(ψ y 2n+1 ), 1 (40) and where M T n is the transposed of the matrix of (36). This means that the recurrence of the r n s n s is basically the adjoint of the recurrence (36) of the (e n, g n, h n, k n ) s Remark that E n D n+1 = (ϕ y 2n+1 )(ψ y 2n+1 )I. So, (40) is ρ n 1 = D 1 n M T n E 1 n ρ n, g n D n ρ n 1 = g n M T n E 1 n ρ n = g n+1 E 1 n ρ n (from (36)), or (ϕ y 2n+1 )(ψ y 2n+1 )g n D n ρ n 1 = g n+1 D n+1 ρ n : g n+1 D n+1 ρ n = (ϕ y 1 )(ψ y 1 )(ϕ y 3 )(ψ y 3 ) (ϕ y 2n+1 )(ψ y 2n+1 )g 0 D 0 ρ 1 so that, for any choice of r n and s n (p n or q n ),

22 Marseille-Luminy July 2007 Elliptic lattices. 7 Linear difference equations. 22 (ψ y 2n+1 )g n+1 r n+1 s n +(ϕ y 2n+1 )h n+1 r n s n+1 +e n+1 r n+1 s n+1 +(ϕ y 2n+1 )(ψ y 2n+1 )k n+1 r n s n = (ϕ y 1 )(ψ y 1 )(ϕ y 3 )(ψ y 3 ) (ϕ y 2n+1 )(ψ y 2n+1 ) [(ψ y 1 )g 0 r 0 s 1 + (ϕ y 1 )h 0 r 1 s 0 + e 0 r 0 s 0 + (ϕ y 1 )(ψ y 1 )k 0 r 1 s 1 ] With the two choices (r, s) = (p, p) and (q, p): (ψ y 2n+1 )g n+1 p n+1 (ϕ)p n (ψ)+(ϕ y 2n+1 )h n+1 p n (ϕ)p n+1 (ψ)+e n+1 p n+1 (ϕ)p n+1 (ψ)+(ϕ y 2n+1 )(ψ y 2n+1 )k n+1 p n (ϕ)p n (ψ) = e 0 (ϕ y 1 )(ψ y 1 )(ϕ y 3 )(ψ y 3 ) (ϕ y 2n+1 )(ψ y 2n+1 ), (ψ y 2n+1 )g n+1 q n+1 (ϕ)p n (ψ)+(ϕ y 2n+1 )h n+1 q n (ϕ)p n+1 (ψ)+e n+1 q n+1 (ϕ)p n+1 (ψ)+(ϕ y 2n+1 )(ψ y 2n+1 )k n+1 q n (ϕ)p n (ψ) = (ϕ y 1 )h 0 q 1 (ϕ)(ϕ y 1 )(ψ y 1 )(ϕ y 3 )(ψ y 3 ) (ϕ y 2n+1 )(ψ y 2n+1 ), Multiply the first equation by q n (ϕ), the second one by p n (ϕ), and subtract: [q n (ϕ)p n+1 (ϕ) p n (ϕ)q n+1 (ϕ)][(ψ y 2n+1 )g n+1 p n (ψ) + e n+1 p n+1 (ψ)] = [q n (ϕ)e 0 p n (ϕ)(ϕ y 1 )h 0 q 1 (ϕ)](ϕ y 1 )(ψ y 1 )(ϕ y 3 )(ψ y 3 ) (ϕ y 2n+1 )(ψ y 2n+1 ), and using the Casorati relation (25) (ψ y 2n+1 )g n+1 p n (ψ) + e n+1 p n+1 (ψ) = [q n (ϕ)e 0 + p n (ϕ)h 0 ]X n+1, (41a) where X n+1 = (ψ y 1)(ψ y 3 ) (ψ y 2n+1 ) (ϕ y 0 )(ϕ y 2 ) (ϕ y 2n ). Similarly, with (r, s) = (p, p) and (p, q), (ϕ y 2n+1 )h n+1 p n (ϕ) + e n+1 p n+1 (ϕ) = [q n (ψ)e 0 + p n (ψ)g 0 ]Xn+1, conj (41b) where X conj n+1 = (ϕ y 1)(ϕ y 3 ) (ϕ y 2n+1 ) is the conjugate of the algebraic function (ψ y 0 )(ψ y 2 ) (ψ y 2n ) X n Difference equation for the denominator p n. Here, e 0 (x) 0. Take (41a) at ψ 1 (y): (y y 2n+1 )g n+1 (ψ 1 (y)))p n (y)+e n+1 (ψ 1 (y))p n+1 (y) = h 0 (ψ 1 (y))p n (ϕ(ψ 1 (y)))x n+1 (ψ 1 (y)), and (41b) at ϕ 1 (y): (y y 2n+1 )h n+1 (ϕ 1 (y))p n (y)+e n+1 (ϕ 1 (y))p n+1 (y) = g 0 (ϕ 1 (y))p n (ψ(ϕ 1 (y)))x conj n+1 (ϕ 1 (y)), so, p n+1 (y) = h 0(ψ 1 (y))p n (ϕ(ψ 1 (y)))x n+1 (ψ 1 (y)) (y y 2n+1 )g n+1 (ψ 1 (y)))p n (y) e n+1 (ψ 1 (y)) = g 0(ϕ 1 (y))p n (ψ(ϕ 1 (y)))x conj n+1 (ϕ 1 (y)) (y y 2n+1 )h n+1 (ϕ 1 (y))p n (y) e n+1 (ϕ 1 (y))

23 Marseille-Luminy July 2007 Elliptic lattices. 7 Linear difference equations. 23 possible only if h 0 (ψ 1 (y))p n (ϕ(ψ 1 (y)))x n+1 (ψ 1 (y)) e n+1 (ψ 1 (y)) At y = some y m : = g 0(ϕ 1 (y))p n (ψ(ϕ 1 (y)))x conj n+1 (ϕ 1 (y)) e n+1 (ϕ 1 (y)) [ gn+1 (ψ 1 (y))) e n+1 (ψ 1 (y)) h n+1(ϕ 1 (y)) e n+1 (ϕ 1 (y)) ] (y y 2n+1 )p n (y) h 0 (x m 1 )p n (y m 1 )X n+1 (x m 1 ) e n+1 (x m 1 ) g 0(x m )p n (y m+1 )X conj n+1 (x m ) e n+1 (x m ) = [ gn+1 (x m 1 ) e n+1 (x m 1 ) h n+1(x m ) e n+1 (x m ) ] (y m y 2n+1 )p n (y m ) Remark that X n+1 (x m 1 ) and X conj n+1(x m ) have the same numerator (y m y 1 )(y m y 3 ) (y m y 2n+1 ), so, 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 ), Therefore, R n (x) = =? If D means p n (x) (x y 0 )(x y 2 ) (x y 2n ) satisfies g 0 (x m ) e n+1 (x m ) R n(y m+1 ) h 0(x m 1 ) e n+1 (x m 1 ) R n(y m 1 ) ] [ hn+1 (x m ) e n+1 (x m ) g n+1(x m 1 ) e n+1 (x m 1 ) p n (y m ) (y m y 1 )(y m y 3 ) (y m y 2n 1 ), then (D (rdp))(y) = At y = some y m : (D p)(y) = p(ψ 1 (y)) p(ϕ 1 (y)), ψ 1 (y) ϕ 1 (y) r(ψ 1 (y)) p(y) p(ϕ(ψ 1 (y))) r(ϕ 1 (y)) p(ψ(ϕ 1 (y))) p(y) y ϕ(ψ 1 (y)) ψ(ϕ 1 (y)) y ψ 1 (y) ϕ 1 (y) (D (rdp))(y m ) = r(x m 1 ) p(y m) p(y m 1 ) y m y m 1 r(x m ) p(y m+1) p(y m ) y m+1 y m x m 1 x m

24 Marseille-Luminy July 2007 Elliptic lattices. 8 Hypergeometric? 24 r(x m )(y m y m 1 ) match if r(x m 1 )(y m+1 y m ) = g 0(x m )e n+1 (x m 1 ). We already encountered a function e n+1 (x m )h 0 (x m 1 ) w(x m ) satisfying a similar difference equation, from Pearson s equation (x m x m 1 )Y 2 (y m ) = g 0(x m 1 ) w(x m 1 ) h 0 (x m 1 ) (x m 1 x m 2 )Y 2 (y m 1 ). So, r(x m )e n+1 (x m ) (y m+1 y m )g 0 (x m ) = w(x m ) Y 2 (y m )(x m x m 1 ). (D p n (y m ) (rd)) (y m y 0 )(y m y 2 ) (y m y 2n ) p n (y m ) 1 = x m 1 x m r(x (y m y 0 )(y m y 2 ) (y m y 2n ) p n (y m 1 ) (y m 1 y 0 )(y m 1 y 2 ) (y m 1 y 2n ) m 1) y m y m 1 p n (y m+1 ) (y r(x m ) m+1 y 0 )(y m+1 y 2 ) (y m+1 y 2n ) p n (y m ) (y m y 0 )(y m y 2 ) (y m y 2n ) y m+1 y m r(x m )e n+1 (x m ) = (y m+1 y m )(x m 1 x m )g 0 (x m ) [ [ ] h0 (x m 1 ) p n (y m ) e n+1 (x m 1 ) (y m y 0 )(y m y 2 ) (y m y 2n ) p n (y m 1 ) (y m 1 y 0 )(y m 1 y 2 ) (y m 1 y 2n ) + 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 ) p n (y m ) (y m y 0 )(y m y 2 ) (y m y 2n ) Therefore, R n (x) = p n (x) (x y 0 )(x y 2 ) (x y 2n ) satisfies (D r(x m )e n+1 (x m ) (rd))r n (y m ) = (y m+1 y m )(x m 1 x m )g 0 (x m ) [ h0 (x m 1 ) e n+1 (x m 1 ) g [ 0(x m ) e n+1 (x m ) + hn+1 (x m ) e n+1 (x m ) g n+1(x m 1 ) e n+1 (x m 1 ) 8. Hypergeometric expansions. ] (ym y 0 )(y m y 2 ) (y m y 2n ) (y m y 1 )(y m y 3 ) (y m y 2n 1 ) From: David R. Masson: The last of the hypergeometric continued fractions, Report-no: OP-SF 12 Sep Dharma P. Gupta; David R. Masson: Contiguous relations, continued fractions and orthogonality Trans. Amer. Math. Soc. 350 (1998), This article is available free of charge Building blocks: ] R n (y m )

25 Marseille-Luminy July 2007 Elliptic lattices. 8 Hypergeometric? 25 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). Indeed, (ϕ(x) y 0 )(ϕ(x) y 1 ) (ϕ(x) y n 1 ) and (ψ(x) y 0 )(ψ(x) y 1 ) (ψ(x) y n 1 ) both vanish at x = x 0, x 1,..., x n 2, and similarly for the {x k }s and the {y k }s. The common denominator is (ϕ(x) y 1 )(ψ(x) y 1 )ϕ(x) y 2 )(ψ(x) y 2 ) = [F (x, y 1 )F (x, y 2 ) F (x, y n )]/Xn 2 (x) = Y 2 (y 1) Y 2 (y n)(x x 0)(x x 1) 2 (x x n 1) 2 (x x n)/x 2 (x) 2, and the numerator is [(ψ n (y y n 1 )ψ n 1 + )(ϕ n (y y n )ϕn 1 + ) (ϕ n (y y n 1 )ϕ n 1 + )(ψ n (y y n)ψ n 1 + )]/(ψ ϕ) = (y 0 + y n 1 y 1 y n )ϕn 1 ψ n 1 +, a symmetric polynomial of degree 2n 2 vanishing at x = x 0, x 1,..., x n 2 and x = x 1, x 2,..., x n 1. The constant C n is found through particular values of x, either x 1 or x n 1 : (y 1 y 1 ) (y 1 y n 1 ) (x 1 x 0 C n = )(x 1 x 1 ) (x 1 x n ) (y 1 y 1)(y 1 y 2) (y 1 y n ) X 2 (x 1 )(x 1 x 0 )(x 1 x 1 ) (x 1 x n 2 ) = (y n y 0 )(y n y 1 ) (y n y n 2 ) (x n 1 x 0 )(x n 1 x 1 ) (x n 1 x n ) (y n y 1)(y n y 2) (y n y n ) X 2 (x n 1 )(x n 1 x 0 )(x n 1 x 1 ) (x n 1 x n 2 ) (Of course, C 0 = 0). Also, (ϕ(x) y 0 )(ϕ(x) y 1 ) (ϕ(x) y n 1 ) (ϕ(x) y 1)(ϕ(x) y 2) (ϕ(x) y n) + (ψ(x) y 0)(ψ(x) y 1 ) (ψ(x) y n 1 ) (ψ(x) y 1)(ψ(x) y 2) (ψ(x) y n) = D n (x) (x x 0)(x x 1 ) (x x n 2 ) (x x 0 )(x x 1 ) (x x n ), where D n is a polynomial of degree 2. Now let us consider the difference equation (23) of the elliptic logarithm with f(x) = N (x y 0 )(x y 1 ) (x y k 1 ) γ k (x y 0 1)(x y 2) (x y k ), as we know that the poles are the y s and that f in interpolated at y 0, y 1,... (cf. Zhedanov [51]). Here, a(x) = (x x 0 )(x x N ), c = 0, and d is a constant time X 2 in (22): (x x 0 )(x x N ) Df(x) = x N X 2 (x) x 0. (x x N) N k=1 γ k C k (x x 0 )(x x 1 ) (x x k 2 ) (x x 1 )(x x 2 ) (x x k ) = x N x 0. Use x x N = x N x k (x x x k 1 x k 1 ) + x k 1 x N (x x k x k 1 x k ): k N k=0 [ x N γ k C ] x k x k x N (x x0 )(x x 1 ) (x x k 1 ) k + γ x k 1 x k+1 C k+1 k x k x k+1 (x x 1 )(x x 2 ) (x x k ) = x N x 0.

26 Marseille-Luminy July 2007 Elliptic lattices. 8 Hypergeometric? 26 So, γ 1 C 1 x 0 x N x 0 x 1 = x N x 0, γ k = x k 1 x k (x 0 x N )(x 1 x N ) (x k 1 x N ) C k (x 0 x N )(x 1 x N ) (x k = 1,, N. k 1 x N ), p n f q n vanishes at x = y 0, y 1, y 2n. Let n p n (x) = δ j (x y 0 )(x y 1 ) (x y j 1 ) 0. We shall manage to represent p n (x)f(x) as a polynomial of degree n (which will be q n ) plus a sum of terms (x y 0)(x y 1 ) (x y k+n 1 ) (x y 1)(x y 2) (x y k ), k = 1, 2,..., N, with vanishing coefficients when k = 1, 2,..., n. x y 0 At n = 1, p 1 (x) = α 0 x + β 0 which interpolates f(x) f(y 0 ) at x = y 1 and y 2 α 0 y 2 + β 0 = whence α 0 = 1 γ 1 γ 1 γ 2 y 1 y 1 y 2 y 2 γ 1 + γ 2 y 2 y 1 y 2 y 2 α 0 y 1 + β 0 = y 1 y 1 γ 1 = (x 1 x 0)(x 2 x N ) (x N x 0)X 2 (x 1 ), y 2 y 0 = y 2 y 0 (y 2 y 0 )(y 2 y 1 ) γ 1 + γ y 2 y 1 2 (y 2 y 1 )(y 2 y 2 ) y 2 y 1 γ 1 + γ 2 y 2 y 1 y 2 y 2 p 1 (x)f(x) = γ 0 (α 0 x + β 0 ) + γ 1 (α 0 x + β 0 ) x y 0 + γ x y 1 2 (α 0 x + β 0 ) (x y 0)(x y 1 ) (x y 1)(x y 2) + Use α 0 x + β 0 = αy k + β 0 y k y k (x y k) + αy k + β 0 y k y (x y k ): k α 0 y 1 + β 0 α 0 y 1 + β 0 (x y 0 )(x y 1 ) α 0 y 2 + β 0 p 1 (x)f(x) = γ 0 (α 0 x + β 0 ) + γ 1 (x y y 1 y 1 0 ) +γ 1 +γ y 1 }{{} y 1 x y 1 2 y 2 y 2 q 1 (x) α 0 y 2 + β 0 (x y 0 )(x y 1 )(x y 2 ) +γ 2 + y 2 y 2 (x y 1)(x y 2) and we have to check that γ 1 α 0 y 1 + β 0 y 1 y 1 + γ 2 α 0 y 2 + β 0 y 2 y 2 = α 0 y 1 + β 0 α 0 y 2 + β 0 γ 1 α 0 + γ 1 + γ y 1 2 y 1 y 2 y 2 = γ 1(y 2 y 2) γ 2 (y 1 y 1) γ 1 (y 2 y 2) + γ 2 (y 2 y 1 ) 1 + γ y 2 y 1 2 γ 1 (y 2 y 2) + γ 2 (y 2 y 1 ) = 0. (x y 0 )(x y 1 ) x y 1

27 Marseille-Luminy July 2007 Elliptic lattices. 9 References. 27 p 1 f(x) q 1 (x) = N k=2 [ α 0 y k γ + β ] 0 α 0 y k+1 + β 0 (x y0 )(x y 1 ) (x y k ) k y k y + γ k+1 k y k+1 y k+1 (x y 1)(x y 2) (x y k ) For a general n, we represent the unknown denominator as p n (x) = n δ j (x y 0 )(x y 1 ) (x y j 1 )(x y j+1) (x y n) j=0. Then, in each term of p n (x)f(x) = n j=0 δ jf(x)(x y 0 )(x y 1 ) (x y j 1 )(x N y j+1 ) (x y n ), we expand f as f(x) = (x y j )(x y j+1 ) (x y j+k 1 ) γ k,j (x y 1 )(x y 2 ) (x y k ), so p n (x)f(x) = n j=0 δ j N k=0 0 γ k,j (x y 0 )(x y 1 ) (x y j+k 1 ) (x y 1)(x y 2) (x y k ) 9. References [1] N.I. Akhiezer, Elements of the Theory of Elliptic Functions, translated from the 2 nd Russian edition (Nauka, Moscow, 1970), Transl. Math. Monographs 79, A.M.S., Providence, [2] Andrews George E.; Askey Richard; Roy Ranjan, Special Functions, Cambridge university press, Encyclopedia of mathematics and its applications. 71, [3] A.I. Aptekarev, Asymptotic properties of polynomials orthogonal on a system of contours, and periodic motions of Toda lattices, Mat. Sb. 125 (167) (1984) (Russian) = Math. USSR Sb. 53 (1986) [4] A.I. Aptekarev, H. Stahl, Asymptotics of Hermite-Pad polynomials, in: A.A. Gonchar, E.B. Saff (Eds.), Progress in Approximation Theory, Springer, Berlin, 1992, pp [5] R. Askey, J. Wilson, Some basic hypergeometric orthogonal polynomials that generalize Jacobi polynomials, Mem. A. M. S. 54 no. 319, pp [6] G.A. Baker, Jr., Essentials of Padé Approximants, Ac. Press, new York, [7] C. Brezinski, History of Continued Fractions and Padé Approximants, Springer, Berlin, [8] B.M. Brown, W.D. Evans, M.E.H. Ismail, The Askey-Wilson polynomials and q-sturm-liouville problems, Math. Proc. Camb. Philos. Soc. 119 (1996) [9] V. P. Burskii, A. S. Zhedanov, The Dirichlet and the Poncelet problems, Reports of RIAM Symposium No.16ME-S1 Physics and Mathematical Structures of Nonlinear Waves Proceedings of a symposium held at Chikushi Campus, Kyushu Universiy, Kasuga, Fukuoka, Japan, November 15-17, No 24.pdf [10] Vladimir P. BURSKII and Alexei S. ZHEDANOV, Dirichlet and Neumann Problems for String Equation, Poncelet Problem and Pell-Abel Equation, Symmetry, Integrability and Geometry: Methods and Applications Vol. 2 (2006), Paper 041, 5 pages arxiv:math.ap/ v1 12 April 2006 [11] J.S. Geronimo, W. Van Assche, Orthogonal polynomials with asymptotically periodic recurrence coefficients, J. Approx. Theory 46 (1986), [12] J.S. Geronimo, W. Van Assche, Orthogonal polynomials on several intervals via a polynomial mapping, Trans. Amer. Math. Soc. 308 (1988), [13] J. Gilewicz, Approximants de Padé, Lect. Notes Math. 667, Springer, Berlin, [14] Gruenbaum, F.Alberto; Haine, Luc: On a q-analogue of Gauss equation and some q-riccati equations, in Ismail, Mourad E. H. (ed.) et al., Special functions, q-series and related topics. Providence, RI: American Mathematical Society, Fields Inst. Commun. 14, (1997).