Transcendence Certificates for D-finite Functions

Although in theory we can decide whether a given D-finite function is transcendental, transcendence proofs remain a challenge in practice. Typically, transcendence is certified by checking certain incomplete sufficient conditions. In this paper we propose an additional such condition which catches some cases on which other tests fail.


INTRODUCTION
An algebraic function is a quantity for which there are polynomials 0 , . . ., , not all zero, such that 0 ( ) + 1 ( ) + • • • + ( ) = 0.A D-finite function is a quantity for which there are polynomials 0 , . . ., , not all zero, such that 0 ( ) + 1 ( ) ′ + • • • + ( ) ( ) = 0.As recognized by Abel, every algebraic function is also D-finite, and it is not hard to construct a differential equation from a known polynomial equation.The other direction is much more difficult, as a given differential equation may or may not have any algebraic solutions.The problems of finding out whether a given differential equation has some (nonzero) algebraic solutions, and finding out whether a given power series solution of a given differential equation is algebraic can be reduced to the problem of finding out whether a given differential equation has only algebraic solutions, using operator factorization [24] or minimization techniques [10], respectively.
The problem to decide whether a given differential equation admits only algebraic solutions has received a lot of attention since the 19th century, when Schwarz, Klein, Fuchs and others studied the problem for equations with = 2 [15], but even this special case was not fully understood until Baldassari and Dwork [1] gave a complete decision procedure in 1979.Only a year later, Singer [21] offered an algorithm that applies to equations of arbitrary order .His algorithm is, however, only of theoretical interest, as it relies on solving a nonlinear system of algebraic equations whose number of variables is determined by a group-theoretic bound involving the term (49 ) 2 .This is far from feasible, even for = 2.However, in practice, for small orders, the bound can be refined, leading to more practical algorithms.This has been done for order 2 [19,22], order 3 [22,23] and orders 4 and 5 [13].The problem remains difficult beyond those known cases.If a differential equation has only algebraic solutions, their minimal polynomials are not difficult to find.One way is to compute a truncated power series solution of the differential equation and then use linear algebra or Hermite-Padé approximation [2] to find a candidate annihilating polynomial.From the first terms of a series solution, we can reliably detect annihilating polynomials of degrees , with ( + 1)( + 1) < .The correctness of such a candidate can be checked by computing the differential equation satisfied by the solution of the candidate equation and comparing it with the input equation.If they do not match, or if no candidate equation is found, repeat the procedure with a higher truncation order and higher degrees , .Eventually, the correct minimal polynomial will be found.
In Sect. 4 we give an alternative method which can decide for a given whether all solutions are algebraic with a minimal polynomial of degree at most , regardless of the degree of the polynomial coefficients of the minimal polynomial.This method has the advantage that need not be guessed in advance, but it still requires a guess for .We are thus led to the question how we can detect with a reasonable amount of computation time that a differential equation has at least one transcendental solution.There are indeed several things that are worth trying.For example, if a differential equation has a logarithmic or an exponential singularity, it cannot only have algebraic solutions.This test was applied for example in order to prove transcendence of the generating function for Kreweras walks with interacting boundaries [9].Another popular test is to determine the asymptotic behaviour of the series coefficients of a solution of the differential equation.If it is not of the form with ∈ Q \ {−1, −2, −3, . . .}, this also proves the presence of a transcendental solution [14].A third possibility is to use arbitrary precision arithmetic [18,20] to compute eigenvalues of monodromy matrices for the differential equation.If there is an eigenvalue that is not a root of unity, there must be a transcendental solution.A fourth idea is to exploit that an algebraic power series ∈ Q[[ ]] must be globally bounded, i.e., there must be nonzero integers , such that ( ) ∈ Z[[ ]].If a given differential operator has a series solution that is not globally bounded, then it cannot only have algebraic solutions.As a fifth approach, we can investigate the -curvature of the differential equation [6,7] and resort to a conjecture of Grothendieck according to which the -curvature is zero for almost all primes if and only if the differential equation has only algebraic solutions.A nice account on this approach was recently given by Bostan, Caruso, and Roques [5].Another idea is to try to prove transcendence via the criterion of Harris and Sibuya [16], which says that for a D-finite function , the reciprocal 1/ is D-finite as well if and only if the logarithmic derivative ′ / is algebraic.Finally, there are powerful criteria for certain special differential equations, e.g., the criterion of Beukers and Heckman for testing algebraicity of a hypergeometric differential equation [3].
All these tests have limitations.The first four tests only provide a sufficient condition for the existence of transcendental solutions, but there are equations with transcendental solutions on which all three tests fail.In addition, for the fourth test, even if we find a solution that looks like it is not globally bounded, it can be difficult to prove that it really is not.A limitation of the -curvature test is the quantifier "almost all": if we encounter a prime (or several primes) for which the -curvature is nonzero, this is strong evidence in favor of a transcendental solution, but there remains a small chance that the prime(s) were just unlucky.The criterion of Harris and Sibuya reduces the problem of proving that ′ / is transcendental to the problem of proving that 1/ is not D-finite, which is typically more difficult.In fact, this criterion is more valuable in the other direction: to prove that 1/ is not D-finite, it suffices to prove that ′ / is not algebraic.The obvious limitation of the criterion of Beukers and Heckman is that it only applies to hypergeometric functions.
In view of this situation, additional sufficient conditions for transcendental solutions that can be tested with reasonable computational cost are of interest.Ideally, such tests should also provide some artifacts that can serve as witness for the existence of transcendental solutions.We propose the term transcendence certificate for such artifacts.For example, a logarithmic or exponential singularity can be viewed as such a transcendence certificate.Observe that the algorithms such as Singer's mentioned earlier do not provide any transcendence certificates but will just report "no algebraic solution" as output.
The purpose of this paper is to introduce a transcendence certificate based on the following classical fact about algebraic functions: P 1. [4,25] Every non-constant algebraic function must have at least one pole.
With our new test, we are able to prove the existence of transcendental solutions for some equations that have no logarithmic singularities, no series solutions with illegal coefficient asymptotics, and whose monodromy matrices have just roots of unity as eigenvalues.We also wish to point out that our approach is applicable to differential equations of any order.
An algebraic function field = [ ]/ is a field extension of the rational function field of finite degree, where is an irreducible polynomial in [ ].For every ∈ ∪ {∞}, the element ∈ can be identified with any of the deg ( ) many roots of the minimal polynomial in the field of Puiseux series at ; we call them the expansions of at .
A Puiseux series is said to be integral if its starting exponent is nonnegative, i.e., if the corresponding function does not have a pole at the expansion point.The element of is called integral at ∈ ∪ {∞} if all its Puiseux series expansions at are integral.In order to extend the definition of integrality to other elements of , note that for every expansion of we have a field homomorphism ℎ : → (( ( − ) 1/ )) (or ℎ : → (( −1/ )) if = ∞) which maps to .Now ∈ is called integral at if for all expansions of the series ℎ ( ) is integral.The element is called (globally) integral if it is integral at every ∈ (but not necessarily at infinity).The set of all integral elements of forms a free [ ]-submodule of , and a basis of this module is called an integral basis of .We say that an element of is completely integral if it is integral at every ∈ ∪ {∞}.According to Proposition 1, the completely integral elements of are precisely the elements of .
Let denote the usual derivation with respect to , i.e., ( ) = ′ , which turns = ( ) or = ( )[ ]/ into differential fields.An element of a differential field is called a constant if ( ) = 0; these constants always form a subfield of .A linear differential operator is an expression of the form = 0 + 1 + • • • + with 0 , . . ., ∈ .If ≠ 0, we call ord( ) = = deg ( ) the order of the operator.The operator is called monic if = 1.The set of all linear differential operators will be denoted by [ ]; it forms a non-commutative ring in which the multiplication is governed by the Leibniz rule = + 1.An operator is called irreducible if it cannot be written as = 1 • 2 with ord( 1 ) ≥ 1 and ord( 2 ) ≥ 1.Every differential field is a [ ]-left-module via the action An element of a differential field is called a solution of an operator ∈ [ ] if • = 0.The set of all solutions of in a differential field is denoted by ( ).It is always a vector space over the constant field of and hence called the solution space of .If the constant field of is , then the dimension of ( ) in is bounded by the order of , but in general it is smaller.We say that has only algebraic solutions if there is a differential field = [ ]/ such that the solution space ( ) in has dimension ord( ).If is an irreducible operator then either all its solutions are algebraic or none of them (except for the zero solution) [21,Prop. 2.5].
] is an operator of order , we call ∈ a singularity of if it is a pole of one of the rational functions 0 / , . . ., −1 / .The point ∞ is called a singularity if, after the substitution ↦ → −1 , the origin 0 becomes a singularity.If ∈ ∪ {∞} is not a singularity of , then has linearly independent Puiseux series solutions at , and they are all integral.
The notion of integrality for differential operators is defined in a similar way as discussed above for algebraic field extensions = [ ]/ .Throughout this paper, we consider only operators which have a basis of Puiseux series solutions at every point ∈ ∪ {∞}.For such an operator ∈ [ ], we have the module [ ]/ where denotes the left ideal is not a ring but only a (left) [ ]-module.In this module, the equivalence class [1] has the property , so [1] can be considered as a solution of in [ ]/ , very much like the element ∈ is a root of .Similar as for algebraic function fields, we can associate [1] ∈ [ ]/ with any solution of in a Puiseux series field (( ( − ) 1/ )) or (( −1/ )).The association of [1] with extends to [ ]/ by mapping an equivalence class [ ] to the series • .The notions of integrality can now be defined like before: for every Puiseux series solution of at , the series Note that in the last two items it suffices to consider points that are singularities of or poles of some of the coefficients of .For any fixed and , these are only finitely many.Also recall that we restrict our attention to operators which have a basis of Puiseux solutions, so that the quantifier "for all Puiseux series solutions" in the definitions above is equivalent to "for all solutions".The set of all integral elements in [ ]/ forms a free [ ]left-module, and a basis of this module is called an integral basis of [ ]/ .An integral basis { 1 , . . ., } is called normal at infinity if there are integers 1 , . . ., ∈ Z such that { 1 1 , . . ., } is a basis of the ( ) ∞ -left-module of all elements of [ ]/ which are integral at infinity.Here, ( ) ∞ refers to the ring of all rational functions / with deg ≤ deg .Integral bases which are normal at infinity always exist, and they can be computed [12,17].
Finally, we recall some fundamental facts about operators.The adjoint * of an operator ∈ [ ] is defined in such a way that for any two operators , ∈ [ ] we have ( + ) * = * + * and ( ) * = * * .We have * = − and * = for all ∈ .Moreover, ord( * ) = ord( ) for every ∈ [ ].The least common left multiple of two operators , ∈ [ ], denoted by lclm( , ), is defined as the unique monic operator of lowest order which has both and as right factor.Its key feature is that whenever is a solution of and is a solution of , then + is a solution of lclm( , ).For the efficient computation of the least common left multiple, see [8].There is a similar construction for multiplication.
The symmetric product ⊗ of two operators , ∈ [ ] is defined as the unique monic operator of lowest order such that whenever is a solution of and is a solution of , then is a solution of ⊗ (regardless of the differential field to which and belong).As a special case, the th symmetric power of an operator ∈ [ ] is defined as ⊗ = ⊗ • • • ⊗ .For the efficient computation of the symmetric powers, see [11].
By construction, we have ( ) + ( ) ⊆ (lclm( , )), and in general, the inclusion is proper.However, if dim ( ) = ord( ) and dim ( ) = ord( ), then we have ( ) + ( ) = (lclm( , )), i.e., the least common multiple cannot have any extraneous solutions.Likewise, if dim ( ) = ord( ) and dim ( ) = ord( ), the solution space of the symmetric product ⊗ is generated by all products with ∈ ( ) and ∈ ( ).These facts were shown by Singer [21] in the context of complex functions, and again using more abstract machinery in the book of van der Put and Singer [24].

PSEUDOCONSTANTS
Let ∈ [ ] be a linear differential operator.As mentioned before, if has a logarithmic or exponential singularity, it follows immediately that does not only have algebraic solutions and we may view the singularity as a transcendence certificate.We continue to exclude this case from consideration, i.e., we continue to assume that has no logarithmic or exponential singularity at any point in ∪ {∞}.In other words, we assume that has a basis of Puiseux series solutions at every point.
but not a constant.
We will say for short that " has a [pseudo]constant" if [ ]/ contains a [pseudo]constant.
and let [ ] ∈ [ ]/ .Let be an extension of such that the solution space ( ) of in has dimension ord( ).
The set of all constants forms a -vector space of dimension at most ord( ).
Conversely, let be the order of and be the representative of order at most − 1 of [ ] .Assume that • is a constant for all ∈ ( ), i.e., • ( • ) = 0.This means that ( ) ⊂ ( • ).Since ( • ) has dimension at most and ( ) has dimension , it follows that ( ) = ( • ).This implies that and • are equal up to an invertible factor in , and therefore that which in turn forces = 0 in contradiction to the assumption that [ ] is not zero.
(3) It is clear that the constants form a -vector space.In order to prove the bound on the dimension, consider a ∈ [ ] with ord( It is clear that is uniquely determined and that the function which maps every constant [ ] to the corresponding is -linear and injective.Now = implies ( ) * = ( ) * , so * * = * * , so − * = * .Since 1 is a solution of the left hand side, it must be a solution of the right hand side, so 0 = ( * ) • 1 = * • , so ∈ ( * ).We have thus constructed an injective -linear map from the space of all constants to the solution space of * in .Since the dimension of the latter is at most ord( ), the claim follows.
If [ ] is a constant, then it is completely integral, but unlike in the case of algebraic functions, the converse is not true in general.This means that pseudoconstants may exist.
. All its solutions are integral at every place including infinity, therefore [1] is completely integral.However, , so it is not a constant.Alternatively, one can observe that has a non-constant solution, and therefore [1] cannot be a constant.So [1] is a pseudoconstant.
In view of Prop. 1, we can regard pseudoconstants as transcendence certificates.

P
. For a contradiction, assume that has only algebraic solutions.Let be an algebraic extension of such that the solution space ( ) in has dimension ord( ).Since algebraic functions are closed under application of linear operators, • is algebraic for all ∈ ( ).Since [ ] is completely integral, • does not have a pole at any ∈ ∪ {∞}.By Prop. 1, this implies that • is constant.Therefore, by Prop.3, [ ] is a constant, which is a contradiction.

E
6. Consider the operator + 1 48 , annihilating the function 1/6 ( − 1) 13/24 2 1 7 8 , 5 6 ; 7  6 ; .The operator is irreducible, and therefore all its solutions have the same nature.By Schwarz' classification and closure properties, they must be transcendental, but let us ignore this argument for the sake of the example.
The singularities of the operator are 0, 1 and ∞, and a basis of solutions at each singularity is given by Therefore, [1] is a pseudoconstant, and thus the operator has no nonzero algebraic solution.
As noted in the introduction, we could also compute the monodromy matrices of around 0, 1 and ∞.If one of them was not a root of unity, this would give another proof of transcendence.However, numeric computations suggest that all eigenvalues are roots of unity in this example.More precisely, the monodromy group around 0 is generated by two matrices 1 and 2 with The operator is irreducible, and therefore all its solutions have the same nature.has the pseudoconstant [ ] , with where ( ) and ( ) are certain polynomials of degree 3 and 6 respectively, with coefficients in Q.So all the solutions of are transcendental.
For operators with at most 3 singularities, the nature of the solutions and the existence of pseudoconstants are determined by the initial exponents of the solutions.Indeed, the operator is then uniquely determined up to a scalar factor by its singularities and initial exponents.Changing the position of the singularities is equivalent to applying a rational change of variables by a Möbius transform, which preserves the nature of the solutions and the pseudoconstants.
This property does not hold for operators with more singularities, as the next example shows.The initial exponents are the same as those in Example 7, but the position of the singularities differ.Unlike the operator in Example 7, the operator does not admit a pseudoconstant.Note that using the technique described in [22], it can be proven that the operator does nonetheless admit only transcendental solutions.

E 9.
In order to illustrate that this proof technique works for operators of any order, we provide1 an operator of order 6 as well as a transcendence certificate.The operator has singularities at 0, 1, . . ., 6 as well as ∞, with the following exponents: (∞) −1 0 1 2 3 4 There are at least two ways to search for pseudoconstants for a given .The first one uses integral bases.It is shown in Lemma 8 of [12] that a basis of the -vector space of all completely integral elements of [ ]/ is given by { : = 1, . . ., ; = 0, . . ., } whenever { 1 , . . ., } is an integral basis that is normal at infinity and 1 , . . ., ∈ Z are such that { 1 1 , . . ., } is a local integral basis at infinity.This motivates the following algorithm.

P .
It is clear that the algorithm is correct if it does not return ⊥.It remains to show that has no pseudoconstant if the algorithm does return ⊥.In view of the remarks before the algorithm, every completely integral element of [ ]/ , and thus in particular every pseudoconstant, is a -linear combination of the .But if all the were constants, then, since the constants also form a -vector space, so would be all their linear combinations.Therefore, if there are pseudoconstants at all, there must be one among the .
An implementation of Algorithm 10 is available in the latest version of the SageMath package ore_algebra2 .Otherwise, in an environment where no functionality for computing integral bases is available, we can use linear algebra to search for pseudoconstants by brute force.This has the advantage of being conceptually more simple, but the disadvantage that we cannot easily recognize the absence of pseudoconstants.Let 1 , . . ., ∈ be the singularities of , and assume that ∞ is not a singularity.At each singularity , let ∈ Q be the smallest exponent appearing in one of the solutions at .Let = ( − 1 ) max(0,⌈− 1 / ⌉ ) • • • ( − ) max(0,⌈− / ⌉ ) , so that [ ] is globally integral.
For each singularity , choose a bound ∈ N on the degree of the denominator of a local integral basis at , and let = 1 + • • • + .We form the ansatz with unknowns , .Evaluating it at all solutions at 1 , . . ., , ∞ gives series whose coefficients are linear combinations of the unknowns , , and setting those coefficients with negative valuations to 0 yields a system of linear equations to solve.Each solution is an operator which is completely integral.However, if no non-zero solution is found, or if all solutions are constants, this is not enough to conclude that the operator does not have a pseudoconstant.It could just mean that the guessed bounds on the denominator were too conservative.
If does not have a pseudoconstant, we could try to apply some transformation to that does not change the nature of the solutions of but may affect the existence of pseudoconstants.For example, applying a gauge transform to does not change the nature of its solutions.However, gauge transforms do not affect the existence of pseudoconstants either.Indeed, let ∈ [ ] be a linear operator, ∈ [ ] be another one and ′ be the gauge transform of such that • does not have a pole for any ∈ ( ), and there exists an ∈ ( ) such that • is not a constant.By definition, this implies that [ ] is a pseudoconstant in [ ]/ .In conclusion, gauge transforms are not strong enough to create pseudoconstants.We will see next that we may have more success with other operations.

SYMMETRIC POWERS
Symmetric powers are useful for proving identities among D-finite functions and they find applications in algorithms for factoring operators [24].They can also be used to decide for a given operator and a given ∈ N whether all solutions of are algebraic functions of degree at most .For, if is an algebraic solution of with a minimal polynomial ∈ [ ] of degree , then has distinct solutions 1 , . . ., in an algebraic closure ¯ of and we can write = ( − 1 ) • • • ( − ).The solutions 1 , . . ., of are conjugates of , and since has coefficients in , we have • ( ) = ( • ) = 0 for every automorphism that fixes .Therefore, 1 , . . ., are also solutions of .For every , the th coefficient of = ( − 1 ) • • • ( − ) is the ( − )th elementary symmetric polynomial of 1 , . . ., and therefore an element of ⊗ ( − ) .As the coefficients of belong to = ( ), they must show up among the rational solutions of ⊗ ( − ) .This observation motivates the following algorithm.
3 form an ansatz + =1 =1 , , − with undetermined coefficients , 4 substitute a truncated series solution of into the ansatz, equate coefficients, and solve the resulting system for the undetermined coefficients , .5 if the system has no solution, return ⊥. 6 let be the polynomial corresponding to one of the solutions of the linear system.7 if all roots of are solutions of , return 8 otherwise, go back to step 4 and try again with a higher truncation order.
Compared to the guess-and-prove approach mentioned in the introduction, the algorithm above has the advantage that only one of the degrees of the minimal polynomials has to be guessed.
Algorithm 12 indicates that symmetric powers know something about algebraicity of solutions.The next result points in the same direction.It says that the symmetric powers of an operator are larger if has a transcendental solution.
(2) Since 2 is a right factor of , we have 1 and among the solutions of .If there is also at least one transcendental solution , then the solution space of ⊗ contains all elements 1 1 2 3   with 1 , 2 , 3 ∈ N such that 1 + 2 + 3 = , and the transcendence of implies that they are all linearly independent over .As these are +2 = Ω( 2 ) many, the claim follows again from dim ( ⊗ ) = ord( ⊗ ).
This theorem provides yet another heuristic test for the existence of transcendental solutions: simply compute ⊗ for the first few and see how their orders grow.As the theorem only makes a statement for asymptotically large , looking at specific values of will not allow us to make any definite conclusion, but it can provide convincing evidence.

E
14. Consider the operators The operator 1 is the annihilator of the roots of so it only has algebraic solutions.The operator 2 is the lclm of the operator from Example 6 and 2 , so it has a transcendental solution and it has 2 as a right factor.The order of the symmetric powers of the operators is growing as follows: The assumption on having 2 as a right factor in the second part of the theorem cannot be dropped, as can be seen for example with = 2 − 1, whose solutions are exp( ) and exp(− ).The solution space of ⊗ is spanned by the terms exp( ( − ( − ))) for ∈ {0, . . ., }, and therefore has dimension +1 = O( ).More generally, for any operator of order ≤ 2, the order of ⊗ is bounded by + −1 ≤ + 1.The divisibility condition says that 1 and are solutions of , and in order to have in addition a transcendental solution, the order of must be at least 3.If does not have 2 as a right factor, apply the theorem to lclm( , 2 ) instead of .Note that has only algebraic solutions if and only if lclm( , 2 ) has only algebraic solutions.More generally, if is any operator that has only algebraic solutions, then has only algebraic solutions if and only if lclm( , ) has only algebraic solutions.This is because, as remarked at the end of Sect.2, the least common multiple does not have any extraneous solutions.Nevertheless, as we show next, there is no hope that lclm( , ) could have any pseudoconstants if not already has any.

P
. Let [ ] be a completely integral element of [ ]/ .Let be an extension of such that ( ) ⊆ has dimension ord( ).
The additional property that is not a constant similarly propagates to at least one of the summands.
In view of this negative result, it is remarkable that taking symmetric products can produce pseudoconstants.For example, the function considered in Example 6 is a product of an algebraic function and a hypergeometric function. .This is the hypergeometric function appearing in Example 6.
The operator does not have a pseudoconstant.However, the operator ⊗5 does have the pseudoconstant [ ( − 1) 3 ].By Theorem 18 below, this implies that all nonzero solutions of are transcendental.
The presence of rational exponents in 1/6 ( − 1) 13/24 means that it does not qualify as a pseudoconstant with our definition.However, considering symmetric powers allows us to clear those denominators.First, observe that the lowest exponents of the solutions of ⊗ are − 1 6 at 0, − 13 24 at 1 and 5 6 at infinity.We are looking for a pseudoconstant of the form [ ( − 1) ] with , integers.Multiplying by such an element adds to the exponent at 0, to the exponent at 1, and subtracts + from the exponent at infinity.The complete integrality condition thus translates into the following inequalities: The solutions, for in {1, . . ., 6}, are represented in Figure 1.The smallest value of for which there is an integer solution is 5, and we recover the pseudoconstant Let ∈ [ ] be a differential operator.Suppose that for some ∈ N the symmetric power ⊗ has a pseudoconstant.Then has at least one transcendental solution.

P
. The solution space of ⊗ is spanned by all products of solutions of .The existence of a pseudoconstant in [ ]/ ⊗ proves that at least one solution of ⊗ is transcendental, and therefore at least one solution of is transcendental.
In other words, a pseudoconstant for ⊗ can be viewed as a transcendence certificate for .As shown by the previous examples, such a certificate may exist even if itself does not have pseudoconstants.So it is worthwhile to search for pseudoconstants of symmetric powers.As shown by the following theorem, we cannot increase our chances to find a pseudoconstant any further by adding some rational solutions to the solution space of .
By Lemma 15, if [ ] is a pseudoconstant, then there exists ∈ {1, . . ., } such that [ ] ⊗ ⊗ ⊗ ( − ) is also a pseudoconstant.This means that for every Puiseux series solution of at some point ∈ ∪ {∞} and every solution ∈ ( ) of we have that • ( − ) is integral, and that for at least one and one , the quantity • ( − ) is not a constant.Fixing one such solution ∈ ( ) \{0} of , it follows that − is a completely integral element of [ ]/ ⊗ and that [ − ] ⊗ is not a constant.Thus ⊗ has the pseudoconstant [ − ] ⊗ .
We have not been able to answer the following question: 20.Is it true that for every operator with at least one transcendental solution there exists an ∈ N such that ⊗ has a pseudoconstant?
If the answer to Question 20 is yes, then this fact in combination with Alg. 12 would yield a new decision procedure for the existence of transcendental solutions.We could simply search in parallel for = 1, 2, 3, . . .for an algebraic solution of of degree and a pseudoconstant of ⊗ .Exactly one of these parallel threads would have to terminate after a finite number of steps.
A natural idea to prove the existence of pseudoconstants of ⊗ for sufficiently large is to show that the linear system, which emerges from a search for pseudoconstants via the linear algebra approach, has more variables than equations for sufficiently large .Unfortunately, this does not seem to be the case: indeed, if ( ) is the order of ⊗ , the ansatz (1) has Θ( ( )) undetermined coefficients.As for the number of equations, it is equal to the number of series coefficients to set to zero: for each series solution ( ∈ {1, . . ., ( )}), the valuation of ( ) can be as low as − , for a total of Θ( ( )) equations.
The following example can perhaps be considered as some piece of empirical evidence that the answer to Question 20 is no.On the other hand, we can show (Prop.23) that for an operator with only algebraic solutions there is always an such that ⊗ has a constant (but of course no pseudoconstant), and this could be considered as some piece of evidence that the answer to Question 20 may be yes. .Thanks to Schwarz' classification, we know that the operator has no algebraic solutions.However, an exhaustive search using integral bases could not find a completely integral element for ⊗ for any ≤ 6, and a heuristic search using linear algebra could not find one for any ≤ 30.

P
. Since has only algebraic solutions, also ⊗ has only algebraic solutions.Moreover, ⊗ has at least one nonzero rational function solution (e.g., the product of all the conjugates of some algebraic solution of ).If is a solution of ⊗ , then so are all the conjugates of , because ⊗ has coefficients in .The solution space of the minimal order annihilating operator of is generated by and its conjugates and therefore a right factor of ⊗ .
Let 1 be a solution of ⊗ which does not belong to span( ), and let 1 be a minimal order annihilating operator of 1 .For = 1, 2, . . ., let be a solution of ⊗ which does not belong to span( ) + ( 1

CONCLUSION
We propose the notion of a transcendence certificate for any kind of artifact whose existence implies that a given differential operator has at least one transcendental solution.Simple transcendence certificates are logarithmic and exponential singularities.Pseudoconstants introduced in Def. 2 can also serve as transcendence certificates.We have given examples of operators that have no logarithmic or exponential singularities but that do have pseudoconstants.
We have also given examples of operators that have no pseudoconstants even though they have transcendental solutions.To such operators, we can try to apply transformations that preserve the existence of transcendental solutions but may lead to the appearance of pseudoconstants.In particular, as shown in Sect.4, it can happen that an operator has no pseudoconstants but some symmetric power ⊗ of does.A pseudoconstant of ⊗ suffices to certify the existence of a transcendental solution of .An open question (Question 20) is whether the existence of transcendental solutions of implies the existence of an such that ⊗ has pseudoconstants.We would be very interested in an answer to this question.
There are further possibilities to transform an operator with no pseudoconstants to one that may have some.For example, we could try to exploit that the composition of a D-finite function with an algebraic function is always D-finite.If is D-finite and is algebraic, then • is algebraic if and only if is algebraic, thus a pseudoconstant for an annihilating operator of • could serve as a transcendence certificate for an annihilating operator of .Note that unlike the transformations considered in this paper, the composition can not only remove singularities but also create new ones.We have not found an example where this process reveals new pseudoconstants.
In another direction, we could try to weaken the requirements of Def. 2. According to our definition, [ ] is a pseudoconstant if every local solution of is such that • has nonnegative valuation.For a transcendence certificate, it would suffice to have one global solution of (a complex function defined on a Riemann surface) which is not constant and has no pole.If we relax Def. 2 accordingly, it may be that additional operators would have pseudoconstants.However, we would no longer know how to decide the existence of pseudoconstants for a given operator.
As predicted by the theorem, for 1 the growth is linear, and for 2 the growth is at least quadratic (cubic).

P 23 .
If ∈ [ ] has only algebraic solutions and is such that all the solutions of have a minimal polynomial of degree at most , then ⊗ has a nonzero constant.
The linear operator which annihilates only the hypergeometric function (without the algebraic function multiplier) does not have a pseudoconstant.If the given operator has no pseudoconstants, we can thus ask whether there is an operator with only algebraic solutions such that ⊗ has pseudoconstants.Of course, as long as nobody tells us how to choose , this observation is not really helpful.What we can easily do however is to multiply the solutions of with each other.It turns out that this is sometimes sufficient.