classical orthogonal polynomials if it is the solution of a di erence equation of the ... in the discrete case, in terms
Konrad-Zuse-Zentrum für Informationstechnik Berlin Takustr. 7, D-14195 Berlin - Dahlem
Wolfram Koepf Dieter Schmersau
Representations of Orthogonal Polynomials
Fachbereich Mathematik und Informatik der Freien Universitat Berlin
Preprint SC 97{06 (February 1997)
Representations of Orthogonal Polynomials Wolfram Koepf Dieter Schmersau
[email protected]
Abstract: Zeilberger's algorithm provides a method to compute recurrence and di erential equations from given hypergeometric series representations, and an adaption of Almquist and Zeilberger computes recurrence and di erential equations for hyperexponential integrals. Further versions of this algorithm allow the computation of recurrence and di erential equations from Rodrigues type formulas and from generating functions. In particular, these algorithms can be used to compute the di erential/di erence and recurrence equations for the classical continuous and discrete orthogonal polynomials from their hypergeometric representations, and from their Rodrigues representations and generating functions. In recent work, we used an explicit formula for the recurrence equation of families of classical continuous and discrete orthogonal polynomials, in terms of the coecients of their di erential/di erence equations, to give an algorithm to identify the polynomial system from a given recurrence equation. In this article we extend these results be presenting a collection of algorithms with which any of the conversions between the di erential/di erence equation, the hypergeometric representation, and the recurrence equation is possible. The main technique is again to use explicit formulas for structural identities of the given polynomial systems.
1 Structural Formulas for Classical Orthogonal Polynomials A family
y(x) = pn(x) = kn xn + : : : (n 2 N := f0; 1; 2; : : : g; kn 6= 0) (1) of polynomials of degree exactly n is a family of classical continuous orthogonal polynomials 0
if it is the solution of a di erential equation of the type
(x) y00 (x) + (x) y0 (x) + n y(x) = 0
(2)
where (x) = ax + bx + c is a polynomial of at most second order and (x) = dx + e is a polynomial of rst order. Since one demands that pn (x) has exact degree n, by equating the highest coecients of xn in (2) one gets 2
n = (an(n 1) + dn) :
(3)
Similarly a family pn (x) of polynomials of degree exactly n, given by (1), is a family of discrete classical orthogonal polynomials if it is the solution of a di erence equation of the type
(x) ry(x) + (x) y(x) + n y(x) = 0 ; 1
(4)
where
y(x) = y(x + 1) y(x) and ry(x) = y(x) y(x 1) denote the forward and backward di erence operators, respectively, and (x) = ax + bx + c and (x) = dx + e are again polynomials of at most second and of rst order, respectively. Again, (3) follows. Since r = r, (4) can also be written in the equivalent form 2
(x) + (x) y(x) (x) ry(x) + n y(x) = 0 ;
and replacing x by x + 1 we arrive at
(x + 1) + (x + 1) y(x) + (x + 1) y(x) + n y(x + 1) = 0 :
(5)
2
It can be shown (see e.g. [14] or [16]) that any solution pn (x) of either (2) or (4) satis es a recurrence equation (n 2 N ; p 0)
pn (x) = (An x + Bn) pn (x) Cn pn (x) +1
or equivalently
1
0
x pn (x) = an pn (x) + bn pn (x) + cn pn (x)
(7)
n bn = B An ;
(8)
+1
with
(6)
1
an = A1 ; n
1
cn = ACn : n
In [14] (compare [16], Eqs. (5) and (10)) we showed that the coecients An ; Bn ; and Cn are given by the explicit formulas An = kkn ; +1
n
Bn = 2 b (nd( +a n2 +a nd) ( da ) 2 ae (+ 2da+n2) a ) kkn n
+1
and
(an + d 2a)n(4ca b )+4a c ab + ae 4acd + db bed + d c (an + d 2a)n kn Cn = k (d 2a +2an) (2an 3a + d) (2an a + d) n in the continuous case, and by the formulas 2
2
2
2
2
2
2
+1 1
An = kkn ; +1
n
Bn = n ( d +( 22abn) ( d2+aa+nd ) (ad) ++ 2e a( dn ) 2 a ) kkn
+1
and
Cn =
n
( n 1 ) ( d + an a ) ( a n d d b a d + a n 2 a n +4 c a + a +2 ea b ) d b e + d c + a e ( d a +2 a n) ( d(a+2n +adn 23aa))n( 2 a n 2 a + d) kkn n 2
2
2
2
+1
2
2
1
2
2
2
in the discrete case, in terms of the coecients a; b; c; d and e of the given di erential/di erence equation. Orthogonal polynomials satisfy further structure equations. One of those is given by the derivative/di erence rules (see e.g. [14])
(x) p0n(x) = n pn (x) + n pn(x) + n pn (x) +1
and
1
(x) rpn (x) = n pn (x) + n pn(x) + n pn (x) +1
or
(n 2 N := f1; 2; 3; : : : g) ;
(n 2 N) ;
1
(10)
(n 2 N ) ;
(x) + (x) pn (x) = Sn pn (x) + Tn pn(x) + Rn pn (x) +1
(9)
1
(11)
respectively. Here
Sn = n ; Tn = n n ; Rn = n : (12) In [14] we showed that the coecients n ; n ; and n are given by the explicit formulas n = a n kkn ; n a db) n = (nd( +a n2 +a nd) ( da ) (22ae + 2an) +1
and
ca b )+ ae + d c bed)(an + d a)(an + d 2a)n kn
n = ((n 1)(an + d (da)(4 2a +2an) (2an 3a + d) (2an a + d) k 2
2
2
n
2
1
in the continuous case, and by the formulas
n = a n kkn ; n
+1
n = n ( d + a n a ) (( 22 aa nn d 2 aa d+ d d) (b d++2 2e aa n )2 a n + 2 a n ) 2
and
2
2
n = (n 1)(d + an a)(and db ad + a n 2a n +4ca + a +2ea b ) dbe + d c + ae kn a) (an + d 2a) n (d a +2(dan+)an (d +2an 3a) (2an 2a + d) kn 2
2
2
2
2
2
2
1
in the discrete case, respectively. Now, we develop further structural identities. Taking the derivative in (2), we get
0 = (x) p000n (x) + (x) + 0 (x) p00n(x) + n + 0 (x) p0n (x) = (ax + bx + c) p000n (x) + (d + 2a) x + (e + b) p00n (x) + (n + d) p0n (x) ; 2
hence y(x) := p0n (x) satis es a di erential equation (a0 x + b0 x + c0 ) y00 (x) + (d0 x + e0 ) y0 (x) + 0n y(x) 2
3
2
of the same type with
a0 = a ; b0 = b ; c0 = c ; d0 = d + 2a ; and e0 = e + b :
(13)
From this we deduce that the equation
x p0n(x) = n p0n (x) + n p0n(x) + n p0n (x) ; +1
(14)
1
namely a recurrence equation for p0n (x), is valid, and from (13) it follows that
n = an(a; b; c; d + 2a; e + b) ; n = bn(a; b; c; d + 2a; e + b) ; and
n = cn(a; b; c; d + 2a; e + b) ; where an (a; b; c; d; e); bn (a; b; c; d; e); and cn (a; b; c; d; e); are given by (8) and the explicit formulas for An ; Bn and Cn . Similarly in the discrete case, applying to (4), we get for y(x) := pn(x)
0 = (x + 1) (x) ry(x) + (x + 1) + (x) y(x) + n + (x) y(x) = (ax + bx + c) ry(x) + (d + 2a) x + d + e + a + b y(x) + (n + d) y(x) ; 2
hence y(x) := pn(x) satis es a di erence equation (a0 x + b0 x + c0 ) ry(x) + (d0 x + e0 ) y(x) + 0n y(x) 2
of the same type with
a0 = a ; b0 = b ; c0 = c ; d0 = d + 2a ; and e0 = d + e + a + b :
(15)
From this we deduce that the equation
x pn(x) = n pn (x) + n pn(x) + n pn (x) ; +1
1
(16)
namely a recurrence equation for pn (x), is valid, and from (15) it follows that
n = an (a; b; c; d + 2a; d + e + a + b) ; n = bn(a; b; c; d + 2a; d + e + a + b) ; and
n = cn(a; b; c; d + 2a; d + e + a + b) ; where an (a; b; c; d; e); bn (a; b; c; d; e); and cn (a; b; c; d; e); are given by (8) and the explicit formulas for An ; Bn and Cn . To obtain a derivative rule for y(x) := p0n (x), we take the derivative of (9) to get (x) p00n(x) + 0 (x) p0n (x) = n p0n (x) + n p0n(x) + n p0n (x) : +1
1
Applying (14) to replace x p0n(x) results in a derivative rule of the form
(x) p00n (x) = a0n p0n (x) + b0n p0n(x) + c0n p0n (x) : +1
1
4
(17)
Similarly in the discrete case a di erence rule of the form
(x) rpn (x) = a0n pn (x) + b0n pn(x) + c0n pn (x) +1
1
(18)
can be obtained for y(x) := y(x). Finally we substitute (17) in the di erential equation. This gives
a0n p0n (x) + b0n p0n(x) + c0n p0n (x) + (x) p0n (x) + n pn(x) = 0 ; +1
1
and replacing x p0n (x) by (14), again, we obtain an equation of the type
pn (x) = abn p0n (x) + bbn p0n(x) + cbn p0n (x) ; +1
1
(19)
in the continuous case, and a similar procedure gives
pn(x) = abn pn (x) + bbn pn(x) + cbn pn (x) +1
1
(20)
in the discrete case. Note that in the discrete case also corresponding equations concerning r are valid. We note in passing that our development shows by simple algebraic arguments that whenever pn(x) is a polynomial system of degree exactly n, satisfying a di erential/di erence equation of type (2)/(4), a recurrence equation of type (6) and a derivative/di erence rule of type (9)/(10) then the system p0n (x) (pn (x)) is again such a system. This has nothing to do with orthogonality. Indeed, in our further development it will become rather important that in the continuous case the powers xn and in the discrete case the falling factorials +1
+1
xn := x (x 1) (x n + 1) = (x n + 1)n = ( 1)n ( x)n which by no means become orthogonal families, have these properties. To deduce the coecients n ; n ; n , a0n ; b0n ; c0n , and abn ; bbn ; cbn , we can follow the above instructions, or we apply the following method: Substituting
pn (x) = kn xn + kn0 xn + kn00 xn + : : : 1
2
in the di erential/di erence equation and equating the coecients of xn determines n , while equating the coecients of xn and xn gives kn0 , and kn00 , respectively, in terms of kn . These values can be substituted in pn (x). Next, we substitute pn (x) in the proposed equation, and equate again the three highest coecients successively to get the three unknowns in terms of a; b; c; d; e; n; kn ; kn , and kn by linear algebra. These computations can be easily carried out by a computer algebra system, e.g. by Maple. With few seconds of computation time, we get Theorem 1 For the solutions of (2) and (4), the relations (14), (17), (19), and (16), (18), (20), respectively, are valid. The coecients n ; n ; n , a0n ; b0n ; c0n , and abn ; bbn ; cbn , are given by 1
1
2
+1
n kn ; n = n + 1 k n
+1
n = 2( db n+(2aann+) (dd a2) a++d 2( ba n )e ) ; 5
n = ( ( n 1 ) ( (adn +2da +a2)a( n4 c) a( 2 ab n) + 3aae ++d d) ( c2 a nb e da) +n (da)n + d a ) kkn ; n a0n = a nn(n+ 1 1) kkn ; n b0n = ( (nd +12)a( an n) (+d d )2( a2 e+a2 a dn b) ) ; d ) ( a n + d a ) n kn ; c0n = ( ( n 1 ) ( a n +(dd a2)a(+4 c2aa n )b ()2+aane +3 ad +cd ) b( 2e da )n( a na + + d) kn abn = n +1 1 kkn ; n 2 e a d b bbn = (d + 2an)(d 2a + 2an) ; 2
2
2
2
1
+1
2
2
2
2
1
+1
cbn = ( ( n ( d1 ) (2aan++2da n )a )((24acna 3ba )++da) e( 2+a nd c a +b edd) ) a n kkn n 2
2
2
2
in the continuous case, and
n kn ; n = n + 1 k n
n =
1
+1
n = n ( d + 2(a2+a n2 b ) 2( da ++ ad n) ( d a+) 2 adn()e a b ) ;
(n 1) (d + an a) (and db ad + a n 2a n +4ca + a +2ea b ) dbe + d c + ae (d + an a) n kn ; (d a + 2an) (d + 2an 3a) (2an 2a + d) kn a0n = a nn(n+ 1 1) kkn ; n b0n = ( n 1 ) ( a n + d )((22aann d 2 aa+d d ) d( db ++ 22 eaan ) 2 a n + 2 a n ) ; 2
2
2
2
2
2
2
2
1
+1
2
2
2
c0n = (n 1)(d + an a)(and db ad + a n 2a n +4ca + a +2ea b ) dbe + d c + ae (d + an a)(an + d)n kn ; ( d a + 2 a n ) ( d + 2 a n 3 a ) ( 2 a n 2 a + d ) kn abn = n +1 1 kkn ; n bbn = 2 a n ( d + a n a ) d b + a d d + 2 e a ; (2an 2a + d)(d + 2an) 2
2
2
2
2
2
2
2
1
+1
2
cbn = (n 1) (d + an a) (and db ad + a n 2a n +4ca + a +2ea b ) dbe + d c + ae kn an (d a + 2an) (d + 2an 3a) (2an 2a + d) kn in the discrete case. 2 2
2
2
2
6
2
1
2
2
2
Note that (19) gives an immediate formula for the antiderivative of a continuous orthogonal polynomial in terms of its neighbors, so that de nite integrals can easily be computed, whereas (20) gives an immediate formula for the antidi erence of a discrete orthogonal polynomial in terms of its neighbors, so that de nite sums can easily be computed. As a direct consequence of Theorem 1 we have the following representations. The de nition of the continuous and discrete families will be given in x 3 and x 5. Corollary 1 The classical continuous orthogonal polynomials have the following antiderivative representations: Z Hn(x) dx = 2 (n1+ 1) Hn (x) +1
(see e.g. [24], (5.5.10)),
Z
Ln (x) dx = Ln (x) Ln (x) (
)
(
(
)
)
+1
(see e.g. [25], VI (1.14)),
Z
+ ) Bn (x) dx = (n + 1) (2n2 +(n ++1 1) (2n + + 2) Bn (x) + (2n + ) (24n + + 2) Bn (x) + (n + ) (2n +2 n) (2n + + 1) Bn (x) ; (
(
)
)
+1
(
)
(
)
1
Z
Cn (x) dx = 2 (n 1+ ) Cn (x) dx 2 (n 1+ ) Cn (x) dx +1
1
(see e.g. [25], V (7.15)),
Z
+ + 1) ; (x) dx P Pn ; (x) dx = (2n + +2 ( n ++ 1) n (2n + + + 2) 2 ( ) ; + (2n + + ) (2n + + + 2) Pn (x) dx 2 (n + ) (n + ) ; (x) dx P n (n + + ) (2n + + ) (2n + + + 1) (
(
)
)
+1
(
)
(
)
1
(see [12], Theorem 6). The classical discrete orthogonal polynomials have the following antidi erence representations:
X
X X
x
x
c (x) ; cn (x) = n + 1 n (
(
)
)
+1
knp (x; N ) = knp (x; N ) p knp (x; N ) ; ( )
( )
( )
+1
mn ; (x) = ( 1)(n + 1) mn ; (x) 1 mn ; (x) ; x X N ) (n + N ) t (x; N ) ; tn(x; N ) = 2 (2n1+ 1) tn (x; N ) 12 tn(x; N ) + (n 2 (2 n + 1) n x (
(
)
)
(
)
+1
+1
1
7
X x
hn ; (x; N ) = (2n + +n ++ 1)+(2 n++1 + + 2) hn ; (x; N ) 2n + 2n + 2n + 2n + N + N + + + h ; (x; N ) n (2n + + ) (2n + + + 2) ) (n + ) (n N ) (n + + + N ) h ; (x; N ) : + ((nn++ + ) (2n + + ) (2n + + + 1) n (
(
)
)
+1
2
2
(
(
)
)
1
Proof: Using the representations for a^n , ^bn and c^n of Theorem 1 with the particular values for a; b; c; d; e and kn of the families (see e.g. [1], [19]) give the results. 2 Note that the representations for a^n , ^bn and c^n of Theorem 1, if applied to Pn (x) = xn or Pn (x) = xn, respectively, yields the simple results Z xn dx = n +1 1 xn ; and X n 1 n x = n + 1x ; +1
+1
x
respectively. The latter is equivalent to the well-known identity m X n+k
k
=0
k
m mXn X = n1! (k + n)n = n1! kn = (n +1 1)! kn k k n m+1 1 = (n + 1)! (n + m)n = n + m : +
=0
=
+1
k k
m n
=
+
n
=
+1
The polynomial system
Kn ; (
)
n (x)= x + 1+ n F 1 x n n 1
1
! 1 n =( 1) F 2
1+
0
n; x +
1+
! ; (21)
which was given in [14], is not orthogonal, but Theorem 1 is still applicable, and we get X ; Kn (x) = (n1+ 1) Kn ; (x) Kn ; (x) : x (
(
)
)
(
)
+1
2 Connection Coecients In this section we would like to consider the problem to determine connection coecients between di erent polynomial systems. Here we assume that Pn (x) = kn xn + : : : (n 2 N ) denotes a family of polynomials of degree exactly n and Qm (x) = km xm + : : : (n 2 N ) denotes a family of polynomials of degree exactly m. Then for any n 2 N a relation of the type 0
0
0
Pn (x) =
n X
m
Cm (n) Qm (x) ;
(22)
=0
is valid, and the coecients Cm (n) (n 2 N ; m = 0; : : : ; n) are called the connection coef cients between the systems Pn (x) and Qm (x). For simplicity we assume that Cm (n) are de ned for all integers n; m and that Cm (n) = 0 outside the above n m-region. 0
8
The connection coecients between many of the classical orthogonal polynomial systems had been determined by di erent kinds of methods (see e.g. [24], [10], [20]) until Askey and Gasper [6] used recurrence equations to prove the positivity of the connection coecients between certain instances of the Jacobi polynomials. In a series of papers ([21]{[22], [3]) Ronveaux et. al. recently used such a method more systematically. Here we will present an algorithmic approach di erent from theirs. Hence, the main idea is to determine recurrence equations for Cm (n). Since Cm (n) depends on two parameters m and n, many mixed recurrence equations are valid as we shall see. The most interesting recurrence equations are those which leave one of the parameters xed. We will determine those recurrence equations, hence pure recurrence equations w.r.t. m and n. The success of this method will heavily depend on whether or not these recurrence equations are of lowest order, i.e., whether or not no recurrence equations of lower order for Cm (n) are valid. In cases when the order of the resulting recurrence equation is one, it de nes a hypergeometric term which can be given explicitly in terms of shifted factorials (or Pochhammer symbols) (a)k = a(a + 1) (a + k 1) = (a + k)= (a) using the initial value Cn (n) = kn =k n . We will see that there are many instances for this situation. Note that w.l.o.g. we could assume that kn = km 1, i.e., that Pen (x) and Qe m (x) are monic systems with connection coecients Cem (n), because if Pn (x) and Qm (x) have leading coecients kn and km , respectively, then their connection coecients Cm (n) are given by
Cm (n) = kn Cem (n) : km
In the last section we have already solved a rather special connection problem: (19)/(20) expresses the connection between the polynomial systems Pn (x) = pn(x) and Qm (x) = p0m (x) or Qm (x) = pm (x), respectively. In this case the connection coecients turn out to be rather simple: almost all of them (namely all with m < n 2) are zero. Now, we consider the generic case. We assume that Pn (x) is a polynomial system given by (2)/(4) with (x) = ax + bx + c, and (x) = dx + e, and that Qm (x) is a polynomial system given by (2)/(4) with (x) = ax + bx + c, and (x) = dx + e. We know then that both Pn (x) and Qm (x) satisfy a recurrence equation (7) whose coecients an (a; b; c; d; e), bn(a; b; c; d; e), and cn(a; b; c; d; e) were given explicitly in the last section. Note that we will denote all coecients connected with Qm (x) by dashes. Hence we have x Pn (x) = an Pn (x) + bn Pn(x) + cn Pn (x) and x Qm (x) = am Qm (x) + bm Qm(x) + cm Qm (x) ; all of an ; bn ; cn ; am ; bm ; cm given explicitly. In three steps, we will now derive three independent recurrence equations for Cm (n). First we consider the term x Pn (x) (see e.g. [22]). Using the de ning equation of Cm (n), and the two recurrence equations for Pn (x) and Qm (x), we get x Pn(x) = an Pn (x) + bn Pn (x) + cn Pn (x) +1
+1
2
2
+1
1
+1
1
n X
+1
=
m n X
1
=0
=
m
an Cm (n + 1) Qm (x) + bn Cm (n) Qm (x) + cn Cm(n 1) Qm (x)
Cm(n) x Qm (x)
=0
9
=
n X
Cm(n) am Qm (x) + bm Qm(x) + cm Qm (x) : +1
m
1
=0
By appropriate index shifts, we can equate the coecient of Qm (x) to get the \cross rule"
an Cm(n + 1) + bn Cm (n) + cn Cm(n 1) = am Cm (n) + bm Cm(n) + cm Cm (n) : (23) 1
1
+1
+1
To deduce a second cross rule in terms of the same variables Cm (n +1); Cm (n); Cm (n 1); Cm (n) and Cm (n), we examine the term x Pn0 (x) (or x Pn (x) in the discrete case). Using both recurrence equations for the derivatives/di erences 1
+1
x Pn0 (x) = n Pn0 (x) + n Pn0 (x) + n Pn0 (x) +1
and
1
x Q0m (x) = m Q0m (x) + m Q0m (x) + m Q0m (x) +1
(or analogously
1
x Pn(x) = n Pn (x) + n Pn(x) + n Pn (x) +1
and
1
x Qm (x) = m Qm (x) + m Qm (x) + m Qm (x) +1
1
in the discrete case), we get
x Pn0 (x) = n Pn0 (x) + n Pn0 (x) + n Pn0 (x) =
n X
m n X
+1
1
n Cm (n + 1) Q0m (x) + n Cm(n) Q0m (x) + n Cm (n 1) Q0m (x)
=0
=
m n X
=0
=
m
Cm (n) x Q0m (x)
Cm (n) m Q0m (x) + m Q0m(x) + m Q0m (x) : +1
1
=0
Again, by appropriate index shifts, we can equate the coecient of Qm (x) to get the cross rule
n Cm (n + 1) + n Cm (n) + n Cm (n 1) = m Cm (n) + m Cm(n) + m Cm (n) (24) 1
1
+1
+1
(and the same result in the discrete case). In a similar way the cross rule
abn Cm(n + 1) + bbn Cm (n) + cbn Cm (n 1) = abm Cm (n) + bbm Cm(n) + cbm Cm (n) (25) 1
1
+1
+1
can be obtained. It turns out, however, that this relation is linearly dependent from (23) and (24), and hence does not yield new information. Now, we specialize a little. First, we consider the continuous case. To obtain reasonably simple results, we assume furthermore that (x) = (x). We consider the term (x) Pn0 (x). Then, using both derivative rules
(x) Pn0 (x) = n Pn (x) + n Pn (x) + n Pn (x) +1
1
10
and
(x) Q0m (x) = m Qm (x) + m Qm (x) + m Qm (x) ; +1
1
we get (x) Pn0 (x) = n Pn (x) + n Pn (x) + n Pn (x) n X
=
m n X
+1
1
( n Cm (n + 1) Qm (x) + n Cm (n) Qm (x) + n Cm (n 1) Qm (x))
=0
=
m n X
Cm (n) (x) Q0m (x)
=
=0
Cm (n) m Qm (x) + m Qm (x) + m Qm (x) : +1
m
1
=0
Again, by appropriate index shifts, this results in the cross rule n Cm (n +1)+ n Cm (n)+ n Cm(n 1) = m Cm (n)+ m Cm(n)+ m Cm (n) : (26) To obtain a pure recurrence equation w.r.t. m, from the three cross rules (23), (24), and (26) by linear algebra we eliminate the variables Cm (n +1) and Cm (n 1); and to obtain a pure recurrence equation w.r.t. n, we eliminate the variables Cm (n) and Cm (n). For simplicity we consider the monic case. Theorem 2 Let Pn(x) be a monic polynomial system given by the di erential equation (2) with (x) = ax + bx + c, and (x) = dx + e, and let Qm (x) be a monic polynomial system given by (2) with (x) = (x), and (x) = dx + e. Then the relation (22) is valid, Cm (n) satisfying the recurrence equation (m n)(am +d a + an)(d + 2am)(d + a + 2am)(d + 3a + 2am) ( d + 2 a m + 2 a ) Cm (n) + ( d b n d + 2 d a m b + d b d + 2 d a m b + 2 d e n a + dde + 2ddbm mbd ed 4a m e m abd + bnda 2eda 4a me 4edam + 2m a e + 2ea n 2ea n mabd + 2mdea + 2 m e a b n d a)( d + 2 a m + 2 a ) ( m + 1 ) ( d + a + 2 a m ) ( d + 3 a + 2 a m ) Cm (n) ( d + 2 a m ) ( m + 1 ) ( a m 2 a + a n d + d ) ( a m + a n + a + d ) (a b m 4 a m c 8 a m c + 2 a m b 4 a d m c + m b d 4 a d c a e + a b c d + b e d 4 a c + b d)( m + 2 ) Cm (n) = 0 with respect to m, with initial values Cn (n) = 1; Cn (n) 0. Furthermore the recurrence equation (d + 2an) (d a + 2an)(d + 2an + 2a)(d + a + 2an)( m +n + 2) ( d + a m + a + a n ) Cm (n + 2) + ( d a + 2 a n ) ( d + a + 2 a n ) ( n + 2 ) ( d + 2 a n ) ( 2ead bd n 2ma e + 2m a e + 2a en 4ea n + 2dbd 2dea + mbda bd + 2eda 4eadn + 2edan abdn bdna + 2a en m abd 4ea n dmbd + 2danb + 2dan b +2ddbn + 2dmea e d + d d e)Cm (n + 1) + ( d + 2 a n + 2 a ) ( n + 2 ) ( n + 1 ) ( a n a m + d d ) ( a n + a m a + d ) ( b e d a e d c 4 a c n d 4 a c n + a b n + n b d ) Cm (n) = 0 with respect to n is valid. 1
1
+1
+1
1
+1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
+1 2
2
2
2
2
2
2
2
2
2
2
+2
+1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
11
2
2
2
2
2
Proof: Using the explicit representations given in the last section in combination with (23), (24), and (26), and elimination of Cm (n+1) and Cm (n 1); or Cm (n) and Cm (n), respectively, yields the results. 2 Note that the recurrence equation given in Theorem 2 reduces to two terms, and hence can be represented by hypergeometric terms, for the connection between Laguerre polynomials (Pn (x) = Ln (x); Qm (x) = Lm (x)), and between the Gegenbauer polynomials (Pn (x) = Cn (x); Qm (x) = Cm (x)). We will consider these and more cases by another method in x 4. Now, let's switch to the discrete case. There are two possibilities to obtain a relation similar to (26). Replacing the derivative by r, the same argument gives (26), again, valid for (x) = (x). If (x) + (x) = (x) + (x), we can replace the derivative by , and adopt the above argument to get the relation 1
(
)
(
+1
)
Sn Cm (n+1)+ Tn Cm (n)+ Rn Cm(n 1) = S m Cm (n)+ T m Cm(n)+ Rm Cm (n) : (27) 1
1
+1
+1
Hence we get
Theorem 3 Let Pn(x) be a monic polynomial system given by the di erence equation (4)
with (x) = ax + bx + c, and (x) = dx + e, and let Qm (x) be a monic polynomial system given by (4) with (x) = (x), and (x) = dx + e. Then the relation (22) is valid, Cm (n) satisfying the recurrence equation 2
(d + 2am + 2a) (d + 3a +2am)(d + a + 2am)(d + 2am)( m +n) ( a n a + d + a m ) Cm (n) ( d + 2 a m + 2 a ) ( d + 3 a + 2 a m ) ( d + a + 2 a m ) ( m + 1 ) ( add + an d + 2ead 2a em + bd m 2na m 2a nm a nd +a n d and 2eadm + 4ea m + badm + 2a m n + adbm +2a mdn 4a m 2ad m 4a dm + dnd 2a m d a md d ma 2a em + 2a mn +andd 2a m 2a en d d dbd 2a m 2a md +2dmand + 2a m nd + 4a em + 4aemd 2aend + 2a ndm 2a mdn 3amdd dm ad 2dbmd 2am bd 2ambd 5dm a + 2ea n d e d d m d + e d + d b n d + a n b d a n b d) Cm (n) + ( m + 1 ) ( d + 2 a m ) ( d + a m + a + a n ) ( m + 2 )(4 a c m + 2 a e m b d m b a m +2eadm badm + 4dcam + 2a d + 4a m 2adbm + 4a m db m +ad m + 2a dm + 6a md + 2d ma + 4a em + 6a m +4dca 2b am + 8a cm + 2a e b a + a m + a + 4a c db bad + d c +ad bd + ae dbe + 2ead +6dm a ) ( a m 2 a d + a n + d ) Cm (n) = 0 2
2
2
2
2
2
2
3
3
2
3
2
2
2
2
2
2
3
2
2
2
2
2
3
2
2
2
3
2
2
3
2
3
2
2
2
2
3
4
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
+1
2
2
2
2
2
2
2
2
2
2
3
2
2
2
3
2
2
2
2
3
2
2
3
2
2
2
2
4
3
3
3
2
2
2
2
2
2
+2
with respect to m, with initial values Cn (n) = 1; Cn (n) 0. Furthermore the recurrence equation +1
( d + 2 a n + 2 a ) ( n + 2 )( n a b d b n d b e + 2 a n e + 4 n a c + 2 d n a + d a n d b n + a n + d c + a e d a n b + 2 d a n e + 4 d c n a) ( n + 1 ) ( a n a + d + a m ) ( a m d + a n + d ) Cm (n) ( d + 2 a n ) ( d a + 2 a n ) ( d + a + 2 a n ) ( n + 2 )( 2 e a d 2 n a m + e d d 2
3
2
2
2
2
2
2
3
2
4
2
2
2
2
2
3
12
2
2
+ 2a nm +amd am d + d nd 2ea m + an dd 2a m n 2a mdn d bn a m d andb + 2a n e + 2a mn + 3a nd + 7a n d +andd + 2ead dmd + 4dn a 4a en ed + 2a ne + d d + 2d an +2dbd + 2a n + ad bd dan b + a md + 2dane + 4a n + 2a n 2dmand 2a m nd + 2a em + 2aemd 4aend 2aed + 2a ndm 2a mdn amdd dbmd am bd + ambd 4ea n + 2 d b n d + 2 a n b d + 2 a n b d + 3 a n d ) Cm (n + 1) + (d + 2an) (d + 2an + 2a)(d a + 2an)(d + a + 2an)( m + n + 2) ( d + a m + a + a n ) Cm (n + 2) = 0 with respect to n is valid. Next let (x) + (x) = (x) + (x), hence a = a, b = b + f , c = c + g, d = d f , e = e g for some constants f; g. Then the relation (22) is valid, Cm (n) satisfying the recurrence equation ( d +f 2am)( d + f 2am 2a) ( d + f a 2am)( d + f 3a 2am) ( m + n ) ( a n a + d + a m ) Cm (n) ( d + f 2 a m 2 a ) ( d + f a 2 a m ) ( d + f 3 a 2 a m ) ( m + 1 )(2 e a m 2 a m n d + 2 a g m + 2 e a d a d bd + d bn + d an + a n d 2a n e + 2a ne a nd + dan b andb 2dane 2aef dan f + 2a m n + 2anmd 2a m am bd ambd 7a m d 3a md 3am d 4amd + f m 4a m 2a m amf b 2a mf n +2a n dm + 2a f mn 2a mn + 2a n m md + f ban +2dg an + 2a m nd + d n + 2a em + a nf dbnf d nf a n f an bf 2ga n + 2ga n 2mf and + mf d + 3f ad 2f a 2f d + 2f d + 4am f d + 2deam + d g + 5a mf + 9a m f md b f ed f gd mf d + 8amf d + f e 2aemf + 2adg m 2af gm 4a dm + 4a f m + df b + f bm 3am f + 2a m g am f b 6f am + f ) Cm (n) ( d + f 2 a m ) ( m + 1 )(4 e a m + 8 a c m 2 b a m + 4a gm + 2ead + 4dca db + d c + ad bd b a + ae + 2a e + 2a d + 4a c + a dbe bad 2aef + 6a m + 4a m am bd 2ambd + 6a m d +6a md + am d + 2amd + 4a m + a m b am + 4a cm 2amf b + 2a em mf d 3f ad + f a + f d f d 3am f d + 2deam + d g 6a mf 6a m f md b + 2dg a f ed 2f ga f gd + mf d 6amf d + f e 2aemf + ag + 4adcm + 2adg m 4af cm 2af gm + f c 2df c 2aeg + dbg + f be f bg + 2a dm 2a f m db m + f b m + f bm + am f +2a m g + f b am f b af b + 2 f a m 2 a f + f b + 2 a g 4 a f c)( m + 2 ) ( d + f a m a n a ) ( a m 2 a + f + a n ) Cm (n) = 0 with respect to m, with initial values Cn (n) = 1; Cn (n) 0. Furthermore the recurrence equation ( n + 2 ) ( d + 2 a n + 2 a )(2 d n a + a n + d c + a e d b e d b n d b n 3
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
3
3
2
2
3
3
3
2
2
2
3
2
3
2
2
2
4
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
3
2
2
2
2
3
3
3
2
2
3
3
3
2
2
4
3
2
2
2
2
2
2
3
2
2
2
3
2
2
2
2
2
3
2
2
2
2
2
2
2
2
2
2
2
3
2
2
2
2
2
2
2
2
2
3
2
2
2
2
2
2
3
2
2
3
2
2
2
2
2
2
2
2
+1
2
2
2
2
2
2
2
3
2
2
2
2
3
2
2
2
2
2
2
2
2
2
2
2
3
2
3
2
2
2
3
3
2
2
2
2
4
2
2
2
2
2
2
2
2
2
3
2
2
2
2
2
2
2
2
2
2
2
2
2
+2
+1
3
2
3
4
2
13
2
2
2
2
3
n a b + d a n + 2 a n e + 4 n a c d a n b + 4 d c n a + 2 d a n e) ( n + 1 ) ( a n a + d + a m ) ( a m + f + a n ) Cm (n) ( d a + 2 a n ) ( d + a + 2 a n ) ( d + 2 a n ) ( n + 2 )( 2 e a m 2 a m n + d + 4 d n a 2 e a d + 2 a n + ad + bd + 4a n + 2a n + d bn + 3d an +4and + 7a n d 2a n e 2a ne + 3a nd + dan b + andb 2dane + 2aef dan f 2a m n 2anmd am bd +ambd a m d + a md am d + 2a mf n 2a n dm + 2a f mn + 2a mn + 2a n m md 2f ban df an + 4dg an 2a m nd +d n + 2a em 2dbnf d nf 2an bf + 4ga n + 4ga n + 2mf and + mf d + mf bd f d + 2deam + d g m d b + 2 d g a f e d + a m f d 2 a e m f 2 d f b) Cm (n + 1) ( d a + 2 a n ) (d + a + 2an)(d + 2an) (d +2an + 2a)( m + n + 2)( d + f am an a) Cm (n + 2) = 0 2
2
2
2
2
2
2
2
2
2
2
3
2
3
2
3
2
2
2
2
3
2
2
3
2
2
2
4
2
2
2
2
2
2
2
2
3
2
2
2
2
2
2
2
2
3
3
2
2
2
2
3
3
3
2
2
2
2
2
2
2
2
3
2
2
2
2
2
2
with respect to n is valid. 2 Note that the recurrence equation for (x) = (x) given in Theorem 3 reduces to two terms, and hence can be represented by hypergeometric terms, for the connection between Charlier polynomials (Pn (x) = cn (x); Qm (x) = cm (x)), between Meixner polynomials (Pn (x) = mn ; (x); Qm (x) = mm; (x)), and between Krawchouk polynomials (Pn (x) = knp (x; N ); Qm (x) = kmp (x; M )), We will consider these and more cases by another method in x 6. (
(
(
)
)
( )
)
( )
( )
3 Hypergeometric Representations: Continuous Case
Note that by Pen ; (x); Cen (x); Le n (x); He n (x); Ben (x) we denote the monic Jacobi, Gegenbauer, Laguerre, Hermite and Bessel polynomials. Their non-monic counterparts have the standardizations (see [1], (22.3), and [2]; Al-Salam denotes the Bessel polynomials by Yn (x)) (
(
)
)
(
)
(
system
kn
; (x) Ln (x) Hn (x) Bn (x) P C n (x) n 2n + + n n n n n n (
1
n
2
)
(
)
n
(
)
n
(
2
!
)
1)
n
!
(
2
(
)
+
+1)
n
2
We get
Theorem 4 Let Pn(x) be a monic polynomial system given by the di erential equation (2)
with (x) = ax + bx + c, and (x) = dx + e. Then the power series coecients Cm (n) given by n X Pn (x) = Cm (n) xm (28) 2
satisfy the recurrence equation
m
=0
(m n)(an + d a + am)Cm (n)+(m +1)(bm + e)Cm (n)+ c(m +1)(m +2)Cm (n) = 0 : (29) +1
+2
In particular, if c = 0, then the recurrence equation (m n)(an + d a + am)Cm (n) + (m + 1)(bm + e)Cm (n) = 0 +1
14
is valid, and we have the hypergeometric representation
Pn (x) =
e d a n db an a d n 4ba a n a n 2
valid for a 6= 0, or
2
F
n; d
1
e b
2
e b n Pn(x) = b d F 1
n
valid for a = 0; b 6= 0, or nally
e n
Pn (x) = d
1
F
1)
!
n d e b x ; b
1
0
! a x ; b
n a a
+(
(30) (31)
!
n d e x ;
(32)
valid for a = 0; b = 0. Therefore, the classical continuous orthogonal polynomials and their monic counterparts have the following hypergeometric power series representations: ! n+ n; n + + + 1 1 x ; Pn (x) = F 2 n +1 ! 2n + + x 1 n n; n 2 F = n 2 2n 1 x ! n+ n; n + + + 1 1 + x n = ( 1) F 2 n +1 ! 2n + + x + 1 n n; n 2 = F ; n 2 2n 1 + x ! n= 2 ; n= 2 + 1 = 2 1 ; Cen (x) = xn F n +1 x ! n xn n=2; n=2 + 1=2 1 ( ) 2 n F ; Cn (x) = n! n + 1 x ! ! n; n n 1 ; n n F x = x Le n (x) = (1 + )n ( 1) F x 1+ ! ! n+ n n; n 1 n ( x ) Ln (x) = n F x ; x = n! F 1+ ! n= 2 ; n= 2 + 1 = 2 1 ; He (x) = xn F (
)
2
1
2
2
)
(
1
)
1
1
2
1
1
1
2
2
(
2
1
2
2
1
2
1
0
0
x ! n= 2 ; n= 2 + 1 = 2 1 ; Hn(x) = 2n xn F x ! ! n n; n + + 1 n x 2 n Ben (x) = (n + + 1) F 2 = x F 2n x2 : n ! ! n; n + + 1 x n 2 ( n + + 1) n n Bn (x) = F x F 2 = : 2n 2n x n
2
0
2
(
)
(
)
2
2
2
0
2
0
0
1
1
1
1
These results are all particular cases of the recurrence equation (29). 15
Proof: Substituting the power series (28) into the di erential equation, and equating the coecients yields the recurrence equation (29). For c = 0 this recurrence equation degenerates to a two-term recurrence equation, and hence establishes the hypergeometric representations (30){(32), using the initial value Cn (n) = 1. A shift in the x-variable then generates the representations for the Jacobi polynomials. The two points of development x = 1 and x = 1 correspond to the zeros of (x). Note that some of the hypergeometric representations correspond to each other by changing the direction of summation. The other representations follow by substituting the particular parameters a; b; c; d, and e into the recurrence equation (29), and using the initial value Cn (n) = kn (or Cn (n) = 1 in the monic case). 2 We would like to mention that the recurrence equation (29) carries the complete information about the hypergeometric representations given in the theorem. The method described results in four di erent hypergeometric representations for the Jacobi polynomials. Many more hypergeometric representations exist, but the algorithmic procedure presented nds power series representations only. For example, the representation (see e.g. [1] (22.5.45)) ! n + x 1 n n; n x + 1 ; F Pn (x) = n 2 + 1 x 1 cannot be discovered by this method. The method was able to nd hypergeometric series representations with point of development x = 0 for the Gegenbauer polynomials which are speci c Jacobi polynomials, but failed in the Jacobi case, though. One might ask whether such a representation exists. This question can be completely answered by an algorithm of Petkovsek [18]. Petkovsek's algorithm nds all hypergeometric term solutions of holonomic recurrence equations, i.e., homogeneous linear recurrence equations with polynomial coecients. Using the recurrence equation (29), an application of Petkovsek's algorithm proves that the Jacobi polynomials do not generally have a hypergeometric series representation at the origin. Note that the method of the last section, although more complicated, does also give the recurrence equation (29), and hence the above results. 1
(
2
)
2
1
0
4 Power Representations Whereas in the last section we considered the speci c connection coecient problem for Qm (x) = xm, in this section the opposite problem, having Pn(x) = xn, is studied. In many applications, one wants to develop a given polynomial in terms of a given orthogonal polynomial system. In this case handy formulas for the powers xn are very welcome.
Theorem 5 Let Qm(x) be a monic polynomial system given by the di erential equation (2)
with (x) = ax + bx + c, and (x) = dx + e. Then the coecients Cm (n) of the power representations n X xn = Cm (n) Qm (x) 2
satisfy the recurrence equation
m
=0
( n m ) ( d + 2 a m ) ( d + 3 a + 2 a m ) ( d + a + 2 a m ) ( d + 2 a m + 2 a ) Cm (n) 2
16
+ (de + bd + 2dbm + 2am b + 2amb + 2ean dbn)(d + 2am + 2a) ( m + 1 ) ( d + 3 a + 2 a m ) ( d + a + 2 a m ) Cm (n) ( m + 2 )( 4 a c m + ab m + 2ab m 4acmd 8a cm + mb d ae d c + bed 4a c 4 a c d + a b + b d)( a m + a n + a + d ) ( m + 1 ) ( d + 2 a m ) Cm (n) = 0 : 2
2
+1
2
2
2
2
2
2
2
2
(33)
2
2
2
+2
If c = 0, then the recurrence equation ( n m ) ( d+2 a m ) ( d+a+2 a m ) Cm (n) + ( m+1 ) ( b m+e ) ( a m+n a+d ) Cm (n) = 0 (34) +1
is valid, and we get (a b 6= 0)
e Cm(n) = db n
b a
a n
!n ( n)m d a d m a m a m 4a e an : d b +
2
b m
(35)
2
+
a
m! m
Therefore, the following representations for the powers in terms of the classical continuous orthogonal polynomials are valid: n ( + + 2m + 1) ( + + m + 1) X ; ( + m + 1) ( + + n + m + 2) ( n)m Pm (x)
(1 x)n = 2n ( + n + 1)
(
)
m
=0
(see e.g. [20], 136, Eq. (2), or [17], x 5.2.4; note the essential misprint in this formula!), (1 + x)n = 2n ( + n + 1)
n X
m
=0
xn =
bX n= c 2
k
+ m + 1) ( n) P ; (x) ; ( 1)m ( ( ++ m++2m1)+ (1) +( + m m + n + m + 2) (
( n=2)k ( n=2 + 1=2)k ( n )k 1 k Ce (x) ; n k ( n=2 =2)k ( n=2 =2 + 1=2)k k! 4 2
=0
bX n= c ( n=2 =2 + 1)k ( n )k ( 1)k C (x) xn = ( )n! 2n n k ( n=2 =2)k k! n k bX n= c n + 2k C (x) n ! = 2n n k k k! ( )n k 2
2
=0
2
2
+1
=0
(see e.g. [20], 144, Eq. (36), or [17], x 5.3.4),
xn = (1 + )n
n X ( n)m
( 1)m Le m (x) ; (1 + ) m ! m m (
)
=0
n ( n) n X m L (x) = n! X n + ( 1)m L (x) m m n m m (1 + )m m (see e.g. [20], 118, Eq. (2), or [17], x 5.5.4), bn= c n X ( n=2)k ( n=2 + 1=2)k e
xn = (1 + )n
(
)
(
=0
x =
=0
2
k
=0
k!
17
Hn k (x) ; 2
)
)
xn =
bX n= c ( 2
k
n= c 1 n=2)k ( n=2 + 1=2)k H (x) = n! bX n k n k n k! 2 2 k k! (n 2k)! Hn k (x) 2
2
=0
2
2
=0
(see e.g. [20], 110, Eq. (4), or [17], x 5.6.4), n
xn = ( ( +2)2)
n ( n) ( =2 + 1) ( =2 + 3=2) X m m m 2m Be (x) : =0
n
xn = ( ( +2)2) n
m (
(n + 2 + )m m!
n m
)
n ( n) ( + 1) ( =2 + 3=2) X m m m B (x)
m m (n + 2 + )m ( =2 + 1=2)m m! n X ( n)m ( + m + 1) B (x) = ( 2)n (2m + + 1) m ! (n + m + + 2) m m (
)
=0
(
)
=0
(see [2], (7.5); note the essential misprint in this formula!; compare [20], 150, Eq. (7)). Proof: In x 2 it was shown how one obtains three essentially di erent cross rules for the connection coecients between Pn (x) and Qm (x). We modify this method here. For Qm (x), we have the di erential equation
(x) Q00m (x) + (x) Q0m (x) + m Qm (x) = 0 with (x) = ax + bx + c, and the derivative rule 2
(x) Q0m (x) = m Qm (x) + m Qm (x) + m Qm (x) ; and it is easily seen that our current Pn (x) = xn satis es any of the derivative rules (x) Pn0 (x) = a n Pn (x) + b n Pn (x) + c n Pn (x) : Hence in our situation, we get the two cross rules (23) with an = 1, bn = cn = 0 Cm (n + 1) = am Cm (n) + bm Cm(n) + cm Cm (n) and (25) with abn = 1=(n + 1), bbn = cbn = 0 +1
1
+1
1
1
1
+1
+1
(36) (37)
1 C (n + 1) = ab C (n) + bb C (n) + cb C (n) (38) m m m m m m n+1 m which we had deduced in x 2. Using the derivative rule (36), we obtain the third cross rule 1
1
+1
+1
a n Cm (n +1)+ b n Cm(n)+ c n Cm(n 1) = m Cm (n)+ m Cm(n)+ m Cm (n) : (39) 1
1
+1
+1
To receive the recurrence equation (33), we use Theorem 1 writing the cross rules in terms of a; b; c; d; and e, only. Then by linear algebra we eliminate the variables Cm (n +1) and Cm (n 1) to obtain a pure recurrence equation w.r.t. m. (Similarly by elimination of the variables Cm (n) and Cm (n) a pure recurrence equation w.r.t. n is obtained.) A shift by one gives (33). If c = 0, then the recurrence equation has still three terms, unfortunately. But since for c = 0 in neither of the three cross rules (37){(39) the variable Cm (n 1) does occur, we can do a similar elimination, this time eliminating the variables Cm (n+1) and Cm (n), leading to the 1
+1
1
18
rst order recurrence equation (34). Hence the hypergeometric representation (35) follows. The power representations for the Jacobi, Laguerre and Bessel polynomials are special cases thereof. In the case of Hermite and Gegenbauer polynomials, (33) contains only the two terms Cm (n) and Cm (n), which leads to the desired representations. 2 Note that, again, the recurrence equation (34) carries the complete information about the hypergeometric representations given in the theorem. As an immediate consequence of the above theorem, we get the following connection coecient results. Corollary 2 The following connection relations between the classical orthogonal polynomials are valid: n X n + + 1) (n + m + + + 1) (2m + + + 1) ((m Pn ; (x) = + + 1) (n + + + 1) m (m + + + 1) ( )n m ; (n + m + + + 2) (n m)! Pm (x) ; +2
(
)
=0
(
)
(see e.g. [4], (13)),
Pn ; (x) = (
)
n X
n + + 1) (n + m + + + 1) ( 1)n m (2m + + + 1) ((m + + 1) (n + + + 1) m (m + + + 1) ( )n m P ; (x) (n + m + + + 2) (n m)! m =0
(
)
(see e.g. [6], (2.8)), bX n= c ( ) (n 2k + ) (k + ) (n k + ) C (x) n k ( ) ( ) m k! (n k + + 1) (see e.g. [5], (3.42)), n ( ) X n m L (x) Ln (x) = ( n m )! m m (see e.g. [20], 119, Eq. (2)), 2
Cn (x) =
2
=0
(
)
(
)
=0
n ( n) ( +1) ( =2+3=2) (n + +1) n ( )n X m m m m ( 1)m B (x) Bn (x) = ( 1)( +2) m ( n +2+ ) ( = 2+1 = 2) ( +1 n ) n m m m m! m n X ( n)m ( + m +1) (n + +1)m ( +1) B (x) ; ( 1)m (2m + +1) m = ! (n + m + + 2) (m n + + 1) m m (
(
)
=0
(
=0
(see [2], (8.2); note the essential misprint in this formula!). Proof: We want to nd the coecients Cm (n) in the relation (22)
Pn(x) =
n X
m
Cm(n) Qm (x) :
=0
19
)
)
Combining
X
Pn (x) =
Aj (n) xj
and
j 2Z
yields the representation
Pn (x) =
XX
xj =
X
Bm (j ) Qm (x)
m2Z
Aj (n) Bm (j ) Qm (x) ;
j 2Zm2Z
and interchanging the order of summation gives
Cm (n) =
X
Aj (n) Bm (j ) :
j 2Z
Similarly, if (as in the Gegenbauer case)
Pn (x) =
X
Aj (n) xn
j 2Z
one gets
j
Dm (n) =
with
and
2
Pn(x) =
X
xj =
X
Bm (j ) Qj
m (x)
2
m2Z
Aj (n) Bm j (n 2j )
j 2Z n X m
Dm (n) Qn
m (x) :
2
=0
Since the summand F (j; m; n) := Aj (n) Bm (j ) turns out to be a hypergeometric term with respect to (j; m; n), i.e., the term ratios F (j + 1; m; n)=F (j; m; n), F (j; m + 1; n)=F (j; m; n), and F (j; m; n + 1)=F (j; m; n) are rational functions, Zeilberger's algorithm ([26], [11], see e.g. [9]) applies and nds recurrence equations for Cm (n) with respect to m and n. In all cases considered, Zeilberger's algorithm nds recurrence equations of rst order with respect to m (as well as for n). The given representations follow then from the initial value 2 Cn(n) = kn=kn . For some applications, it is important to know the rate of change in the direction of the parameters of the orthogonal systems, given in terms of the system itself. By a limiting process, these parameter derivative representations can be obtained from the results of Corollary 2. Corollary 3 The following representations for the parameter derivatives of the classical orthogonal polynomials are valid:
; 1 @ P ; (x) = nX Pn (x) + n @ m + +1+m+n + + 1 + 2 m ( + m + 1)n m P ; (x) n m ( + + m + 1)n m m 1
(
)
(
)
=0
(
(see [7], Theorem 3), nX
2n m
)
2m + +
m @ Pe ; (x) = + + 1 + 2 m ( + m + 1)n m Pe ; (x) ; n m @ m n m 2n + + + + 1 + m + n ( + + m + 1)n m 1
(
)
=0
(
n
20
)
; 1 @ P ; (x) = nX Pn (x) + n @ m + +1+m+n + m + 1)n m P ; (x) ( 1)n m + n+ 1m+ 2 m ( (+ + m + 1)n m m 1
(
)
(
)
=0
(
(see [7], Theorem 3), nX
)
2m + +
2)n m
m @ Pe ; (x) = + + 1 + 2 m ( + m + 1)n m Pe ; (x) ; ( n @ n m 2n + + + + 1 + m + n ( + + m + 1)n m m m 1
(
)
(
n
=0
2 (1 + m) 2 @ C (x) = nX + 2 + m + n Cn (x) @ n (2 + m ) (2 + 1 + 2 m ) m nX 2 (1 + ( 1)n m ) ( + m) + Cm (x) (2 + m + n ) ( n m ) m 1
=0
1
=0
(see [12], Theorem 10),
@ Ce (x) = nX 2m @ n m 1
n
+1
=0
bX n= c
( )m n! (1 + ( 1)n m ) ( + m) Ce (x) ( )n m! (2 + m + n) (n m) m
n!
2
=
k
=1
k 4k (n
( + n 2k)
2
n 2k + Ce (x) ; 2k)! k (n k + ) n k 2
@ L (x) = nX 1 L (x) ; m @ n m n m 1
(
)
(
)
=0
(see [12], Theorem 10),
@ Le (x) = nX ( 1)n m n! Le (x) ; @ n n m m! m m 1
(
)
(
)
=0
@ B (x) = nX 1 Bn (x) + n @ m +n+m+1 ! ( 1)n m 2m(n+ m+) 1 ( + m +n1) n 1
=0
m
@ Be (x) = nX n! ( 2)n m n @ ( n m ) ( + n + m + 1) ( + 2 m + 2) n m
m! Bm(x) ;
e
1
2
=0
Proof: Given the connection relation
Pn (x) =
n X m
Cm (n; ; ) Pm (x) ;
=0
21
2
m
1
m! Bm (x) :
)
we build the di erence quotient n C (n; ; ) Pn (x) Pn (x) = X m Pm (x) Pn (x)
nX C (n; ; ) m Pm (x) ; = Cn (n ; ; ) 1 Pn (x) + m m
=0
1
=0
so that with !
@ P (x) = lim Cn(n; ; ) 1 P (x) + nX lim Cm(n; ; ) P (x) (40) n m ! @ n m ! 2 since the systems Pn (x) are continuous w.r.t. . This gives the results. Note that for monic polynomials (and moreover if kn does not depend on as in the Laguerre 1
=0
case) the rst limit in (40) equals zero. Hence the parameter derivative representations are simplest in such a case.
5 Hypergeometric Representations: Discrete Case By hn ; (x; N ) and Qn (x; ; ; N ) we denote two commonly used standardizations of the Hahn polynomials (see [19], and [23]), and by mn ; (x), knp (x; N ) and cn (x) the Meixner, Krawchouk and Charlier polynomials are denoted, respectively. They have the standardizations system hn ; (x; N ) Qn (x; ; ; N ) mn ; (x) knp (x; N ) cn (x) Kn ; (x) + + 2n n n n n kn 1 n Nn n n n (
)
(
(
)
(
(
(
+
+
)
(
+1)
)
)
1
+1)
( )
(
( )
(
1
)
)
(
)
1
!
The polynomials tn (x; N ) := hn ; (x; N ) are the discrete Chebyshev polynomials. The polynomials Kn ; (x) given in (21), are not orthogonal, but satisfy the di erence equation ry(x) + ( x + ) y(x) + n y(x) = 0 : (0 0)
(
)
The monic counterparts of the discrete systems will be denoted by he n ; (x; N ), Qe n (x; ; ; N ), ten (x; N ), me n ; (x), kenp (x; N ) and cen (x), respectively. Observe that therefore by ehn we do not denote the Hahn-Eberlein polynomials ehn; (x; N ) as in [19]. In the continuous case, we looked for power series representations, i.e., we set Qm (x) = xm . The corresponding choice in the discrete case is a representation in terms of the falling factorial Qm(x) = xm := x (x 1) (x m + 1) = (x m + 1)m = ( 1)m ( x)m : We get Theorem 6 Let Pn(x) be a monic polynomial system given by the di erence equation (4) with (x) = ax + bx + c, and (x) = dx + e. Then the series coecients Cm (n) given by (
(
)
( )
(
)
)
(
)
2
Pn (x) =
n X
m
=0
22
Cm (n) xm
(41)
satisfy the recurrence equation ( a n + a m a + d ) ( n m ) Cm (n) (42) + ( m + 1 ) ( a n 2 a m a n a m + n d 2 d m b m d e ) Cm (n) ( m + 1 ) ( m + 2 ) ( a m + 2 a m + d m + b m + a + d + b + c + e ) Cm (n) = 0 : 2
2
+1
2
+2
If c = 0, then the recurrence equation ( n m ) ( a m + d + a n a ) Cm (n) ( m + 1 ) ( a m + m b + m d + e ) Cm (n) = 0 (43) 2
+1
is valid, and we have the hypergeometric representation
Pn(x) =
b d p b d 2 ae b d p b d 2 1n a a d d na
d
+ +
a
( + )
( + )
2
2
+ +
(
a n
a n d x; n 1 + a p ae b d b d 2 ae ; a
)
a
valid for a 6= 0, or
2
+
4
2
ae
4
2
2
0 n; F@b d pb d2 3
+
4
(
)
4
2
n
1 n 4
1 1A ;
(44)
! e b n n; x d Pn(x) = b + d 1+ d ; F e n b d b+d valid for a = 0; b + d = 6 0, or nally ! e n n; x b Pn(x) = b F e ; 2
(45)
1
+
2
(46)
0
valid for a = 0; d = b. Therefore, the classical discrete orthogonal polynomials and their monic counterparts have the following hypergeometric series representations:
hn ; (
)
n (x; N ) = ( n1)! ( + 1)n (N n)n F 3
(see e.g. [19], p. 54, Table 2.4),
ehn ; (
)
)n (x; N ) = (1(1++ n)n+(1 + N )n F 3
!
2
n; x; n + 1 + + 1 + 1; 1 N
!
n; x; n + 1 + + 1 ; + 1; 1 N
2
!
n; x; n + 1 tn(x; N ) = ( n)n F 1 ; 1; 1 N ! n; x; n + 1 n ! (1 N ) n 1 ; ten(x; N ) = (1=2) 4n F 1 ; 1 N n ! n; x; n + 1 + + Qn (x; ; ; N ) = F 1 + 1; N 1)n (N
3
3
3
2
23
2
2
(see e.g. [23], 1.5)
Qe n (x; ; ; N ) = ehn ; (x; N + 1) ; (
mn ; (
)
(x) = ( )n F 2
)
n; x 1 1
1
(see e.g. [19], p. 54, Table 2.4),
me n ; (
)
knp
n F (x) = ( )n 1 2
( )
1)n N
(x; N ) = (
n
2
F
!
( )
2
cn (
)
(x) = F 2
cen
)
(x) = (
)n
!
1
0
2
n; x 1 ; N p
n; x 1
(see e.g. [19], p. 54, Table 2.4), (
n; x 1 N p
1
(see e.g. [19], p. 54, Table 2.4),
kenp (x; N ) = ( N )n pn F
F
0
!
n; x 1 1 ;
1
pn
!
!
!
n; x 1 :
These results are all particular cases of the recurrence equation (43). Proof: Substituting the series (41) into the di erence equation, and equating the coecients of the falling factorials yields the recurrence equation (42) which had been obtained by Lesky [16]. This conversion can be easily done using a computer algebra system by bringing the given di erence equation into the form (5), expanding it, and replacing any occurrence of y(x) by (m + 1) Cm , any occurrence of a product x y(x) by Cm + m Cm and any occurrence of a shift y(x + 1) by Cm + (m + 1) Cm since +1
1
+1
xm = m xm ; x xm = xm + m xm ; and (x + 1)m = xm + m xm : 1
+1
1
Iteratively for all nonnegative integers j; k any of the terms xj k y(x) and xj y(x + k) can be replaced by these rules. Note that this method can also be applied for higher order di erence equations with polynomials coecients. Di erent from the continuous case, the recurrence equation (42) does not degenerate to a two-term recurrence equation for c = 0. To get (43), nevertheless, we must use a di erent approach. One possibility is to apply Petkovsek's algorithm to the recurrence equation (42), leading to (43). Another possibility is to modify the method which will be used in the next section to deduce representations of the falling factorials in terms of discrete orthogonal systems. This method yields (43) directly. 24
As soon as (43) is deduced, the initial value Cn (n) = 1 gives the hypergeometric representations (44){(46) which include all other representations by substituting the particular parameters a; b; c; d, and e. 2 We would like to mention that, again, a single recurrence equation, (42), carries the complete information about the hypergeometric representations given in the theorem. Note furthermore, that the radicals in (44) do only occur by the representation used: the radical factors come in pairs whose product is radical-free. Note that the computation which gives (44), answers a question raised by Koornwinder [15]. For more examples of this type see [13]. Our method was able to nd hypergeometric series representations for the particular case c = 0. This is the most important situation since all the classical discrete orthogonal families are of this type, corresponding to the fact that their discrete support has zero as left boundary point (see e.g. [19], Tables 2.1{2.3). By construction, all the series representations determined have an upper parameter x. The question remains, however, whether or not such a hypergeometric series representation might be valid for c 6= 0, too. In general, the answer is no. Petkovsek's algorithm shows that the recurrence equation (42) does not generally have a hypergeometric term solution. Note that the hypergeometric representation (21) for Kn ; (x) is not of this type, and cannot be obtained by the given method. By Petkovsek's algorithm there is no representation (41) with a hypergeometric term Cm (n) for these polynomials. (
)
6 Falling Factorial Representations Whereas in the last section we considered the speci c connection coecient problem for Qm (x) = xm , in this section the opposite problem, having Pn(x) = xn , is studied.
Theorem 7 Let Qm(x) be a monic polynomial system given by the di erence equation (4)
with (x) = ax + bx + c, and (x) = dx + e. Then the coecients Cm (n) of the falling factorial representations n X (47) xn = Cm (n) Qm (x) 2
m
=0
satisfy the recurrence equation
( 2 m a + a + d ) ( 2 m a + 3 a + d ) ( 2 m a + 2 a + d ) ( 2 m a + d ) ( n m ) Cm (n) + ( 2 m a + a + d ) ( 2 m a + 3 a + d ) ( 2 m a + 2 a + d ) ( m + 1 )(2 m n a 2 m a + m ad + 2m ab + 2mna + 2mnad 2ma mad + 2mab + md + 2 m d b + n a d + 2 n a e n d b a d + d b + d e) Cm (n) +( m + 1 ) ( 2 m a + d )(m a + 4 m a + 2 m a d + 6 m a + 6 m a d + 4m a c + 2m a e + m ad m adb m ab + 4ma + 6ma d + 8ma c + 4ma e + 2mad 2madb + 4madc + 2made 2mab md b mdb + a + 2a d +4a c + 2a e + ad adb + 4adc + 2ade ab + ae d b + d c d b d b e)( m + 2 ) ( m a + n a + a + d ) Cm (n) = 0 : (48) 2
2
2
2
2
2
2
2
2
+1
4
2
2
2
2
2
3
2
2
2
3
3
2
2
2
2
2
3
2
3
2
3
2
2
2
2
2
2
2
+2
25
2
2
2
2
2
2
3
2
2
If c = 0, then 0 = (d + a + 2 a m) (d + 2 a m) ( n + m) Cm (n) (a n + d + a m) (m + 1) (a m + m d + m b + e) Cm (n) : 2
(49)
+1
Therefore, the following representations for the falling factorials in terms of the classical discrete orthogonal polynomials are valid:
xn =
n ( + 1) (1 N ) ( 1)n (1 + + + 2m) ( n) (1 + + ) X n n m m h ; (x; N ) ; (
m
=0
xn =
( + + 2)n (1 + + ) (n + 2 + + )m ( + 1)m (1 N )m m
)
n ( + 1) (1 N ) ( 1)n ( n) ( =2 + =2 + 1) ( =2 + =2 + 3=2) 4m X n n m m m he ; (x; N ) ;
m (n + 2 + + )m ( + 1)m (1 N )m m! m n (1 + ) ( N ) ( 1)n ( + + 1 + 2m) ( n) (1 + + ) X m m Q (x; ; ; N ) n n xn = ( + + 2) ( + + 1) ( n + 2 + + ) m ! m n m m (compare [8], (4.2){(4.3)), (
( + + 2)n
=0
)
=0
n ( n) (1 + 2m) )n ( 1)n X m tm (x; N ) ; xn = (1 N n+1 ( n + 2) m (1 N )m m =0
n ( n)m (3=2)m 4m te (x; N ) ; )n ( 1)n X xn = (1 N m n+1 m (n + 2)m (1 N )m m! =0
n ( 1)n ( )n X
n
( n)m ; mm (x) ; ( )m m! m n m n ( 1)n ( )n ( n)m ; X me m (x) ; xn = ( )m m! m n ( 1)n ( N ) pn m ( n) X n m k p (x; N ) ; xn = m ( N )m m n ( 1)n ( N ) pn m ( n) X n m ke p (x; N ) ; xn = m ( N ) m ! m m n n ( n) X m c (x) ; n x = m m! m n ( 1)n ( )n m ( n) X m ce (x) : xn = m m ! m
xn =
1
(
)
=0
1
(
)
=0
( )
=0
( )
=0
(
)
=0
(
)
=0
Proof: In x 2 it was shown how one obtains three essentially di erent cross rules for the connection coecients between Pn (x) and Qm (x). We modify this method here. For Qm (x), we have the di erence equation
(x) rQm(x) + (x) Qm (x) + m Qm(x) = 0 26
with (x) = ax + bx + c, and the di erence rule (11) ( (x) + (x)) Qm (x) = m Qm (x) + ( m m ) Qm (x) + m Qm (x) ; and it is easily seen that our current Pn (x) = xn satis es any of the di erence rules 2
+1
1
( (x) + (x)) Pn (x) = a n Pn (x) + n a (2n 1) + b + d Pn (x) + n (n 1)(a(n 1) + b + d) + c + e Pn (x) : +1
(50)
1
Hence in our situation, we get the two cross rules (23) with an = 1, bn = n, cn = 0 (51) Cm (n + 1) = am Cm (n) + bm Cm(n) + cm Cm (n) and (25) with abn = 1=(n + 1), bbn = cbn = 0 1 a C (n + 1) = ab C (n) + bb C (n) + cb C (n) (52) m m m m m m n+1 n m which we had deduced in x 2. Using the di erence rule (50), we obtain the third cross rule 1
1
1
+1
1
+1
+1
+1
a n Cm (n + 1) + n a (2n 1) + b + d Cm (n) + n (n 1)(a(n 1) + b + d) + c + e Cm (n 1) (53) = m Cm (n) + m Cm (n) + m Cm (n) : To receive the recurrence equation (48), we use Theorem 1 writing the cross rules in terms of a; b; c; d; and e, only. Then by linear algebra we eliminate the variables Cm (n +1) and Cm (n 1) to obtain a pure recurrence equation w.r.t. m. (Similarly by elimination of the variables Cm (n) and Cm (n) a pure recurrence equation w.r.t. n is obtained.) A shift by one gives (48). If c = 0, then the recurrence equation has still three terms, unfortunately. For c = 0, we nd a fourth cross rule to eliminate one more variable in the following way. Since the (second) di erence rule (10) (x) rQm (x) = m Qm (x) + m Qm (x) + m Qm (x) is valid, we can use the fact that for c = 0 any of the di erence rules (x) rPn (x) = a n Pn (x) + n (a n + b) Pn (x) ; and therefore the fourth cross rule a n Cm (n + 1) + n (a n + b) Cm (n) = m Cm (n) + m Cm (n) + m Cm (n) (54) is valid. Eliminating the variables Cm (n+1), Cm (n), and Cm (n) from the four cross rules (51){(54) gives the rst order recurrence equation (49). This leads to the desired representations. 2 Whereas we admit that all the shifted factorial representations of the theorem essentially were known ([8], [22]), our presentation uni es this development. In particular, the recurrence equation (49) carries the complete information about the falling factorial representations given in the theorem. Petkovsek's algorithm proves, again, that for the family Kn ; (x) there is no representation (47) with a hypergeometric term Cm (n). As an immediate consequence of Theorem 7, we get the following connection coecient results. 1
1
1
+1
+1
+1
+1
1
+1
1
1
1
+1
1
(
27
)
+1
Corollary 4 The following connection relations between the classical discrete orthogonal polynomials are valid:
hn ; (x; N ) = (
)
n ( ) (1 N ) ( + 1) X n n n
(2 + + )n n!
m
=0
(1 + + )m (n + 1 + + )m hm ; (x; N ) ; (55) ( +( ++1++1)2m) (1 N() n() m+1) m m ( +2+ n + )m ( n + +1)m n X ( + 1)n (1 N )n (1 + + )n ( )n he n ; (x; N ) = n (
(
)
)
(2 + + )n ( =2 + =2 + 1=2)n ( =2 + =2 + 1)n 4 2 + =2 + 1)m ( =2 + =2 + 3=2)m 4m he ; (x; N ) ; ( n(1)m (nN+) 1 (+ ++1) )m( ( = m m m + 2 + n + )m ( n + + 1)m m! n ( + 1) (1 N ) ( ) ( 1)n X n n n ; hn (x; N ) = (2 + +
) n ! n m m ( +( ++ 1++1)2m) (1 (N )n)m( (1++1) +( +)m (n++n 1++2) +( )m ( 1)n + 1) hm ; (x; N ) ; m m m m n X ( + 1)n (1 N )n (1 + + )n ( )n ( 1)n he n ; (x; N ) = n m (2 + + )n ( =2 + =2 + 1=2)n ( =2 + =2 + 1)n 4 m ( =2 + =2 + 3=2)m ( 4)m he ; (x; N ) ; ( n)m(( n++ 1++n ++2) )m(1( =2N+) =( 2 ++ 1) m 1)m ( n + 1)m m! m m ehn ; (x; N ) = (56) bX n= c ( n=2)k ( (n 1)=2)k ( )k ((N n)=2)k (N n +1)=2)k ( n 1=2)k he ; (x; N ) ; n k (1=4 =2 n=2)k ( n + 1=2 )k ( n=2 1=4 =2)k k! 4k k + 1)n ( + 1=2)n (2 + 1)n hn ; (x; N ) = (( + (57) 1)n ( + 1=2)n (2 + 1)n N n n n n n k bX n= c N n (
) 4 k k k k k k k h ; (x; N ) ; ( n=2)k ( =2 n=2 1=4)k ( n + 1=2)k ( n=2 + 1=2)k k! n k k 2)n (2 + 1)n Qn(x; ; ; N ) = (( ++ 11==2) (58) n (2 + 1)n bX n= c ( n=2)k ( (n 1)=2)k ( )k (3=4 =2 n=2)k ( n 1=2)k Q (x; ; ; N ) ; ( n=2)k ( =2 n=2 1=4)k ( n +1=2)k ( n=2+1=2)k k! n k k n ( ) ( 1)n X n Qn(x; ; ; N ) = (2 + + )n m ( + + 1 + 2m) ( n)m (1 + + )m (n + 1 + + )m ( 1)m Q (x; ; ; N ) m ( + + 1) ( + 2 + n + )m (1 + n)m m! (compare [8], (4.1), (4.5)), n X ( )n ( + 1)n Qn (x; ; ; N ) = m ( + 1)n (2 + + )n m
=0
(
(
)
)
=0
(
(
)
)
=0
(
(
)
)
2
(
)
2
=0
(
+1
2
2
2
)
3
2
2
2
4
2
2
1
+1
2
2
(
)
2
=0
2
2
=0
=0
=0
28
( + + 1 + 2m) ( n)m (1 + + )m ( + 1)m (n + 1 + + )m Q (x; ; ; N ) ( + + 1) ( + 1)m ( + + n + 2)m ( n + 1)m m! m (compare [8], (4.1), (4.5)),
mn ; (x) = (
n X ( )n ( n)m m ; (x) ( n + 1 ) m! m
)
(
m
=0
(compare [8], (5.5)),
me n ; (x) = (
n n X
)
m
1
m
=0
)
m
( )n ( n)m e ; ( n + 1 )m m! mm (x) ; (
)
( 1) m n n X ( n ) m
; ( )n ( ) m! mm ; (x) mn (x) = ( 1) m m (
(
)
)
=0
(compare [8], (5.4)),
me n ; (
)
n X
(x) =
m
=0
knp (x; N ) = ( )
n ( 1) ( 1)
m
( )n ( n)m ( 1)m m e m ; (x) ; ( )m m! (
n X
)
m (p q)n m ( N )nn !(( nN)m) ( 1) kmq (x; N ) m m ( )
=0
(compare [8], (5.11)),
n X
m (p q)n m ( N )(n (N n) )mm(! 1) kemq (x; N ) ; m m n pn m (M N ) ( n) X n m k p (x; M ) knp (x; N ) = m n ! ( N M n + 1) m m (compare [8], (5.12)),
kenp (x; N ) = ( )
( )
=0
( )
( )
=0
n pn m (M N ) ( n) X n m ke p (x; M ) ; (N M n + 1) m! m
kenp (x; N ) = ( )
( )
m
m n X
=0
cn (x) = (
)
m ( 1)n n ( )n m ( mn!)m cm (x) m ( )
=0
(compare [8], (5.16)),
cen (x) = (
)
n X
( 1)m ( )n m ( mn!)m cem (x) ; m ( )
=0
; n Kn (x) = (
)
n
m
n ( n)m X
(1
m!
m
X n ( n)m f ; (x) = K =0
(1
n (
)
m!
n m
=0
29
n m )
+
n m )
+
Km ; (x) ; (
)
fm ; (x) : K (
)
Proof: Combining the representations
Pn (x) =
X
Aj (n) xj
xj =
and
j 2Z
X m2Z
Bm(j ) Qm (x) ;
and using Zeilberger's algorithm, the method of Corollary 2 yields the results. The connection relations for the polynomials Kn ; (x) cannot be obtained by this method. Here Theorem 2 leads straightforwardly to the result. 2 Although besides (56){(58) the connection results were essentially known ([8], [3]), our development gives a uni ed treatment of them and makes new results like (56){(58) easily accessible. Note that some of the representations are rather complicated. We suggest the idea to use the notation p fq for the summand of p Fq , i.e. (
)
!
!
1 X upper upper x = f x; k : p q p Fq lower lower k With this notation, (55) could be written in the standardized hypergeometric notation ! ; 1 N; +1 ; hn (x; N ) = f 1; n 2+ + =0
(
)
3
1
n X
!
n; 1 + + ; n + 1 + + ; =2 + =2 + 3=2; 1 f 1; m hm ; (x; N ) : 1 N; +1; =2+ =2+1=2; +2+ n + ; n + +1 (
4
m
=0
)
5
Finally, we deduce the parameter derivatives for the classical discrete orthogonal polynomials. Corollary 5 The following representations for the parameter derivatives of the classical discrete orthogonal polynomials are valid:
; @ h ; (x; N ) = nX 1 hn (x; N ) + @ n m + +m+n+1 ( 1)n m ( + +1+2m) (1 N + m)n m ( +1+ m)n (n m) ( + + 1 + m)n m 1
(
)
(
)
=0
m h ; m (
)
(x; N ) ;
@ he ; (x; N )= nX ( 1)n m ( + +1+2m) (1 N + m)n m ( +1+ m)n m n! eh ; (x; N ) ; m @ n ( + +1+2m) n m m! m ( + + m + n +1) (n m) 1
(
)
(
2
=0
)
2
@ Q (x; ; ; N ) = nX 1 1 @ n + + m + n + 1 + m + 1 Qn(x; ; ; N ) + m ( + + 1 + 2m) ( + 1 + m)n m n! Q ( x ; ; ; N ) ; m (n m) ( + 1 + m)n m ( + + 1 + m)n m m! 1
=0
; 1 @ h ; (x; N ) = nX hn (x; N ) + @ n m + +m+n+1 + + 1 + 2m (1 N + m)n m ( + 1 + m)n n m ( + + 1 + m)n m 1
(
)
(
)
=0
30
m h ; m (
)
(x; N ) ;
(1 N + m)n m ( +1+ m)n m n! eh ; (x; N ) ; + +1+2m @ eh ; (x; N )= nX m @ n ( + + m + n +1) ( n m ) ( + +1+2m) n m m! m 1
(
)
(
2
=0
)
2
@ Q (x; ; ; N ) = nX 1 Qn (x; ; ; N ) + n @ m + +m+n+1 n! ( 1)n m ( + + 1 + 2m) Q ( x ; ; ; N ) ; n m ( + + 1 + m)n m m! m @ ; n( 1 + + n) ; n
; @ mn (x) = (1 ) mn (x) (1 ) mn (x) 1
=0
(
(
)
)
(
)
1
(see e.g. [12], Theorem 9),
n(1 n) ; @ ; @ me n (x) = (1 ) me n (x) ; (
(
)
)
1
2
@ m ; (x) = nX n! mm ; (x) n @ m ! ( n m ) m 1
(
)
(
)
=0
(see [12], Theorem 10),
@ me ; (x) = nX n m n! me m ; (x) ; @ n 1 m ! ( n m ) m 1
(
)
(
)
=0
@ k p (x; N ) = ( 1 + n N ) k p (x; N ) n @p n ( )
( )
1
(see e.g. [12], Theorem 9),
@ ke p (x; N ) = n ( 1 + n N ) ke p (x; N ) ; n @p n @ c (x) = n c (x) n c (x) @ n n n ( )
( )
1
(
(
)
)
(
)
1
(see e.g. [12], Theorem 9),
@ ce (x) = n ce (x) ; n @ n @ K ; (x) = nX n m n! K ; (x) ; m @ n m (n m) m! (
(
)
)
1
1
(
1
)
(
)
=0
nX
@ f ; n! f ; @ Kn (x) = m (n m) m! Km (x) : 1
(
)
(
)
=0
Proof: If the derivative is taken with respect to a variable occurring as an argument rather than as a parameter in the hypergeometric representation, its representation can be easily obtained from the derivative rule of the generalized hypergeometric function, and the chain rule. In those cases, the representations need at most two neighboring polynomials. The other cases can be handled similarly to Corollary 3. 2 31
7 Conclusion Here, we want to recall the algorithms to convert between the di erent types of representations: 1. Hypergeometric Representation ! Recurrence Equation: Zeilberger's algorithm 2. Hypergeometric Representation ! Di erence/Di erential Equation: Zeilberger's/Almkvist-Zeilberger's algorithm 3. Di erence/Di erential Equation ! Recurrence Equation: Theorem 1 4. Di erence/Di erential Equation ! Hypergeometric Representation: method of x 3 and x 5 5. Recurrence Equation ! Di erence/Di erential Equation: Algorithms 1 and 2 in [14] 6. Recurrence Equation ! Hypergeometric Representation: combination of methods 5 and 4
References [1] Abramowitz, M. and Stegun, I. A.: Handbook of Mathematical Functions. Dover Publ., New York, 1964. [2] Al-Salam, W.A.: The Bessel polynomials. Duke Math. J. 24, 1957, 529{545. [3] Area, I., Godoy, E., Ronveaux, A. and Zarzo, A.: Minimal recurrence relations for connection coecients between classical orthogonal polynomials: discrete case. Preprint. [4] Askey, R.: Dual equations and classical orthogonal polynomials. J. Math. Anal. Appl. 24, 1968, 677{685. [5] Askey, R. and Fitch, J.: Integral representations for Jacobi polynomials and some applications. J. Math. Anal. Appl. 26, 1969, 411{437. [6] Askey, R. and Gasper, G.: Jacobi polynomial expansions of Jacobi polynomials with non-negative coecients. Proc. Camb. Phil. Soc. 70, 1971, 243{255. [7] Frohlich, J.: Parameter derivatives of the Jacobi polynomials and the Gaussian hypergeometric function. Integral Transforms and Special Functions 2, 1994, 252{266. [8] Gasper, G.: Projection formulas for orthogonal polynomials of a discrete variable. J. Math. Anal. Appl. 45, 1974, 176{198. [9] Graham, R.L., Knuth, D.E. and Patashnik, O.: Concrete Mathematics. A Foundation for Computer Science. Addison-Wesley, Reading, Massachussets, second edition, 1994. [10] Hua, L.K.: Harmonic Analysis of Functions of Several Complex Variables in the Classical Domains. Translations of Mathematical Monographs Vol. 6, Amer. Math. Soc., Providence, R.I., 1963. 32
[11] Koepf, W.: Algorithms for m-fold hypergeometric summation. Journal of Symbolic Computation 20, 1995, 399{417. [12] Koepf, W.: Identities for families of orthogonal polynomials and special functions. Integral Transforms and Special Functions, 1997, to appear. [13] Koepf, W.: On a problem of Koornwinder. Konrad-Zuse-Zentrum Berlin (ZIB), Preprint SC 96-52, 1996. [14] Koepf, W. and Schmersau, D.: Algorithms for classical orthogonal polynomials. KonradZuse-Zentrum Berlin (ZIB), Preprint SC 96-23, 1996. [15] Koornwinder, T. H.: Hypergeometric series evaluation by Zeilberger's algorithm. In: Open Problems, ed. by Walter van Assche. J. of Comput. and Appl. Math. 48, 1993, 225{243. [16] Lesky, P.: U ber Polynomlosungen von Di erentialgleichungen und Di erenzengleichungen zweiter Ordnung. Anzeiger der Osterreichischen Akademie der Wissenschaften, math.naturwiss. Klasse 121, 1985, 29{33. [17] Magnus, W., Oberhettinger, F. and Soni, R.P.: Formulas and Theorems for the Special Functions of Mathematical Physics. Springer, Berlin{Heidelberg{New York, 1966. [18] Petkovsek, M.: Hypergeometric solutions of linear recurrences with polynomial coecients. J. Symbolic Comp. 14, 1992, 243{264. [19] Nikiforov, A.F., Suslov, S.K. and Uvarov, V.B.: Classical orthogonal polynomials of a discrete variable. Springer-Verlag, Berlin{Heidelberg{New York, 1991. [20] Rainville, E.D.: Special Functions. The MacMillan Co., New York, 1960. [21] Ronveaux, A., Zarzo, A. and Godoy, E.: Recurrence relations for connection coecients between two families of orthogonal polynomials. J. Comp. Appl. Math. 62, 1995, 67{73. [22] Ronveaux, A., Belmehdi, S., Godoy, E. and Zarzo, A.: Recurrence relation approach for connection coecients. Applications to classical discrete orthogonal polynomials. Centre de Recherches Mathematiques, CRM Proceedings and Lecture Notes 9, 1996, 319{335. [23] Koekoek, R. und Swarttouw, R. F.: The Askey-scheme of hypergeometric orthogonal polynomials and its q analogue. Report 94{05, Technische Universiteit Delft, Faculty of Technical Mathematics and Informatics, Delft, 1994. Electronic version: http://www.can.nl/~renes/index.html [24] Szego, G.: Orthogonal Polynomials. Amer. Math. Soc. Coll. Publ. Vol. 23, New York City, 1939. [25] Tricomi, F.G.: Vorlesungen u ber Orthogonalreihen. Grundlehren der Mathematischen Wissenschaften 76, Springer-Verlag, Berlin{Gottingen{Heidelberg, 1955. [26] Zeilberger, D.: A fast algorithm for proving terminating hypergeometric identities. Discrete Math. 80, 1990, 207{211. 33
