Integral D-finite Functions Manuel Kauers Institute for Algebra - - PowerPoint PPT Presentation

Integral D-finite Functions

Manuel Kauers

Institute for Algebra Johannes Kepler University Linz, Austria joint work with Christoph Koutschan

1

Integral

p0(x)f(x) + p1(x)f′(x) + · · · + pr(x)f(r)(x) = 0

D-finite Functions

Manuel Kauers

Institute for Algebra Johannes Kepler University Linz, Austria joint work with Christoph Koutschan

1

• Integral

p0(x)f(x) + p1(x)f′(x) + · · · + pr(x)f(r)(x) = 0

D-finite Functions

Manuel Kauers

Institute for Algebra Johannes Kepler University Linz, Austria joint work with Christoph Koutschan

1

Z · · · · · · Q

Integral

p0(x)f(x) + p1(x)f′(x) + · · · + pr(x)f(r)(x) = 0

D-finite Functions

Manuel Kauers

Institute for Algebra Johannes Kepler University Linz, Austria joint work with Christoph Koutschan

1

k[x] · · · · · · k(x) Z · · · · · · Q

Integral

p0(x)f(x) + p1(x)f′(x) + · · · + pr(x)f(r)(x) = 0

D-finite Functions

Manuel Kauers

Institute for Algebra Johannes Kepler University Linz, Austria joint work with Christoph Koutschan

1

Ok[x] · · · · · · k(x) k[x] · · · · · · k(x) Z · · · · · · Q

Integral

p0(x)f(x) + p1(x)f′(x) + · · · + pr(x)f(r)(x) = 0

D-finite Functions

Manuel Kauers

Institute for Algebra Johannes Kepler University Linz, Austria joint work with Christoph Koutschan

1

??? · · · · · · D-finite functions Ok[x] · · · · · · k(x) k[x] · · · · · · k(x) Z · · · · · · Q

Integral

p0(x)f(x) + p1(x)f′(x) + · · · + pr(x)f(r)(x) = 0

D-finite Functions

Manuel Kauers

Institute for Algebra Johannes Kepler University Linz, Austria joint work with Christoph Koutschan

1

• An element of Q is called integral if its denominator is 1 or −1.

2

• An element of Q is called integral if its denominator is 1 or −1.
• An element of k(x) is called integral if its denominator is in k.

2

• An element of Q is called integral if its denominator is 1 or −1.
• An element of k(x) is called integral if its denominator is in k.

Note: This is the case if and only if the element has no poles.

2

• An element of Q is called integral if its denominator is 1 or −1.
• An element of k(x) is called integral if its denominator is in k.

Note: This is the case if and only if the element has no poles.

• An element of k(x) is called integral if its monic (!) minimal

polynomial belongs to k[x][y].

2

• An element of Q is called integral if its denominator is 1 or −1.
• An element of k(x) is called integral if its denominator is in k.

Note: This is the case if and only if the element has no poles.

• An element of k(x) is called integral if its monic (!) minimal

polynomial belongs to k[x][y]. Note: This is the case if and only if the corresponding function has no poles on any of its branches.

2

• An element of Q is called integral if its denominator is 1 or −1.
• An element of k(x) is called integral if its denominator is in k.

Note: This is the case if and only if the element has no poles.

• An element of k(x) is called integral if its monic (!) minimal

polynomial belongs to k[x][y]. Note: This is the case if and only if the corresponding function has no poles on any of its branches. Example: √ x and

3

√ x2 are integral but √ 1/x is not.

2

• An element of Q is called integral if its denominator is 1 or −1.
• An element of k(x) is called integral if its denominator is in k.

Note: This is the case if and only if the element has no poles.

• An element of k(x) is called integral if its monic (!) minimal

polynomial belongs to k[x][y]. Note: This is the case if and only if the corresponding function has no poles on any of its branches. Example: M = y2 − x √ x and

3

√ x2 are integral but √ 1/x is not.

2

• An element of Q is called integral if its denominator is 1 or −1.
• An element of k(x) is called integral if its denominator is in k.

Note: This is the case if and only if the element has no poles.

• An element of k(x) is called integral if its monic (!) minimal

polynomial belongs to k[x][y]. Note: This is the case if and only if the corresponding function has no poles on any of its branches. Example: M = y2 − x √ x and M = y3 − x2

3

√ x2 are integral but √ 1/x is not.

2

• An element of Q is called integral if its denominator is 1 or −1.
• An element of k(x) is called integral if its denominator is in k.

Note: This is the case if and only if the element has no poles.

• An element of k(x) is called integral if its monic (!) minimal

polynomial belongs to k[x][y]. Note: This is the case if and only if the corresponding function has no poles on any of its branches. Example: M = y2 − x √ x and M = y3 − x2

3

√ x2 are integral but M = y2 − 1

x

√ 1/x is not.

2

• An element of Q is called integral if its denominator is 1 or −1.
• An element of k(x) is called integral if its denominator is in k.

Note: This is the case if and only if the element has no poles.

• An element of k(x) is called integral if its monic (!) minimal

polynomial belongs to k[x][y]. Note: This is the case if and only if the corresponding function has no poles on any of its branches. Example: M = y2 − x √ x and M = y3 − x2

3

√ x2 are integral but M = xy2 − 1 √ 1/x is not.

2

Consider the field K = k(x) 3 √ x2 .

3

Consider the field K = k(x) 3 √ x2 . It is the k(x)-vector space generated by 1,

3

√ x2, (

3

√ x2)2.

3

Consider the field K = k(x) 3 √ x2 . It is the k(x)-vector space generated by 1,

3

√ x2, (

3

√ x2)2. Let O be the set of all integral elements of K.

3

Consider the field K = k(x) 3 √ x2 . It is the k(x)-vector space generated by 1,

3

√ x2, (

3

√ x2)2. Let O be the set of all integral elements of K. This is a k[x]-module, but it is not generated by 1,

3

√ x2, (

3

√ x2)2.

3

Consider the field K = k(x) 3 √ x2 . It is the k(x)-vector space generated by 1,

3

√ x2, (

3

√ x2)2. Let O be the set of all integral elements of K. This is a k[x]-module, but it is not generated by 1,

3

√ x2, (

3

√ x2)2. In fact, 1

x

3

√ x4 ∈ O \

• k[x] + k[x]

3

√ x2 + k[x](

3

√ x2)2 .

3

Consider the field K = k(x) 3 √ x2 . It is the k(x)-vector space generated by 1,

3

√ x2, (

3

√ x2)2. Let O be the set of all integral elements of K. This is a k[x]-module, but it is not generated by 1,

3

√ x2, (

3

√ x2)2. In fact, 1

x

3

√ x4 ∈ O \

• k[x] + k[x]

3

√ x2 + k[x](

3

√ x2)2 . It can be shown that

• 1,

3

√ x2,

1 x

3

√ x4 is a module basis of O.

3

Consider the field K = k(x) 3 √ x2 . It is the k(x)-vector space generated by 1,

3

√ x2, (

3

√ x2)2. Let O be the set of all integral elements of K. This is a k[x]-module, but it is not generated by 1,

3

√ x2, (

3

√ x2)2. In fact, 1

x

3

√ x4 ∈ O \

• k[x] + k[x]

3

√ x2 + k[x](

3

√ x2)2 . It can be shown that

• 1,

3

√ x2,

1 x

3

√ x4 is a module basis of O. Such a basis is called an integral basis for K.

3

Classical Problem: Given an irreducible polynomial M ∈ k(x)[y], find an integral basis for K = k(x)[y]/M, i.e., a k[x]-module basis for the set O of all integral elements of K.

4

Classical Problem: Given an irreducible polynomial M ∈ k(x)[y], find an integral basis for K = k(x)[y]/M, i.e., a k[x]-module basis for the set O of all integral elements of K. Classical Algorithms:

• Trager’s algorithm
• van Hoeij’s algorithm

4

Classical Problem: Given an irreducible polynomial M ∈ k(x)[y], find an integral basis for K = k(x)[y]/M, i.e., a k[x]-module basis for the set O of all integral elements of K. Classical Algorithms:

• Trager’s algorithm – based on ideal arithmetic
• van Hoeij’s algorithm

4

Classical Problem: Given an irreducible polynomial M ∈ k(x)[y], find an integral basis for K = k(x)[y]/M, i.e., a k[x]-module basis for the set O of all integral elements of K. Classical Algorithms:

• Trager’s algorithm – based on ideal arithmetic
• van Hoeij’s algorithm – based on series arithmetic

4

Classical Problem: Given an irreducible polynomial M ∈ k(x)[y], find an integral basis for K = k(x)[y]/M, i.e., a k[x]-module basis for the set O of all integral elements of K. Classical Algorithms:

• Trager’s algorithm – based on ideal arithmetic
• van Hoeij’s algorithm – based on series arithmetic

4

Classical Problem: Given an irreducible polynomial M ∈ k(x)[y], find an integral basis for K = k(x)[y]/M, i.e., a k[x]-module basis for the set O of all integral elements of K. Classical Algorithms:

• Trager’s algorithm – based on ideal arithmetic
• van Hoeij’s algorithm – based on series arithmetic

Key Fact:

• An element a ∈ K is integral if and only if all its Puiseux

series expansions at all places have nonnegative valuation.

4

Example: M = (25

16x3 + 2x4) − x3 y − (2x + 1)y2 + y3

5

Example: M = (25

16x3 + 2x4) − x3 y − (2x + 1)y2 + y3

−3 −2 −1 1 2 3 −3 −2 −1 1 2 3

5

Example: M = (25

16x3 + 2x4) − x3 y − (2x + 1)y2 + y3

−3 −2 −1 1 2 3 −3 −2 −1 1 2 3

5

Example: M = (25

16x3 + 2x4) − x3 y − (2x + 1)y2 + y3

x = 0 y 1st sol 1 + 2x + · · · 2nd sol

5 4x3/2 − 9 20x5/2 + · · ·

3rd sol − 5

4x3/2 + 9 20x5/2 + · · ·

5

Example: M = (25

16x3 + 2x4) − x3 y − (2x + 1)y2 + y3

x = 0 y y2 1st sol 1 + 2x + · · · 1 + 4x + · · · 2nd sol

5 4x3/2 − 9 20x5/2 + · · · 25 16x3 + · · ·

3rd sol − 5

4x3/2 + 9 20x5/2 + · · · 25 16x3 + · · ·

5

Example: M = (25

16x3 + 2x4) − x3 y − (2x + 1)y2 + y3

x = 0 y y2 y2 − y 1st sol 1 + 2x + · · · 1 + 4x + · · · 2x + · · · 2nd sol

5 4x3/2 − 9 20x5/2 + · · · 25 16x3 + · · · 25 16x3 + · · ·

3rd sol − 5

4x3/2 + 9 20x5/2 + · · · 25 16x3 + · · · 25 16x3 + · · ·

5

Example: M = (25

16x3 + 2x4) − x3 y − (2x + 1)y2 + y3

x = 0 y y2

1 x(y2 − y)

1st sol 1 + 2x + · · · 1 + 4x + · · · 2 + · · · 2nd sol

5 4x3/2 − 9 20x5/2 + · · · 25 16x3 + · · · 25 16x2 + · · ·

3rd sol − 5

4x3/2 + 9 20x5/2 + · · · 25 16x3 + · · · 25 16x2 + · · ·

5

Example: M = (25

16x3 + 2x4) − x3 y − (2x + 1)y2 + y3

x = 0 y y2

1 x(y2 − y)

1st sol 1 + 2x + · · · 1 + 4x + · · · 2 + · · · 2nd sol

5 4x3/2 − 9 20x5/2 + · · · 25 16x3 + · · · 25 16x2 + · · ·

3rd sol − 5

4x3/2 + 9 20x5/2 + · · · 25 16x3 + · · · 25 16x2 + · · ·

The element 1

x(y2 − y) is integral but does not belong to

k[x] + k[x]y + k[x]y2.

5

Example: M = (25

16x3 + 2x4) − x3 y − (2x + 1)y2 + y3

x = 0 y y2

1 x(y2 − y)

1st sol 1 + 2x + · · · 1 + 4x + · · · 2 + · · · 2nd sol

5 4x3/2 − 9 20x5/2 + · · · 25 16x3 + · · · 25 16x2 + · · ·

3rd sol − 5

4x3/2 + 9 20x5/2 + · · · 25 16x3 + · · · 25 16x2 + · · ·

The element 1

x(y2 − y) is integral but does not belong to

k[x] + k[x]y + k[x]y2. In fact, an integral basis is given by

• 1, y, 1

x(y2 − y)

• .

5

General outline of the algorithm:

6

General outline of the algorithm: 1 Start with the basis (b0, b1, . . . , bd−1) = (1, y, . . . , yd−1).

6

General outline of the algorithm: 1 Start with the basis (b0, b1, . . . , bd−1) = (1, y, . . . , yd−1). 2 If there exists an integral element a which does not belong to k[x]b0 + · · · + k[x]bd−1, say of degree i with respect to y, replace bi by a and goto 2.

6

General outline of the algorithm: 1 Start with the basis (b0, b1, . . . , bd−1) = (1, y, . . . , yd−1). 2 If there exists an integral element a which does not belong to k[x]b0 + · · · + k[x]bd−1, say of degree i with respect to y, replace bi by a and goto 2. 3 Otherwise, return b0, . . . , bd−1 and stop.

6

General outline of the algorithm: 1 Start with the basis (b0, b1, . . . , bd−1) = (1, y, . . . , yd−1). 2 If there exists an integral element a which does not belong to k[x]b0 + · · · + k[x]bd−1, say of degree i with respect to y, replace bi by a and goto 2. 3 Otherwise, return b0, . . . , bd−1 and stop. Facts:

6

General outline of the algorithm: 1 Start with the basis (b0, b1, . . . , bd−1) = (1, y, . . . , yd−1). 2 If there exists an integral element a which does not belong to k[x]b0 + · · · + k[x]bd−1, say of degree i with respect to y, replace bi by a and goto 2. 3 Otherwise, return b0, . . . , bd−1 and stop. Facts:

• Existence of an element a can be decided by making a

suitable ansatz, equating coefficients in the Puiseux series, and solving a linear system.

6

General outline of the algorithm: 1 Start with the basis (b0, b1, . . . , bd−1) = (1, y, . . . , yd−1). 2 If there exists an integral element a which does not belong to k[x]b0 + · · · + k[x]bd−1, say of degree i with respect to y, replace bi by a and goto 2. 3 Otherwise, return b0, . . . , bd−1 and stop. Facts:

• Existence of an element a can be decided by making a

a =

1 x−α(β0b0 + · · · + βi−1bi−1 + bi)

suitable ansatz, equating coefficients in the Puiseux series, and solving a linear system.

6

General outline of the algorithm: 1 Start with the basis (b0, b1, . . . , bd−1) = (1, y, . . . , yd−1). 2 If there exists an integral element a which does not belong to k[x]b0 + · · · + k[x]bd−1, say of degree i with respect to y, replace bi by a and goto 2. 3 Otherwise, return b0, . . . , bd−1 and stop. Facts:

• Existence of an element a can be decided by making a

a =

1 x−α(β0b0 + · · · + βi−1bi−1 + bi)

suitable ansatz, equating coefficients in the around α Puiseux series, and solving a linear system.

6

General outline of the algorithm: 1 Start with the basis (b0, b1, . . . , bd−1) = (1, y, . . . , yd−1). 2 If there exists an integral element a which does not belong to k[x]b0 + · · · + k[x]bd−1, say of degree i with respect to y, replace bi by a and goto 2. 3 Otherwise, return b0, . . . , bd−1 and stop. Facts:

• Existence of an element a can be decided by making a

a =

1 x−α(β0b0 + · · · + βi−1bi−1 + bi)

suitable ansatz, equating coefficients in the around α Puiseux series, and solving a linear system.

• Termination of the algorithm can be shown by considering a

certain polynomial whose degree decreases whenever some bi is replaced by an a.

6

General outline of the algorithm: 1 Start with the basis (b0, b1, . . . , bd−1) = (1, y, . . . , yd−1). 2 If there exists an integral element a which does not belong to k[x]b0 + · · · + k[x]bd−1, say of degree i with respect to y, replace bi by a and goto 2. 3 Otherwise, return b0, . . . , bd−1 and stop. Facts:

• Existence of an element a can be decided by making a

a =

1 x−α(β0b0 + · · · + βi−1bi−1 + bi)

suitable ansatz, equating coefficients in the around α Puiseux series, and solving a linear system.

• Termination of the algorithm can be shown by considering a

disc(b0, . . . , bd−1) certain polynomial whose degree decreases whenever some bi is replaced by an a.

6

\end{old} \begin{new}

7

Consider a linear differential operator L = p0 + p1∂ + · · · + pr∂r ∈ k(x)[∂].

8

Consider a linear differential operator L = p0 + p1∂ + · · · + pr∂r ∈ k(x)[∂]. For any α, such an operator L admits a fundamental system of generalized series solutions of the form exp

• P((x − α)1/s)
• (x − α)νQ
• (x − α)1/s, log(x − α)
• for some s ∈ N, P ∈ k[t], ν ∈ k, and Q ∈ k[[u]][v].

8

Consider a linear differential operator L = p0 + p1∂ + · · · + pr∂r ∈ k(x)[∂]. For any α, such an operator L admits a fundamental system of generalized series solutions of the form exp

• P((x − α)1/s)
• (x − α)νQ
• (x − α)1/s, log(x − α)
• for some s ∈ N, P ∈ k[t], ν ∈ k, and Q ∈ k[[u]][v].

We restrict the attention to operators L where P = 0 and s = 1 for all its solutions.

8

Consider a linear differential operator L = p0 + p1∂ + · · · + pr∂r ∈ k(x)[∂]. For any α, such an operator L admits a fundamental system of generalized series solutions of the form exp

• P((x − α)1/s)
• (x − α)νQ
• (x − α)1/s, log(x − α)
• for some s ∈ N, P ∈ k[t], ν ∈ k, and Q ∈ k[[u]][v].

We restrict the attention to operators L where P = 0 and s = 1 for all its solutions.

8

Consider a linear differential operator L = p0 + p1∂ + · · · + pr∂r ∈ k(x)[∂]. For any α, such an operator L admits a fundamental system of generalized series solutions of the form exp

• P((x − α)1/s)
• (x − α)νQ
• (x − α)1/s, log(x − α)
• for some s ∈ N, P ∈ k[t], ν ∈ k, and Q ∈ k[[u]][v].

We restrict the attention to operators L where P = 0 and s = 1 for all its solutions. We call such a series integral if ν ≥ 0.

8

Consider a linear differential operator L = p0 + p1∂ + · · · + pr∂r ∈ k(x)[∂]. For any α, such an operator L admits a fundamental system of generalized series solutions of the form exp

• P((x − α)1/s)
• (x − α)νQ
• (x − α)1/s, log(x − α)
• for some s ∈ N, P ∈ k[t], ν ∈ k, and Q ∈ k[[u]][v].

We restrict the attention to operators L where P = 0 and s = 1 for all its solutions. We call such a series integral if ν ≥ 0. (Well, ν might not be real; see the paper for a more careful definition.)

8

Consider the algebra A = k(x)[∂]/L, where L = k(x)[∂]L is the left ideal generated by L.

9

Consider the algebra A = k(x)[∂]/L, where L = k(x)[∂]L is the left ideal generated by L. This is a k(x)-vector space generated by 1, ∂, . . . , ∂r−1.

9

Consider the algebra A = k(x)[∂]/L, where L = k(x)[∂]L is the left ideal generated by L. This is a k(x)-vector space generated by 1, ∂, . . . , ∂r−1. An element q0 + q1∂ + · · · + qr−1∂r−1 ∈ A is called integral if for every series solution f of L at any α ∈ k the corresponding series q0 f + q1 f′ + · · · + qr−1 f(r−1) is integral.

9

Consider the algebra A = k(x)[∂]/L, where L = k(x)[∂]L is the left ideal generated by L. This is a k(x)-vector space generated by 1, ∂, . . . , ∂r−1. An element q0 + q1∂ + · · · + qr−1∂r−1 ∈ A is called integral if for every series solution f of L at any α ∈ k the corresponding series q0 f + q1 f′ + · · · + qr−1 f(r−1) is integral. Fact: The integral elements of A form a k[x]-submodule of A.

9

Consider the algebra A = k(x)[∂]/L, where L = k(x)[∂]L is the left ideal generated by L. This is a k(x)-vector space generated by 1, ∂, . . . , ∂r−1. An element q0 + q1∂ + · · · + qr−1∂r−1 ∈ A is called integral if for every series solution f of L at any α ∈ k the corresponding series q0 f + q1 f′ + · · · + qr−1 f(r−1) is integral. Fact: The integral elements of A form a k[x]-submodule of A. Want: A k[x]-module basis of this module.

9

Example: L = 2x(2x − 1)∂2 − (4x2 + 1)∂ + (2x + 1).

10

Example: L = 2x(2x − 1)∂2 − (4x2 + 1)∂ + (2x + 1). x = 0 1 1st sol 1 + x + 1

2x2 + · · ·

2nd sol x1/2 + · · ·

10

Example: L = 2x(2x − 1)∂2 − (4x2 + 1)∂ + (2x + 1). x = 0 1 ∂ 1st sol 1 + x + 1

2x2 + · · ·

1 + x + · · · 2nd sol x1/2 + · · ·

1 2x−1/2 + · · ·

10

Example: L = 2x(2x − 1)∂2 − (4x2 + 1)∂ + (2x + 1). x = 0 1 ∂ x∂ 1st sol 1 + x + 1

2x2 + · · ·

1 + x + · · · x + x2 + · · · 2nd sol x1/2 + · · ·

1 2x−1/2 + · · · 1 2x1/2 + · · ·

10

Example: L = 2x(2x − 1)∂2 − (4x2 + 1)∂ + (2x + 1). x = 0 1 ∂ x∂ 1st sol 1 + x + 1

2x2 + · · ·

1 + x + · · · x + x2 + · · · 2nd sol x1/2 + · · ·

1 2x−1/2 + · · · 1 2x1/2 + · · ·

1 and x∂ are integral elements of k(x)[∂]/L, but ∂ is not.

10

Example: L = 2x(2x − 1)∂2 − (4x2 + 1)∂ + (2x + 1). x = 1/2 1 x∂ 1st sol

1 2 + (x− 1 2) + · · · 1 4 + 3 4(x− 1 2) + · · ·

2nd sol 1 + (x− 1

2) + · · · 1 2 + 3 2(x− 1 2) + · · ·

10

Example: L = 2x(2x − 1)∂2 − (4x2 + 1)∂ + (2x + 1). x = 1/2 1 x∂ 2x∂ − 1 1st sol

1 2 + (x− 1 2) + · · · 1 4 + 3 4(x− 1 2) + · · · 1 2(x− 1 2) + · · ·

2nd sol 1 + (x− 1

2) + · · · 1 2 + 3 2(x− 1 2) + · · ·

2(x− 1

2) + · · ·

10

Example: L = 2x(2x − 1)∂2 − (4x2 + 1)∂ + (2x + 1). x = 1/2 1 x∂

1 2x−1

• 2x∂ − 1
• 1st sol

1 2 + (x− 1 2) + · · · 1 4 + 3 4(x− 1 2) + · · · 1 4 + · · ·

2nd sol 1 + (x− 1

2) + · · · 1 2 + 3 2(x− 1 2) + · · ·

1 + · · ·

10

Example: L = 2x(2x − 1)∂2 − (4x2 + 1)∂ + (2x + 1). x = 1/2 1 x∂

1 2x−1

• 2x∂ − 1
• 1st sol

1 2 + (x− 1 2) + · · · 1 4 + 3 4(x− 1 2) + · · · 1 4 + · · ·

2nd sol 1 + (x− 1

2) + · · · 1 2 + 3 2(x− 1 2) + · · ·

1 + · · ·

1 2x−1

• 2x∂ − 1
• is an integral element of k(x)[∂]/L, but does not

belong to k[x] + k[x] x∂

10

Main result: The idea of van Hoeij’s algorithm carries over from the algebraic case to the D-finite case.

11

Main result: The idea of van Hoeij’s algorithm carries over from the algebraic case to the D-finite case. In view of ALGEBRAIC ⊆ D-FINITE, our version may be viewed as a generalization of van Hoeij’s original algorithm.

11

Why should we care?

12

Why should we care?

• Trager uses integral bases to integrate algebraic functions.

12

Why should we care?

• Trager uses
• integral bases to integrate algebraic functions.

12

Why should we care?

• Trager uses
• integral bases to
• integrate algebraic functions.

12

Why should we care?

• Trager uses integral bases to integrate algebraic functions.
• Is there a similar integration algorithm for D-finite functions?

12

Why should we care?

• Trager uses integral bases to integrate algebraic functions.
• Is there a similar integration algorithm for D-finite functions?

Maybe.

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• Example. L = 2x(2x − 1)∂2 − (4x2 + 1)∂ + (2x + 1).

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• Example. L = 2x(2x − 1)∂2 − (4x2 + 1)∂ + (2x + 1).

An integral basis of k(x)[∂]/L is

• 1,

1 2x−1(2x∂ − 1)

• .

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• Example. L = 2x(2x − 1)∂2 − (4x2 + 1)∂ + (2x + 1).

An integral basis of k(x)[∂]/L is

• 1,

1 2x−1(2x∂ − 1)

• .

For a solution y of L, let ω0 := 1 · y, ω1 :=

1 2x−1(2x∂ − 1) · y and

consider f = (4x2 + 37x − 11)ω0 − (28x3 − 40x2 + x + 1)ω1 4(x − 1)2x2 .

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• Example. L = 2x(2x − 1)∂2 − (4x2 + 1)∂ + (2x + 1).

An integral basis of k(x)[∂]/L is

• 1,

1 2x−1(2x∂ − 1)

• .

For a solution y of L, let ω0 := 1 · y, ω1 :=

1 2x−1(2x∂ − 1) · y and

consider f = (4x2 + 37x − 11)ω0 − (28x3 − 40x2 + x + 1)ω1 4(x − 1)2x2 . Then a Hermite-reduction-like calculation can find f = ∂ · (11 + 4x)ω0 + 5(2x − 1)ω1 8(1 − x)2x2 + 0

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• Example. L = 2x(2x − 1)∂2 − (4x2 + 1)∂ + (2x + 1).

An integral basis of k(x)[∂]/L is

• 1,

1 2x−1(2x∂ − 1)

• .

For a solution y of L, let ω0 := 1 · y, ω1 :=

1 2x−1(2x∂ − 1) · y and

consider f = (4x2 + 37x − 11)ω0 − (28x3 − 40x2 + x + 1)ω1 4(x − 1)2x2 . Then a Hermite-reduction-like calculation can find

• f =

(11 + 4x)ω0 + 5(2x − 1)ω1 8(1 − x)2x2

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• Example. L = 2x(2x − 1)∂2 − (4x2 + 1)∂ + (2x + 1).

An integral basis of k(x)[∂]/L is

• 1,

1 2x−1(2x∂ − 1)

• .

For a solution y of L, let ω0 := 1 · y, ω1 :=

1 2x−1(2x∂ − 1) · y and

consider f = (4x2 + 37x − 11)ω0 − (28x3 − 40x2 + x + 1)ω1 4(x − 1)2x2 . Then a Hermite-reduction-like calculation can find

• f =

(11 + 4x)ω0 + 5(2x − 1)ω1 8(1 − x)2x2 So far, we have not worked out whether this works in general.

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