Lazy Hermite Reduction and Creative Telescoping for Algebraic Functions
aa r X i v : . [ c s . S C ] F e b Lazy Hermite Reduction and CreativeTelescoping for Algebraic Functions ∗ Shaoshi Chen a,b , Lixin Du a,b,c , Manuel Kauers ca KLMM, Academy of Mathematics and Systems Science,Chinese Academy of Sciences,Beijing 100190, China b School of Mathematical Sciences,University of Chinese Academy of Sciences,Beijing 100049, China c Institute for Algebra, Johannes Kepler University,Linz, A4040, Austria [email protected], [email protected]@jku.at
February 16, 2021
Abstract
Bronstein’s lazy Hermite reduction is a symbolic integration techniquethat reduces algebraic functions to integrands with only simple poles with-out the prior computation of an integral basis. We sharpen the lazy Her-mite reduction by combining it with the polynomial reduction to solve thedecomposition problem of algebraic functions. The sharpened reduction isthen used to design a reduction-based telescoping algorithm for algebraicfunctions in two variables.
The integration problem for algebraic functions has a long history that can betraced back at least to the work of Euler and others on elliptic integrals [3]. In1826, Abel initiated the study of general integrals of algebraic functions, whichare now called abelian integrals [2]. Liouville in 1833 proved that if the integral ∗ S. Chen was partially supported by the NSFC grants 11871067, 11688101, the Fundof the Youth Innovation Promotion Association, CAS, and the National Key Research andDevelopment Project 2020YFA0712300. L. Du was supported by the NSFC grant 11871067and the Austrian FWF grant P31571-N32. M. Kauers was supported by the Austrian FWFgrant P31571-N32.
1f an algebraic function is an elementary function, then it must be the sum ofan algebraic function and a linear combination of logarithms of algebraic func-tions (see [29, Chapter IX] for a detailed historical overview). In 1948, Rittpresented an algebraic approach to the problem of integration in finite terms inhis book [34]. Based on Liouville’s theorem and some developments in differen-tial algebra [35], Risch in 1970 finally solved this classical problem by giving acomplete algorithm [33]. After Risch’s work, more efficient algorithms have beengiven due to the emerging developments in symbolic computation [20, 36, 11, 13].Passing from indefinite integration to definite integration with parameters, thecentral problem shifts to finding linear differential equations satisfied by the in-tegrals of algebraic functions with parameters. In this direction, the first workwas started by Picard in [32] and later a systematical method was developedby Manin in his work on proving the function-field analogue of Mordell’s con-jecture [30]. In the 1990s, another powerful method was developed by Almkvistand Zeilberger [4] by including the trick of differentiating under the integral signin the framework of creative telescoping [39].In a given differential ring R with the derivation D , one can ask two fun-damental problems: one is the integrability problem , i.e., deciding whether agiven element f ∈ R is of the form D ( g ) for some g ∈ R , if such a g exists,we say that f is integrable in R ; another is the decomposition problem , i.e., de-composing a given element f ∈ R into the form D ( g ) + r with g, r ∈ R and r is minimal in certain sense and r = 0 if and only if f is integrable in R . Foralgebraic functions, the integrability problem was studied by Liouville in his twomemoirs [27, 28]. Hermite reduction [22] solves the decomposition problem forrational functions. Trager in his thesis [36] extended Hermite reduction to thealgebraic case. His algorithm requires the computation of an integral bases inthe beginning. In order to avoid this expensive step, Bronstein [10] introducedthe lazy Hermite reduction that partially solves the decomposition problem.The first contribution in this paper is a sharpened version of the lazy Hermitereduction. We combine the lazy Hermite reduction with a further reduction,namely the polynomial reduction, in order to solve the decomposition problemcompletely.When the differential ring R is equipped with two derivations D , D , onecan also consider the creative telescoping problem: for a given element f ∈ R ,decide whether there exist c , . . . , c r ∈ R , not all zero, such that D ( c i ) = 0 forall i ∈ { , . . . , r } and c r D r ( f ) + · · · + c f = D ( g ) for some g ∈ R. The operator L = c r D r + · · · + c if exists is called a telescoper for f . Forevery algebraic function there exists such a telescoper, and many constructionalgorithms have been developed in [30, 38, 16, 9, 15]. Our second contributionis an adaption of the reduction-based approach from [15] using the sharpenedlazy Hermite reduction.The remainder of this paper is organized as follows. First we recall Bron-stein’s idea of lazy Hermite reduction in Section 2. Instead of an integral basis,2t uses a so-called “suitable basis” of the function field. In Section 3 we havea closer look at these bases. After developing the polynomial reduction in Sec-tion 4, we will present the telescoping algorithm for algebraic functions basedon the sharpened lazy Hermite reduction in Section 5. We conclude our pa-per by some experimental comparisons among several telescoping algorithms inSection 6. Trager’s generalization of Hermite reduction to algebraic functions works as fol-lows [36, 21, 12, 15]. Let K be a field of characteristic zero and m ∈ K ( x )[ y ]be an irreducible polynomial over K ( x ). Then A = K ( x )[ y ] / h m i is an algebraicextension of K ( x ). When there is no ambiguity, we also use y to represent theelement y + h m i in A , which can be viewed as a root of m . Let W = ( ω , . . . , ω n )be an integral basis of A . Let f = uv d P nk =1 a k ω k ∈ A with d > u, v, a , . . . , a n ∈ K [ x ] such that gcd( u, v ) = gcd( v, v ′ ) = gcd( v, a , . . . , a n ) = 1.We seek b , . . . , b n , c , . . . , c n ∈ K [ x ] such that for g = v d − P nk =1 b k ω k and h = uv d − P nk =1 c k ω k such that f = g ′ + h. The g in this equation can be found by solving a certain linear system over K [ x ] / h v i , and once g is known, h can be computed as h = f − g ′ . Let e ∈ K [ x ]and M ∈ K [ x ] n × n be such that eW ′ = M W , and assume (without loss ofgenerality) that e | uv . Then the coefficient vector b = ( b , . . . , b n ) of v d − g satisfies b (cid:0) uve − M − ( k − uv ′ I n (cid:1) ≡ ( a , . . . , a n ) mod v (2.1)and using that W is an integral basis, it can be shown that this linear systemhas a unique solution, see [36, 15] for further details. Applying the processrepeatedly, we can eliminate all multiple poles from the integrand, i.e., we canfind g and h such that f = g ′ + h and h = q − P nk =1 p k ω k for some polynomials p , . . . , p n , q with q squarefree.If W is not an integral basis, the linear system (2.1) may or may not have aunique solution. Example 1.
Let m = y − x and f = y ( x +1) x .1. For W = ( x, xy ) we have e = 2 x and M = (cid:18) (cid:19) . Applying the reduc-tion to f = x +1) x xy leads to the linear system b (cid:18) − x + 1) 00 − ( x + 1) (cid:19) ≡ (0 ,
2) mod x which has a unique solution. . For W = ( x, y ) we have e = 2 x and M = (cid:18) (cid:19) . Applying the reductionto f = x ( x +1) y leads to the linear system b (cid:18) − ( x + 1) (cid:19) ≡ (0 ,
2) mod x, which is solvable, but not uniquely.3. For W = ( x, ( x +1) y ) we have e = 2 x ( x +1) and M = (cid:18) x + 1) 00 3 x + 1 (cid:19) .Applying the reduction to f = 1 x ( x + 1) ( x + 1) y leads to the linear system b (cid:18) − x − ( x + 1) (cid:19) ≡ (0 ,
2) mod x ( x + 1) which has no solutions. Note that none of the bases above in the example above is an integral basis.However, all the bases consist of integral elements and have the property that e is squarefree. Bronstein [10] calls such a basis a suitable basis and observes thatwhenever we apply Hermite reduction to a suitable basis and find that the linearsystem (2.1) has no solution, then we can construct from any unsolvable systeman integral element of A that does not belong to the K [ x ]-module generated bythe elements of W . We can then replace W by a suitable basis of an enlarged K [ x ]-module which also includes A and proceed with the reduction. Example 2.
We continue the previous example.1. For W = ( x, xy ) , no basis update is needed because the linear system hasa unique solution.2. For W = ( x, y ) , the right kernel element (1 , of the matrix in the linearsystem translates into the new integral element x +1 , which does not belongto the K [ x ] module generated by x and y in A . A basis of the modulegenerated by x , y , and x + 1 is (1 , y ) .3. For W = ( x, ( x + 1) y ) , from the lack of solutions of the linear system itcan be deduced that xy is an integral element not belonging to the modulegenerated by x and y in A . A basis of the module generated by x , y , and xy is ( x, y ) . Starting from a basis W consisting of integral elements, there can be at mostfinitely many basis updates before we reach an integral basis. Therefore, it takesat most finitely many basis updates (and possibly fewer than needed for reaching4n integral basis) to complete the reduction process. This variant of Hermitereduction, which avoids the potentially expensive computation of an integralbasis at the beginning, is called lazy Hermite reduction. Its final result is asuitable basis ˜ W of A and g, h ∈ A such that f = g ′ + h and the coefficients of h with respect to ˜ W are rational functions with a squarefree common denominator.In the examples above, we may find ˜ W = (1 , y ), g = − yx , and h = − yx ( x +1) .One of the key features of Hermite reduction is that we can decide theintegrability problem. For example, if we write a rational function f ∈ K ( x ) inthe form f = g ′ + h for some g, h ∈ K ( x ) where h has a squarefree denominatorand numerator degree less than denominator degree, then f admits an integral in K ( x ) if and only if h = 0. Trager generalizes this criterion to algebraic functionsas follows. By a change of variables, he first ensures that the integrand f has adouble root at infinity. Then he performs Hermite reduction with respect to anintegral basis that is normal at infinity. If this gives g, h such that f = g ′ + h ,then f is integrable in A if and only if h = 0 [36, 15].Unfortunately, this criterion does not extend to the lazy version of Hermitereduction. Even if we produce a double root of the integrand at infinity andmake the suitable basis normal at infinity (which amounts to a local integralbasis computation that we actually would prefer to avoid altogether), a nonzeroremainder h does not imply that f is not integrable. Example 3.
Let m = y − x and f = yx . For W = (1 , ( x + 1) y ) we have e = 2 x ( x + 1) and M = (cid:18) x + 1 (cid:19) . The lazy Hermite reduction finds g = − x ) y x and h = y x . Here f = (cid:0) − y x (cid:1) ′ is integrable and has a root oforder ≥ at infinity, but the remainder h is nonzero. Somewhat surprisingly, Bronstein does not address this issue in his re-port [10]. In the following sections, we develop a fix using the technique ofpolynomial reduction.
Let A = K ( x )[ y ] / h m i with m ∈ K [ x, y ] being an irreducible polynomial over K ( x ). Let ¯ K be the algebraic closure of K . If n = deg y ( m ), there are n distinctsolutions in the field ¯ K hh x − a ii := [ r ∈ N \{ } ¯ K (( ( x − a ) /r ))of formal Puiseux series around a ∈ ¯ K . There are also n distinct solutions inthe field ¯ K hh x − ii := [ r ∈ N \{ } ¯ K (( x − /r ))of formal Puiseux series around ∞ . For a fixed a ∈ ¯ K ∪{∞} , let y , . . . , y n be all n roots of m in ¯ K hh x − a ii (or ¯ K hh x − ii if a = ∞ ). There are n distinct K ( x )-embeddings σ , . . . , σ n of A into ¯ K hh x − a ii (or ¯ K hh x − ii if a = ∞ ) such that5 i ( f ( y )) = f ( y i ) for any f ∈ A . Then for each a ∈ ¯ K ∪ {∞} , we can associate f ∈ A with n series σ i ( f ) for i = 1 , . . . , n . Moreover, if we equip the fields A ,¯ K hh x − a ii and ¯ K hh x − ii with natural differentiations with respect to x , theneach embedding σ i is a differential homomorphism, i.e., σ i ( f ′ ) = σ i ( f ) ′ for any f ∈ A .A nonzero Puiseux series around a ∈ ¯ K can be written in the form P = X i ≥ c i ( x − a ) r i , where c i ∈ ¯ K, c = 0 and r i ∈ Q . The valuation map ν a on ¯ K hh x − a ii is definedby ν a ( P ) = r if P is nonzero and ν a (0) = ∞ . Replacing x − a by x , we get the valuation map ν ∞ on ¯ K hh x − ii . A series P in ¯ K hh x − a ii or ¯ K hh x − ii is called integral if its valuation is nonnegative. The value function val a : A → Q ∪ {∞} is defined by val a ( f ) := n min i =1 ν a ( σ i ( f ))for any f ∈ A . An element f ∈ A is called (locally) integral at a ∈ ¯ K ∪ {∞} ifval a ( f ) ≥
0, i.e., every series associated to f (around a ) is integral. The element f ∈ A is called (globally) integral if val a ( f ) ≥ a ∈ ¯ K (“at all finiteplaces”), i.e., f is locally integral at every a ∈ ¯ K . A basis of the K [ x ]-moduleof all integral elements of A is called an integral basis of A . The elements of A that are locally integral at some point a ∈ K form a K ( x ) a module, where K ( x ) a is the subring of K ( x ) consisting of all rational functions which do nothave a pole at a . In the case a = ∞ , K ( x ) a is the ring of all rational functions p/q with deg x ( p ) ≤ deg x ( q ). A basis of the K ( x ) a -module of locally integralelements of A is called a local integral basis at a .For a series P ∈ ¯ K hh x − a ii , the smallest positive integer r such that P ∈ ¯ K (( ( x − a ) /r )) is called the ramification index of P . If r >
1, the series P issaid to be ramified . For an element f ∈ A , a point a ∈ ¯ K is called a branchpoint of f if one of the series associated to f around a is ramified.Let W = ( ω , . . . , ω n ) be a K ( x )-vector space basis of A . Throughout thissection, let e ∈ K [ x ] and M = (( m i,j )) ni,j =1 ∈ K [ x ] n × n be such that eW ′ = M W and gcd( e, m , , m , , . . . , m n,n ) = 1. As already mentioned in the previoussection, W is a called a suitable basis if e is squarefree and ω i ’s are integralfor each i . Every integral basis is a suitable basis, see [15, Lemma 3]. Now weexplore some further properties of such bases. Lemma 4.
Let W be an integral basis of A . Let e ∈ K [ x ] and M ∈ K [ x ] n × n be such that eW ′ = M W . If a ∈ ¯ K is a root of e , there exists ω ∈ W such that a is a branch point of ω .Proof. Let W = ( ω , . . . , ω n ) and M = (( m i,j )) ni,j =1 ∈ K [ x ] n × n , and let a be a6oot of e . By gcd( e, m , , . . . , m n,n ) = 1, there is i ∈ { , . . . , n } such that ω ′ i = 1 e n X j =1 m i,j ω j , where a is not a common root of m i, , . . . , m i,n .From the above expression of ω ′ i , we get ω ′ i does not belong to the modulegenerated by W over K [ x ]. Since W is a local integral basis at a , it follows that ω i is not locally integral at a . Thus val a ( ω ′ i ) < a were not a branch point of ω i , then all Puiseux series at a associatedto ω i were power series, and then all Puiseux series at a associated to ω ′ i werepower series, implying val a ( ω ′ i ) ≥
0. As we have seen before that val a ( ω ′ i ) < a must be a branch point.In order to give a converse of Lemma 4, we consider the series associatedto an algebraic function. For a ramified Puiseux series P = P i ≥ c i ( x − a ) r i ∈ ¯ K hh x − a ii , let δ ( P ) := min { r i | r i ∈ Q \ Z , i ≥ } be the minimal fractional exponent of P . Define δ ( P ) = ∞ if the series P isnot ramified. Then δ ( P ′ ) = δ ( P ) −
1. Similar as the valuation of a series, thefunction δ satisfies δ ( P + Q ) ≥ min { δ ( P ) , δ ( Q ) } for any P, Q ∈ ¯ K hh x − a ii . Lemma 5.
Let W be a K ( x ) -vector space basis of A . Let e ∈ K [ x ] and M ∈ K [ x ] n × n be such that eW ′ = M W . Let a ∈ ¯ K . If there exists some ω ∈ W suchthat a is a branch point of ω , then a is a root of e .Proof. Suppose that a is a branch point of some ω ∈ W . It implies that for suchan element ω , there is a K ( x )-embedding σ of A into the field of Puiseux series(around a ) such that the series σ ( ω ) is ramified. Let r = min { δ ( σ ( ω )) | ω ∈ W } .Then r ∈ Q \ Z . Choose an element ω i ∈ W such that δ ( σ ( ω i )) = r . Then σ ( ω i )must be ramified. After differentiating the series σ ( ω i ), its minimal fractionalexponent decreases strictly by 1. This means δ ( σ ( ω i ) ′ ) = δ ( σ ( ω i )) − r − . Let M = (( m i,j )) ni,j =1 ∈ K [ x ] n × n . Since σ is a differential homomorphismand a K ( x )-embedding, we have σ ( ω i ) ′ = σ ( ω ′ i ) = 1 e n X j =1 m i,j σ ( ω j ) . After multiplying by a polynomial, the minimal fractional exponent of a serieswill not decrease. So if a is not a root of e , then δ ( σ ( ω i ) ′ ) ≥ n min j =1 δ ( m i,j σ ( ω j )) ≥ n min j =1 δ ( σ ( ω j )) = r. This leads to a contradiction. Thus a must be a root of e .7e now show that the polynomial e does not depend on the choice of thebasis of A but only on the K [ x ]-submodule it generates. Let U and V be two K ( x )-vector space bases of A . Let e u , e ν ∈ K [ x ] and M u , M ν ∈ K [ x ] n × n besuch that e u U ′ = M u U and e ν V ′ = M ν V . Suppose that U and V generate thesame submodule of A over K [ x ]. Then there exists a matrix T ∈ K [ x ] such that U = T V and T is an invertible matrix over K [ x ]. Taking derivatives, we get U ′ = (cid:18) T ′ T − + T e ν M ν T − (cid:19) U = 1 e u M u U. Since
T, T − ∈ K [ x ] n × n , we have e u divides e ν . Similarly the fact that V = RU with R = T − ∈ K [ x ] n × n implies that e ν divides e u . Thus e u = e ν when e u , e ν are monic. Lemma 6.
Let W be a suitable basis and U be an integral basis of A . Let W = T U with T ∈ K [ x ] n × n . Let e ∈ K [ x ] and M ∈ K [ x ] n × n be such that eW ′ = M W . If a ∈ ¯ K is a root of det( T ) , then a is a root of e . That means if W is not a local integral basis at a ∈ ¯ K , then a is a root of e .Proof. First we shall make a change of bases. Consider the Smith normal formof the matrix T . This means there are two matrices P, Q ∈ K [ x ] n × n such thatboth P and Q are invertible over K [ x ] and P T Q = Λ for some diagonal matrixΛ ∈ K [ x ] n × n . Then W = P − Λ Q − U and hence P W = Λ( Q − U ). Replacing P W and Q − U by W and U , respectively, we may assume W = T U with T = diag( r , . . . , r n ) ∈ K [ x ] n × n . This operation does not change themodule generated by W or U , respectively. Note that det( P T Q ) and det( T )are equal up to a unit in K . It suffices to prove the result for such special bases W and U .Let e u ∈ K [ x ] and M u = (( a i,j )) ni,j =1 ∈ K [ x ] n × n be such that e u U ′ = M u U .Differentiating both sides of W = T U yields that W ′ = (cid:18) T ′ T − + T e u M u T − (cid:19) W = 1 e M W. (3.1)Substituting T = diag( r , . . . , r n ) and M u = (( a i,j )) ni,j =1 , we get the the i -thdiagonal entry of e M is r ′ i r i + a i,i e u .Let a ∈ ¯ K be a root of det( T ). Since det( T ) = r r · · · r n , there is i ∈{ , , . . . , n } such that a is a root of r i . If a is not a root of e u , then a must be apole of the entry r ′ i r i + a i,i e u . Since e is a common multiple of the denominator ofall the entries of e M , we have a is a root of e . Now suppose that a is a root of e u . Since U is an integral basis, Lemma 4 implies that there is u ∈ U such that a is a branch point of u . Then one of the series associated to u is ramified. Inother words, such a series has at least one fractional exponent. Since W = T U for some invertible matrix T in K ( x ) n × n , there is ω ∈ W such that one of theseries associated to ω is also ramified. Therefore, a is a branch point of ω . ByLemma 5, a is root of e . 8ince the inverse of the matrix T is T − = T ) T ∗ , where T ∗ is the adjointmatrix of T , the least common multiple of the denominator of the entries of T − is bounded by det( T ). By investigating Equation (3.1), we see a possibleroot of e must come from det( T ) and e u . Combining the last paragraph in theargument of Lemma 6, we get those roots of det( T ) and e u are exactly rootsof e . In particular, when W is a suitable basis, the corresponding polynomial e is the squarefree part of the product det( T ) e u . Therefore, if the submodulegenerated by a suitable basis is larger, then e is smaller. Polynomial reduction is a postprocessing step for Hermite reduction which wasfirst introduced for hyperexponential terms [6] and has later been formulated foralgebraic and fuchsian D-finite functions [15, 17]. For the latter cases, like forTrager’s criterion, integral bases that are normal at infinity are employed. Ourgoal in this section is to relax this requirement to suitable bases, so as to obtaina version of polynomial reduction which can serve as a natural continuationof the lazy Hermite reduction process and provides the feature that the finalremainder is zero if and only if the integrand is integrable.Let h ∈ A be the remainder of lazy Hermite reduction (see Section 2) withrespect to a suitable basis W = ( ω , . . . , ω n ). Then h can be written in the form h = n X i =1 h i de ω i (4.1)with h i , d ∈ K [ x ] such that gcd( d, e ) = gcd( h , . . . , h n , d ) = 1 and d is square-free. If h is integrable in A , we shall prove that d is a constant in K . When W is an integral basis, this result was proved in [15, Lemma 9]. The followinglemma is a local version. Lemma 7.
Let h ∈ A be in the form (4.1) . If h is integrable in A and W is alocal integral basis at a ∈ ¯ K , then d has no root at a .Proof. Suppose h is integrable in A . Then there exists H = P ni =1 b i ω i ∈ A with b i ∈ K ( x ) such that h = H ′ . It suffices to show that every b i has no pole at a .Otherwise H has a pole at a , because W is a local integral basis at a . Then h has at least a double pole at a . This contradicts the fact that d, e are squarefree.Thus d has no root at a . Theorem 8.
Let h ∈ A be in the form (4.1) . If h is integrable in A , then d isin K .Proof. Suppose that h is integrable in A . In order to show d is a constant, weshow that for any a ∈ ¯ K , a is not a root of d . If W is a local integral basisat a , then the conclusion follows from Lemma 7. Now we assume that W is nota local integral basis at a . By Lemma 6, we know that a is a root of e . Sincegcd( d, e ) = 1, it follows that a is not a root of d .9o further reduce the lazy Hermite remainder, we give an upper bound forthe denominator of its integral if h is integrable in A . This bound does notdepend on the integrand h , but only depends on the discriminant of a suitablebasis in its representation.Recall that the discriminant of a tuple W = ( ω , . . . , ω n ) of elements of A isdefined by the determinantDisc( W ) = det(Tr( ω i ω j )) , where Tr is the trace map from A to K ( x ). If the ω i ’s are integral functions,then their traces are polynomials, and thus the discriminant is a polynomial. If W = T U where T ∈ K [ x ] n × n is the change of basis matrix, then Disc( W ) =Disc( U ) det( T ) . Lemma 9.
Let h ∈ A be in the form (4.1) . Let U be an integral basis of A andlet T ∈ K [ x ] n × n be such that W = T U . If h is integrable in A , i.e., there exists u ∈ K [ x ] and q = ( q , . . . , q n ) ∈ K [ x ] n such that h = n X i =1 q i u ω i ! ′ and gcd( q , . . . , q n , u ) = 1 , then u divides det( T ) . Hence u divides Disc( W ) / Disc( U ) .Proof. Since h has a squarefree denominator with respect to W and W = T U with T ∈ K [ x ] n × n , it follows that h has a squarefree denominator with respectto U . This means h is also a remainder with respect to the integral basis U .Assume that h is integrable in A . Then Lemma 7 implies that h = ( P ni =1 a i u i ) ′ with a i ∈ K [ x ]. Write U = r RW with r ∈ K [ x ], R = (( b i,j )) ni,j =1 ∈ K [ x ] n × n and gcd( r, b , , b , , . . . , b n,n ) = 1. Then n X i =1 a i u i = n X i =1 a i r n X j =1 b i,j ω j = n X j =1 r n X i =1 a i b i,j ! ω j . Thus u divides r because a i , b i,j ∈ K [ x ] and two antiderivatives of h only differby a constant.Since W = T U , we have r R = T − = T ) T ∗ . So r divides det( T ) andhence u also divides det( T ). Moreover, Disc( W ) = Disc( U ) det( T ) . Since u i ’sand ω i ’s are integral elements, the discriminants Disc( W ) , Disc( U ) are polyno-mials in K [ x ]. Therefore det( T ) divides the polynomial Disc( W ) / Disc( U ), sodoes u .If we already know that W is an integral basis, then the quotient Disc( W ) / Disc( U )is a unit in K . By Lemma 9, we see u is a constant. If no integral basis is avail-able, then from the condition u divides Disc( W ), an upper bound of u can bechosen as the product over p ⌊ r/ ⌋ where p runs through the irreducible factorsof Disc( W ) and r is the multiplicity of p in Disc( W ).10 xample 10. Let m = y − x and h = yx .1. For W = (1 , y ) , we have Disc( U ) = 4 x . So we can choose u = 1 . In fact, W is an integral basis and h = (2 y ) ′ .2. For W = (1 , ( x + 1) y ) , we have Disc( W ) = 4( x + 1) x . So we can choose x + 1 as an upper bound of u . In fact, h = ( x +1 x + 1) y ) ′ . If h is integrable in A , we shall reduce h to zero, otherwise we hope toremove all possible integrals whose denominators are bounded by u . Beforethat we write h as two parts with denominators d and e , respectively. Bythe extended Euclidean algorithm, there are polynomials r i , s i ∈ K [ x ] suchthat h i = r i d + s i e and deg x ( r i ) < deg x ( d ). Then the lazy Hermite reductionremainder h decomposes as h = n X i =1 h i de ω i = n X i =1 r i d ω i + n X i =1 s i e ω i . (4.2)Our second goal is to confine the s i ’s to a finite-dimensional vector spaceover K . This is a generalization of the polynomial reduction in [15]. In thisprocess, we shall rewrite the second term of the remainder h in (4.2) withrespect to another basis. This new basis is used to perform the polynomialreduction and obtain the following additive decomposition. Definition 11.
Let f be an element in A . Let W and V be two K ( x ) -vectorspace bases of A . Let e, a ∈ K [ x ] and M, B ∈ K [ x ] n × n be such that eW ′ = M W and aV ′ = BV . Suppose that f can be decomposed into f = g ′ + 1 d P W + 1 a QV, (4.3) where g ∈ A , d ∈ K [ x ] is squarefree and gcd( d, e ) = 1 , P, Q ∈ K [ x ] n with deg x ( P ) < deg x ( d ) and Q ∈ N V , which is a finite-dimensional K -vector space.The decomposition in (4.3) is called an additive decomposition of f with respectto x if it satisfies the condition that P, Q are zero if and only if f is integrablein A . Given an algebraic function, its additive decomposition always exists forsome integral basis W and some basis V which is a local integral basis at infinity,see [15, Theorem 14]. We shall show below that we can always find an additivedecomposition with respect to certain suitable bases.Let V be a K ( x )-vector space basis of A , and let a ∈ K [ x ] and B =(( b i,j )) ni,j =1 ∈ K [ x ] n × n be such that aV ′ = BV . We do not require thatgcd( a, b , , b , , . . . , b n,n ) = 1. Let u ∈ K [ x ] and p ∈ K [ x ] n . A direct calcu-lation yields that (cid:16) pu V (cid:17) ′ = (cid:16) pu (cid:17) ′ V + pu V ′ = aup ′ − au ′ p + upBu a V. (4.4)This motivates the following definition.11 efinition 12. For a given polynomial u ∈ K [ x ] , let the map φ V : K [ x ] n → u − K [ x ] n be defined by φ V ( p ) = 1 u ( aup ′ − au ′ p + upB ) for any p ∈ K [ x ] n . We call φ V the map for polynomial reduction with respectto u and V , and call the subspace im( φ V ) = { φ V ( p ) | p ∈ K [ x ] n } ⊆ u − K [ x ] n the subspace for polynomial reduction with respect to u and V . Note that, by construction, if q = φ V ( p ), then qa V = ( pu V ) ′ . So qa V isintegrable.We can always view an element of K [ x ] n (resp. K [ x ] n × n ) as a polynomial in x with coefficient in K n (resp. K n × n ). In this sense, we use the notation lt( · )for the leading term of a vector (resp. matrix). For example, if p ∈ K [ x ] n is ofthe form p = p ( r ) x r + · · · + p (1) x + p (0) , p ( i ) ∈ K n , where p ( r ) = 0, then deg x ( p ) = r , lt( p ) = p ( r ) x r . Let { e , . . . , e n } be thestandard basis of K n . Then the K [ x ]-module K [ x ] n viewed as K -vector spaceis generated by S := { e i x j | ≤ i ≤ n, j ∈ N } . Definition 13.
Let N V be the K -subspace of K [ x ] n generated by { t ∈ S | t = lt( p ) for all p ∈ im( φ V ) ∩ K [ x ] n } Then K [ x ] n = (im( φ V ) ∩ K [ x ] n ) ⊕ N V . We call N V the standard complement of im( φ V ) . For any p ∈ K [ x ] n , there exists p ∈ K [ x ] n and p ∈ N V such that pa V = (cid:16) p u V (cid:17) ′ + p a V. This decomposition is called the polynomial reduction of p with respect to u and V . If u = 1 , then the polynomial reduction with respect to u falls back to thesituation discussed in [15]. Proposition 14.
Let a ∈ K [ x ] and B ∈ K [ x ] n × n be such that aV ′ = BV , asbefore. If deg x ( B ) ≤ deg x ( a ) − , then N V is a finite dimensional K -vectorspace.Proof. Consider the map ˜ φ V : K [ x ] n → K [ x ] n defined by˜ φ V ( p ) = aup ′ − au ′ p + upB p ∈ K [ x ] n . Then ˜ φ V ( p ) = u φ V ( p ). It is easy to check that im( φ V ) ∩ K [ x ] n and im( ˜ φ V ) ∩ u K [ x ] n are isomorphic as K -vector spaces via the multi-plication by u . Considering the codimension of subspaces in K [ x ] n over K , wehave the formuladim K ( N V ) = codim K (im( φ V ) ∩ K [ x ] n )= codim K (cid:16) im( ˜ φ V ) ∩ u K [ x ] n (cid:17) ≤ codim K (cid:16) im( ˜ φ V ) (cid:17) + codim K (cid:0) u K [ x ] n (cid:1) . Let µ := deg x ( a ) − ℓ := deg x ( u ) and s := deg x ( p ). Since deg x ( aup ′ ) =deg x ( au ′ p ) = s + ℓ + µ , we havedeg x ( ˜ φ V ( p )) ≤ s + ℓ + max { µ, deg x ( B ) } . By an argument analogous to [15, Proposition 12], we distinguish two casesdeg x ( B ) < µ and deg x ( B ) = µ , and get the codimension of im( ˜ φ V ) is finite.Since K [ x ] n / ( u K [ x ] n ) ∼ = ( K [ x ] /u K [ x ]) n , the codimension of u K [ x ] n is alsofinite.The condition deg x ( B ) ≤ deg x ( a ) − V is a local integral basisat infinity, but may not hold for an arbitrary basis. So we introduce a weakerbasis that satisfies the degree condition. This is an analogue of suitable basisat infinity. Definition 15.
A basis V of A is called suitable at infinity if for a ∈ K [ x ] and B ∈ K [ x ] n × n such that aV ′ = BV we have deg x ( B ) < deg x ( a ) . There always exists a basis which is suitable at infinity. We can find sucha basis as follows. Start from an arbitrary K ( x )-basis V = ( v , . . . , v n ) of thefunction field A . We can make its elements v , . . . , v n integral at infinity byreplacing each v i by x − τ v i for a sufficiently large τ ∈ N . Consider a ∈ K [ x ]and B ∈ K [ x ] n × n be such that aV ′ = BV . If deg x ( B ) < deg x ( a ), we are done.If not, consider a row b in B with deg x ( b ) ≥ deg x ( a ) and set v = xa − bV .Then v is integral at infinity (because at infinity differentiating increases thevaluation) but it does not belong to the K ( x ) ∞ -module generated by V (becauseof deg x ( b ) ≥ deg x ( a )). Therefore the K ( x ) ∞ -module generated by V and v isstrictly larger than the K ( x ) ∞ -module generated by V . Replace V by a basis ofthis enlarged module, and update a and B such that aV ′ = BV . If we now havedeg x ( B ) < deg x ( a ), we are done, otherwise repeat the process just described.The iteration will terminate because with every update of V the K ( x ) ∞ -module generated by it gets enlarged, and since all these modules are containedin the module of elements of A that are integral at infinity, after at most finitelymany updates V will be a local integral basis at infinity. At least then, thedesired degree condition must hold, because otherwise there would be an inte-gral element which is not a K ( x ) ∞ -linear combination of the basis elements, incontradiction to the basis being integral at infinity.13 heorem 16. Let f be an element in A . Let V be a basis of A which is suitableat infinity. Then there exists a suitable basis W of A such that f admits anadditive decomposition in (4.3) with respect to the bases W and V .Proof. We present a constructive proof to show the existence of an additivedecomposition of f . After performing the lazy Hermite reduction on f , we get f = ˜ g ′ + 1 d P W + 1 e U W where W is a suitable basis, P = ( r , . . . , r n ) ∈ K [ x ] n and U = ( s , . . . , s n ) ∈ K [ x ] n with r i , s i introduced in (4.2). Let W = b CV for some b ∈ K [ x ] and C ∈ K [ x ] n × n . Let a ∈ K [ x ] and B ∈ K [ x ] n × n be such that aV ′ = BV .Multiplying a and B by some polynomial, we may assume that a is a commonmultiple of e and b . Rewriting the remainder in terms of the new basis V , weget 1 e U W = 1 eb U CV = 1 a ˜ U V, (4.5)for some ˜ U ∈ K [ x ] n . By Theorem 9, if f is integrable, there exist ˜ u ∈ K [ x ] and R ∈ K [ x ] n such that 1 e U W = (cid:18) RW ˜ u (cid:19) ′ = (cid:18) RCVu (cid:19) ′ , (4.6)where u = ˜ ub . Next, we apply the polynomial reduction with respect to thepolynomial u and decompose ˜ U into φ V ( ˜ U ) + ˜ U with ˜ U , ˜ U ∈ K [ x ] n and˜ U ∈ N V . Then we have 1 a ˜ U V = (cid:18) u ˜ U V (cid:19) ′ + 1 a ˜ U V. (4.7)We then get the decomposition (4.3) by setting g = ˜ g + u ˜ U V and Q = ˜ U .Assume that f is integrable. Then Theorem 9 implies that d ∈ K . Sincedeg x ( P ) < deg x ( d ), we have P = 0. Combining (4.7), (4.5) and (4.6) yieldsthat 1 a QV = 1 a ˜ U V − (cid:18) u ˜ U V (cid:19) ′ = (cid:18) u ˜ QV (cid:19) ′ , (4.8)where ˜ Q = RC − ˜ U . So Q = φ V ( ˜ Q ) ∈ im( φ V ) ∩ K [ x ] n . Since im( φ V ) ∩ K [ x ] ∩ N V = { } , it follows that Q = 0. Example 17.
We continue with Example 3 by applying the polynomial reduc-tion to the lazy Hermite remainder h = x ( x +1) ( x + 1) y . Since Disc( W ) =4( x + 1) x , we choose u = x + 1 . Note that W is already a suitable basisat infinity. The map for the polynomial reduction with respect to W and u is φ ( p ) = u ( eup ′ − eu ′ p + epM ) for any p ∈ K [ x ] n . Then h = ( x +1) x + 1) y ) ′ reduces to . Reduction-Based Telescoping
Lazy Hermite reduction in combination with the polynomial reduction just de-scribed can be used for deciding whether a given algebraic function admits analgebraic integral. Most algebraic functions don’t. The next question of inter-est may then be whether the algebraic function at hand can be deformed insome way to a related one that is algebraically integrable. Creative telescopingproduces such a deformation. It is applicable to functions f which besides theintegration variable x involve some other parameter t . The task of creative tele-scoping is to find a nonzero operator L ( t, D t ) such that L ( t, D t ) · f is integrable.Such an operator is called a telescoper for f .Several techniques are known for computing such a telescoper. The so-calledreduction based approach is one of them, and it has attracted a lot of attentionin recent years. It was first proposed for rational functions [5]. Given f = p/q ∈ K ( x ) with K = C ( t ), we can use Hermite reduction to find g i , h i ∈ K ( x )( i = 0 , , , . . . ) such that D it f = D x g i + h i . If c , . . . , c r ∈ K are such that c h + · · · + c r h r = 0, then( c + · · · + c r D rt ) · f = D x · ( c g + · · · + c r g r ) , so the operator L = c + · · · + c r D rt is a telescoper for f .Some recomputation can be avoided by observing that instead of D it f we mayas well integrate D t h i − , because D x and D t commute. With this optimization,reduction based telescoping can be summarized as follows. Algorithm 18.
Input: a function f depending on x and t ;Output: a telescoper for f find g , h such that f = g ′ + h . for r = 1 , , , . . . do: if h , . . . , h r − are linearly dependent over K return c + c D t + · · · + c r − D r − t with c i not all zero and P r − i =0 c i h i = 0 . find g r , h r such that D t h r − = g ′ r + h r The termination of this procedure can be secured in two ways. One way isto ensure that the remainders h , h , . . . belong to a finite dimensional K -vectorspace. Then the remainders must eventually become linearly dependent. Thesecond way is to ensure that the map which sends f to h such that f = g ′ + h for some g is K -linear and has the set of all integrable elements as its kernel.It is then guaranteed that the procedure will not miss a telescoper, so it willterminate because we know that for every algebraic function there does exist atelescoper.Besides for rational functions, both arguments have been worked out forvarious larger classes of functions [6, 14, 17, 7, 37], including the class of algebraicfunctions [15]. The version for algebraic functions is uses Trager’s Hermitereduction followed by a polynomial reduction, both steps requiring an integralbasis of the function field. Using Theorem 16, we will argue that reduction basedtelescoping also works with lazy Hermite reduction and the variant of polynomial15eduction developed in the previous section, with the obvious advantage thatno integral bases computation is needed.For doing so, we must take into consideration that lazy Hermite reductiontakes a suitable basis as input but may return the result with respect to anadjusted suitable basis. Let W , W , . . . denote the suitable bases with respectto which the result of the i th call to lazy Hermite reduction is returned. Bysupplying W i − as input to the i th call, we can ensure that the K [ x ]-modulegenerated by W i − is contained in the K [ x ]-module generated by W i , for every i .Therefore, if W r = W r − for some r , we can rewrite all remainders h , . . . , h r − in the bases W r and V without introducing new denominators. (Note thatno update is required for V .) In order to meet the conditions specified inTheorem 16, it may be necessary to rerun the polynomial reduction on the newrepresentations of the old remainders. The termination of the algorithm thenfollows via the second way indicated above. In order to also justify terminationin the first way, it suffices to observe that we can keep V fixed throughoutthe computation, so the termination follows directly from the finite dimensionof N V . For the paper [16] in 2012, we have created a collection of about 100 integra-tion problems, mostly originating from applications in combinatorics [31, 8],and we have compared the performance of six different approaches, includ-ing Chyzak’s algorithm [18, 19, 26] as implemented by Koutschan [23, 25],Koutschan’s ansatz-based method [24], as well as the method based on residueelimination we proposed in [16]. The result of the evaluation was somewhatinconclusive. Most algorithms outperformed the other algorithms at least forsome examples. At the same time, the timing differences can be significant.For the present paper, we have evaluated the performance of reduction basedcreative telescoping using lazy Hermite reduction on the benchmark set from2012. The runtime was taken on the same computer (a 64bit Linux serverwith 24 cores running at 3GHz; in 2012 it had 100G RAM, meanwhile it wasupgraded to 700G). The new experiments were performed with a more recentversion of Mathematica (Mathematica 12.1.1). Timings and code are availableon our website [1].An interesting observation is that in all examples of the collection (at leastthose which finished within the specified time limit of 30h), the lazy Hermitereduction procedure never encountered a situation where the linear system (2.1)did not have a unique solution so that a basis change would have been required.In almost all cases, we also found that no basis change was needed before enter-ing the reduction process in order to turn to a suitable basis. Examples wherethe default basis was not suitable and needed to be adjusted generally tookmuch more time than examples where the default basis was already suitable.In comparison to the earlier methods, the inclusion of reduction based cre-ative telescoping does not change the general observation that no algorithm16s clearly superior to the others. The new approach is faster than the otherapproaches on some of the examples, while on others it is much worse.
Example 19.
Consider the rational function f ( x, y, t ) = (1 − x − y − t + ( xy + xt + yt )) − . The computation of an annihilating operator for res x,y f ( x, y, tx y ) is completedin about 4.5 seconds by our new approach. The other techniques need at leasttwice as long and up to 500 seconds. In this example, the telescoper has order 6and coefficients of degree 15. On the other hand, computing an annihilatingoperator for res x,y f ( x, y, tx y ) takes about half an hour using the reduction basedapproach, while all other approaches all need less than 90 seconds. Here thetelescoper has order 4 and coefficients of degree 11. It is not clear why thealgorithms perform so differently on such closely related input. In view of this diverse and unpredictable performance of the various al-gorithms, it is advantageous to have several independent techniques available.Having more techniques at our disposal increases the chances that a hard in-stance arising from an application can be completed by at least one of them.
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