Parallel Telescoping and Parameterized Picard--Vessiot Theory

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by  Shaoshi Chen

Parallel Telescoping and Parameterized
Picard–Vessiot Theory
∗
4
1
Shaoshi Chen, Ruyong Feng, and Ziming Li
0
2 KLMM, AMSS, Chinese Academy of Sciences
n Beijing 100190, (China)
a schen,[email protected],
J
[email protected]
9
1
Michael F. Singer

]
C Department of Mathematics,
S North Carolina State University
s. Raleigh, NC 27695-8205 (USA)
c
[email protected]
[

1 January 21, 2014
v
6
6
6
4 Abstract
.
1 Parallel telescoping is a natural generalization of differential creative-
0 telescoping for single integrals to line integrals. It computes a linear
4 ordinary differential operator L, called a parallel telescoper, for several
1 multivariate functions, such that the applications of L to the functions
:
v yield antiderivatives of a single function. We present a necessary and
i sufficient condition guaranteeing the existence of parallel telescopers for
X
differentially finite functions, and develop an algorithm to compute min-
r imal ones for compatible hyperexponential functions. Besides computing
a
annihilatorsofparametriclineintegrals,weusetheparalleltelescopingfor
determiningGaloisgroupsofparameterizedpartialdifferentialsystemsof
first order.
1 Introduction
Theproblemoffinding lineardifferentialequationswithpolynomialcoefficients
for parametric integrals has a long history. It at least dates back to Picard [27]
who proved the existence of such linear differential equations for integrals of
∗S.C. R.F. and Z.L. weresupported by two NSFC grants (91118001, 60821002/F02) and
a973project(2011CB302401), M.F.S.wassupportedbytheNSFgrantCCF-1017217.
1
algebraic functions involving parameters. His result has been generalized to
higher-dimensionalcasesandled to Gauss–Maninconnections[21, 22, 14]. The
key for obtaining such linear differential equations is the method of creative
telescoping, whichwasfirstformulatedas analgorithmictoolbyZeilbergerand
his collaborators in 1990s [34, 35, 33]. The method enables us to prove a large
amount of combinatorial identities in an automatic way [26]. For more recent
developments, see the survey article [16].
Given a function f(t,x) described by two linear differential equations with
polynomialcoefficientsintandx,themethodofdifferentialcreative-telescoping
[1]findsalineardifferentialoperatorLin∂/∂twithpolynomialcoefficientsint
such that L(f) = ∂g/∂x, where g is usually a linear combination of partial
derivatives of f over the field of rational functions in t and x. The operator L
is called a telescoper for f, and the function g is called a certificate of L. They
can be used to evaluate parametric integrals of f with respect to x.
Recently, a connectionhas been revealedbetweenthe method ofdifferential
creative-telescoping and Galois theory of parameterized differential equations
in [8, 2, 9, 28, 12]. Considera first-orderpartialdifferential systemof the form:
∂Y ∂Y
=f , ..., =f , (1)
1 n
∂x ∂x
1 n
where f ,...,f are rational functions in t, x ,...,x satisfying compatibility
1 n 1 n
conditions. Its parameterized Galois group can be determined by constructing
alinearordinarydifferentialoperatorLin∂/∂twithpolynomialcoefficientsint
such that
∂g ∂g
L(f )= ,...,L(f )=
1 n
∂x ∂x
1 n
for a single rationalfunction g. The operator L will be referred as to a parallel
telescoper for f , ..., f with respect to x , ..., x . Parallel telescopers may
1 n 1 n
also be used to evaluate parametric line integrals in the same manner as we do
for single integrals by classical creative-telescoping.
In this paper, we present a necessary and sufficient condition guarantee-
ing the existence of parallel telescopers for differentially finite functions (see
Definition 2). The condition can easily be verified if the given functions are
hyperexponential. We develop an algorithmto compute a paralleltelescoper of
minimal order for compatible hyperexponential functions. The algorithm can
be used for constructing parallel telescopers for non-compatible ones, although
its output may not be of minimal order. We also show how to determine the
Galois group of a differential system of the form (1) by parallel telescoping.
The rest of the paper is organized as follows. In Section 2, we review the
notion of differentially finite elements. In Section 3, we study the existence
of parallel telescopers. We present an algorithm in Section 4 for constructing
minimal parallel telescopers for hyperexponential functions. In Section 5, par-
allel telescoping is applied to determine Galois groups of parameterized partial
differential systems of first order.
2
2 Differentially finite elements
Let k be an algebraically closed field of characteristic zero. Assume that δ is
i
the usual partialderivative withrespect to x onthe field k(x ,...,x )for all i
i 1 n
with 1 i n. For brevity, we set x = (x ,...,x ). Over the differential
1 n
≤ ≤
field (k(x), δ ,...,δ ) there is a noncommutative algebra k(x) D ,...,D
1 n 1 n
{ } h i
whose commutation rules are
D D =D D and D f =fD +δ (f)
i j j i i i i
for all i,j 1,...,n and f k(x). The algebra is also called the ring of
∈ { } ∈
differential operators associated to k(x). The commutation rules imply the
following fact:
Fact 1. LetL k(x) D ,...,D and[D ,L]=D L LD for someiwith1
1 n i i i
∈ h i − ≤
i n.
≤
(i) [D ,L]=0 if and only if L is free of x .
i i
(ii) If L is free of D , then so is [D ,L].
i i
(iii) If L is in k[x] D ,...,D , then the degree of [D ,L] in x is less than
1 n i i
h i
that of L.
Due to the noncommutativity of k(x) D ,...,D , we make a convention
1 n
h i
that ideals, vector spaces, modules and submodules are all left ones in this
paper.
Let M be a module over k(x) D ,...,D . For an operator L in the
1 n
h i
ring k(x) D ,...,D and h M, the scalar product of L and h is denoted
1 n
h i ∈
by L(h). We say that L is an annihilator of h if L(h) = 0. The set of all
annihilators of h is denoted by ann(h), which is an ideal in k(x) D ,...,D .
1 n
h i
Definition 2. Let h be an element of a module over the ring k(x) D ,...,D .
1 n
h i
We say that h is differentially finite (abbreviated as D-finite) over k(x) if
ann(h) k(x) D = 0 for all i with 1 i n.
i
∩ h i6 { } ≤ ≤
It is straightforward to see that h is D-finite if and only if the submodule
generated by h is a finite-dimensional linear space over k(x). It follows that,
if h ,...,h are D-finite elements in a module over k(x) D ,...,D , so is
1 m 1 n
h i
every element in the submodule generated by h ,...,h .
1 m
When a module consists of functions in x ,...,x , its D-finite elements are
1 n
called D-finite functions, which are ubiquitous in combinatorics as generating
functions. D-finite functions were first systematical investigated by Stanley
in [30]. Their important algebraic properties have been revealed by Lipshitz
in [19, 20]. We recall a lemma in [19, Lemma 3], which is the starting point of
our study on parallel telescopers.
3
Lemma 3 (Lipshitz,1988). If h is a D-finiteelement in a module over the ring
k(x) D ,...,D , then
1 n
h i
ann(h) k(x ,...,x ,x ,...,x ) D ,D = 0
1 i−1 i+1 n i j
∩ h i6 { }
for all i,j 1,...,n with i=j.
∈{ } 6
The next lemma allows one to remove redundant variables.
Lemma 4. Let h be a D-finite element in a module over k(x) D ,...,D . If
1 n
h i
D (h)=D (h)= =D (h)=0
m+1 m+2 n
···
forsomem 1,...,n 1 ,thenhisalsoaD-finiteelementoverk(x ,...,x ).
1 m
∈{ − }
Proof. BythedefinitionofD-finiteelements,itsufficestoshowthattheintersec-
tionofann(h)andk(x ,...,x ) D isnontrivialforalliwith1 i m. Sup-
1 m i
h i ≤ ≤
posethecontrary. Then,withoutlossofgenerality,wemayfurthersupposethat
every nonzero annihilator of h in k[x] D involves x . Among those annihila-
1 n
h i
tors,wechooseone,sayP,whosedegreeinx isminimal. ByFact1(i),[D ,P]
n n
is nonzero. By Fact 1 (ii), it belongs to k(x ,...,x ) D . Since both D (h)
1 n 1 n
h i
andP(h)areequaltozero,[D ,P]isalsoanonzeroannihilatorofhink[x] D .
n 1
h i
By Fact 1 (iii), it has degree in x less than that of P, a contradiction.
n
3 Parallel Telescopers
Inthis section,we define the notionof paralleltelescopers forseveralmultivari-
ate functions in a module-theoretic setting, and study under what conditions
parallel telescopers exist for D-finite elements.
3.1 Definition of parallel telescopers
In order to define parallel telescopers, we introduce a new indeterminate t,
and extend the field k(x) to k(t,x), which is denoted by K, and set ∆ =
δ ,δ ,...,δ , where δ stands for the usual partial derivative with respect
t 1 n t
{ }
to t on K. Moreover, let us denote by R the ring K D ,D ,...,D of linear
t 1 n
h i
differential operators. The notions such as D-finite elements and annihilators
carry over naturally to K and R.
Definition 5. Let f ,...,f be in an R-module. A nonzero operator L
1 n
∈
k(t) D is called a parallel telescoper for f , ..., f with respect to x if there
t 1 n
h i
exists an element g in the submodule generated by f , ..., f over R, such that
1 n
L(t,D )(f )=D (g) for all 1 i n.
t i i
≤ ≤
The element g is called a certificate of L with respect to x.
By Definition 5, the parallel telescopers for f , ..., f and zero form an
1 n
ideal in the ring k(t) D . The ideal is principal, since k(t) D is a left Eu-
t t
h i h i
clidean domain. A generator of the ideal is called a minimal parallel telescoper
for f ,...,f with respect to x.
1 n
4
3.2 Existence of parallel telescopers
We derive a necessary and sufficient condition on the existence of parallel tele-
scopersforD-finiteelements. Tothisend,weneedadifferentialanalogueof[26,
Thm. 6.2.1].
Lemma 6. Let h be an element of an R-module. If h is D-finite over K, then,
for every i 1,...,n , there exists a nonzero operator
∈{ }
L k(t,x ,...,x ,x ,...,x ) D
i 1 i−1 i+1 n t
∈ h i
and an element g in the submodule generated by h such that L (h)=D (g ).
i i i i
Proof. Let N be the submodule generated by h over R.
By Lemma 3, h has a nonzero annihilator in K D ,D , which is free of x .
t i i
h i
Among all of the x -free and nonzero annihilators for h, we choose one, say P ,
i i
whose degree in D is minimal. If the degree d of P in D is equal to zero,
i i i i
then the lemma holds by taking L = P and g = 0. Assume that d > 0. We
i i i i
can always write
P =L +D Q , (2)
i i i i
where L is in the ring k(t,x ,...,x ,x ,...,x ) D , and Q is in the ring
i 1 i−1 i+1 n t i
h i
k(t,x ,...,x ,x ,...,x ) D ,D whose degree in D is less than d .
1 i−1 i+1 n t i i i
h i
Setg :=Q (h). SinceP (h)=0andg isinN,itremainstoshowthatL is
i i i i i
nonzeroin(2). SupposethatL =0. ThenD (g )=0,which,togetherwiththe
i i i
D-finiteness of g , implies that g is D-finite overk(t,x ,...,x ,x ,...,x )
i i 1 i−1 i+1 n
by Lemma 4. Thus, there exists a nonzero linear operator R in the ring
i
k[t,x ,...,x ,x ,...,x ] D such that R (g ) = 0. It follows that the
1 i−1 i+1 n t i i
h i
product R Q is also a nonzero and x -free annihilator of h. But it has degree
i i i
in D less than d , which contradicts the minimality assumption for the degree
i i
of P in D .
i i
To study the existence of parallel telescopers, we introduce the notion of
compatible elements with respect to x.
Definition 7. The elements f ,...,f of an R-module are said to be com-
1 n
patible with respect to x if the compatibility conditions D (f ) = D (f ) for
i j j i
all 1 i<j n hold.
≤ ≤
The following lemma shows that the compatibility conditions are sufficient
for the existence of parallel telescopers for D-finite elements.
Lemma 8. Let f ,...,f be elements of an R-module. If they are D-finite
1 n
over K and compatible with respect to x, then there exists a parallel telescoper
for f , ..., f with respect to x.
1 n
Proof. Set x = (x ,...,x ), and set R := k(t,x ) D ,D ,...,D for
m 1 m m m t 1 m
h i
all m with 1 m n.
≤ ≤
We proceed by induction on n. If n=1, then f has a telescoper in k(t) D
1 t
h i
with respect to x by Lemma 6. Assume that the lemma holds for any n 1
1
−
5
elements that are both D-finite over k(t,x ) and compatible with respect
n−1
to x .
n−1
Assume that f ,f ,...,f are D-finite over k(t,x ) and compatible with
1 2 n n
respect to x . Denote by N the submodule generated by f , ..., f over R .
n 1 n n
By Lemma 6, there exists a nonzero operator L in k(t,x ) D such that
n n−1 t
h i
L (f )=D (g ) for some g N. (3)
n n n n n
∈
Withoutlossofgenerality,wefurtherassumethatL in(3)isofminimaldegree
n
in D and is monic with respect to D .
t t
First, we show that L belongs to k(t) D . For all i with 1 i n 1,
n t
h i ≤ ≤ −
we set L = [D ,L ], which belongs to k(t,x ) D by Fact 1 (ii), and has
i i n n−1 t
h i
degreeinD lessthanthatofL ,becauseL ismonicwithrespecttoD . Note
t n n t
thatL D (f )=L D (f )=D L (f ),inwhichthefirstequalityisimmediate
n i n n n i n n i
fromthecompatibilityconditionD (f )=D (f ),andthesecondfromthefact
i n n i
that L is free of x . Thus, L (f ) = D L (f ) D L (f ), which, together
n n i n i n n n n i
−
with (3), implies that
L (f )=D D (g ) D L (f )=D f˜ , (4)
i n i n n n n i n i
− (cid:16) (cid:17)
where f˜ := D (g ) L (f ) for i = 1,...,n 1. Since f˜ belongs to N,
i i n n i i
− −
we see that L = 0, for otherwise, L would not be a nonzero operator that
i n
satisfies (3) and has minimal degree in D by (4). Thus, L is free of x by
t n i
Fact 1 (i). Accordingly, L k(t) D . Moreover, L =0 and (4) imply
n t i
∈ h i
D (f˜)=0 for all i with 1 i n 1. (5)
n i
≤ ≤ −
Next,weapplytheinductionhypothesistof˜,...,f˜ . Sincef ,...,f are
1 n−1 1 n
D-finite over k(t,x ), so is g , and so is f˜ for all i with 1 i n 1. By (5)
n n i
and Lemma 4, f˜ is D-finite over k(t,x ). Moreover, f˜≤,...≤,f˜ − are com-
i n−1 1 n−1
patible with respect to x because f , ..., f are compatible with respect
n−1 1 n−1
tox andbecauseL is free ofx . Therefore,there existanonzeroopera-
n−1 n n−1
tor L˜ k(t) D and an element g˜ in the submodule generated by f˜,...,f˜
t 1 n−1
∈ h i
over R such that
n−1
L˜ f˜ =D (g˜) for i 1,...,n 1 and D (g˜)=0. (6)
i i n
(cid:16) (cid:17) ∈{ − }
The first equality in (6) is due to the induction hypothesis, and the second due
to (5). Moreover, g˜ belongs to N.
At last, we verify that L˜L is a parallel telescoper for f , ..., f . Set g =
n 1 n
L˜(g ) g˜. It belongs to N because both g and g˜ do. For i 1,...,n 1 ,
n n
− ∈ { − }
L˜L (f )=L˜ D (g ) f˜ by the definitionoff˜ in(4). ItfollowsfromL˜D =
n i i n i i i
(cid:16) − (cid:17)
D L˜ and the first equality of (6) that, for all i 1,...,n 1 ,
i
∈{ − }
L˜L (f )=D L˜(g ) D (g˜)=D L˜(g ) g˜ =D (g).
n i i n i i n i
− (cid:16) − (cid:17)
6
Applying L˜L to f , we get
n n
L˜L (f )=L˜D (g )=D L˜(g ) =D (g),
n n n n n n n
(cid:16) (cid:17)
in which the first equality follows from (3) and the last from the second one
in (6). Therefore, L˜L is indeed a parallel telescoper for f ,...,f with respect
1 n
to x .
n
The next theorem is a necessaryand sufficient condition on the existence of
parallel telescopers for D-finite elements.
Theorem 9. Let f ,...,f be D-finite elements of an R-module. Then they
1 n
have a parallel telescoper with respect to x if and only if there exists a nonzero
operator P k(t) D such that
t
∈ h i
P(D (f ) D (f ))=0 for all 1 i<j n. (7)
i j j i
− ≤ ≤
Proof. Assume that f ,...,f have a parallel telescoper P with respect to x.
1 n
Then there exists an element g in the submodule generated by f , ..., f such
1 n
that P(f ) = D (g) for all i with 1 i n. Since D D (g) = D D (g), we
i i i j j i
≤ ≤
have P(D (f ) D (f ))=0.
i j j i
−
Conversely, assume that there exists a nonzero operator P k(t) D such
t
∈ h i
that
P(D (f ) D (f ))=0 for all 1 i<j n.
i j j i
− ≤ ≤
Then P(f ),...,P(f ) are compatible, because P is free of x. So there is a
1 n
parallel telescoper L for P(f ),...,P(f ) by Lemma 8. Therefore, LP is a
1 n
parallel telescoper for f , ..., f with respect to x.
1 n
4 Hyperexponential case
LetE beadifferentialfieldextensionof(K,∆). Thesetofextendedderivations
onE is alsodenotedby ∆. The derivations in∆ areassumedto commute with
each other. Furthermore, we assume the subfield of constants in E is k.
For an element h E and an operator L R of the form
∈ ∈
L= a DiDj1 Djn
i,j1,...,jn t 1 ··· n
i,j1,X...,jn≥0
with a K, we define the application of L to h as
i,j1,...,jn ∈
L(h)= a δi δj1 δjn(h).
i,j1,...,jn t◦ 1 ◦···◦ n
i,j1,X...,jn≥0
Then E is an R-module whose multiplication is the application of an operator
in R to an element of E.
A nonzero element h E is said to be hyperexponential over K if the
∈
logarithmic derivative δ(h)/h belongs to K for all δ ∆. Hyperexponential
∈
functions are D-finite elements. In fact, the submodule generated by several
hyperexponential functions over R is the linear space spanned by them. Two
hyperexponential functions are said to be similar if their ratio belongs to K.
7
4.1 Determining the existence
Thenextpropositionallowsonetodeterminetheexistenceofparalleltelescopers
for hyperexponential functions.
Proposition 10. Let h E be hyperexponential over K. Then ann(h)
∈ ∩
k(t) D = 0 if and only if the logarithmic derivative of h with respect to t is
t
h i6 { }
of the form
δ (p)
t +r for some p k(x)[t] and r k(t). (8)
p ∈ ∈
Proof. Assume that δ (h)/h is of the form (8). Since p is a polynomial in t
t
over k(x), there exists a nonzero operator L in k(t) D annihilating p. It is
t
h i
easy to verify that (D r)(h/p)=0. Therefore, h is annihilated by a nonzero
t
−
operatorink(t) D . SuchanoperatoristhesymmetricproductofLandD r.
t t
h i −
Conversely,assumethatthereexistsanonzeroelementL ann(h) k(t) D .
t
∈ ∩ h i
Then δ (h)/h is a rational solution of the Riccati equation associated to L,
t
although it does not have to be in k(t). By formula (4.3) in [31, page 107],
δ (h) δ (P) R
t t
= +Q+ ,
h P S
whereP,Q,R andS arepolynomialsin t overthe algebraicclosureofk(x), the
rootsofS aresingularpoints ofL,andthe rootsofP arenonsingularones(see
also [5, Theorem 1]). Moreover, one can assume that deg (R) < deg (S) and
t t
thatS is monic. Since the singularpoints ofLareink,the coefficients ofS are
ink aswell. FollowingthealgorithmforcomputingrationalsolutionsofRiccati
equations described in [5, 4.3] or [31, Exercise 4.10], we see that R belongs
§
to k[t]. The same conclusion holds for Q by the algorithm in [5, 4.2], as Q is
§
constructedby analyzing the pole of the associatedRiccati equationat infinity.
Set r = Q+R/S, which belongs to k(t), and set s = δ (h)/h r, which is
t
−
in k(t,x) and equal to δ (P)/P. Thus, the linear differential equation δ (Y)=
t t
sY has a polynomial solution P. Since s belongs to k(t,x), the equation must
have a polynomial solution p in k(x)[t], which implies that δ (p)/p = s. Then,
t
the logarithmic derivative ∂h/∂t is of the form (8).
One can decide if the logarithmic derivative δ (h)/h in Proposition 10 is of
t
the form (8) by computing its squarefree partial fraction decomposition with
respect to t. A more efficient way is to apply Algorithm WeakNormalizer in [6,
6.1]toδ (h)/h,whichdeliversapolynomialpink(x)[t]suchthatthedifference
t
§
of δ (h)/h and δ (p)/p belongs to k(t) if and only if δ (h)/h is of the form (8).
t t t
Let h ,...,h be hyperexponential functions. Then h , ..., h have a par-
1 n 1 n
allel telescoper with respect to x if and only if, for every pair i,j with 1 i<
≤
j n, there exists a nonzero operator P k(t) D such that
i,j t
≤ ∈ h i
P (D (h ) D (h ))=0. (9)
i,j i j j i
−
This is because the least common left multiple of the P can be taken as
i,j
the operator P in (7) of Theorem 9. For each pair (h ,h ), there are three
i j
8
cases to be considered: (i) If D (h ) = D (h ), then set P = 1. (ii) If h is
i j j i i,j i
similar to h , then the difference D (h ) D (h ) is hyperexponential. So we
j i j j i
−
can find P by Proposition 10. (iii) If h is not similar to h , then (9) implies
i,j i j
thatbothP (D (h ))andP (D (h ))areequaltozero. Proposition10isalso
i,j i j i,j j i
applicable to the last case.
Example 11. Consider the hyperexponential functions
t(x +t+t2u) ((t+1)2+x x +t(x 1))u tx
1 1 2 1 1
h = , h = − − ,
1 2
u(t+x )√t u(t+x )√t
1 2
where u:=t+x +x . A direct calculation yields
1 2
1
h:=D (h ) D (h )=
2 1 1 2
− −√t
The logarithmic derivative of h in t belongs to k(t). Then P :=2tD +1 is the
t
operator in k D such that P(h) = 0. So h and h have a parallel telescoper
t 1 2
h i
with respect to x and x by Proposition 10.
1 2
4.2 Computing minimal parallel telescopers
This subsectionis devotedto computing minimal paralleltelescopers. First,we
present a recursive algorithm, named ParaTele, for hyperexponential functions
thatarebothcompatibleandsimilar. Next, weshowthatthe algorithmcanbe
easily adapted to compute minimal parallel telescopers for merely compatible
hyperexponential functions.
Algorithm ParaTele: Given compatible functions r h, ..., r h, where h is hy-
1 n
perexponential over K and r , ..., r are rational functions in K, compute a
1 n
minimal parallel telescoper L(t,D ) for r h,...,r h with respect to x and a
t 1 n
certificate g of L.
1. Compute a minimal telescoper L for r h with certificate g by the algo-
n n n
rithms in [1, 4].
2. If n=1, then return (L ,g ); otherwise, set
n n
f˜ :=D (g ) L (r h) for i=1,...,n 1.
i i n n i
− −
3. Run ParaTele for functions f˜,...,f˜ to get (L˜,g˜), where L˜ is of min-
1 n−1
imal order and g˜ is in the submodule generated by f˜’s over the ring
i
k(t,x ,...,x ) D ,...,D .
1 n−1 1 n−1
h i
4. Return L:=L˜L and g :=L˜(g ) g˜.
n n
−
Note that some of the rational functions r , ..., r in Algorithm ParaTele
1 n
may be equalto zero. So the input consists of either zero orsimilar hyperexpo-
nentialfunctions. This guaranteesthatthe recursioninstep3 canbe executed.
9
It follows from the proof of Lemma 8 that Algorithm ParaTele always com-
putes aparalleltelescoper. Toshowits minimality,weneedalemmathatplays
a similar role for hyperexponential functions as Lemma 4 for D-finite ones.
Recall that x = (x ,...,x ) and R denotes k(t,x ) D ,D ,...,D ,
m 1 m m m t 1 m
h i
where m=1,...,n.
Lemma 12. Let h ,...,h be hyperexponential elements of E. Assume that,
1 m
for all i with 1 i m,
≤ ≤
D (h )= =D (h )=0. (10)
m+1 i n i
···
Let N and N be the submodule generated by h , ..., h over R and R ,
m 1 m m
respectively. If there exists a nonzero operator T k(t) D and a N such
t
∈ h i ∈
that
T(h )=D (a) for all i with 1 i m,
i i
≤ ≤
then there exists b N such that
m
∈
T(h )=D (b) for all i with 1 i m.
i i
≤ ≤
In other words, T is a parallel telescoper for h , ..., h with respect to x .
1 m m
Proof. Withoutlossofgenerality,assumethat h ,...,h isamaximallinearly
1 ℓ
{ }
independentsubsetof h ,...,h overK. Thena= ℓ a h forsomea
{ 1 m} j=1 j j j ∈
K, because N is the linear space spanned by h1, ..., hPℓ over K. Hence,
ℓ ℓ
T(h )= D (a h )= (δ (a )+a r )h , (11)
i i j j i j j i,j j
Xj=1 Xj=1
where r stands for the logarithmic derivative δ (h )/h and i ranges from 1
i,j j i i
to m. Then there exist s 1,...,ℓ and w k(t,x ) such that
i ∈{ } i,si ∈ m
T(h )=w h for all i 1,...,m .
i i,si si ∈{ }
Infact,s canbeanyintegerbetween1andℓandw mustbezeroifT(h )=0;
i i,si i
and s is unique if T(h ) is nonzero by Proposition 4.1 in [18]. Thus, (11) can
i i
be rewritten as
ℓ
w h = (δ (a )+a r )h .
i,s s i j j i,j j
Xj=1
By the linear independence of h , ..., h , T(h )=D (a) is equivalent to
1 ℓ i i
δ (a )+a r =w ,
i si s i,si i,si
 (12)
 δ (a )+a r =0 for j 1,...,m with j=s .
i j j i,j i
∈{ } 6

Let ξ ,...,ξ k be such that b = a (x ,ξ ,...ξ ) is well-defined
m+1 n i i m m+1 n
∈
foralli with1 i ℓ. Then(12)stillholdsifwe replacea byb fori=1,...,
i i
≤ ≤
10