Integral P-Recursive Sequences
aa r X i v : . [ c s . S C ] F e b Integral P-Recursive Sequences ∗ Shaoshi Chen a,b , Lixin Du a,b,c , Manuel Kauers c , Thibaut Verron 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, [email protected]
February 10, 2020
Abstract
In an earlier paper, the notion of integrality known from algebraicnumber fields and fields of algebraic functions has been extended to D-finite functions. The aim of the present paper is to extend the notionto the case of P-recursive sequences. In order to do so, we formulate ageneral algorithm for finding all integral elements for valued vector spacesand then show that this algorithm includes not only the algebraic and theD-finite cases but also covers the case of P-recursive sequences.
Singularities play an essential role in algorithms for analyzing recurrence ordifferential equations, and for symbolic summation and integration. The “local”behaviour at a singularity typically gives rise to severe restrictions of the possible“global” shape of a solution, and such restrictions are exploited in the design ofalgorithms for finding such solutions. It is therefore important to have access to ∗ S. Chen was supported by the NSFC grants 11871067, 11688101 and the Fund of the YouthInnovation Promotion Association, CAS. L. Du was supported by the NSFC grant 11871067and the Austrian FWF grant P31571-N32. M. Kauers was supported by the Austrian FWFgrants F5004 and P31571-N32. T. Verron was supported by the Austrian FWF grant P31571-N32. k = C ( x ) be the field of rational functions in x over a field C and K = k ( α ) be an algebraic extension of k . Every element of K has aminimal polynomial m ∈ C [ x ][ y ]. An element of K is called integral if all itsseries expansions only involve terms with nonnegative exponents. The integralelements of K form a C [ x ]-submodule of K , which somehow plays the role in K that Z plays in Q . An integral basis of K is a k -vector space basis of K whichat the same time is a C [ x ]-module basis of the module of integral elements.Trager [15, 2, 1, 3] used integral bases in his integration algorithm for al-gebraic functions. This was one of the motivations for introducing the notionof integral D-finite functions [12], which were then used not only for integra-tion [4] but also for solving differential equations in terms of hypergeometricseries [10, 9]. Also for D-finite functions, integrality is defined in terms of theexponents appearing in the series expansions. The goal of the present paper isto introduce a notion of integrality for the recurrence case. Our hope is thatthis work will subsequently be useful for the development of new summationalgorithms.A major difference between the differential case and the shift case is the factthat singularities are no longer isolated points α ∈ C . Instead, as pointed outfor instance in [17], singularities should be viewed as orbits α + Z ∈ C/ Z con-sisting of some α ∈ C together with all elements of C that have integer distanceto α . Instead of certain kinds of series solutions at α of differential operatorsor algebraic equations, we have to consider certain kinds of sequence solutions α + Z → C of a recurrence operator. This makes the matter considerably moretechnical.We proceed in two stages. In the first stage (Sections 2 and 3), we give ageneral formulation of the algorithm proposed by van Hoeij for algebraic func-tion fields [16] and adapted to D-finite functions by Kauers and Koutschan [12].The general formulation applies to arbitrary valued vector spaces, and we iden-tify the computational assumptions on which the correctness and terminationarguments of the algorithms are based. In Section 4, we show how it indeedgeneralizes the previous algorithms. In the second stage (Section 5), we showhow the general setting developed in Sections 2 and 3 can be applied to the shiftcase. In this section, we recall basic terminologies about valuations on fields andvector spaces from [8, 18, 14]. Let k be a field of characteristic zero and Γ bea totally ordered abelian group, written additively, and let Γ ∞ = Γ ∪ {∞} inwhich α + ∞ = ∞ + α = ∞ for all α ∈ Γ ∞ and β < ∞ for all β ∈ Γ. A mapping ν : k → Γ ∞ is called a valuation on k if for all a, b ∈ k ,( i ) ν ( a ) = ∞ if and only if a = 0; 2 ii ) ν ( ab ) = ν ( a ) + ν ( b );( iii ) ν ( a + b ) ≥ min { ν ( a ) , ν ( b ) } .The pair ( k, ν ) is called a valued field and ν ( k \ { } ) ⊆ Γ is called the valuegroup of ν . The set O ( k,ν ) := { a ∈ k | ν ( a ) ≥ } forms a subring of k that iscalled the valuation ring of ν . Example 1.
A typical example of a valued field is the field of rational functions.Let C be a field of characteristic 0 and Γ = Z . For any irreducible p ∈ C [ x ] and f ∈ C ( x ) \ { } , we can always write f = p m a/b for some m ∈ Z and a, b ∈ C [ x ] with gcd( a, b ) = 1 and p ∤ ab . The valuation ν p ( f ) of f at p isdefined as the integer m . Set ν p (0) = ∞ . Then ( C ( x ) , ν p ) is a valued field with O ( C ( x ) ,ν p ) = { f ∈ C ( x ) | ν p ( f ) ≥ } being a local ring with its maximal idealgenerated by p . The valuation ν ∞ defined by ν ∞ ( f ) = deg x ( b ) − deg x ( a ) for any f = a/b ∈ C ( x ) is called the valuation at ∞ . Any valuation ν on the field C ( x ) is either ν ∞ or ν p for some irreducible p ∈ C [ x ] (see [5, Chapter 1, §
3] in thelanguage of places). When p = x − z with z ∈ C , we will write ν z instead of ν p .For z ∈ C , the field of formal Laurent series C (( x − z )) admits a valuation ν ( z ) ,defined as ν ( z ) (cid:16)P i ≥ n c i ( x − z ) i (cid:17) = n , where c n = 0 . Any r ∈ C ( x ) admits arepresentation r L in C (( x − z )) with ν z ( r ) = ν ( z ) ( r L ) . Definition 2.
Let V be a vector space over a valued field ( k, ν ) . A map val : V → Γ ∞ is called a value function on V if for all x, y ∈ V and a ∈ k , ( i ) val( x ) = ∞ if and only if x = 0 ; ( ii ) val( ax ) = ν ( a ) + val( x ) ; ( iii ) val( x + y ) ≥ min { val( x ) , val( y ) } .The pair ( V, val) is called a valued vector space over k . An element x ∈ V issaid to be integral if val( x ) ≥ . Remark 3.
Let U be any subspace of a valued vector space ( V, val) . Then therestriction of val on U is also a value function on U , which makes ( U, val) avalued vector space. Proposition 4.
Let ( k, ν ) be a valued field and ( V, val) be a valued vector spaceover k . The set O ( V, val) ⊆ V of all integral elements in V forms an O ( k,ν ) -module.Proof. For any a, b ∈ O ( k,ν ) and x, y ∈ O ( V, val) , we haveval( ax + by ) ≥ min { val( ax ) , val( by ) } = min { ν ( a ) + val( x ) , ν ( b ) + val( y ) } . Since ν ( a ) , ν ( b ) ≥ x ) , val( y ) ≥
0, we have val( ax + by ) ≥
0. So ax + by ∈ O ( V, val) . 3 k -vector space basis of a valued vector space ( V, val) which is at the sametime an O ( k,ν ) -module basis of O ( V, val) is called a (local) integral basis withrespect to val. Assume that the module O ( V, val) has a local integral basis { x , . . . , x r } and x = a x + · · · + a r x r ∈ V . Then val( x ) ≥ ν ( a i ) ≥ i = 1 , . . . , r . When does a local integral basis exist and howto construct such a basis are the main problems we study in this paper. Valuefunctions and integral bases for algebraic functions fields have been extensivelystudied both theoretically [5, 7, 14] and algorithmically [15, 17, 16] and havealso been extended to the D-finite case [12]. Example 5. (See [14, Example 3.3]) Any finite dimensional k -vector space canbe equipped with a valuation. More precisely, let V be a vector space over avalued field ( k, ν ) of dimension r . Let { B , . . . , B r } be a basis of V . Take values γ , . . . , γ r in Γ and define val : V → Γ ∪ {∞} by for all a , . . . , a r ∈ k , val r X i =1 a i B i ! = min { γ + ν ( a ) , . . . , γ r + ν ( a r ) } . It is easy to check that val is a value function on V . Example 6.
Let C be an algebraically closed field of characteristic 0, k = C ( x ) and ν z be the valuation of k at z ∈ C as in Example 1. Then ( k, ν z ) is a valuedfield. Let K = k ( β ) with β being algebraic over C ( x ) . Any nonzero element B ∈ K can be expanded as a Puiseux series of the form B = X i ≥ c i ( x − z ) r i , where c i ∈ C with c = 0 and r i ∈ Q with r < r < · · · . The value function val z : K → Q ∪ {∞} is then defined by val z ( B ) = r for nonzero B ∈ K and val z (0) = ∞ . In this setting, O ( K, val z ) is a free C [ x ] -module. Example 7.
Let C be a field with characteristic 0, and consider a linear dif-ferential operator L = ℓ + · · · + ℓ r D r ∈ C ( x )[ D ] with ℓ r = 0 . The quotientmodule V = C ( x )[ D ] / h L i is a C ( x ) -vector space with , D, . . . , D r − as a basis.Its element is a solution of L . If z ∈ C is a so-called regular singular pointof L [11], then there are r linearly independent solutions in the C -vector spacegenerated by C [[[ x − z ]]] := [ ν ∈ C ( x − z ) ν C [[ x − z ]][log( x − z )] . Following [12], we construct a value function val z on V as follows. Firstchoose a function ι : C/ Z × N → C with ι ( ν + Z , j ) ∈ ν + Z for every ν ∈ C and j ∈ N , with ι ( ν + Z , j ) + ι ( ν + Z , j ) − ι ( ν + ν + Z , j + j ) ≥ or every ν , ν ∈ C and j , j ∈ N , and with ι ( Z ,
0) = 0 . This function picksfrom each Z -equivalence class in C a canonical representative.Using this auxiliary function, the valuation val z ( t ) of a term t := ( x − z ) ν + i log( x − z ) j is the integer ν + i − ι ( ν + i, j ) , and the valuation val z ( f ) of a series f ∈ C [[[ x − z ]]] is the minimum of the valuations of all the termsappearing in it (with nonzero coefficients). The valuation of is defined as ∞ .The value function val z ( · ) : V → Z ∪ {∞} is then defined as the smallestvaluation of a series B · f , when f runs through all solutions of L . We nowcheck that the function val z is indeed a value function. ( i ) Let B ∈ V . Clearly if B = 0 , val α ( B ) = ∞ for all α ∈ ¯ C . Conversely,assume that val α ( B ) = ∞ , then by definition val α ( B · f ) = ∞ and so B · f = 0 for all f ∈ Sol α ( L ) , which implies that the dimension of thesolution space of B is at least r . But the order of B is less than r , andthe dimension of the solution space of a nonzero operator cannot exceedits order, so it follows that B = 0 . ( ii ) For any a ∈ C ( x ) ⊆ ¯ C [[[ x − α ]]] and f ∈ ¯ C [[[ x − α ]]] , the valuation of af is the sum of the valuations of a and f by definition. Then for any B ∈ V , val α ( aB ) = min f ∈ Sol α ( L ) { val α ( aB · f ) } , which is then equal to ν α ( a ) + val α ( B ) . ( iii ) By val α (( B + B ) · f )) ≥ min { val α ( B · f ) , val α ( B · f ) } for all f ∈ Sol α ( L ) ,we have val α ( B + B ) ≥ min(val α ( B ) , val α ( B )) for B , B ∈ V . When Γ = Z , the valued field ( k, ν ) can be endowed with a topology. Wesummarize here the relevant constructions, more details can be found in [13,Chapter 2]. For a ∈ k , let | a | = e − ν ( a ) . The properties of the valuation ensurethat | · | is an absolute value, called the ν -adic absolute value. This absolutevalue defines a topology on k , in which elements are “small” if their valuationis “large”.Recall that a sequence of elements ( c n ) ∈ k N is said to be Cauchy if foreach ǫ >
0, there exists N ∈ N such that for every m, n > N , | c m − c n | < ǫ ,or, equivalently, if for each M ∈ Z , there exists N ∈ N such that for every m, n > N , ν ( c m − c n ) > M . The field k is said to be complete if every Cauchysequence is convergent.The completion of k is a minimal field extension k ν which is complete. Itcan be constructed as follows. As a set, let k ν be the set of all Cauchy sequencesin k , modulo the equivalence relation ( c n ) ≡ ( d n ) ⇔ ( c n − d n ) converges to 0 atinfinity. The field k is contained in k ν via the constant sequences. Ring oper-ations on k extend to k ν component-wise, and make k ν a field. The valuationon k extends to k ν by taking the limit of the valuations of the terms of thesequences, we use the same letter ν for that valuation.An important feature of the topology on k and k ν is that the ν -adic absolutevalue is ultrametric: it satisfies the stronger triangular condition | a + b | ≤ | a | , | b | ). In particular, any series P ∞ n =0 a n with a n ∈ k ν and | a n | → k ν . Example 8.
The completion of C ( x ) w.r.t. the valuation ν z is C (( x − z )) , andits completion w.r.t. ν ∞ is C ((1 /x )) . These definitions extend naturally to a valued k -vector space. Just like inthe case of fields, the hypotheses (i) and (iii) of Definition 2 ensure that we candefine a norm on V by setting || v || = e − val( v ) . This turns V into a topologicalvector space: addition and scalar multiplication are continuous.Part (ii) of Definition 2 further ensures that || cv || = | c | · || v || for c ∈ k , v ∈ V . In particular, if a sequence ( a n ) n ∈ N in k converges to 0, then ( a n v ) n ∈ N converges to 0 in V .More generally, if B , . . . , B r ∈ V and ( a (1) n ) , . . . , ( a ( r ) n ) are sequences in k converging to a (1) ∞ , . . . , a ( r ) ∞ , respectively, then the sequence ( a (1) n B + · · · + a ( r ) n B r )in V converges to a (1) ∞ B + · · · + a ( r ) ∞ B r .Let V ν be the k ν -vector space obtained from scalar extension of V . If V isfinite dimensional and B , . . . , B r is a basis, V ν can be seen as the k ν -vector spacegenerated by B , . . . , B r , identifying its elements with elements of V wheneverpossible, and it is the completion of V with respect to the above topology. Remark 9.
The inequality dim k ν V ν ≤ dim k V always holds, but it may happenthat the inequality is strict. For example, consider C (( x )) as a C ( x ) -vectorspace, with valuation ν = ν , and let V be a r -dimensional sub-vector space of C (( x )) . Then V ν = C (( x )) has dimension over C (( x )) . In this section, we present a general algorithm for computing local and globalintegral bases of valued vector spaces and conditions on the termination of thisalgorithm.
Given a valued field ( k, ν ), a basis of a k -vector space V of dimension r , anda value function val on V , our goal is to compute a local integral basis of V if it exists. The algorithm described below is based on the algorithm given byvan Hoeij [16] for computing integral bases of algebraic function fields. It alsocovers the adaption by Kauers and Koutschan to D-finite functions [12]. Forsimplicity, we restrict to the case Γ = Z .For the algorithm to apply in the general setting, we need to make thefollowing assumptions.A. arithmetic in k and V is constructive, and ν and val are computable.B. we know an element x ∈ k with ν ( x ) = 1.6. for any given B , . . . , B d ∈ V , we can find α , . . . , α d − in k such thatval( α B + · · · + α d − B d − + B d ) > α i ’s exist.D. the completion V ν of V has dimension r .The algorithm is then as follows. Algorithm 10.
INPUT: a k -vector space basis B , . . . , B r of V OUTPUT: a local integral basis of V w.r.t. val for d = 1 , . . . , r , do: replace B d by x − val( B d ) B d . while there exist α , . . . , α d − ∈ k such that val( α B + · · · + α d − B d − + B d ) > , choose such α , . . . , α d − . replace B d by x − ( α B + · · · + α d − B d − + B d ) . return B , . . . , B r . Theorem 11.
Alg. 10 is correct.Proof.
We show by induction on d that for every d = 1 , . . . , r , the outputelements B , . . . , B d form a local integral basis for the subspace of V generatedby the input elements B , . . . , B d . From the updates in lines 2 and 5, it isclear that the output elements generate the same subspace, so the only claimto be proven is that they are also module generators for the module of integralelements.For d = 1, line 2 ensures that val( B ) = 0, and no further change is goingto happen in the while loop. When val( B ) = 0, then the integral elements ofthe subspace generated by B are precisely the elements uB for u ∈ k with ν ( u ) ≥
0, so B is an integral basis.Now assume that d is such that B , . . . , B d − is an integral basis, and let B d ∈ V . After executing line 2, we may assume val( B d ) ≥
0. After termination of thewhile loop, we know that there are no α , . . . , α d − ∈ k such that val( α B + · · · + α d − B d − + B d ) >
0. Let α , . . . , α d ∈ k be such that A = α B + · · · + α d B d is an integral element. We have to show that ν ( α i ) ≥ i = 1 , . . . , d .We cannot have ν ( α d ) <
0, otherwise, val( α − d A ) >
0, which would contra-dict the termination condition of the while loop. Thus ν ( α d ) ≥
0. But then,val( α d B d ) ≥
0, so A − α d B d is also integral. Since A − α d B d is in the k -subspacegenerated by B , . . . , B d − and the latter is an integral basis by induction hy-pothesis, it follows that ν ( α i ) ≥ i = 1 , . . . , d − V . The second proof requires an additional assumption but has theadvantage of providing a bound for the number of iterations of the loop.7 heorem 12. Alg. 10 terminates.Proof.
Assume that for some d ∈ { , . . . , r } , the loop does not terminate. Let B d,i be the value of B d before entering the i th iteration, and let ˜ B d,i = x i B d,i .For all i , val( B d,i ) = 0 and val ˜ B d,i = i . For all i ∈ N , there exists a j,i ∈ C for j ∈ { , . . . , d − } such that˜ B d,i = x i B d,i − + d − X j =0 a j,i B j = x ˜ B d,i − + x i d − X j =0 a j,i B j and B d,i has valuation 0 at α . We can unroll the sum as˜ B d,i = B d, + d − X j =0 i − X k =0 x k a j,k ! B j . Viewing this equality in V ν and taking the limit as i → ∞ yields˜ B d, ∞ := lim i →∞ ˜ B d,i = B d, + d − X j =0 ∞ X k =0 x k a j,k ! B j . Furthermore, ˜ B d, ∞ has valuation ∞ , so it is zero and B d, = − d − X j =0 ∞ X k =0 x k a j,k ! B j in V ν .But by hypothesis (D), V ν has dimension r , so B , . . . , B r must be linearlyindependent over k ν too. This is a contradiction, so the loop terminates.The second termination proof is more explicit. It depends on a generalizationof what is called discriminant in fields of algebraic numbers or functions. Definition 13.
Let ( V, val) be a valued vector space of finite dimension r overa valued field ( k, ν ) with the value group Γ . Let x ∈ k be such that ν ( x ) = 1 and B V denote the set of all bases of V . A map Disc : B V → Γ is called a discriminant function on V if for every basis B , . . . , B r of V , we have ( i ) γ := Disc( { B , . . . , B r } ) ≥ if all the B i ’s are integral in V ( ii ) if there exist α , . . . , α d − ∈ k with d ≤ r such that val V ( ˜ B d ) > d min i =1 { val V ( B i ) } , where ˜ B d := α B + · · · + α d − B d − + B d , then Disc( { B , . . . , x − ˜ B d , . . . , B r } ) < γ. heorem 14. Let ( V, val V ) be a valued vector space of finite dimension r overa valued field ( k, ν ) with the value group Z . If ν is surjective and there exists adiscriminant function Disc : B V → Z , Alg. 10 terminates.Proof. Since ν is surjective, there exists x ∈ k such that ν ( x ) = 1. Let { B , . . . , B r } be any basis of V over k . We may always assume that val V ( B i ) = 0by replacing B i by x − val V ( B i ) B i for all i . It suffices to show that Alg. 10 termi-nates on { B , . . . , B r } . Let γ = Disc( { B , . . . , B r } ) ∈ N . At any intermediatestep of Alg. 10, B , . . . , B r are always integral and form a basis of V . If α i ’sexist in the while loop, γ decreases strictly. So there are at most γ times ofbasis updating, which implies that Alg. 10 terminates. In a next step, we seek integral bases with respect to several valuations simulta-neously. Instead of a single valuation val : V → Z ∪ {∞} , we have a set of valua-tions ν z : k → Z ∪ {∞} ( z ∈ Z ) and a set of value functions val z : V → Z ∪ {∞} ( z ∈ Z ) and want to find a vector space basis B , . . . , B r of V that is also an O ( k,ν z ) -module basis of O ( V, val z ) for every z ∈ Z . The idea is to apply Alg. 10repeatedly. In order to make this work, we impose the following additionalassumptions:(B ′ ) for every z ∈ Z we know an element x z ∈ k with ν z ( x z ) = 1 and ν ζ ( x z ) = 0for all ζ ∈ Z \ { z } (C ′ ) for every z ∈ Z and any given B ,. . . ,B d ∈ V , we can compute α ,. . . ,α d − ∈ k with ν ζ ( α i ) ≥ i and all ζ ∈ Z \ { z } such thatval z ( α B + · · · + α d − B d − + B d ) > , or prove that no such α i ’s exist.(D ′ ) for every z ∈ Z , the completion V ν z of V has dimension r .E. we know a finite set Z ⊆ Z and a basis B , . . . , B r of V that is an integralbasis for all z ∈ Z \ Z .Under these circumstances, we can proceed as follows. Algorithm 15.
INPUT: a k -vector space basis B , . . . , B r of V which is anintegral basis for all z ∈ Z \ Z OUTPUT: an integral basis for all z ∈ Z for all z ∈ Z , do: apply Alg. 10 to B , . . . , B r , using ν z , val z and x z in place of ν, val , and x ,and ensuring in step 3 that ν ζ ( α i ) ≥ for all i and all ζ ∈ Z . replace B , . . . , B r by the output of Alg. 10. return B , . . . , B r . Theorem 16.
Alg. 15 is correct. roof. We only have to show that one application of Alg. 10 does not de-stroy the integrality properties arranged in earlier calls. To see that this isthe case, consider the effects of steps 2 and 5 with respect to a value functionother than val z . If val ζ is such a function, then by the assumption on x z , wehave ν ζ ( x z ) = 0, so B , . . . , B d − , B d and B , . . . , B d − , x ez B d generate the same O ( k,ν z ) -module. Hence this step is safe. Likewise, by the assumptions on the α i chosen in step 5, { B , . . . , B d − , B d } and { B , . . . , B d − , B d + P d − i =1 α i B i } generate the same O ( k,ν z ) -module. So this step is safe too. We shall discuss one more refinement. In applications, we typically have k =¯ C ( x ) where C is a field and ¯ C is an algebraic closure of C , with the usualvaluation ν z for z ∈ ¯ C (see Example 1). For this valuation, x z = x − z is acanonical choice.For theoretical purposes it is advantageous to work with vector spaces over k ,but computationally it would be preferable to work with coefficients in C ( x )rather than ¯ C ( x ). It is therefore desirable to ensure that the basis elementsreturned by Alg. 15 have coefficients in C ( x ) with respect to the input basis.Note that in this setting, we have the following properties: Lemma 17.
1. For every automorphism σ : ¯ C → ¯ C leaving C fixed, for ev-ery z ∈ Z , and for every u ∈ ¯ C ( x ) , we have ν z ( u ) = ν σ ( z ) ( σ ( u )) , where σ ( u ) is the element of ¯ C ( x ) obtained by applying σ to the coefficients of u .2. For every u ∈ ¯ C ( x ) \ { } , and for every z ∈ Z , u admits a unique Laurentseries expansion u = c z ( x − z ) ν z ( u ) + ( x − z ) ν z ( u )+1 r with c z ∈ ¯ C \ { } and ν z ( r ) ≥ . The constant c z in item 2 is called the leading coefficient of u .The second property of the lemma ensures that the coefficients α ,. . . ,α d − ∈ ¯ C ( x ) from (C) and (C ′ ) can be chosen in ¯ C . Indeed, we can replace α i byits leading coefficient if ν z ( α i ) = 0 and by zero otherwise, because whenever α , . . . , α d − ∈ ¯ C ( x ) is a solution and β , . . . , β d − ∈ ¯ C ( x ) are arbitrary with ν z ( β i ) ≥ i , then also α + β , . . . , α d − + β d − is a solution.If we restrict α , . . . , α d − to ¯ C , then there can be at most one solutionwhenever we seek a solution in step 3 of Alg. 10, because the difference of any twodistinct solutions would be a nontrivial ¯ C -linear combination of B , . . . , B d − ,and by the invariant of the outer loop, B , . . . , B d − already form an integralbasis of the k -subspace they generate.We shall adopt the following last assumption, stating that we can apply σ on V :F. We know a basis B , . . . , B r as in (E) such that for every automorphism σ : ¯ C → ¯ C fixing C , and for all α , . . . , α r ∈ k , we have val z ( α B + · · · + α r B r ) = val σ ( z ) ( σ ( α ) B + · · · + σ ( α r ) B r ).10sing this assumption, it can further be shown that the unique elements α , . . . , α d − ∈ ¯ C from (C ′ ) must in fact belong to C ( z ) (if they exist at all). Thisis because if some α i were in ¯ C \ C ( z ), then there would be some automorphism σ : ¯ C → ¯ C fixing C ( z ) but moving α i , and (F) would imply that σ ( α ) , . . . , σ ( α d )would be another solution to (C ′ ), in contradiction to the uniqueness.In order to ensure that the output elements of Alg. 15 are C ( x )-linear com-binations of the input elements, we adjust Alg. 10 as follows. Let G be theGalois group of C ( z ) over C . In step 2, instead of replacing B d by x − val z ( B d ) z ,we replace B d by (cid:18) Y σ ∈ G σ ( x z ) − val z ( B d ) (cid:19) B d . Note that Q σ ∈ G σ ( x z ) = Q σ ∈ G σ ( x − z ) is the minimal polynomial of z in C [ x ].In step 5 of Alg. 10, we choose α , . . . , α d − ∈ C ( z ) (if there are any), andinstead of replacing B d by x − z ( α B + · · · + α d − B d − + α d B d ) (with α d = 1),we replace B d by A := d X i =1 (cid:18) X σ ∈ G σ (cid:18) α i x z (cid:19)(cid:19) B i . Proposition 18.
When the steps 2 and 5 of Alg. 10 are adjusted as indicated,Alg. 15 returns an integral basis of V whose elements are C ( x ) -linear combina-tions of the input elements.Proof. By Galois theory, Q σ ∈ G σ ( x z ) = Q σ ∈ G σ ( x − z ) ∈ C ( x ) and ˜ α i := P σ ∈ G σ ( α i / ( x − z )) ∈ C ( x ) for every i . Therefore, all updates in the modi-fied Alg. 10 replace certain basis elements by C ( x )-linear combinations of basiselements.It remains to show that the output is an integral basis for all z ∈ Z . To seethis, we have to check the effect of Alg. 10 concerning val z and concerning val ζ for ζ ∈ Z \ { z } . For the latter, we distinguish the case when ζ is conjugate to z and when it is not.By part 1 of Lemma 17, for all ζ ∈ Z that are not conjugate to z we have ν ζ (˜ α i ) ≥ i = 1 , . . . , d − ν ζ (˜ α d ) = 0. Therefore, B , . . . , B d − and A generate the same O ( k,ν ζ ) -module as B , . . . , B d − and B d , for every ζ ∈ Z thatis not conjugate to z . This settles the case when ζ is not conjugate to z .Next, observe that val z ( x − z ( α B + · · · + α d B d )) ≥ x z , α , . . . , α d . Moreover, by part 1 of Lemma 17, ν z ( σ ( x − z )) = ν σ − ( z ) ( x − z ) =0 for every σ ∈ G \ { id } , and ν z ( σ ( α i )) = ν σ − ( z ) ( α i ) ≥ ν ζ ( α i ) ≥ ζ . Therefore val z ( σ ( x − z )( σ ( α ) B + · · · + σ ( α d ) B d ) ≥ σ ∈ G \{ id } .It follows that val z ( A ) ≥ max σ ∈ G val z (cid:16) d X i =1 σ (cid:18) α i x − z (cid:19) B i (cid:17) ≥ . Moreover, since α d = 1 and val σ ( z ) ( x z ) = 0 for all σ = id, we have that B , . . . , B d − and A generate the same O ( k,ν z ) -module as B , . . . , B d − and x − z ( α B + · · · + α d B d ). This settles the concern about val z .11inally, if ζ is conjugate to z , say ζ = σ ( z ) for some automorphism σ ∈ G ,then val ζ ( A ) = val ζ ( σ ( A )) = val z ( A ) ≥ A is a C ( x )-linear combination of the original basis elements. So A belongs to the O ( k,ν ζ ) -module of all integral elements (w.r.t. val ζ ) of the subspace generatedby B , . . . , B d in V , so we are not making the module larger than we should.Conversely, the old B d belongs to the O ( k,ν ζ ) -module generated by B , . . . , B d − and A , so by updating B d to A , the module generated by B , . . . , B d does notbecome smaller.Informally, what happens by taking the sums over the Galois group is thatthe algorithm working locally at z simultaneously works at all its conjugates.If for a certain z , the set Z contains z as well as its conjugates, it is fair (andadvisable) to discard all the conjugates from Z and only keep z . More precisely,the whole process requires only knowing the minimal polynomial of z in C [ x ],so for applications where the set Z is computed as the set of roots of somepolynomial p ∈ C [ x ], the algorithms can proceed with the factors of p insteadof all its roots. We will see below how the algorithms in [16, 12] for computing integral basesare special cases of the general formulation in Section 3.Let C be a computable subfield of C and k = ¯ C ( x ) with a valuation ν z for z ∈ ¯ C . The value function val z on V = k ( β ) with β ∈ C ( x ) is defined inExample 6 and on V = ¯ C ( x )[ D ] / h L i is defined in Example 7. We show that theassumptions imposed on value functions in Section 3 are fulfilled in the algebraicand D-finite settings. Note that (B), (C), (D) are subsumed in (B ′ ), (C ′ ), (D ′ ),respectively.(A) It is assumed that C is a computable field, so it is clear that arithmetic in¯ C ( x ) and V are computable, and that ν z on ¯ C ( x ) is also computable. Thevalue functions val z for algebraic and D-finite functions are computablesince we can determine first few terms of Puiseux or generalized seriessolutions by algorithms in [11, 6].(B ′ ) For every z ∈ Z , we can take x z = x − z such that ν z ( x z ) = 1 and ν ζ ( x z ) = 0 for all ζ ∈ Z \ { z } .(C ′ ) Done in [12, Section 4].(D ′ ) Clear.(E) In the algebraic case, we can choose as Z the set of singularities of β ∈ C ( x ) which is clearly a finite set. In the D-finite case, we can choose as Z the set of zeros of ℓ r which are the only possible singularities by [12,Lemma 9]. 12F) If α and ¯ α are conjugates, let σ be an element of the Galois group of¯ C/C such that ¯ α = σ ( α ). In particular σ ( L ) = L and σ ( B ) = B . Forall i ∈ { , . . . , r } , σ ( f α,i ) ∈ C [[[ x − ¯ α ]]] is a solution of σ ( L ) = L . Since σ is an automorphism, the σ ( f α,i ) form a fundamental system of L in¯ C [[[ x − ¯ α ]]]. For all i ∈ { , . . . , r } , B · σ ( f α,i ) = σ ( B ) · σ ( f α,i ) = σ ( B · f α,i ),and the equality of the valuations follows. In the algebraic case, thisequality follows from the property of Duval’s rational Puiseux series (seethe remarks on [6, page 120]).The termination of the general algorithm 10 in the algebraic and D-finitecases have been shown in [16, 12] by using classical discriminants and generalizedWronskians. The discriminant functions in these cases can be taken as thecompositions of the valuation ν z and these functions. More precisely, for a basis B , . . . , B r of V = k ( β ), the discriminant function Disc in the algebraic settingis defined as Disc( { B , . . . , B r } ) = val z (det(Tr( B i B j ))) , where Tr is the trace map from V to C ( x ). If B , . . . , B r are integral, thendet(Tr( B i B j )) ∈ ¯ C [ x ] and hence Disc( { B , . . . , B r } ) ∈ N . Let P ∈ C ( x )[ y ]be the minimal polynomial for β and β , . . . , β r ∈ C ( x ) be the roots of P .Then V ≃ C ( x )[ y ] / h P i . So for each i , there exists a unique Q i ∈ C ( x )[ y ] withdeg y ( Q B ) < r such that B i = Q i ( β ). It is well-known thatdet(Tr( B i B j )) = det( Q i ( β j )) . (1)If there exist a , . . . , a d − ∈ k such that˜ B d = 1 x − z ( a B + · · · + a d − B d − + B d )is integral, then the formula (1) implies thatDisc( { B , . . . , ˜ B d , . . . , B r } ) = Disc( { B , . . . , B r } ) − . So Disc is indeed a discriminant function on k ( β ). In the case of D-finite func-tions, for any basis B = { B , . . . , B r } of V = ¯ C ( x )[ D ] / h L i and fundamentalseries solutions b , . . . , b r ∈ ¯ C [[[ x − z ]]] of L , the generalized Wronskian isdefined as wr L,z ( B ) := det((( B i · b j )) ri,j =1 ) ∈ ¯ C [[[ x − z ]]] . The discriminant function Disc can be defined as the valuation of wr
L,z ( B ) at z . By the proof of Theorem 18 in [12], Disc is indeed a discriminant functionon C ( x )[ D ] / h L i . We now turn to recurrence operators. We consider the Ore algebra C ( x )[ S ]with the commutation rule Sx = ( x + 1) S . We fix an operator L = ℓ +13 S + · · · + ℓ r S r ∈ C ( x )[ S ] with ℓ , ℓ r = 0, and we consider the vector space V = ¯ C ( x )[ S ] / h L i , where h L i = ¯ C ( x )[ S ] L . The operator L acts on a sequence f : α + Z → ¯ C through ( L · f )( z ) := ℓ ( z ) f ( z ) + · · · + ℓ r ( z ) f ( z + r ) for all z ∈ α + Z . This action turns ¯ C α + Z into a (left) C [ x ][ S ]-module, but not to a(left) C ( x )[ S ]-module, because a sequence f : α + Z → ¯ C cannot meaningfullybe divided a polynomial which has a root in α + Z . In order to obtain a C ( x )[ S ]-module, consider the space ¯ C (( q )) α + Z of all sequences f : α + Z → ¯ C (( q )) whoseterms are Laurent series in a new indeterminate q , and define the action of L = ℓ + · · · + ℓ r S r ∈ C ( x )[ S ] on a sequence f : α + Z → ¯ C (( q )) through( L · f )( z ) := ℓ ( z + q ) f ( z ) + · · · + ℓ r ( z + q ) f ( z + r ) for all z ∈ α + Z . Note thatno ℓ i ∈ C ( x ) can have a pole at z + q for any z ∈ α + Z when α ∈ ¯ C and q ¯ C .For a fixed operator L = ℓ + · · · + ℓ r S r ∈ C [ x ][ S ] with ℓ , ℓ r = 0, theset Sol( L ) := { f : α + Z → ¯ C (( q )) : L · f = 0 } is a ¯ C (( q ))-vector space ofdimension r . Indeed, a basis b , . . . , b r is given by specifying the initial values b i ( α + j ) = δ i,j for i, j = 1 , . . . , r and observing that the operator L uniquelyextends any choice of initial values indefinitely to the left as well as to theright. The reason is again that q ¯ C implies ℓ ( z + q ) , ℓ r ( z + q ) = 0 for every z ∈ α + Z , so there is no danger that computing a certain sequence term b i ( z )from b i ( z + 1) , . . . , b i ( z + r ) or from b i ( z − , . . . , b i ( z − r ) could produce adivision by zero. Instead of a division by zero, we can only observe a divisionby q .The valuation ν q ( a ) of a nonzero Laurent series a ∈ ¯ C (( q )) is the smallest n ∈ Z such that the coefficient [ q n ] a of q n in a is nonzero. We further define ν q (0) = + ∞ . For a nonzero solution f : α + Z → ¯ C (( q )) of an operator L ∈ C [ x ][ S ], we will be interested in how the valuation changes as z ranges through α + Z . As we have noticed, there can be occasional divisions by q as we extend f towards the left or the right, so ν q ( f ( z )) can go up and down as z movesthrough α + Z . In fact, it can go up and down arbitrarily often, as the solution f : Z → ¯ C (( q )), f ( z ) = 1+ q +( − z of the operator L = S − z is a root of ℓ we can have ν q ( f ( z )) < min { ν q ( f ( z + 1)) , . . . , ν q ( f ( z + r )) } , and only when z is a root of ℓ r ( x − r ) we can have ν q ( f ( z )) < min { ν q ( f ( z − , . . . , ν q ( f ( z − r )) } . Since the nonzero polynomials ℓ , ℓ r have at most finitely many roots in α + Z ,we can conclude that bothlim inf n →−∞ ν q ( f ( α + n )) and lim inf n → + ∞ ν q ( f ( α + n ))are well-defined for every solution f : α + Z → ¯ C (( q )) of L . Their differencevg f := lim inf n → + ∞ ν q ( f ( α + n )) − lim inf n →−∞ ν q ( f ( α + n ))is called the valuation growth of f . Considerations about the valuation growthare used for example in algorithms for finding hypergeometric solutions [17].14n our context, solutions with negative valuation growth are troublesome,because we want to define the valuation of a residue class B ∈ ¯ C ( x )[ S ] / h L i at z in terms of the valuations of the sequence terms ( B · b )( z ) ∈ ¯ C (( q )), where b runs through Sol( L ). When b ∈ Sol( L ) has negative valuation growth, thenwe can have ν q (( B · b )( z )) < z , which makes it hard tomeet assumption (E). Moreover, if all solutions have positive valuation growth,we have ν q (( B · b )( z )) > z , which is also in conflict withassumption (E). In order to circumvent this problem, we let Z ⊆ ¯ C be suchthat for each orbit α + Z with Z ∩ ( α + Z ) = ∅ and for which L has a solution in¯ C (( q )) α + Z with nonzero valuation growth, the set Z ∩ ( α + Z ) has a (computable)right-most element. We then define the value function val z : V → Z ∪ {∞} byval z ( B ) := min b ∈ Sol( L ) (cid:16) ν q (( B · b )( z )) − lim inf n →∞ ν q ( b ( z − n )) (cid:17) . We use the convention ∞ − ∞ = ∞ . Proposition 19. val z is a value function for every z ∈ Z .Proof. We check the conditions of Def. 2.( i ) If B = 0, then B · b is the zero sequence for every b ∈ Sol( L ), so ν q (( B · b )( z )) = ∞ for all n ∈ Z .Conversely, let B ∈ ¯ C ( x )[ S ] be such that val z ([ B ]) = ∞ . We may assumethat the order of B is less than r , so that [ B ] = 0 is equivalent to B = 0.By val z ([ B ]) = ∞ we have ν q (( B · b )( z )) = ∞ for all b ∈ Sol( L ), i.e.,( B · b )( z ) = 0 for all b ∈ Sol( L ).If b , . . . , b r is a basis of Sol( L ), then the matrix M = (( b j ( z + i − ri,j =1 ∈ ¯ C (( q )) r × r is regular. Now if B were nonzero and β k S k is a nonzero term appear-ing in B , then multiplying the k th row of M by β k and adding suitablemultiples of other rows to the k th row, we obtain a matrix whose k th rowis 0, because ( B · b )( z ) = · · · = ( B · b r )( z ) = 0. On the other hand, thedeterminant of this matrix is equal to β k det( M ) = 0, so B cannot benonzero.( ii ) Clear by ν q (( uf )( z )) = ν q ( u ) + ν q ( f ( z )) for all u ∈ ¯ C (( q )) and f ∈ ¯ C (( q )) z + Z .( iii ) Clear by ν q ((( B + B ) · u )( z )) = ν q (( B · u )( z )+( B · u )( z )) ≥ min( ν q (( B · u )( z )) , ν q (( B · u )( z ))) for all u ∈ ¯ C (( q )) z + Z .Next, we show that we can meet the computability assumptions of Section 3.Note again that (B), (C), (D) are subsumed in (B ′ ), (C ′ ), (D ′ ), respectively.15A) It is assumed that C is a computable field, so it is clear that arithmeticin ¯ C ( x ) and V are computable, and that ν z is computable. We show thatval z is computable as well.Let ζ ∈ z + Z be such that all roots of ℓ ℓ r contained in z + Z are to theright of ζ , and consider the basis b , . . . , b r of Sol( L ) in ¯ C (( q )) z + Z definedby the initial values b j ( ζ + i −
1) = δ i,j ( i, j = 1 , . . . , r ). We shall provethat for all η ∈ z + Z ,val η ( B ) = r min j =1 ν q (( B · b j )( η )) . Since we can compute ( B · b j )( η ) for any j = 1 , . . . , r and η ∈ z + Z , thisimplies that val η is computable. In particular, val z is then computable.We have min ri =1 ν q ( b j ( ζ + i − j = 1 , . . . , r by construction,and in fact lim inf n → + ∞ ν q ( b j ( ζ − n )) = 0 for j = 1 , . . . , r , because at noposition ζ − n the valuation can be smaller than the minimum valuationof its r neighbors to the right or than the minimum valuation of its r neighbors to the left, due to the lack of roots of ℓ ℓ r in the range underconsideration.Let now b = c b + · · · + c r b r for coefficients c , . . . , c r ∈ ¯ C (( q )). Let v := min rj =1 ν q ( c j ). Assume that v = 0, and let j be such that ν q ( c j ) = 0.Then for all η ∈ z + Z , ν q ( b ( η )) ≥ r min j =1 ν q ( b j ( η ))and ν q (( B · b )( η )) ≥ min rj =1 ν q (( B · b j )( η )).Furthermore, by construction of the basis of b j ’s, for all i ∈ { , . . . , r } , b ( ζ + i −
1) = c i , so min ri =1 ν q ( b ( ζ + i − ℓ ℓ r left of ζ , it implies thatlim inf n → + ∞ ν q ( b ( ζ − n )) = 0 . It follows from the above that ν q (( B · b )( η )) − lim inf n → + ∞ ν q ( b ( η − n )) ≥ r min j =1 ν q (( B · b j )( η )) . Assume now that v = 0. In that case, consider q − v b = q − v c b + · · · + q − v c r b r , with min rj =1 ν q ( q − v c j ) = 0. From the above, ν q (( B · q − v b )( η )) − lim inf n → + ∞ ν q ( q − v b ( η − n )) ≥ r min j =1 ν q (( B · b j )( η )) . η ∈ z + Z we have ν q ( q − v b ( η )) = ν q ( b ( η )) − v and ν q (( B · q − v b )( η )) = ν q (( q − v B · b )( η ))= ν q (( B · b )( η )) − v, it still holds that ν q (( B · b )( η )) − lim inf n → + ∞ ν q ( b ( η − n )) ≥ r min j =1 ν q (( B · b j )( η )) , so that indeed val η ( B ) = min rj =1 ν q (( B · b j )( η )).(B ′ ) We can take x z = x − z .(C ′ ) Let B , . . . , B d ∈ C ( x )[ S ] / h L i be given. We can then compute v :=min di =1 val z ( B i ) and we can find the required α , . . . , α d − ∈ ¯ C by equatingthe coefficients of q n for n ≤ v in the linear combination α ( B · b j )( z ) + · · · + α d − ( B d − · b j )( z ) + ( B d · b j )( z ) to zero and solving the resultinginhomogeneous linear system for α , . . . , α d − .(D ′ ) Clear.(E) First we shall prove that if α + Z does not contain a root of ℓ ℓ r , then B = { , S, . . . , S r − } is an integral basis for all z ∈ Z ∩ α + Z . For such z ,consider the basis b , . . . , b r of Sol( L ) ⊆ ¯ C (( q )) α + Z with b j ( z + i −
1) = δ i,j ( i, j = 1 , . . . , r ). By the discussion of (A), for any operator A ∈ V , wehave val z ( A ) = r min j =1 ν q (( A · b j )( z )) . Let A = p + · · · + p r − S r − be an operator in V = ¯ C ( x )[ S ] / h L i . Bythe construction of the basis b j ’s, for all j = { , . . . , r } , ( A · b j )( z ) = p j ( x + q − z ). It imples that r min j =1 ν q (( A · b j )( z )) = r − min j =0 ν z ( p j ) . So A is integral if and only if ν z ( p j ) ≥ j and B is an integral basisat z . Since ℓ ℓ r can have at most finitely many roots, we have restrictedthe required subset Z to finitely many orbits α + Z . In each of theseorbits, there is a natural bound for Z to the left after lack of roots of ℓ ℓ r by the similar argument as above. If L has a solution with nonzerovaluation growth, then the bound to the right is given by the choice of Z .Now suppose all solutions of L in ¯ C (( q )) α + Z have zero valuation growth.Let ζ ∈ α + Z be such that all roots of ℓ ℓ r are contained to the left. Foreach z = ζ + n with n ≥
0, choosing the basis b j ( z + i −
1) = δ i,j ( i, j =1 , . . . , r ), we getlim inf n → + ∞ ν q ( b j ( z + n )) = r min i =1 ν q ( b j ( z + i − j = 1 , . . . , r . Then lim inf n → + ∞ ν q ( b j ( z − n )) = 0. For any operator A ∈ V , it again follows that val z ( A ) = min rj =1 ν q (( A · b j )( z )) and hence B is an integral basis at such a point z for the same reason.(F) We can take any basis of V = ¯ C ( x )[ S ] / h L i whose basis elements belongto C ( x )[ S ] / h L i , for example 1 , S, . . . , S r − .If z, ˜ z ∈ ¯ C are conjugates, let σ be an element of the Galois group of ¯ C over C that maps z to ˜ z . Then for every solution f ∈ ¯ C (( q )) z + Z of L also σ ( f ) ∈ C (( q )) ˜ z + Z is a solution of L , because L has coefficients in C , so σ ( L ) = L .Since we have σ (( α + · · · + α r − S r − )( f ))= ( σ ( α ) + · · · + σ ( α r − ) S r − )( σ ( f ))for any α , . . . , α r − ∈ ¯ C ( x ), it follows thatval z ( α + · · · + α r − S r − ) ≥ val ˜ z ( σ ( α ) + · · · + σ ( α r − ) S r − ) . Equality follows by exchanging z and ˜ z .We now define the discriminant function in the shift setting. For each α ∈ Z ,by the item (A), we can choose a basis b , . . . , b r of Sol( L ) such that val α ( B ) =min rj =1 ν q (( B · b j )( α )). For any k -vector space basis B = { B , . . . , B r } of V =¯ C ( x )[ S ] / h L i , we can takeDisc α ( B ) := ν q (det(((( B i · b j )( α ))) ri,j =1 )) ∈ Z . It is well-defined since the matrix (( B i · b j )( α )) = ( p i,ℓ ) · ( b j ( α + ℓ − B i = P rj =1 p i,ℓ S ℓ − with p i,ℓ ∈ ¯ C ( x ). If B i ’s are integral for α ,then ν q (( B i · b j )( α )) ≥ i, j = 1 , . . . , r . It follows that Disc α ( B ) ≥ B d by ( x − α ) − A d with A d = α B + · · · + α r − B d − + B d suchthat val α ( A d ) > min di =1 val α ( B i ), the discriminant is replaced by Disc α ( B ) − Example 20.
Let L = ( x + 1) + ( x − S + ( x + 1) S . For every α / ∈ Z , wehave that { , S, S } is a local integral basis for V = C ( x )[ S ] / h L i at α + Z . Forthe orbit Z , choosing b j ( − i −
1) = δ i,j for i, j = 1 , , , we obtain a basis ofthe solution space in C (( q )) Z : n · · · − − · · · b ( n ) · · · − q q ( q − q +1 · · · b ( n ) · · · − q − · · · b ( n ) · · · − q +2 q q − q +2 q ( q +1) · · · hen val α ( B ) = min j =1 ν q (( B · b j )( α )) for any operator B ∈ V and α ∈ Z . Sincethe solution b has negative valuation growth, for a global integral basis the set Z has to be bounded on the right in the orbit Z . Take Z = C \ { , , . . . } . At α = 0 ,we have is locally integral, but S, S are not since val ( S ) = val ( S ) = − .However, xS, xS are locally integral. By our algorithm, we can find a localintegral basis at : (cid:26) , x − x + 1 x S, − x + S (cid:27) . Using such a basis as an input, continue to find all locally integral elements at α = − . Similarly replace B = − x + S by ( x + 1) B since val ( B ) = − .This operation does change the local integrality at Z \ {− } , because x + 1 isinvertible in the localization of C [ x ] at any z = − . So the output local integralbasis at α = − is also a global integral basis for Z : (cid:26) , x − x + 1 x S, − x + 2 x + − x − x ( x + 1) S + 1 x + 1 S (cid:27) . References [1] Manuel Bronstein. The lazy Hermite reduction. Technical Report 3562,INRIA, 1998.[2] Manuel Bronstein. Symbolic integration tutorial. ISSAC’98, 1998.[3] Shaoshi Chen, Manuel Kauers, and Christoph Koutschan. Reduction-basedcreative telescoping for algebraic functions. In
Proceedings of ISSAC’16 ,pages 175–182, 2016.[4] Shaoshi Chen, Mark van Hoeij, Manuel Kauers, and Christoph Koutschan.Reduction-based creative telescoping for fuchsian D-finite functions.
Jour-nal of Symbolic Computation , 85:108–127, 2018.[5] Claude Chevalley. Introduction to the Theory of Algebraic Functions ofOne Variable. AMS Mathematical Surveys, 1951.[6] Dominique Duval. Rational Puiseux expansions.
Compositio Mathematica ,70(2):119–154, 1989.[7] Antonio J. Engler and Alexander Prestel.
Valued Fields . Springer, 2005.[8] L´aszl´o Fuchs. Vector spaces with valuations.
Journal of Algebra , 35(1-3):23–38, 1975.[9] Erdal Imamoglu.
Algorithms for Solving Linear Differential Equations withRational Function Coefficients . PhD thesis, Florida State University, 2017.[10] Erdal Imamoglu and Mark van Hoeij. Computing hypergeometric solutionsof second order linear differential equations using quotients of formal so-lutions and integral bases.
Journal of Symbolic Computation , 83:254–271,2007. 1911] Edward L. Ince.
Ordinary Differential Equations . Dover, 1926.[12] Manuel Kauers and Christoph Koutschan. Integral D-finite functions. In
Proceedings of ISSAC’15 , pages 251–258, 2015.[13] Jean-Pierre Serre. Local fields.
Graduate Texts in Mathematics , 1979.[14] Jean Pierre Tignol and Adrian R. Wadsworth.
Value Functions on SimpleAlgebras, and Associated Graded Rings , volume 3733 of
Springer Mono-graphs in Mathematics . Springer-Verlag, Switzerland, 2015.[15] Barry M. Trager.
Integration of Algebraic Functions . PhD thesis, MIT,1984.[16] Mark van Hoeij. An algorithm for computing an integral basis in an al-gebraic function field.
Journal of Symbolic Computation , 18(4):353–363,1994.[17] Mark van Hoeij. Finite singularities and hypergeometric solutions of linearrecurrence equations.
Journal of Pure and Applied Algebra , 139:109–131,1999.[18] Guangxing Zeng. Valuations on a module.