Telescopers for Rational andAlgebraic Functions via Residues

# Telescopers for Rational and Algebraic Functions via Residues

## Abstract

We show that the problem of constructing telescopers for functions of  variables is equivalent to the problem of constructing telescopers for algebraic functions of  variables and present a new algorithm to construct telescopers for algebraic functions of two variables. These considerations are based on analyzing the residues of the input. According to experiments, the resulting algorithm for rational functions of three variables is faster than known algorithms, at least in some examples of combinatorial interest. The algorithm for algebraic functions implies a new bound on the order of the telescopers.

Symbolic Integration, Creative Telescoping
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3

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I.1.2Computing MethodologiesSymbolic and Algebraic Manipulation[Algorithms]

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Algorithms

## 1 Introduction

The problem of creative telescoping is to find, for a given “function” in several variables , , linear differential operators involving only the and derivations with respect to the , and some other “functions” such that

 L(f)=Dx1(g1)+⋯+Dxm(gm),

where  denotes the derivative with respect to . The main motivation for computing such operators (called “telescopers” for ) is that, under suitable technical assumptions on and the domain , these operators have the definite integral

 F(t1,…,tn)=∫Ωf(t1,…,tn,x1,…,xm)dx1⋯dxm

as a solution. Once differential operators for have been found, other algorithms can next be used for determining possible closed forms, or asymptotic information, or recurrence equations for the series coefficients of .

There are general algorithms for computing telescopers when the input  is holonomic [25, 15, 24, 20, 9] as well as special-purpose algorithms designed for restricted input classes [25, 26, 5]. The focus in the present paper is on two such restricted input classes: rational and algebraic functions of several variables. Our first result is that an algorithm for computing telescopers for rational functions of  variables directly leads to an algorithm for computing telescopers for algebraic functions of  variables and vice versa (Section 2). Our second result is a new algorithm for creative telescoping of algebraic functions of two variables (Section 3), which, by the equivalence, also implies a new algorithm for creative telescoping of rational functions of three variables. The algorithm for algebraic functions is mainly interesting because it implies a new bound on the order of the telescoper in this case (Theorem 15), while the implied algorithm for rational functions is mainly interesting because at least for some examples it provides an efficient alternative to other methods (Section 4).

For a precise problem description, let  be a field of characteristic zero, and  be the field of rational functions in  and  over . Let  denote the  variables . The algebraic closure of a field  will be denoted by . The usual derivations  and  are denoted by  and , respectively. Let be the ring of linear differential operators in  with coefficients in . Then we are interested in the following two problems:

###### Problem 1.

Given , find a nonzero operator  such that

 L(f)=Dx1(g1)+⋯+Dxm(gm)for some~{}gj∈k(t,x).

Such an  is called a telescoper for , and the rational functions  are called certificates of .

###### Problem 2.

Given , find a nonzero operator  such that

 L(α)=Dx1(β1)+⋯+Dxm−1(βm−1) for % some~{}βj∈¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯k(t,^xm).

Such an  is called a telescoper for , and the algebraic functions  are called certificates of .

Both the equivalence of these two problems and the new algorithm for Problem 2 (when ) are based on the general idea of eliminating residues in the input. As an introduction to this approach, let us consider the problem of finding a telescoper and certificate for a rational function in two variables, that is, given a rational function , we want to find a nonzero such that for some . We may consider as an element of , where , and as such we may write

 f=p+n∑i=1mi∑j=1αi,j(x−βi)j, (1)

where , the are the roots in of the denominator of and the are in . We refer to the element as the residue of  at . Using Hermite reduction, one sees that a rational function is of the form for some if and only if all residues of are zero. Therefore to find a telescoper for it is enough to find a nonzero operator such that has only zero residues. For example assume that has only simple poles, i.e., , and squarefree. We then know that the Rothstein-Trager resultant [23, 19]

 R:=resultantx(a−zDx(b),b)∈K[z]

is a polynomial whose roots are the residues at the poles of . Given a squarefree polynomial in , differentiation with respect to and elimination allow one to construct a nonzero linear differential operator such that annihilates the roots of this polynomial. Applying to each term of (1) one sees that has zero residues at each of its poles. Applying Hermite reduction to allows us to find a such that .

The main idea in the method described above is that nonzero residues are the obstruction to being the derivative of a rational function and one constructs a linear operator to remove this obstruction. Understanding how residues form an obstruction to integrability and constructing linear operators to remove this obstruction will be the guiding principal that motivates the results which follow.

The authors would like to thank Barry Trager for useful discussions and outlining the proof of Proposition 11.

## 2 Telescopers for rationalfunctions

### 2.1 Rational and algebraic integrability

In this section, we give a criterion which decides whether or not  is a telescoper for a rational function in . Again, let . A rational function  is said to be rational integrable with respect to  if  for some . An algebraic function  is said to be algebraic integrable with respect to  if  for some . By taking traces, one can show that if is algebraic integrable with respect to , then an antiderivative of  already exists in the field .

For a rational function , Hermite reduction with respect to  decomposes  into

 f=Dxm(r)+ab, (2)

where and such that and is squarefree with respect to . It is clear that is rational integrable with respect to  if and only if in (2) is rational integrable with respect to . Over the field , one can write a rational function as

 f=p+n∑i=1mi∑j=1αij(xm−βi)j,

where and the  are in . We call the -residue of  at , denoted by .

###### Proposition 3.

Let  and . Then

• if  for some

• for all  with .

###### Proof.

The first assertion follows by observing the effect of  on each term in the partial fraction decomposition of . By Hermite reduction, we can decompose  into

 f=Dxm(r)+n∑i=1αixm−βi.

By the first assertion, either  if  or  if  for all . Applying  to the two sides of the equation above yields

 Dxj(f) =Dxj(Dxm(r))+n∑i=1(Dxj(αi)xm−βi+αiDxj(βi)(xm−βi)2) =Dxm(Dxj(r)−n∑i=1αiDxj(βi)xm−βi)+n∑i=1Dxj(αi)xm−βi.

Then we have either  if  or  if  for all . The second assertion follows.

If  is written as the form in (2), then we have

Therefore, all the -residues of  are roots of the Rothstein-Trager resultant (see [19, 23])

 R:=resultantxm(b,a−zDxm(b))∈K(^xm)[z].
###### Lemma 4.

Let . Then is rational integrable with respect to  if and only if all the -residues of  are algebraic integrable with respect to .

###### Proof.

By the Hermite reduction and partial fraction decomposition, can be written as

 f=Dxm(r)+n∑i=1αixm−βi,

where , and the  are pairwise distinct.

Suppose that all the -residues  of  are algebraic integrable with respect to , i.e., for some . Note that for each  we have

 Dxj(γi,j)xm−βi=Dxj(γi,jxm−βi)+Dxm(γi,jDxj(βi)xm−βi).

Then we get

 αixm−βi=m−1∑j=1Dxj(γi,jxm−βi)+Dxm(m−1∑j=1γi,jDxj(βi)xm−βi).

Therefore, is rational integrable with respect to  by taking

 gj=n∑i=1γi,jxm−βiandgm=r+n∑i=1m−1∑j=1γi,jDxj(βi)xm−βi.

Note that all the  and  are in  because  and  are roots of a polynomial in .

Suppose now that  is rational integrable with respect to , i.e., for some . For any , taking the -residues of  and , respectively, and using Proposition 3 we get

 residuexm(f,βi)=αi=m−1∑j=1Dxj(residuexm(gj,βi)),

which implies that  is algebraic integrable with respect to .

###### Example 5.

Let . Then the -residue of  at  is . Since , is rational integrable with respect to  and . More precisely,

 f=Dx1(x1x1+x2)+Dx2(−x1x1+x2).
###### Example 6.

Let . Then the -residue of  at  is . Since  has no antiderivative in , is not rational integrable with respect to  and .

### 2.2 Equivalence

###### Theorem 7.

Let . Then  is a telescoper for  if and only if  is a telescoper for every -residue of .

###### Proof.

By a similar calculation as in the proof of Proposition 3, we have

 L(residuexm(f,β))=residuexm(L(f),β) (3)

for any  and . If  is a telescoper for , then  for some . By Proposition 3 and Equation (3), for the -residue at any pole  of  with respect to , we have

 L(α)=m−1∑j=1Dxj(residuexm(gj,β)).

So  is a telescoper for . Conversely, assume that  is a telescoper for any -residue of . Note that any -residue of  is of the form , which is algebraic integrable by assumption. Then  is rational integrable by Lemma 4. Therefore, is a telescoper for .

Now we can present an explicit translation between the two telescoping problems by using Theorem 7.

If we can solve Problem 2, then for a rational function , first, we can perform Hermite reduction to decompose  into ; second, we compute the resultant ; finally, we get a telescoper for  by constructing telescopers for all the roots of  in  and taking their least common left multiple.

On the other hand, if we can solve Problem 1, then for an algebraic function  with minimal polynomial , we compute a telescoper  for the rational function . Note that is the -residue of  at . Therefore, is a telescoper for .

###### Example 8.

Consider the rational function

 f=2y(1−x)x(x+1)(x+2)(t+x)(xy−y−t4)1−x(2−x+(x+1)(x+2)(t+x)(xy−y−t4)2).

In order to find a telescoper for , we view as a rational function in with coefficients in and determine its residues in . Write and for the numerator and denominator of . Since is squarefree, the residues of are precisely the roots of the the Rothstein-Trager resultant . In the present example, these are

 t4x−1±1√x(x+1)(x+2)(x+t).

According to Theorem 7, it now suffices to find a telescoper for this algebraic function. This problem is discussed in the following section.

## 3 Telescopers for algebraicfunctions

We showed above how focusing on residues can yield a technique to find telescopers of rational functions by reducing this question to a similar one for algebraic functions. In this section we describe an algorithm to solve this latter problem for algebraic functions of two variables. In what follows, the term “algebraic function” will always refer to functions of two variables  and . When one tries to use residues to solve the problem of finding telescopers for algebraic functions one must deal with several complications. The first is a technical complication. One does not have a global way of expressing a function similar to partial fractions and so must rely on local expansions. This forces one to look at differentials rather than functions in order to define the notion of residue in a manner that is independent of local coordinates. The second complication is a more substantial one. There are differentials having zero residues everywhere that are not of the form , i.e. is not the derivative of an algebraic function. Nonetheless, one knows that there must exist an operator of order equal to twice the genus of the curve associated to such that for some algebraic . This will force us to add an additional step to find our desired telescoper. In Section 3.1, we will gather some facts concerning differentials in function fields of one variable that will be needed in our algorithm. In Section 3.2 we describe the algorithm.

### 3.1 Derivations and Differentials

In this section we review some notation and facts concerning function fields of one variable (cf. [2, 4, 8, 11, 16]). In the previous section the results and calculations depended heavily on the notion of the residue of a rational function of at an algebraic function of . In the present section we shall also need to use the notion of a residue but since we are dealing with algebraic functions instead of rational functions, the appropriate notion is that of a residue of a differential at a place of the associated function field . We will denote this by and refer to the above mentioned books for basic definitions and properties. We note that when , and , then , where and is the place of .

Let be a differential field of charactersitic zero with derivation denoted by (for example, with as above). Let be transcendental over and an algebraic extension of . We may extend the derivation to a derivation on by first letting and then taking the unique extension to . We define a derivation on by letting be zero on , and taking the unique extension of from to . We shall also assume that the constants are precisely . This is equivalent to saying that the minimal polynomial of over is absolutely irreducible (cf. [10]). In [8], Chapter VI, §7, Chevalley shows that can be used to define a map (which we denote again by ) on differentials such that . The map furthermore has the following properties:

1. for any , and

2. for any place of and any differential ,

 residueP(Dxtω)=Dxt(residueP(ω)).

Given we will want to find an operator and an element such that . In terms of differentials, this latter equation may be written as , where .

We shall have occasion to write our field as for some other which is transcendental over and algebraic over and work with the derivation defined in a similar manner as above. We will need to know that if we can find a telescoper with respect to the derivation then we can convert this into a telescoper with respect to . The following lemma and proposition allow us to do this.

###### Lemma 9.

Let and be as above and let be a differential of . For any there exists such that

 (D¯xt)i(ω)−(Dxt)i(ω)=dui. (4)
###### Proof.

Write . Lemma 1 of [16] (see also Lemma 3 in Chapter VI, §7 of [8]) implies that

 D¯xt(ω)−Dxt(ω)=−d(¯αDxt(¯x)). (5)

Letting , we have equation (5) for . One can verify by induction that (5) holds for , where

###### Proposition 10.

Let , ,

 (D¯xt)n+an−1(D¯xt)n−1+…+a0∈K⟨D¯xt⟩,

and such that

 ((D¯xt)n+an−1(D¯xt)n−1+…+a0)(ω)=d¯β.

One can effectively find such that

 ((Dxt)n+an−1(Dxt)n−1+…+a0)(α)=Dx(β).
###### Proof.

From Lemma 9 we have that

 ((D¯xt)n+an−1(D¯xt)n−1+…+a0)(ω) =((Dxt)n(ω)+dun)+an−1((Dxt)n−1(ω)+dun−1) +…+a0ω.

Therefore, taking into account that the belong to ,

 ((Dxt)n+an−1(Dxt)n−1+…+a0)(ω)

which implies the conclusion of the proposition with .

In the algorithm described in the next section, we will consider a differential in and assume that

1. has no poles at any place above the place of at infinity, and

2. the places where does have a pole are all unramified above places of .

We describe below an algorithm that allows one to select an such that and such that satisfies conditions 1. and 2. above with respect to . The algorithm of Section 3.2 can be used to produce a telescoper with respect to and Proposition 10 allows one to convert this telescoper to a telescoper with respect to . In the following proposition, the proof that condition 2. can be fulfilled was outlined to us by Barry Trager [21, 22].

###### Proposition 11.

Let be a differential in . One can effectively find an such that and

1. has no poles at any place above the place of at infinity, and

2. the places where does have a pole are all unramified above places of .

###### Proof.

If 1. does not hold, let be selected so that has no poles above , let

 ¯x=cxx−c.

This change of variables interchanges and the point at infinity, so 1. is now satisfied with respect to and we shall henceforth abuse notation and assume that 1. is satisfied with respect to .

Let be a nonsingular curve that is a model of . The elements of can be considered as functions on . As noted in [21, p. 63], ramification occurs when the line of projection from the curve down to the -axis is tangent to the curve and, for each pole of , there are only a finite number of projection directions that are tangent to the curve at this pole. Therefore for all but finitely many choices of an integer , if we let , will satisfy 2. with respect to . One can refine this argument and produce a finite set of integers that are to be avoided. This is done in the following way.

Let be an indeterminate and consider the field , where and . Let and assume that (after a possible change of ), satisfies a monic polynomial over . The behavior of various objects in when one reduces modulo a prime ideal of is considered in [11, Chapter III, §6]. We shall be interested in reducing modulo ideals of the form , where is an integer. One can effectively calculate an integral basis of the integral closure of in (cf. [12, 21]) and from this a complementary basis ([2, Chapter 5, §2], [4, §22]). In Chapter III §6.2 of [11], Eichler gives a method that will produce a finite set such that for , the set is again an integral basis of the integral closure of in . This method can be refined (and the set slightly increased if need be) so that is also a complementary basis. Expressing in terms of this complementary basis,

 ω=1b(¯x)n∑i=1pi(M,¯x)w′i(M)d¯x,

one sees that will have poles precisely at the zeroes of . If one selects such that is relatively prime to , the discriminant of the integral basis , then will not have poles at ramification points. The finitely many values of that do not satisfy this latter condition are roots of

 S(M)=resultantX( resultantY(b(X+MY),F(X,Y)), resultantY(D(X+MY),F(X,Y))),

where is the minimal polynomial of over .

### 3.2 An Algorithm to Calculate Telescopers for Algebraic Functions

We assume we are given a function field of one variable and a differential in . We shall furthermore assume that satisfies conditions 1. and 2. of Proposition 11. We will describe an algorithm to find , not all zero, and such that

 (an(Dxt)n+an−1(Dxt)n−1+…+a0)(ω)=dβ.

If , then is a telescoper for with certificate . The algorithm has two steps. The first step finds an operator such that applying this operator to results in a differential with only zero residues. The second step finds an operator of order at most twice the genus of and an element such that .

Step 1. We will describe two methods for constructing an operator that annihilates the residues of . The first one requires one to calculate in algebraic extensions of while the second only requires calculations in . Throughout, let be a minimal polynomial of over and let

 ω=αdx=ABdx

for some with no finite poles and .

Method 1. We make no assumptions concerning ramification at the poles but for convenience we do assume that the poles of only occur at finite points. Let be a root of . For any branch of at , we may write

 ω=pa(z)dz,

where for some positive integer and is a Laurent series in with coefficients in . One can calculate the coefficient of in and this will be the residue of at this place. In this way, one can calculate the possible residues of . Let be a Galois extension of containing . Let be the field of -constants in and be a -basis of . Let where is the Wronskian determinant. One sees that is a nonzero linear differential polynomial with coefficients in such that for . Define

 L1(Y)=lclm{Lσ(Y)∣σ∈G},

where is the Galois group of over , denotes the linear differential polynomial resulting from applying to each coefficient of and denotes the least common left multiple. We then have that has coefficients in and annihilates the residues of .

Method 2. We now assume that has poles only at finite places and that there is no ramification at the poles. This implies that at any place corresponding to a pole, we may write for some . Therefore the residue of at this place is

 α−1=(Dx[(x−x0)−i0−1α])x=x0.

This is the key to the following, parts of which in a slightly different form appear in [7].

###### Proposition 12.

Given as above, one can compute a polynomial of degree

 m:=degZ(R)≤degY(F)degx(B∗),

with the square free part of , such that if is a nonzero residue of then . Furthermore, one can compute a nonzero operator such that has residue zero at all places.

###### Proof.

We may write

 αdx=ABdx=A1B1dx+A2B22dx+⋯+AℓBℓℓdx,

where the are regular at finite places and is the squarefree decomposition of . To achieve our goal it is therefore enough to prove the claim for a differential of the form , where is regular at finite places and is squarefree. Following [7], we let be a differential indeterminate and let

 h=(Au−n)(n−1)(n−1)!∈K(x,y)⟨u⟩,

where is the ring of differential polynomials in with coefficients in and denotes -fold differentiation with respect to . Let be a place where has a pole and let and denote the values of and at the place. We note that since is regular at and is not ramified, any derivative of is also regular at (one needs the hypothesis that these places are unramified to make this claim). Taking into account the rules of differentiation, we see that

 h=p(x,y,u,u′,…,u(n−1))q(x)ut,

where , is some positive integer and does not vanish at , i.e. . Let

 ~p=p(x,Y,B′,12B′′,13B(3),…,1nB(n))∈K[x,Y]

and

 ~q=q(x)(B′)t∈K[x].

One then shows, as in [7], that is the residue of at .

The above argument shows that the polynomial

 R=resultantx(resultantY(~p−Z~q,F),B)∈K[Z]

vanishes at the residues of . The degree estimate for follows from the general degree estimate for resultants which states for any that is at most

 degu(S)degv(T)+degv(S)degu(T).

This implies first that the inner resultant in the definition of has -degree at most . (Note that no degree estimates for