What Is a Holonomic Function?

What is a holonomic function?

Christoph Koutschan

Johann Radon Institute for Computational and Applied Mathematics (RICAM) Austrian Academy of Sciences

December 16, 2013 SFB Status Seminar, Strobl Algorithmic and Enumerative Combinatorics What is a holonomic function? Notation: Let K be a computable field of characteristic zero. Preliminaries: A finitely generated left module over the Weyl algebra As(K) is holonomic if it is zero or if it has Bernstein dimension s, i.e., the dimension of the graded associated module with respect to the Bernstein filtration (which filters according to total degree).

Definition 1: A function f(x1, . . . , xs) is holonomic if the left K K As( )-module As( )/ AnnAs(K)(f) is holonomic. In other words, if it is the solution of a maximally overdetermined system of differential equations. Definition 2: A function f is holonomic if the following K(x1, . . . , xs)-vector space is finite-dimensional: O O dimK(x1,...,xs)( / AnnO(f)) < ∞, where denotes the Ore algebra K(x , . . . , x )[D ; 1, d ] ··· [D ; 1, d ]. 1 s x1 dx1 xs dxs

1 / 38 A sequence f(n) is called P-recursive (or P-finite) if it satisfies a linear recurrence equation with polynomial coefficients:

pd(n)f(n + d) + ··· + p1(n)f(n + 1) + p0(n)f(n) = 0,

p0, . . . , pd ∈ K[n] not all zero.

−→ In both cases, only ﬁnitely many initial conditions! −→ Also called holonomic function resp. holonomic sequence.

D-ﬁnite and P-recursive

A function f(x) is called D-finite if it satisfies a linear ordinary differential equation with polynomial coefficients:

(d) 0 pd(x)f (x) + ··· + p1(x)f (x) + p0(x)f(x) = 0, p0, . . . , pd ∈ K[x] not all zero.

2 / 38 −→ In both cases, only ﬁnitely many initial conditions! −→ Also called holonomic function resp. holonomic sequence.

D-ﬁnite and P-recursive

A function f(x) is called D-finite if it satisfies a linear ordinary differential equation with polynomial coefficients:

(d) 0 pd(x)f (x) + ··· + p1(x)f (x) + p0(x)f(x) = 0, p0, . . . , pd ∈ K[x] not all zero.

A sequence f(n) is called P-recursive (or P-finite) if it satisfies a linear recurrence equation with polynomial coefficients:

pd(n)f(n + d) + ··· + p1(n)f(n + 1) + p0(n)f(n) = 0, p0, . . . , pd ∈ K[n] not all zero.

2 / 38 −→ Also called holonomic function resp. holonomic sequence.

D-ﬁnite and P-recursive

A function f(x) is called D-finite if it satisfies a linear ordinary differential equation with polynomial coefficients:

(d) 0 pd(x)f (x) + ··· + p1(x)f (x) + p0(x)f(x) = 0, p0, . . . , pd ∈ K[x] not all zero.

A sequence f(n) is called P-recursive (or P-finite) if it satisfies a linear recurrence equation with polynomial coefficients:

pd(n)f(n + d) + ··· + p1(n)f(n + 1) + p0(n)f(n) = 0, p0, . . . , pd ∈ K[n] not all zero.

−→ In both cases, only ﬁnitely many initial conditions!

2 / 38 D-ﬁnite and P-recursive

A function f(x) is called D-finite if it satisfies a linear ordinary differential equation with polynomial coefficients:

(d) 0 pd(x)f (x) + ··· + p1(x)f (x) + p0(x)f(x) = 0, p0, . . . , pd ∈ K[x] not all zero.

A sequence f(n) is called P-recursive (or P-finite) if it satisfies a linear recurrence equation with polynomial coefficients:

pd(n)f(n + d) + ··· + p1(n)f(n + 1) + p0(n)f(n) = 0, p0, . . . , pd ∈ K[n] not all zero.

−→ In both cases, only ﬁnitely many initial conditions! −→ Also called holonomic function resp. holonomic sequence.

2 / 38 Can express any shift as K(n)-linear combination of Hn and Hn+1: 2n+7 n+3 Hn+4 = n+4 Hn+3 − n+4 Hn+2 2n+7 2n+5 n+2 n+3 = n+4 n+3 Hn+2 − n+3 Hn+1 − n+4 Hn+2 3n2+18n+26 (2n+7)(n+2) = (n+3)(n+4) Hn+2 − (n+3)(n+4) Hn+1 3n2+18n+26 2n+3 n+1 (2n+7)(n+2) = (n+3)(n+4) n+2 Hn+1 − n+2 Hn − (n+3)(n+4) Hn+1 2(2n+5)(n2+5n+5) (n+1)(3n2+18n+26) = (n+2)(n+3)(n+4) Hn+1 − (n+2)(n+3)(n+4) Hn

Example: Harmonic Numbers Pn 1 Example: The harmonic numbers Hn = k=1 k satisfy the recurrence

nHn = (2n − 1)Hn−1 − (n − 1)Hn−2 (n ≥ 2) with initial conditions H0 = 0 and H1 = 1.

3 / 38 2n+7 n+3 = n+4 Hn+3 − n+4 Hn+2 2n+7 2n+5 n+2 n+3 = n+4 n+3 Hn+2 − n+3 Hn+1 − n+4 Hn+2 3n2+18n+26 (2n+7)(n+2) = (n+3)(n+4) Hn+2 − (n+3)(n+4) Hn+1 3n2+18n+26 2n+3 n+1 (2n+7)(n+2) = (n+3)(n+4) n+2 Hn+1 − n+2 Hn − (n+3)(n+4) Hn+1 2(2n+5)(n2+5n+5) (n+1)(3n2+18n+26) = (n+2)(n+3)(n+4) Hn+1 − (n+2)(n+3)(n+4) Hn

Example: Harmonic Numbers Pn 1 Example: The harmonic numbers Hn = k=1 k satisfy the recurrence

nHn = (2n − 1)Hn−1 − (n − 1)Hn−2 (n ≥ 2) with initial conditions H0 = 0 and H1 = 1.

Can express any shift as K(n)-linear combination of Hn and Hn+1:

Hn+4

3 / 38 2n+7 2n+5 n+2 n+3 = n+4 n+3 Hn+2 − n+3 Hn+1 − n+4 Hn+2 3n2+18n+26 (2n+7)(n+2) = (n+3)(n+4) Hn+2 − (n+3)(n+4) Hn+1 3n2+18n+26 2n+3 n+1 (2n+7)(n+2) = (n+3)(n+4) n+2 Hn+1 − n+2 Hn − (n+3)(n+4) Hn+1 2(2n+5)(n2+5n+5) (n+1)(3n2+18n+26) = (n+2)(n+3)(n+4) Hn+1 − (n+2)(n+3)(n+4) Hn

Example: Harmonic Numbers Pn 1 Example: The harmonic numbers Hn = k=1 k satisfy the recurrence

nHn = (2n − 1)Hn−1 − (n − 1)Hn−2 (n ≥ 2) with initial conditions H0 = 0 and H1 = 1.

Can express any shift as K(n)-linear combination of Hn and Hn+1: 2n+7 n+3 Hn+4 = n+4 Hn+3 − n+4 Hn+2

3 / 38 3n2+18n+26 (2n+7)(n+2) = (n+3)(n+4) Hn+2 − (n+3)(n+4) Hn+1 3n2+18n+26 2n+3 n+1 (2n+7)(n+2) = (n+3)(n+4) n+2 Hn+1 − n+2 Hn − (n+3)(n+4) Hn+1 2(2n+5)(n2+5n+5) (n+1)(3n2+18n+26) = (n+2)(n+3)(n+4) Hn+1 − (n+2)(n+3)(n+4) Hn

Example: Harmonic Numbers Pn 1 Example: The harmonic numbers Hn = k=1 k satisfy the recurrence

nHn = (2n − 1)Hn−1 − (n − 1)Hn−2 (n ≥ 2) with initial conditions H0 = 0 and H1 = 1.

Can express any shift as K(n)-linear combination of Hn and Hn+1: 2n+7 n+3 Hn+4 = n+4 Hn+3 − n+4 Hn+2 2n+7 2n+5 n+2 n+3 = n+4 n+3 Hn+2 − n+3 Hn+1 − n+4 Hn+2

3 / 38 3n2+18n+26 2n+3 n+1 (2n+7)(n+2) = (n+3)(n+4) n+2 Hn+1 − n+2 Hn − (n+3)(n+4) Hn+1 2(2n+5)(n2+5n+5) (n+1)(3n2+18n+26) = (n+2)(n+3)(n+4) Hn+1 − (n+2)(n+3)(n+4) Hn

Example: Harmonic Numbers Pn 1 Example: The harmonic numbers Hn = k=1 k satisfy the recurrence

nHn = (2n − 1)Hn−1 − (n − 1)Hn−2 (n ≥ 2) with initial conditions H0 = 0 and H1 = 1.

3 / 38 2(2n+5)(n2+5n+5) (n+1)(3n2+18n+26) = (n+2)(n+3)(n+4) Hn+1 − (n+2)(n+3)(n+4) Hn

Example: Harmonic Numbers Pn 1 Example: The harmonic numbers Hn = k=1 k satisfy the recurrence

nHn = (2n − 1)Hn−1 − (n − 1)Hn−2 (n ≥ 2) with initial conditions H0 = 0 and H1 = 1.

Can express any shift as K(n)-linear combination of Hn and Hn+1: 2n+7 n+3 Hn+4 = n+4 Hn+3 − n+4 Hn+2 2n+7 2n+5 n+2 n+3 = n+4 n+3 Hn+2 − n+3 Hn+1 − n+4 Hn+2 3n2+18n+26 (2n+7)(n+2) = (n+3)(n+4) Hn+2 − (n+3)(n+4) Hn+1 3n2+18n+26 2n+3 n+1 (2n+7)(n+2) = (n+3)(n+4) n+2 Hn+1 − n+2 Hn − (n+3)(n+4) Hn+1

3 / 38 Example: Harmonic Numbers Pn 1 Example: The harmonic numbers Hn = k=1 k satisfy the recurrence

nHn = (2n − 1)Hn−1 − (n − 1)Hn−2 (n ≥ 2) with initial conditions H0 = 0 and H1 = 1.

3 / 38 If f(n) and g(n) are P-recursive then also the following are P-recursive • f(n) + g(n) • f(n) · g(n) • f(an + b) for integers a and b

A sequence is P-recursive iﬀ its generating function is D-ﬁnite.

Closure Properties

If f(x) and g(x) are D-ﬁnite then also the following are D-ﬁnite • f(x) + g(x) • f(x) · g(x) • f(a(x)) if a(x) is algebraic

4 / 38 A sequence is P-recursive iﬀ its generating function is D-ﬁnite.

Closure Properties

If f(x) and g(x) are D-ﬁnite then also the following are D-ﬁnite • f(x) + g(x) • f(x) · g(x) • f(a(x)) if a(x) is algebraic

If f(n) and g(n) are P-recursive then also the following are P-recursive • f(n) + g(n) • f(n) · g(n) • f(an + b) for integers a and b

4 / 38 Closure Properties

If f(x) and g(x) are D-ﬁnite then also the following are D-ﬁnite • f(x) + g(x) • f(x) · g(x) • f(a(x)) if a(x) is algebraic

If f(n) and g(n) are P-recursive then also the following are P-recursive • f(n) + g(n) • f(n) · g(n) • f(an + b) for integers a and b

A sequence is P-recursive iﬀ its generating function is D-ﬁnite.

4 / 38 Ansatz: want to ﬁnd c0, . . . , cd ∈ K[n] such that

0 = cd(n)h(n + d) + ··· + c0(n)h(n)

= cd(n)f(n + d)g(n + d) + ··· + c0(n)f(n)g(n) = cd(n) fd,d1−1f(n + d1 − 1) + ··· + fd,0f(n) × gd,d2−1g(n + d2 − 1) + ··· + gd,0g(n) + ··· + c0(n)f(n)g(n) d −1 d −1 X1 X2 = ri,j(c0, . . . , cd, n)f(n + i)g(n + j) i=0 j=0

All coeﬃcients ri,j must vanish: this yields d1d2 equations for the unknowns c0, . . . , cd. The choice d = d1d2 ensures a solution.

Proof Show that for P-recursive sequences f(n) and g(n) also h(n) = f(n)g(n) is P-recursive. Assume f and g satisfy recurrences of order d1 and d2, respectively.

5 / 38 = cd(n)f(n + d)g(n + d) + ··· + c0(n)f(n)g(n) = cd(n) fd,d1−1f(n + d1 − 1) + ··· + fd,0f(n) × gd,d2−1g(n + d2 − 1) + ··· + gd,0g(n) + ··· + c0(n)f(n)g(n) d −1 d −1 X1 X2 = ri,j(c0, . . . , cd, n)f(n + i)g(n + j) i=0 j=0

All coeﬃcients ri,j must vanish: this yields d1d2 equations for the unknowns c0, . . . , cd. The choice d = d1d2 ensures a solution.

Proof Show that for P-recursive sequences f(n) and g(n) also h(n) = f(n)g(n) is P-recursive. Assume f and g satisfy recurrences of order d1 and d2, respectively. Ansatz: want to ﬁnd c0, . . . , cd ∈ K[n] such that

0 = cd(n)h(n + d) + ··· + c0(n)h(n)

5 / 38 = cd(n) fd,d1−1f(n + d1 − 1) + ··· + fd,0f(n) × gd,d2−1g(n + d2 − 1) + ··· + gd,0g(n) + ··· + c0(n)f(n)g(n) d −1 d −1 X1 X2 = ri,j(c0, . . . , cd, n)f(n + i)g(n + j) i=0 j=0

All coeﬃcients ri,j must vanish: this yields d1d2 equations for the unknowns c0, . . . , cd. The choice d = d1d2 ensures a solution.

0 = cd(n)h(n + d) + ··· + c0(n)h(n)

= cd(n)f(n + d)g(n + d) + ··· + c0(n)f(n)g(n)

5 / 38 d −1 d −1 X1 X2 = ri,j(c0, . . . , cd, n)f(n + i)g(n + j) i=0 j=0

All coeﬃcients ri,j must vanish: this yields d1d2 equations for the unknowns c0, . . . , cd. The choice d = d1d2 ensures a solution.

0 = cd(n)h(n + d) + ··· + c0(n)h(n)

= cd(n)f(n + d)g(n + d) + ··· + c0(n)f(n)g(n) = cd(n) fd,d1−1f(n + d1 − 1) + ··· + fd,0f(n) × gd,d2−1g(n + d2 − 1) + ··· + gd,0g(n) + ··· + c0(n)f(n)g(n)

5 / 38 All coeﬃcients ri,j must vanish: this yields d1d2 equations for the unknowns c0, . . . , cd. The choice d = d1d2 ensures a solution.

0 = cd(n)h(n + d) + ··· + c0(n)h(n)

5 / 38 Proof Show that for P-recursive sequences f(n) and g(n) also h(n) = f(n)g(n) is P-recursive. Assume f and g satisfy recurrences of order d1 and d2, respectively. Ansatz: want to ﬁnd c0, . . . , cd ∈ K[n] such that

0 = cd(n)h(n + d) + ··· + c0(n)h(n)

All coeﬃcients ri,j must vanish: this yields d1d2 equations for the unknowns c0, . . . , cd. The choice d = d1d2 ensures a solution.

5 / 38 • multidimensional sequences f(n1, . . . , ns) (the ni are called discrete variables) • mixed setting: functions in several continuous and discrete variables f(x1, . . . , xs, n1, . . . , nr)

Multivariate Generalization

Try to generalize the ﬁniteness property to

• multivariate functions f(x1, . . . , xs) (the xi are called continuous variables)

6 / 38 • mixed setting: functions in several continuous and discrete variables f(x1, . . . , xs, n1, . . . , nr)

Multivariate Generalization

Try to generalize the ﬁniteness property to

• multivariate functions f(x1, . . . , xs) (the xi are called continuous variables)

• multidimensional sequences f(n1, . . . , ns) (the ni are called discrete variables)

6 / 38 Multivariate Generalization

Try to generalize the ﬁniteness property to

• multivariate functions f(x1, . . . , xs) (the xi are called continuous variables)

• multidimensional sequences f(n1, . . . , ns) (the ni are called discrete variables) • mixed setting: functions in several continuous and discrete variables f(x1, . . . , xs, n1, . . . , nr)

6 / 38 The Legendre polynomials can be deﬁned recursively:

P0(x) = 1

P1(x) = x

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x). (i) Pn+j(x): i, j ≥ 0 .

(3) Pn+1+3(x) = 22 22 222 2 2 2 2 2 (n4(2nn28xx+5)642(x−n−nxnx2++3(3)00+16−nxnnxnx−+2nnx−+18−3n4−+8n(xn+15n−+3(x+6n+22)(2+3))x n+3)−+3)0004 0 000 − 2 PPnn2+2((x(x)x2) +) +− 2 PnP+1(PxPP(n)x+1()(xx()x)) nx+32−1 (n(xn+12+2)(−(x1)22−n+3)1)3 nx+32x−−1P1n+1n+1(nxn) 2 6(4(n2n(n−2x+1)(x+71)(+3nxnnx+2)2+5x+22−xxn)20+nx2−n+18x2+6) −+ 2 2 Pn(x) P (x) − (n(+2)(x(x2−−n1)1)+3)2 (x2nP−n1)+13 (x) n

0 222 00(3)00(3)(4) 0 0 00(3)0002 0 0 0 00 −→(P((−→n(n(xnxx+1++−P−−( 1)P 2)3)nx1)n1)(P1))Px(Pn−xnP)P+1+2n+3)nnisxP+1is(((xx∂x((nD)x)x-ﬁnite)( +) +)x−-ﬁnite + +) 2 (1(2−xP 46 24nxP−xP w.r.t.(n+ w.r.t.x(nn+x+1 3)(5))x 1)(P−xP)(xnnxP−)x(nand)n−x.(+1(−+2()nnx(−(n)+−xnx( =(−)n 1)(ofn1)(+ + 02)(+P+ 1)(n (n rank 3)n 1)n(+nxP+xP+) 2)+n =1)22)+1 3)nP 2)).(P 0Pn(xPxn(n)x(n+1)(x =)+1x =)( =)x 0( = 0 =x) 0) =0 0 = 0 0

Example: Legendre Polynomials Pn(x) This family of (orthogonal) polynomials is a particular solution of the diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0.

(i) Consider the set Pn (x): i ≥ 0 .

(4) Pn (x) =

7 / 38 The Legendre polynomials can be deﬁned recursively:

P0(x) = 1

P1(x) = x

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x). (i) Pn+j(x): i, j ≥ 0 .

(3) Pn+1+3(x) = 22 22 222 2 2 2 2 2 (n4(2nn28xx+5)42(x−n−nxnx2++300+16−nxnnxnx−+2nnx−+18−3n4−+8nxn+15n+3(x+6n+22+3))x +3)−0004 0 0 − PPn2+2((xx2)) +− PnP+1(PxP(n)x+1()x()x) nx+32−1 (n(xn+12+2)(−(x1)22−n+3)1)3 nx+32−1Pn+1n+1(nx) 2 6(4(n2n(n−2x+1)(x+71)(+3nxnnx+2)2+5x+22−xxn)20+nx2−n+18x2+6) −+ 2 2 Pn(x) P (x) − (n(+2)(x(x2−−n1)1)+3)2 (x2nP−n1)+13 (x) n

0 22 00(3)00(3) 0 0 000002 0 0 0 −→(P((−→n(nxnx+1++−P−( 1)P 2)3)nx1)n1)(P)Px(Pn−xnP)+1+2n+3)nisxP+1is(((xx∂x((nD)x)x-ﬁnite)( +) +)x−-ﬁnite + +) 2 (1(2−xP 4 24nxP−xP w.r.t.(n+ w.r.t.x(nn+x+1 3)(5))x 1)P−xP)(nnxP−x(nand)nx.(+1(−+2()nnx−()n+−xnx( =()n 1)(ofn1)(+ + 0+P+ 1)(n (n rank 3)n 1)(+nxP+xP) 2)+n =1)22)+1nP 2)).(P 0n(xPxn()x(n+1)x =)+1 =)( =x 0( = 0x) 0) =0 = 0 0

Example: Legendre Polynomials Pn(x) This family of (orthogonal) polynomials is a particular solution of the diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0.

(i) Consider the set Pn (x): i ≥ 0 .

(4) Pn (x) = 6x (3) (n−2)(n+3) 00 − x2−1 Pn (x) + x2−1 Pn (x)

i 2 (4) (3) 00 (x − 1)Pn (x) + 6xPn (x) − (n − 2)(n + 3)Pn (x) = 0 7 / 38 The Legendre polynomials can be deﬁned recursively:

P0(x) = 1

P1(x) = x

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x). (i) Pn+j(x): i, j ≥ 0 .

Example: Legendre Polynomials Pn(x) This family of (orthogonal) polynomials is a particular solution of the diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0.

(i) Consider the set Pn (x): i ≥ 0 .

(4) Pn (x) = 6x (3) (n−2)(n+3) 00 − x2−1 Pn (x) + x2−1 Pn (x)

7 / 38 The Legendre polynomials can be deﬁned recursively:

P0(x) = 1

P1(x) = x

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x). (i) Pn+j(x): i, j ≥ 0 .

0 222 00(3)00(4) 0 0 (3)0002 0 0 00 −→(P((−→n(n(xnxx+1++−P−−( 1)P 2)3)nx1)n1)(P1))Px(Pn−xnP)P+1+2n+3)nnisxP+1is(((xx∂x((nD))x-ﬁnite)x( +) +x)−-ﬁnite +) + 2 (1(2−xP 62 4nxP−xP w.r.t.(n+ w.r.t.x(nn+x+1 3)5)) 1)(P−xP(xnnxP)x(nand)n−x.(+1−+2()nx(−(n)+xnx( =(−)n 1)(ofn+ + 02)(+P+ 1)( (n rank 3)n 1)n(nxP+xP+)+n =1)22)+1 3)n 2)).(P 0P(xPxnn)(n+1)(x =+1x =)()x 0( = 0 =x)) =0 0 = 0 0

Example: Legendre Polynomials Pn(x) This family of (orthogonal) polynomials is a particular solution of the diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0.

(i) Consider the set Pn (x): i ≥ 0 .

(4) Pn (x) = 6x (3) (n−2)(n+3) 00 − x2−1 Pn (x) + x2−1 Pn (x)

i 2 (3) 00 0 (x − 1)Pn (x) + 4xPn (x) − (n − 1)(n + 2)Pn(x) = 0 7 / 38 The Legendre polynomials can be deﬁned recursively:

P0(x) = 1

P1(x) = x

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x). (i) Pn+j(x): i, j ≥ 0 .

(3) Pn+1+3(x) = 2 2 2 2 2 2 2 2 4(2(nn28x+5)462(xn−xnx2+300(3)+16−nnxnx+2nx−−3n4−+8n(n+15n−+3(xn+22)(2+3))x n+3)−+3)04 0 000 − 2 PPnn+2((x(x)x) +) +− 2 PnP+1PPP(nx+1()(xx()x)) nx+32−1 (n(xn+12+2)(−(x1)22−n+3)1)3 nx+32x−−1P1n+1n+1(nxn) 2nn(n22x+1)(+7nxn2+5x2−x n2+nx2−n+18x2+6) −+ 4(n x+3nx+2xP) n(x) P (x) − (n+2)((x2−n1)+3)2 (x2P−n1)+13 (x) n

Example: Legendre Polynomials Pn(x) This family of (orthogonal) polynomials is a particular solution of the diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0.

(i) Consider the set Pn (x): i ≥ 0 .

(4) Pn (x) = n2x2−n2+nx2−n+18x2+6 00 (x2−1)2 Pn (x) 6(n−1)(n+2)x 0 − (x2−1)2 Pn(x)

7 / 38 The Legendre polynomials can be deﬁned recursively:

P0(x) = 1

P1(x) = x

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x). (i) Pn+j(x): i, j ≥ 0 .

0 22 00(3)(3)(4) 0 00(3)0002 0 0 0 00 −→(P((−→n(nxnx+1++−P−( 1)P 2)3)nx1)n(1)P)Px(Pn−xnP)+1+2n+3)nisxP+1is((x∂x((nDx)x-ﬁnite)() +)x−-ﬁnite + +) (1(2− 46 24nxP−xP w.r.t.(n+ w.r.t.xnn++1 3)(5))x 1)(PxP)(xnnxP−)x(and)n−x.+1(−+2()nx(−(n)−xnx( =(−)n(ofn1)(+ + 02)(++ 1)(n ( rank 3)n 1)n+nP+xP+ 2)+n 1)22)+1 3)nP 2)).(PPn(xPxn(n)x(n+1)(x =)+1x =)( =)x 0( = 0 =x) 0) =0 0 = 0 0

Example: Legendre Polynomials Pn(x) This family of (orthogonal) polynomials is a particular solution of the diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0.

(i) Consider the set Pn (x): i ≥ 0 .

(4) Pn (x) = n2x2−n2+nx2−n+18x2+6 00 (x2−1)2 Pn (x) 6(n−1)(n+2)x 0 − (x2−1)2 Pn(x)

i 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0 7 / 38 The Legendre polynomials can be deﬁned recursively:

P0(x) = 1

P1(x) = x

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x). (i) Pn+j(x): i, j ≥ 0 .

(3) Pn+1+3(x) = 22 22 22 2 2 2 2 (n4(2nn2xx+5)642x−−nxn2++3(3)00+16nxnxnx−2n−+18−3n4+8n(xn+15n−(x+6n+22)(2+3))x n−+3)0004 0 00 − 2 PPnn2+2((x(x)x2) +) +− 2 PnP+1(PxP(n)x+1()(xx()x)) nx+32−1 (n(xn+12+2)(−1)2n+3) nx+32x−−1P1n+1n+1(xn) 2 6(4(2nn−2xx+71)(+3nxnnx+2)+5+2xx) 0 − 2 2 Pn(x) − (n(+2)(x(x2−−n1)1)+3)2 nPn+1(x)

0 222 00(3)00(3)(4) 0 0 00(3)0002 0 0 0 00 −→(P((n(n(xnxx+1++−P−−( 1) 2)3)nx1)1)(P1))PxPn−nP)P+1+2n+3nnisxP+1(((xx∂x((n)x)x-ﬁnite)( +) +)x− + +) 2 (1(2−xP 46 24nxP−xP w.r.t.(n+x(nn+x+1 3)(5))x 1)(P−xP)(xnnxP−)(nand)n−x(+1(−+2()nnx(−(n)+−xnx( =(−)n 1)(ofn1)(+ + 02)(+P+ 1)(n (n rank 3)n 1)n(+nxP+xP+) 2)+n =1)22)+1 3)nP 2)).(P 0Pn(xPxn(n)x(n+1)(x =)+1x =)( =)x 0( = 0 =x) 0) =0 0 = 0 0

Example: Legendre Polynomials Pn(x) This family of (orthogonal) polynomials is a particular solution of the diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0.

(i) Consider the set Pn (x): i ≥ 0 .

(4) Pn (x) =

8x(n2x2−n2+nx2−n+3x2+3) 0 − (x2−1)3 Pn(x) n(n+1)(n2x2−n2+nx2−n+18x2+6) + (x2−1)3 Pn(x)

−→ Pn(x) is D-ﬁnite w.r.t. x. 7 / 38 The Legendre polynomials can be deﬁned recursively:

P0(x) = 1

P1(x) = x

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x). (i) Pn (x): i ≥ 0 .

(4) Pn+3((xx)) = = 22 22 222 2 2 2 2 2 (n4(2nn28xx+5)642(x−n−nxnx2++3(3)00+16−nxnnxnx−+2nnx−+18−3n4−+8n(xn+15n−+3(x+6n+22)(2+3))x n+3)−+3)0004 0 000 − 2 PPnn2+2((x(x)x2) +) +− 2 PnP+1(PxPP(n)x+1()(xx()x)) nx+32−1 (n(xn+12+2)(−(x1)22−n+3)1)3 nx+32x−−1P1n+1n+1(nxn) 2 6(4(n2n(n−2x+1)(x+71)(+3nxnnx+2)2+5x+22−xxn)20+nx2−n+18x2+6) −+ 2 2 Pn(x) P (x) − (n(+2)(x(x2−−n1)1)+3)2 (x2nP−n1)+13 (x) n

Example: Legendre Polynomials Pn(x) This family of (orthogonal) polynomials is a particular solution of the diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0.

(i) Consider the set Pn+j(x): i, j ≥ 0 . j

(3) Pn+1(x) =

7 / 38 The Legendre polynomials can be deﬁned recursively:

P0(x) = 1

P1(x) = x

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x). (i) Pn (x): i ≥ 0 .

(4) Pn+3((xx)) = = 22 22 222 2 2 2 2 2 (n4(2nn28xx+5)62(x−n−nxnx2++3(3)+16−nxnnxnx−+2nnx−+18−3n4−+8n(xn+15n−+3x+6+22)(2)x n+3)−+3)0004 000 − 2 PPnn2+2(x()x2 +) − 2 Pn+1(PxPP(n)x+1()(xx()x)) nx+3−1 ((xn2+2)(−(x1)22−n+3)1)3 n+3x −P1n+1(nxn) 2 6(4(n2n(n−2x+1)(x+71)(+3nxnnx+2)2+5x+22−xxn)20+nx2−n+18x2+6) −+ 2 2 Pn(x) P (x) − (n(+2)(x(x2−−n1)1)+3)2 (x2nP−n1)+13 (x) n

0 222 0000(3)(4) 0 0 00(3)02 0 0 00 −→(P((−→n(n(xnxx+1++−P−−( 1)P 2)3)nx1)n1)(P1))Px(Pn−xnP)P+1+2n+3)nnisxP+1is(((xx∂x((nD)x)x-ﬁnite)( +) +)x−-ﬁnite + +) 2 (1(2−xP 46 2nxP−xP w.r.t.(n+ w.r.t.x(nn+x+1 3)(5))x 1)(P−xP)(xnnxP−)x(nand)n−x.(+1(−+2()nnx(−(n)+−xnx =(−) 1)(ofn1)(+ + 02)(P+ 1)(n (n rankn 1)n(+nx+xP+) 2)+ =1)22) 3)nP 2)).(P 0PnxPn(n)x(n+1(x =)+1x)( =)x 0( = =x) 0) =0 0 = 0 0

Example: Legendre Polynomials Pn(x) This family of (orthogonal) polynomials is a particular solution of the diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0.

(i) Consider the set Pn+j(x): i, j ≥ 0 . j

(3) Pn+1(x) = 4x 00 n(n+3) 0 − x2−1 Pn+1(x) + x2−1 Pn+1(x)

i 2 (3) 00 0 (x − 1)Pn+1(x) + 4xPn+1(x) − n(n + 3)Pn+1(x) = 0 7 / 38 The Legendre polynomials can be deﬁned recursively:

P0(x) = 1

P1(x) = x

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x). (i) Pn (x): i ≥ 0 .

Example: Legendre Polynomials Pn(x) This family of (orthogonal) polynomials is a particular solution of the diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0.

(i) Consider the set Pn+j(x): i, j ≥ 0 . j

(3) Pn+1(x) = 4x 00 n(n+3) 0 − x2−1 Pn+1(x) + x2−1 Pn+1(x)

7 / 38 The Legendre polynomials can be deﬁned recursively:

P0(x) = 1

P1(x) = x

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x). (i) Pn (x): i ≥ 0 .

0 222 (3)00(3)(4) 0 0 00(3)002 0 0 0 00 −→(P((−→n(n(xnxx+1++P−−−( 1)P 2)3)nx1)n1)(P1))Px(n−PxnP)P+1+2+3)nnnisxP+1is(((xx∂x((nD)x)-ﬁnite)x( +) +x)−-ﬁnite +) + 2 (1(2−xP 46 4nxP−xP w.r.t.(n+ w.r.t.x(nn+x 3)(+15))x 1)(P−xP)xn(nxP−)x(nand)n−x.(+1(+2−()nnx(−n()+−xx( =(−)n 1)(ofn1)( + 02)(+P+n (n rank 3)n 1)n(+xP+xP+) 2)n =1)22)+1 3)nP).(P 0Pn(xxn(n)x(+1)(x =)x =)( =)x 0 = 0 =) 0 =0 0 0

Example: Legendre Polynomials Pn(x) This family of (orthogonal) polynomials is a particular solution of the diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0.

(i) Consider the set Pn+j(x): i, j ≥ 0 . j

(3) Pn+1(x) = 4x 00 n(n+3) 0 − x2−1 Pn+1(x) + x2−1 Pn+1(x)

i 2 00 0 (x − 1)Pn+1(x) + 2xPn+1(x) − (n + 1)(n + 2)Pn+1(x) = 0 7 / 38 The Legendre polynomials can be deﬁned recursively:

P0(x) = 1

P1(x) = x

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x). (i) Pn (x): i ≥ 0 .

(4) Pn+3((xx)) = = 22 22 222 2 2 2 2 2 −4(2nnn8x+5)x64(x−n−nxnPxP+(3)00+16−nxn(nxx−+()xnnx ++18)−−4−n(xn+15n−+3(+6n+22)(+3)xPn+3)−+3)004(P0xP()x000)(x(x)) −nx+32−1 P(n(xn+12+2+2)(−(x(1)x22−n)+3)1) +3 nx+32x−2−11nP+1nP+1nn+1n(x) 6(n2n(n2−x+1)(+71)(nxn+2)2+5x2−x n20+nx2−n+18x2+6) − Pn(x) + (n(+2)(x2−n1)+3)2 (x2n−1)3 Pn(x)

Example: Legendre Polynomials Pn(x) This family of (orthogonal) polynomials is a particular solution of the diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0.

(i) Consider the set Pn+j(x): i, j ≥ 0 . j

(3) Pn+1(x) = (n2x2−n2+3nx2−3n+8x2) 0 (x2−1)2 Pn+1(x) 4(n2x+3nx+2x) − (x2−1)2 Pn+1(x)

7 / 38 The Legendre polynomials can be deﬁned recursively:

P0(x) = 1

P1(x) = x

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x). (i) Pn (x): i ≥ 0 .

(4)(3) Pn+1+3((xx)) = = 22 22 222 2 2 2 2 2 (n4(2nn28xx+5)642(x−n−nxnx2++3(3)00+16−nxnnxnx−+2nnx−+18−3n4−+8n(xn+15n−+3(x+6n+22)(2+3))x n+3)−+3)0004 0 000 − 2 PPnn2+2((x(x)x2) +) +− 2 PnP+1(PxPP(n)x+1()(xx()x)) nx+32−1 (n(xn+12+2)(−(x1)22−n+3)1)3 nx+32x−−1P1n+1n+1(nxn) 2 6(4(n2n(n−2x+1)(x+71)(+3nxnnx+2)2+5x+22−xxn)20+nx2−n+18x2+6) −+ 2 2 Pn(x) P (x) − (n(+2)(x(x2−−n1)1)+3)2 (x2nP−n1)+13 (x) n

Example: Legendre Polynomials Pn(x) This family of (orthogonal) polynomials is a particular solution of the diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0.

(i) Consider the set Pn+j(x): i, j ≥ 0 . j

7 / 38 This family of (orthogonal) polynomials is a particular solution of the diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0.

(i) Pn (x): i ≥ 0 .

Example: Legendre Polynomials Pn(x) The Legendre polynomials can be deﬁned recursively:

P0(x) = 1

P1(x) = x

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x). (i) Consider the set Pn+j(x): i, j ≥ 0 . j

7 / 38 This family of (orthogonal) polynomials is a particular solution of the diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0.

(i) Pn (x): i ≥ 0 .

(4)(3) Pn+1((xx)) = = 22 22 222 2 2 2 2 2 (n4n28xx642(x−n−nnx2++3(3)00+16−nxnnxnx−+2nnx−+18−3n4−+8n(xn+15−+3(x+6n2)(2+3))x n+3)−+3)0004 0 000 − 2 Pn 2 ((xx)2) + + 2 P P(PxPPn)+1((xx()x)) x2−1 (n(xn+12+2)(−(x1)22−n+3)1)3 x2x−−1P1n+1n+1(nxn) 2 6(4(n2n(n−2x+1)(x+71)(+3nxnnx+2)2+5x+22−xxn)20+nx2−n+18x2+6) −+ 2 2 Pn(x) P (x) − (n(+2)(x(x2−−n1)1)+3)2 (x2nP−n1)+13 (x) n

0 222 00(3)00(3)(4) 0 0 00(3)0002 0 0 0 00 −→(P((−→n(n(xnxx+1++−P−−( 1)P 2)nx1)n1)(P1))Px(Pn−xnP)P+1+2n)nnisxP+1is(((xx∂x((nD)x)x-ﬁnite)( +) +)x−-ﬁnite + +) 2 (1(2−xP 46 24nxP−xP w.r.t.(n+ w.r.t.x(nn+x+1 3)()x 1)(P−xP)(xnnxP−)x(nand)n−x.(+1(−()nnx(−(n)+−xnx( =(−)n 1)(ofn1)(+ + 02)(+P+ 1)(n (n rank 3)n 1)n(+nxP+xP+) 2)+n =1)2+1 3)nP 2)).(P 0Pn(xPxn(n)x(n)(x =)+1x =) =) 0( = 0 =x 0) 0 0 = 0

Example: Legendre Polynomials Pn(x) The Legendre polynomials can be deﬁned recursively:

P0(x) = 1

P1(x) = x

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x). (i) Consider the set Pn+j(x): i, j ≥ 0 . j

Pn+3(x) = (2n+5)x n+2 n+3 Pn+2(x) − n+3 Pn+1(x)

(n + 3)Pn+3(x) − (2n + 5)xPn+2(x) + (n + 2)Pn+1(x) = 0 7 / 38 This family of (orthogonal) polynomials is a particular solution of the diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0.

(i) Pn (x): i ≥ 0 .

Example: Legendre Polynomials Pn(x) The Legendre polynomials can be deﬁned recursively:

P0(x) = 1

P1(x) = x

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x). (i) Consider the set Pn+j(x): i, j ≥ 0 . j

Pn+3(x) = (2n+5)x n+2 n+3 Pn+2(x) − n+3 Pn+1(x)

7 / 38 This family of (orthogonal) polynomials is a particular solution of the diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0.

(i) Pn (x): i ≥ 0 .

(4)(3) Pn+1((xx)) = = 2 2 22 2 2 2 2 2 (n(2nn28xx+5)642(x−n−nxnx2++3(3)00−nxnnx−+2nnx−+183n−+8(xnn−+3(x+6n+22)(2+3))x n+3)+3)000 0 000 − 2 PPnn2+2((x(x)x2) +) +− 2 PnP+1(xPP()x()(xx)) nx+32−1 (nx+12−(x1)22−1)3 nx+32x−−1P1n+1n+1(nxn) 6(4(n(n−2+1)(x1)(+3nnx+2)2x+22−xxn)20+nx2−n+18x2+6) −+ 2 2 P (x) P (x) − (x(x2−−1)1)2 (x2nP−n1)+13 (x) n

0 222 00(3)00(3)(4) 0 0 00(3)0002 0 0 0 00 −→(P((−→n((nxnxx+1++−P−−( 1)P 3)nx1)n1)(P1))Px(Pn−xPn)P+1n+3)nnisxP+1is(((xx∂(x(nD)x)x-ﬁnite)( +) +)x-ﬁnite− + +) 2 (1(2−xP 46 24xPn−xP w.r.t.(n w.r.t.+x(nn+x+1( 5))x 1)(P−xP)(xnnxP−)x(nand)n−x.((−+2()nnx(−(n)+−xnx( =(−)n 1)(ofn1)(+ + 02)(+P+ 1)(nn (rank 3) 1)n(+nxP+xP) 2)+n =22)+1 3)nP 2)).(P 0n(xPx(n)xn+1)( =)+1x =( =)x 0( 0 =x) 0) = 0 = 0 0

Example: Legendre Polynomials Pn(x) The Legendre polynomials can be deﬁned recursively:

P0(x) = 1

P1(x) = x

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x). (i) Consider the set Pn+j(x): i, j ≥ 0 . j

Pn+3(x) = 4n2x2−n2+16nx2−4n+15x2−4 (n+2)(n+3) Pn+1(x) 2n2x+7nx+5x − (n+2)(n+3) Pn(x)

(n + 2)Pn+2(x) − (2n + 3)xPn+1(x) + (n + 1)Pn(x) = 0 7 / 38 This family of (orthogonal) polynomials is a particular solution of the diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0.

(i) Pn (x): i ≥ 0 .

Example: Legendre Polynomials Pn(x) The Legendre polynomials can be deﬁned recursively:

P0(x) = 1

P1(x) = x

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x). (i) Consider the set Pn+j(x): i, j ≥ 0 . j

Pn+3(x) = 4n2x2−n2+16nx2−4n+15x2−4 (n+2)(n+3) Pn+1(x) 2n2x+7nx+5x − (n+2)(n+3) Pn(x)

7 / 38 This family of (orthogonal) polynomials is a particular solution of the diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0.

(i) Pn (x): i ≥ 0 .

Example: Legendre Polynomials Pn(x) The Legendre polynomials can be deﬁned recursively:

P0(x) = 1

P1(x) = x

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x). (i) Consider the set Pn+j(x): i, j ≥ 0 . j

7 / 38 This family of (orthogonal) polynomials is a particular solution of the diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0.

(i) Pn (x): i ≥ 0 .

222 00(3)00(3)(4) 0 00(3)0002 0 0 0 00 −→(((−→n(n(xxx++−P−− 1)P 2)3)n1)n1)(P1)Px(PnxnP)P+1+2n+3)nnis+1is(((xx∂x((D)x)x-ﬁnite) +) +)−-ﬁnite + + 2 (1(2xP 46 24nxP−xP w.r.t.n+ w.r.t.x(nnx+1 3)(5))x(P−xP)(xnnx−)x(nand)n−x.(+1(−+2)nn(−(n+−xnx((−)n 1)(ofn1)(+ +2)(+P+ 1)(n (n rank 3)n 1)n(+nxP+xP+) 2)+n =1)22)+1 3)nP 2)).(P 0Pn(xPxn(n)x(n+1)(x =)+1x =)( =)x 0( = 0 =x) 0) =0 0 = 0 0

Example: Legendre Polynomials Pn(x) The Legendre polynomials can be deﬁned recursively:

P0(x) = 1

P1(x) = x

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x). (i) Consider the set Pn+j(x): i, j ≥ 0 . j

i 0 0 Pn+1(x) − xPn(x) − (n + 1)Pn(x) = 0 7 / 38 This family of (orthogonal) polynomials is a particular solution of the diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0.

(i) Pn (x): i ≥ 0 .

Example: Legendre Polynomials Pn(x) The Legendre polynomials can be deﬁned recursively:

P0(x) = 1

P1(x) = x

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x). (i) Consider the set Pn+j(x): i, j ≥ 0 . j

7 / 38 This family of (orthogonal) polynomials is a particular solution of the diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0.

(i) Pn (x): i ≥ 0 .

0 222 00(3)00(3)(4) 0 0 00(3)000 0 0 00 −→P((−→(n(xnxx+1+−P−−(P 2)3)nx1)n1)(1))Px(P−xnP)P+2n+3)nnisxP+1is((x∂x((nD)xx-ﬁnite)( +))x−-ﬁnite + +) 2(2−xP 46 24nxPxP w.r.t.(n+ w.r.t.(nn+x+1 3)(5))x 1)(−xP)(xnxP−)xnand)n−.(+1(−+2(nnx((n)+−xnx( =−)n 1)(of1)(+ + 02)(+P 1)(n (n rank 3)nn(+nxP++) 2)+n =1)22)+1 3)P 2)).P 0Pn(Pxn(nx(n+1)(x)+1x =)( =)x( = 0 =x) 0) =0 0 = 0 0

Example: Legendre Polynomials Pn(x) The Legendre polynomials can be deﬁned recursively:

P0(x) = 1

P1(x) = x

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x). (i) Consider the set Pn+j(x): i, j ≥ 0 . j

i 2 0 (n + 1)Pn+1(x) + (1 − x )Pn(x) − (n + 1)xPn(x) = 0 7 / 38 This family of (orthogonal) polynomials is a particular solution of the diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0.

(i) Pn (x): i ≥ 0 .

0 222 00(3)00(3)(4) 0 0 00(3)0002 0 0 0 00 (P((−→n(n(xnxx+1++−−−( 1)P 2)3)x1)n1)P1))P(Pn−xnPP+1+2n+3)nnxP+1is(((xxx((nD)x)x)( +) +)x−-ﬁnite + +) 2 (1(2−xP 46 24nxP−xP(n+ w.r.t.x(nn+x+1 3)(5))x 1)(P−xP)(xnxP−)x(n)n−x.(+1(−+2()nnx(−(n)+−nx( =(−)n 1)n1)(+ + 02)(+P+ 1)(n (n 3)n 1)n(+nxP+xP+) 2)+n =1)2)+1 3)nP 2)(P 0Pn(xPxn(n)x(n+1)(x =)+1x =)( =)x 0( = 0 =x) 0) =0 0 = 0 0

Example: Legendre Polynomials Pn(x) The Legendre polynomials can be deﬁned recursively:

P0(x) = 1

P1(x) = x

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x). (i) Consider the set Pn+j(x): i, j ≥ 0 . j

−→ Pn(x) is ∂-finite w.r.t. n and x (of rank 2). 7 / 38 Again, finitely many (modulo details) initial conditions suffice to fix f.

∂-Finiteness

First attempt at a deﬁnition:

Let f(x1, . . . , xs, n1, . . . , nr) be a function in the continuous variables x1, . . . , xs and in the discrete variables n1, . . . , nr. If there is a ﬁnite set of basis functions of the form

di1 dis ... f(x1, . . . , xs, n1 + j1, . . . , nr + jr) i1 is dx1 dxs with i1, . . . , is, j1, . . . , jr ∈ N such that any shifted partial derivative of f (of the above form) can be expressed as a K(x1, . . . , xs, n1, . . . , nr)-linear combination of the basis functions.

8 / 38 ∂-Finiteness

First attempt at a deﬁnition:

Let f(x1, . . . , xs, n1, . . . , nr) be a function in the continuous variables x1, . . . , xs and in the discrete variables n1, . . . , nr. If there is a ﬁnite set of basis functions of the form

Again, finitely many (modulo details) initial conditions suffice to fix f.

8 / 38 Example 1: The Legendre diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0 translates to the operator 2 2 (x − 1)Dx + 2xDx − n(n + 1). Example 2: The three-term recurrence

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x) translates to the operator 2 (n + 2)Sn − (2n + 3)xSn + (n + 1).

Algebraic Setting Write diﬀerential/diﬀerence equations in operator notation: • shift operator Sv: Svf(v) = f(v + 1) d • partial derivative Dv: Dvf(v) = dv f(v) • q-shift operator Sv,q: Sv,qf(v) = f(qv) • arbitrary operator ∂v: any of the above

9 / 38 Example 2: The three-term recurrence

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x) translates to the operator 2 (n + 2)Sn − (2n + 3)xSn + (n + 1).

Example 1: The Legendre diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0 translates to the operator 2 2 (x − 1)Dx + 2xDx − n(n + 1).

9 / 38 Algebraic Setting Write diﬀerential/diﬀerence equations in operator notation: • shift operator Sv: Svf(v) = f(v + 1) d • partial derivative Dv: Dvf(v) = dv f(v) • q-shift operator Sv,q: Sv,qf(v) = f(qv) • arbitrary operator ∂v: any of the above

Example 1: The Legendre diﬀerential equation 2 00 0 (x − 1)Pn (x) + 2xPn(x) − n(n + 1)Pn(x) = 0 translates to the operator 2 2 (x − 1)Dx + 2xDx − n(n + 1). Example 2: The three-term recurrence

nPn(x) = (2n − 1)xPn−1(x) − (n − 1)Pn−2(x) translates to the operator 2 (n + 2)Sn − (2n + 3)xSn + (n + 1). 9 / 38 More general: 0 Dxa(x) = a(x)Dx + a (x) Sna(n) = a(n + 1)Sn, etc.

Even more general:

∂va = σ(a)∂v + δ(a) where σ is an automorphism and δ a σ-derivation, i.e., δ(ab) = σ(a)δ(b) + δ(a)b.

Operator Algebra Diﬀerential equations and recurrences are translated to skew polynomials. Noncommutative multiplication:

Dxx = xDx + 1,Snn = nSn + Sn, etc.

10 / 38 Even more general:

∂va = σ(a)∂v + δ(a) where σ is an automorphism and δ a σ-derivation, i.e., δ(ab) = σ(a)δ(b) + δ(a)b.

Operator Algebra Diﬀerential equations and recurrences are translated to skew polynomials. Noncommutative multiplication:

Dxx = xDx + 1,Snn = nSn + Sn, etc.

More general: 0 Dxa(x) = a(x)Dx + a (x) Sna(n) = a(n + 1)Sn, etc.

10 / 38 Operator Algebra Diﬀerential equations and recurrences are translated to skew polynomials. Noncommutative multiplication:

Dxx = xDx + 1,Snn = nSn + Sn, etc.

More general: 0 Dxa(x) = a(x)Dx + a (x) Sna(n) = a(n + 1)Sn, etc.

Even more general:

∂va = σ(a)∂v + δ(a) where σ is an automorphism and δ a σ-derivation, i.e., δ(ab) = σ(a)δ(b) + δ(a)b.

10 / 38 In fact, the above notation is a shortcut for

K(v, w, . . . )[∂v; σv, δv][∂w; σw, δw] ···

Example: The operators that we encountered with the Legendre polynomials live in the Ore algebra K K d (x, n)hDx,Sni = (x, n)[Dx; 1, dx ][Sn; σn, 0].

Operator Algebra Such operators form an Ore algebra

K(v, w, . . . )h∂v, ∂w,... i, i.e., multivariate polynomials in the ∂’s with coeﬃcients being rational functions in v, w, . . . , where K is a computable ﬁeld of characteristic 0 (i.e., containing Q).

11 / 38 Example: The operators that we encountered with the Legendre polynomials live in the Ore algebra K K d (x, n)hDx,Sni = (x, n)[Dx; 1, dx ][Sn; σn, 0].

Operator Algebra Such operators form an Ore algebra

In fact, the above notation is a shortcut for

K(v, w, . . . )[∂v; σv, δv][∂w; σw, δw] ···

11 / 38 Operator Algebra Such operators form an Ore algebra

In fact, the above notation is a shortcut for

K(v, w, . . . )[∂v; σv, δv][∂w; σw, δw] ···

Example: The operators that we encountered with the Legendre polynomials live in the Ore algebra K K d (x, n)hDx,Sni = (x, n)[Dx; 1, dx ][Sn; σn, 0].

11 / 38 Eq1 + Eq2 corresponds to P + Q r(v) · Eq1 corresponds to r(v) · P d dv Eq1 corresponds to Dv · P Eq1|v→v+1 corresponds to Sv · P

Let Eq1 and Eq2 be two linear equations for a function f, given by the operators P and Q, i.e., P · f = Q · f = 0. Then • • • • are also valid equations for f. More generally,

P,Q ∈ AnnO f =⇒ P + Q ∈ AnnO f L ∈ O,P ∈ AnnO f =⇒ LP ∈ AnnO f

which states that AnnO f is a left ideal in O.

Annihilating Ideals Let now O be such an Ore algebra. The set AnnO f := P ∈ O : P · f = 0 is called the annihilator of the function f.

12 / 38 Eq1 + Eq2 corresponds to P + Q r(v) · Eq1 corresponds to r(v) · P d dv Eq1 corresponds to Dv · P Eq1|v→v+1 corresponds to Sv · P

Then • • • • are also valid equations for f. More generally,

P,Q ∈ AnnO f =⇒ P + Q ∈ AnnO f L ∈ O,P ∈ AnnO f =⇒ LP ∈ AnnO f

which states that AnnO f is a left ideal in O.

Annihilating Ideals Let now O be such an Ore algebra. The set AnnO f := P ∈ O : P · f = 0 is called the annihilator of the function f.

Let Eq1 and Eq2 be two linear equations for a function f, given by the operators P and Q, i.e., P · f = Q · f = 0.

12 / 38 corresponds to P + Q corresponds to r(v) · P

corresponds to Dv · P corresponds to Sv · P More generally,

P,Q ∈ AnnO f =⇒ P + Q ∈ AnnO f L ∈ O,P ∈ AnnO f =⇒ LP ∈ AnnO f

which states that AnnO f is a left ideal in O.

Annihilating Ideals Let now O be such an Ore algebra. The set AnnO f := P ∈ O : P · f = 0 is called the annihilator of the function f.

Let Eq1 and Eq2 be two linear equations for a function f, given by the operators P and Q, i.e., P · f = Q · f = 0. Then

• Eq1 + Eq2 • r(v) · Eq1 d • dv Eq1 • Eq1|v→v+1 are also valid equations for f.

12 / 38 More generally,

P,Q ∈ AnnO f =⇒ P + Q ∈ AnnO f L ∈ O,P ∈ AnnO f =⇒ LP ∈ AnnO f

which states that AnnO f is a left ideal in O.

Annihilating Ideals Let now O be such an Ore algebra. The set AnnO f := P ∈ O : P · f = 0 is called the annihilator of the function f.

Let Eq1 and Eq2 be two linear equations for a function f, given by the operators P and Q, i.e., P · f = Q · f = 0. Then

• Eq1 + Eq2 corresponds to P + Q • r(v) · Eq1 corresponds to r(v) · P d • dv Eq1 corresponds to Dv · P • Eq1|v→v+1 corresponds to Sv · P are also valid equations for f.

12 / 38 Annihilating Ideals Let now O be such an Ore algebra. The set AnnO f := P ∈ O : P · f = 0 is called the annihilator of the function f.

Let Eq1 and Eq2 be two linear equations for a function f, given by the operators P and Q, i.e., P · f = Q · f = 0. Then

• Eq1 + Eq2 corresponds to P + Q • r(v) · Eq1 corresponds to r(v) · P d • dv Eq1 corresponds to Dv · P • Eq1|v→v+1 corresponds to Sv · P are also valid equations for f. More generally,

P,Q ∈ AnnO f =⇒ P + Q ∈ AnnO f L ∈ O,P ∈ AnnO f =⇒ LP ∈ AnnO f which states that AnnO f is a left ideal in O. 12 / 38 j In other words, if the left ideal of annihilating operators of f

AnnO(f) = {P ∈ O : P · f = 0}

is a zero-dimensional ideal.

Deﬁnition: ∂-Finite Function

Let O = K(v, w, . . . )h∂v, ∂w,... i be an Ore algebra. A function f(v, w, . . . ) is ∂-ﬁnite w.r.t. O if “all its shifts and derivatives” O · f = {P · f : P ∈ O} form a ﬁnite-dimensional K(v, w, . . . )-vector space: dimK(v,w,... ) O/ AnnO(f) < ∞.

13 / 38 In other words, if the left ideal of annihilating operators of f

AnnO(f) = {P ∈ O : P · f = 0}

is a zero-dimensional ideal.

Deﬁnition: ∂-Finite Function

i 13 / 38 j

Deﬁnition: ∂-Finite Function

In other words, if the left ideal of annihilating operators of f

AnnO(f) = {P ∈ O : P · f = 0} is a zero-dimensional ideal.

13 / 38 Gr¨obnerBases

14 / 38 2. This set of functions is closed under many operations. n • addition, e.g., x + Pn(x)

• multiplication, e.g., Pn(x)Pn+1(x) √ 2 • certain substitutions, e.g., P2n+3 x + 1 0 • operator application, e.g., Pn+2(x) 3. These operations (closure properties) can be executed algorithmically. 4. Many elementary and special functions are covered.

Why ∂-Finite Functions?

1. Definition gives rise to a finite data structure. • annihilating ideal of operators (Gröbnerbasis is finite) • finitely many initial values

15 / 38 3. These operations (closure properties) can be executed algorithmically. 4. Many elementary and special functions are covered.

Why ∂-Finite Functions?

1. Definition gives rise to a finite data structure. • annihilating ideal of operators (Gröbnerbasis is finite) • finitely many initial values 2. This set of functions is closed under many operations. n • addition, e.g., x + Pn(x)

• multiplication, e.g., Pn(x)Pn+1(x) √ 2 • certain substitutions, e.g., P2n+3 x + 1 0 • operator application, e.g., Pn+2(x)

15 / 38 4. Many elementary and special functions are covered.

Why ∂-Finite Functions?

15 / 38 Why ∂-Finite Functions?

15 / 38 (Incomplete) List of ∂-Finite Functions ArcCsc, KelvinBei, HypergeometricPFQ, ExpIntegralE, ArcTanh, HankelH2, AngerJ, JacobiP, ChebyshevT, AiryBi, AiryAi, Sinc, Multinomial, CatalanNumber, QBinomial, CosIntegral, ArcSech, SphericalHankelH2, HermiteH, ExpIntegralEi, Beta, AiryBiPrime, SphericalBesselJ, Binomial, ParabolicCylinderD, Erfc, EllipticK, Fibonacci, QFactorial, Cos, Hypergeometric2F1, Erf, KelvinKer, HypergeometricPFQRegularized, Log, Factorial, BesselY, Cosh, CoshIntegral, ArcTan, ArcCoth, LegendreP, LaguerreL, EllipticE, SinhIntegral, Sinh, BetaRegularized, SphericalHankelH1, ArcSin, EllipticThetaPrime, Root, LucasL, AppellF1, FresnelC, LegendreQ, ChebyshevU, GammaRegularized, Erﬁ, HarmonicNumber, BesselI, KelvinKei, ArithmeticGeometricMean, Exp, ArcCot, EllipticTheta, Hypergeometric0F1, EllipticPi, GegenbauerC, ArcCos, WeberE, FresnelS, EllipticF, ArcCosh, Subfactorial, QPochhammer, Gamma, StruveH, WhittakerM, ArcCsch, Hypergeometric1F1, SinIntegral, BesselJ, StruveL, ArcSec, Factorial2, KelvinBer, BesselK, ArcSinh,

HankelH1, Sqrt, PolyGamma, HypergeometricU, AiryAiPrime, Sin, 16 / 38 SphericalBesselY, WhittakerW, Pochhammer, Hypergeometric2F1Regularized, Hypergeometric1F1Regularized, Hypergeometric0F1Regularized, Application 1 Finite Elements

Joint work with Joachim Sch¨oberl and Peter Paule

17 / 38 Problem Setting

Simulate the propagation of electromagnetic waves using the Maxwell equations dH dE = curl E, = − curl H dt dt where H and E are the magnetic and the electric ﬁeld respectively.

Deﬁne basis functions (this is the 2D case):

i (2i+1,0) 2y ϕi,j(x, y) := (1 − x) Pj (2x − 1)Pi 1−x − 1 using the Legendre and Jacobi polynomials.

Problem: Represent the partial derivatives of ϕi,j(x, y) in the basis (i.e., as linear combinations of shifts of the ϕi,j(x, y) itself).

18 / 38 Sketch of the algorithm:

1. Work in the Ore algebra O = Q(i, j, x, y)hSi,Sj,Dxi.

2. Compute a Gr¨obnerbasis G of AnnO ϕi,j(x, y). 3. Choose index sets A and B. 4. Reduce the above ansatz with G and obtain a normal form. 5. Do coeﬃcient comparison with respect to x and y.

6. Solve the resulting linear system for ak,l, bm,n ∈ Q(i, j). 7. If there is no solution, go back to step 3.

Make an Ansatz! More precisely, we need a relation of the form

X d X ak,l(i, j) dx ϕi+k,j+l(x, y) = bm,n(i, j)ϕi+m,j+n(x, y), (k,l)∈A (m,n)∈B

d that is free of x and y (and similarly for dy ).

19 / 38 Make an Ansatz! More precisely, we need a relation of the form

X d X ak,l(i, j) dx ϕi+k,j+l(x, y) = bm,n(i, j)ϕi+m,j+n(x, y), (k,l)∈A (m,n)∈B

d that is free of x and y (and similarly for dy ).

Sketch of the algorithm:

1. Work in the Ore algebra O = Q(i, j, x, y)hSi,Sj,Dxi.

2. Compute a Gr¨obnerbasis G of AnnO ϕi,j(x, y). 3. Choose index sets A and B. 4. Reduce the above ansatz with G and obtain a normal form. 5. Do coeﬃcient comparison with respect to x and y.

6. Solve the resulting linear system for ak,l, bm,n ∈ Q(i, j). 7. If there is no solution, go back to step 3.

19 / 38 Result

With this method, we ﬁnd the relation

d (2i + j + 3)(2i + 2j + 7) dx ϕi,j+1(x, y)+ d 2(2i + 1)(i + j + 3) dx ϕi,j+2(x, y)− d (j + 3)(2i + 2j + 5) dx ϕi,j+3(x, y)+ d (j + 1)(2i + 2j + 7) dx ϕi+1,j(x, y)− d 2(2i + 3)(i + j + 3) dx ϕi+1,j+1(x, y)− d (2i + j + 5)(2i + 2j + 5) dx ϕi+1,j+2(x, y)+ 2(i + j + 3)(2i + 2j + 5)(2i + 2j + 7)ϕi,j+2(x, y)+ 2(i + j + 3)(2i + 2j + 5)(2i + 2j + 7)ϕi+1,j+1(x, y) = 0

d and a similar one for dy ϕi,j(x, y). −→ The use of these previously unknown formulae caused a considerable speed-up in the numerical simulations.

20 / 38 That was nice, but we want (and can) do more. . .

What about integrals

Z b f(x, . . . ) dx a and sums b X f(n, . . . ) n=a

21 / 38 Z b integration F (x) = f(x, y) dy a

d Telescoping: write f(x,n, yk) = gdy(n,g(x, k + y) 1). − g(n, k). Zb b X d Then F (n) = gd(yn,g( kx,+ y) 1)d−yg(n, k) = g(x,n, b)+− 1)g(−x,g a()n,. a). k=aa Creative Telescoping: write dd d cd(nx)fdx(dnf+(x,d, y k)) + +······++cc00((xn))ff((x,n, y k)) = gd(yn,g( kx,+ y) 1). − g(n, k). IntegratingSumming from fromaatotob byieldsyields a a recurrence diﬀerential for equationF (n): for F (x): dd cd(nx)Fdx(dnF+(xd)) + +······++cc00((xn))FF((xn)) = g(x,n, b)+− 1)g(−x,g a()n, a).

Creative Telescoping Method for doing integrals and sums (aka Feynman’s diﬀerentiating under the integral sign)

b X Consider the following summation problem: F (n) = f(n, k) k=a

22 / 38 Z b integration F (x) = f(x, y) dy a

d f(x, y) = dy g(x, y). Z b d dy g(x, y) dy g(x, b) − g(x, a). a Creative Telescoping: write dd d cd(nx)fdx(dnf+(x,d, y k)) + +······++cc00((xn))ff((x,n, y k)) = gd(yn,g( kx,+ y) 1). − g(n, k). IntegratingSumming from fromaatotob byieldsyields a a recurrence diﬀerential for equationF (n): for F (x): dd cd(nx)Fdx(dnF+(xd)) + +······++cc00((xn))FF((xn)) = g(x,n, b)+− 1)g(−x,g a()n, a).

Creative Telescoping Method for doing integrals and sums (aka Feynman’s diﬀerentiating under the integral sign)

b X Consider the following summation problem: F (n) = f(n, k) k=a Telescoping: write f(n, k) = g(n, k + 1) − g(n, k). b X Then F (n) = g(n, k + 1) − g(n, k) = g(n, b + 1) − g(n, a). k=a

22 / 38 Z b integration F (x) = f(x, y) dy a

d f(x, y) = dy g(x, y). Z b d dy g(x, y) dy g(x, b) − g(x, a). a

dd d cd(x) dxd f(x, y) + ··· + c0(x)f(x, y) dy g(x, y). Integrating from a to b yields a diﬀerential equation for F (x): dd cd(x) dxd F (x) + ··· + c0(x)F (x) g(x, b) − g(x, a)

Creative Telescoping Method for doing integrals and sums (aka Feynman’s diﬀerentiating under the integral sign)

b X Consider the following summation problem: F (n) = f(n, k) k=a Telescoping: write f(n, k) = g(n, k + 1) − g(n, k). b X Then F (n) = g(n, k + 1) − g(n, k) = g(n, b + 1) − g(n, a). k=a Creative Telescoping: write

cd(n)f(n + d, k) + ··· + c0(n)f(n, k) = g(n, k + 1) − g(n, k).

Summing from a to b yields a recurrence for F (n):

cd(n)F (n + d) + ··· + c0(n)F (n) = g(n, b + 1) − g(n, a).

22 / 38 b X summation F (n) = f(n, k) k=a f(n, k) = g(n, k + 1) − g(n, k). b X g(n, k + 1) − g(n, k) g(n, b + 1) − g(n, a). k=a

cd(n)f(n + d, k) + ··· + c0(n)f(n, k) g(n, k + 1) − g(n, k).

Summing from a to b yields a recurrence for F (n):

cd(n)F (n + d) + ··· + c0(n)F (n) g(n, b + 1) − g(n, a).

Creative Telescoping Method for doing integrals and sums (aka Feynman’s diﬀerentiating under the integral sign)

Z b Consider the following integration problem: F (x) = f(x, y) dy a

d Telescoping: write f(x, y) = dy g(x, y). Z b d Then F (n) = dy g(x, y) dy = g(x, b) − g(x, a). a Creative Telescoping: write dd d cd(x) dxd f(x, y) + ··· + c0(x)f(x, y) = dy g(x, y). Integrating from a to b yields a diﬀerential equation for F (x): dd cd(x) dxd F (x) + ··· + c0(x)F (x) = g(x, b) − g(x, a)

22 / 38 A reasonable choice for where to look for g is O · f.

d Then the task is to ﬁnd P (n, Sn) = cd(n)Sn + ··· + c0(n) and Q ∈ O such that P − (Sk − 1)Q · f = 0 ⇐⇒ P − (Sk − 1)Q ∈ AnnO(f).

Creative Telescoping, O = K(n, k)hSn,Ski

cd(n)f(n + d, k) + ··· + c0(n)f(n, k) = g(n, k + 1) − g(n, k)

= (Sk − 1) · g(n, k). Where should we look for a suitable g(n, k)? Note that there are trivial solutions like: k−1 X g(n, k) := cd(n)f(n + d, i) + ··· + c0(n)f(n, i) i=0

23 / 38 Creative Telescoping, O = K(n, k)hSn,Ski

cd(n)f(n + d, k) + ··· + c0(n)f(n, k) = g(n, k + 1) − g(n, k)

= (Sk − 1) · g(n, k). Where should we look for a suitable g(n, k)? Note that there are trivial solutions like: k−1 X g(n, k) := cd(n)f(n + d, i) + ··· + c0(n)f(n, i) i=0

A reasonable choice for where to look for g is O · f.

d Then the task is to ﬁnd P (n, Sn) = cd(n)Sn + ··· + c0(n) and Q ∈ O such that P − (Sk − 1)Q · f = 0 ⇐⇒ P − (Sk − 1)Q ∈ AnnO(f).

23 / 38 Then the creative telescoping operator i j CT = S − 3 + (S − 1) + (S − 1) n i n − i − j + 1 j n − i − j + 1

n with CT i,j,n−i−j = 0 implies that

F (n + 1) = 3F (n).

Creative Telescoping (Example 1)

Let F (n) denote the double sum over the trinomial coeﬃcients

n n n n X X n X X n! F (n) = = . i, j, n − i − j i!j!(n − i − j)! j=0 i=0 j=0 i=0

24 / 38 Creative Telescoping (Example 1)

Let F (n) denote the double sum over the trinomial coeﬃcients

n n n n X X n X X n! F (n) = = . i, j, n − i − j i!j!(n − i − j)! j=0 i=0 j=0 i=0

Then the creative telescoping operator i j CT = S − 3 + (S − 1) + (S − 1) n i n − i − j + 1 j n − i − j + 1

n with CT i,j,n−i−j = 0 implies that

F (n + 1) = 3F (n).

24 / 38 The creative telescoping operator

y(1 − x2) yz(1 − y2) (z3 − z)D2 + (3z2 − 1)D + z + D + D z z x xyz − 1 y xyz − 1

that annihilates the integrand, certifies that P (z) satisfies the differential equation

(z3 − z)P 00(z) + (3z2 − 1)P 0(z) + zP (z) = 0.

Creative Telescoping (Example 2)

The lattice Green’s function of the square lattice is given by

Z 1 Z 1 1 P (z) = √ p dx dy. 0 0 (1 − xyz) 1 − x2 1 − y2

25 / 38 Creative Telescoping (Example 2)

The lattice Green’s function of the square lattice is given by

Z 1 Z 1 1 P (z) = √ p dx dy. 0 0 (1 − xyz) 1 − x2 1 − y2

The creative telescoping operator

y(1 − x2) yz(1 − y2) (z3 − z)D2 + (3z2 − 1)D + z + D + D z z x xyz − 1 y xyz − 1 that annihilates the integrand, certifies that P (z) satisfies the differential equation

(z3 − z)P 00(z) + (3z2 − 1)P 0(z) + zP (z) = 0.

25 / 38 • Corresponds to an m-fold summation/integration problem.

• w = w1, . . . , wm are the summation/integration variables.

• v = v1, v2,... are the surviving parameters.

• P (v, ∂v) is called the telescoper.

• The Qi(v, w, ∂v, ∂w) are called the delta parts. • The delta parts can be viewed as certiﬁcates for the correctness of the telescoper. • Research topic: develop fast algorithms to compute it!

Creative Telescoping in Full Generality In general, a creative telescoping operator has the form

P (v, ∂v) + ∆1Q1(v, w, ∂v, ∂w) + ··· + ∆mQm(v, w, ∂v, ∂w)

where ∆i = Swi − 1 or ∆i = Dwi (depending on the problem).

26 / 38 Creative Telescoping in Full Generality In general, a creative telescoping operator has the form

P (v, ∂v) + ∆1Q1(v, w, ∂v, ∂w) + ··· + ∆mQm(v, w, ∂v, ∂w)

where ∆i = Swi − 1 or ∆i = Dwi (depending on the problem).

• Corresponds to an m-fold summation/integration problem.

• w = w1, . . . , wm are the summation/integration variables.

• v = v1, v2,... are the surviving parameters.

• P (v, ∂v) is called the telescoper.

26 / 38 Answer 1: No! In general not for ∂-ﬁnite functions.

Answer 2: Yes! If additionally the function is holonomic.

Combine the two notions: • Use ∂-ﬁniteness for computations. • Use holonomy for justiﬁcations.

Holonomy

Question: Does there always exist such a telescoping operator?

27 / 38 In general not for ∂-ﬁnite functions.

Answer 2: Yes! If additionally the function is holonomic.

Combine the two notions: • Use ∂-ﬁniteness for computations. • Use holonomy for justiﬁcations.

Holonomy

Question: Does there always exist such a telescoping operator?

Answer 1: No!

27 / 38 In general not for ∂-ﬁnite functions.

If additionally the function is holonomic.

Combine the two notions: • Use ∂-ﬁniteness for computations. • Use holonomy for justiﬁcations.

Holonomy

Question: Does there always exist such a telescoping operator?

Answer 1: No!

Answer 2: Yes!

27 / 38 If additionally the function is holonomic.

Combine the two notions: • Use ∂-ﬁniteness for computations. • Use holonomy for justiﬁcations.

Holonomy

Question: Does there always exist such a telescoping operator?

Answer 1: No! In general not for ∂-ﬁnite functions.

Answer 2: Yes!

27 / 38 Combine the two notions: • Use ∂-ﬁniteness for computations. • Use holonomy for justiﬁcations.

Holonomy

Question: Does there always exist such a telescoping operator?

Answer 1: No! In general not for ∂-ﬁnite functions.

Answer 2: Yes! If additionally the function is holonomic.

27 / 38 Holonomy

Question: Does there always exist such a telescoping operator?

Answer 1: No! In general not for ∂-ﬁnite functions.

Answer 2: Yes! If additionally the function is holonomic.

Combine the two notions: • Use ∂-ﬁniteness for computations. • Use holonomy for justiﬁcations.

27 / 38 Diﬀerently stated: f is holonomic if for any (s − 1)-subset E ⊂ x1, . . . , xs,Dx1 ,...,Dxs , |E| = s − 1,

there exists a nonzero element in AnnW(f) that is free of all generators in E.

−→ This is why a creative telescoping operator always exists.

Holonomic Functions

Assume that f(x1, . . . , xs) depends only on continuous variables. Consider the Weyl algebra

W K = [x1, . . . , xs]hDx1 ,...,Dxs i.

Then f is holonomic if the left ideal AnnW(f) has dimension s (which, by Bernstein’s inequality, is the minimum possible).

28 / 38 −→ This is why a creative telescoping operator always exists.

Holonomic Functions

Assume that f(x1, . . . , xs) depends only on continuous variables. Consider the Weyl algebra

W K = [x1, . . . , xs]hDx1 ,...,Dxs i.

Then f is holonomic if the left ideal AnnW(f) has dimension s (which, by Bernstein’s inequality, is the minimum possible).

Diﬀerently stated: f is holonomic if for any (s − 1)-subset E ⊂ x1, . . . , xs,Dx1 ,...,Dxs , |E| = s − 1, there exists a nonzero element in AnnW(f) that is free of all generators in E.

28 / 38 Holonomic Functions

Assume that f(x1, . . . , xs) depends only on continuous variables. Consider the Weyl algebra

W K = [x1, . . . , xs]hDx1 ,...,Dxs i.

Then f is holonomic if the left ideal AnnW(f) has dimension s (which, by Bernstein’s inequality, is the minimum possible).

Diﬀerently stated: f is holonomic if for any (s − 1)-subset E ⊂ x1, . . . , xs,Dx1 ,...,Dxs , |E| = s − 1, there exists a nonzero element in AnnW(f) that is free of all generators in E.

−→ This is why a creative telescoping operator always exists.

28 / 38 −→ This equivalence holds only in the continuous case!

A sequence is deﬁned to be holonomic if its (multivariate) generating function is holonomic.

1 Example: The sequence n2+k2 is ∂-ﬁnite but not holonomic.

∂-Finite and Holonomic Functions

Theorem: The function f(x1, . . . , xs) is holonomic if and only if it is ∂-ﬁnite.

29 / 38 A sequence is deﬁned to be holonomic if its (multivariate) generating function is holonomic.

1 Example: The sequence n2+k2 is ∂-ﬁnite but not holonomic.

∂-Finite and Holonomic Functions

Theorem: The function f(x1, . . . , xs) is holonomic if and only if it is ∂-ﬁnite.

−→ This equivalence holds only in the continuous case!

29 / 38 1 Example: The sequence n2+k2 is ∂-ﬁnite but not holonomic.

∂-Finite and Holonomic Functions

Theorem: The function f(x1, . . . , xs) is holonomic if and only if it is ∂-ﬁnite.

−→ This equivalence holds only in the continuous case!

A sequence is deﬁned to be holonomic if its (multivariate) generating function is holonomic.

29 / 38 ∂-Finite and Holonomic Functions

Theorem: The function f(x1, . . . , xs) is holonomic if and only if it is ∂-ﬁnite.

−→ This equivalence holds only in the continuous case!

A sequence is deﬁned to be holonomic if its (multivariate) generating function is holonomic.

1 Example: The sequence n2+k2 is ∂-ﬁnite but not holonomic.

29 / 38 2. An annihilating ideal is given by its Gr¨obnerbasis (i.e., a ﬁnite set of generators that allows to decide ideal membership and equality of ideals).

3. Integrals and sums are treated by the method of creative telescoping.

4. The output is always given as an annihilating ideal, not as a closed form.

Principia Holonomica

1. Functions and sequences are represented by their annihilating left ideals (and initial values).

30 / 38 3. Integrals and sums are treated by the method of creative telescoping.

4. The output is always given as an annihilating ideal, not as a closed form.

Principia Holonomica

1. Functions and sequences are represented by their annihilating left ideals (and initial values).

2. An annihilating ideal is given by its Gr¨obnerbasis (i.e., a ﬁnite set of generators that allows to decide ideal membership and equality of ideals).

30 / 38 4. The output is always given as an annihilating ideal, not as a closed form.

Principia Holonomica

1. Functions and sequences are represented by their annihilating left ideals (and initial values).

2. An annihilating ideal is given by its Gr¨obnerbasis (i.e., a ﬁnite set of generators that allows to decide ideal membership and equality of ideals).

3. Integrals and sums are treated by the method of creative telescoping.

30 / 38 Principia Holonomica

1. Functions and sequences are represented by their annihilating left ideals (and initial values).

2. An annihilating ideal is given by its Gr¨obnerbasis (i.e., a ﬁnite set of generators that allows to decide ideal membership and equality of ideals).

3. Integrals and sums are treated by the method of creative telescoping.

4. The output is always given as an annihilating ideal, not as a closed form.

30 / 38 Application 2 Special Function Identities

31 / 38 Some Special Function Identities n 2 2 n k 3 X n k + n X nk + n X k = (1) k k k k j k=0 k=0 j=0 (m+ 1 ,−m− 1 ) Z ∞ 1 πP 2 2 (a) dx = m (2) 4 2 m+1 m+ 3 m+ 1 0 (x + 2ax + 1) 2 2 (a + 1) 2 Z ∞ √ −x a/2 a −t a +n e x n!Ln(x) = e t 2 Ja 2 tx dt (3) 0 Z ∞ ∞ ∞ m n −x2 X X Hm(x)Hn(x)r s e √ dx = πe2rs (4) m!n! −∞ m=0 n=0 Z 1 1 πinΓ(n + 2ν)J (a) 2ν− 2 iax (ν) n+ν 1 − x e Cn (x) dx = ν−1 ν (5) −1 2 a n!Γ(ν) √ 2 ∞ n sin z + 2tz X (−t) yn−1(z) = (6) z n! n=0

32 / 38 << HolonomicFunctions.m Annihilator[Exp[-x]*x^(a/2)*n!*LaguerreL[n, a, x], {S[a], S[n], Der[x]}]

{2Sn − 2xDx + (−a − 2n − 2), 2 2 2 2 4x Dx + (4x + 4x)Dx + (−a + 2ax + 4nx + 4x), 2 2 2 2xSa + (2ax + 2x + 2x)Dx + (−a + ax − a + 2nx + 2x)} CreativeTelescoping[Exp[-t]*t^(a/2+n)*BesselJ[a,2*Sqrt[t*x]], Der[t], {S[a], S[n], Der[x]}]

{{−2Sn + 2xDx + (a + 2n + 2), 2 2 2 2 4x Dx + (4x + 4x)Dx + (−a + 2ax + 4nx + 4x), 2 2 2 2xSa + (2ax + 2x + 2x)Dx + (−a + ax − a + 2nx + 2x)}, {−2t, −4tx, −2tx}} −→ The annihilating ideals agree; check a few initial values.

Computer Proof of a Special Function Identity Z ∞ √ −x a/2 a −t a +n e x n!Ln(x) = e t 2 Ja 2 tx dt. 0

33 / 38 CreativeTelescoping[Exp[-t]*t^(a/2+n)*BesselJ[a,2*Sqrt[t*x]], Der[t], {S[a], S[n], Der[x]}]

Computer Proof of a Special Function Identity Z ∞ √ −x a/2 a −t a +n e x n!Ln(x) = e t 2 Ja 2 tx dt. 0 << HolonomicFunctions.m Annihilator[Exp[-x]*x^(a/2)*n!*LaguerreL[n, a, x], {S[a], S[n], Der[x]}]

{2Sn − 2xDx + (−a − 2n − 2), 2 2 2 2 4x Dx + (4x + 4x)Dx + (−a + 2ax + 4nx + 4x), 2 2 2 2xSa + (2ax + 2x + 2x)Dx + (−a + ax − a + 2nx + 2x)}

33 / 38 Computer Proof of a Special Function Identity Z ∞ √ −x a/2 a −t a +n e x n!Ln(x) = e t 2 Ja 2 tx dt. 0 << HolonomicFunctions.m Annihilator[Exp[-x]*x^(a/2)*n!*LaguerreL[n, a, x], {S[a], S[n], Der[x]}]

{{−2Sn + 2xDx + (a + 2n + 2), 2 2 2 2 4x Dx + (4x + 4x)Dx + (−a + 2ax + 4nx + 4x), 2 2 2 2xSa + (2ax + 2x + 2x)Dx + (−a + ax − a + 2nx + 2x)}, {−2t, −4tx, −2tx}}

−→ The annihilating ideals agree; check a few initial values. 33 / 38 Application 3 Enumerative Combinatorics

Joint work with Manuel Kauers and Doron Zeilberger 34 / 38 Some notions from partition theory: • partition of an integer, e.g., 4 = 2 + 1 + 1 • plane partition: two-dimensional arrangement of summands, 2 1 e.g., 1 • Ferrer’s diagram: three-dimensional representation of PPs, 2 1 e.g., = 1 • totally symmetric plane partition (TSPP): Ferrer’s diagram is symmetric w.r.t. rotation and reﬂection • orbit: see the picture

The q-TSPP Conjecture “The Holy Grail of Enumerative Combinatorics”

Last surviving open problem of the classic (Stanley, 1986) A baker’s dozen of conjectures concerning plane partitions

35 / 38 The q-TSPP Conjecture “The Holy Grail of Enumerative Combinatorics”

Last surviving open problem of the classic (Stanley, 1986) A baker’s dozen of conjectures concerning plane partitions Some notions from partition theory: • partition of an integer, e.g., 4 = 2 + 1 + 1 • plane partition: two-dimensional arrangement of summands, 2 1 e.g., 1 • Ferrer’s diagram: three-dimensional representation of PPs, 2 1 e.g., = 1 • totally symmetric plane partition (TSPP): Ferrer’s diagram is symmetric w.r.t. rotation and reﬂection • orbit: see the picture

35 / 38 Andrews-Robbins q-TSPP conjecture (1983):

X Y 1 − qi+j+k−1 q|π/S3| = 1 − qi+j+k−2 π∈T (n) 1≤i≤j≤k≤n

Counting formula is obtained by q → 1 (Stembridge 1995):

Y i + j + k − 1 |T (n)| = i + j + k − 2 1≤i≤j≤k≤n

Let T (n) denote set of TSPPs with largest part ≤ n, e.g. n = 2:

1 · q0 1 · q1 1 · q2 1 · q3 1 · q4

36 / 38 Let T (n) denote set of TSPPs with largest part ≤ n, e.g. n = 2:

1 · q0 1 · q1 1 · q2 1 · q3 1 · q4

Andrews-Robbins q-TSPP conjecture (1983):

X Y 1 − qi+j+k−1 q|π/S3| = 1 − qi+j+k−2 π∈T (n) 1≤i≤j≤k≤n

Counting formula is obtained by q → 1 (Stembridge 1995):

Y i + j + k − 1 |T (n)| = i + j + k − 2 1≤i≤j≤k≤n 36 / 38 where ! i+j−1 i + j − 2 i + j − 1 i ai,j := q + q + (1 + q )δi,j − δi,j+1 i − 1 q i q

where n (1 − qn)(1 − qn−1) ··· (1 − qn−k+1) := k k−1 k q (1 − q )(1 − q ) ··· (1 − q) is the q-binomial coeﬃcient.

The Determinant

Soichi Okada: the q-TSPP conjecture is true if

2 Y 1 − qi+j+k−1 det (ai,j) = =: bn 1≤i,j≤n 1 − qi+j+k−2 1≤i≤j≤k≤n

37 / 38 The Determinant

Soichi Okada: the q-TSPP conjecture is true if

2 Y 1 − qi+j+k−1 det (ai,j) = =: bn 1≤i,j≤n 1 − qi+j+k−2 1≤i≤j≤k≤n where ! i+j−1 i + j − 2 i + j − 1 i ai,j := q + q + (1 + q )δi,j − δi,j+1 i − 1 q i q where n (1 − qn)(1 − qn−1) ··· (1 − qn−k+1) := k k−1 k q (1 − q )(1 − q ) ··· (1 − q) is the q-binomial coeﬃcient.

37 / 38 • cn,j not given explicitly • perfect application for the holonomic systems approach • very large computations • feasible only due to a new algorithm

The Holonomic Ansatz (by Doron Zeilberger) “Pull out of the hat” a holonomic function cn,j and prove

cn,n = 1 (n ≥ 1), n X cn,jai,j = 0 (1 ≤ i < n), j=1 n X bn c a = (n ≥ 1). n,j n,j b j=1 n−1

Then det(ai,j)1≤i,j≤n = bn holds for all n.

38 / 38 The Holonomic Ansatz (by Doron Zeilberger) “Pull out of the hat” a holonomic function cn,j and prove

cn,n = 1 (n ≥ 1), n X cn,jai,j = 0 (1 ≤ i < n), j=1 n X bn c a = (n ≥ 1). n,j n,j b j=1 n−1

Then det(ai,j)1≤i,j≤n = bn holds for all n.

• cn,j not given explicitly • perfect application for the holonomic systems approach • very large computations • feasible only due to a new algorithm

38 / 38