Algebraic independence of Mahler functions via radial asymptotics
AALGEBRAIC INDEPENDENCE OF MAHLER FUNCTIONSVIA RADIAL ASYMPTOTICS
RICHARD P. BRENT, MICHAEL COONS, AND WADIM ZUDILIN
Abstract.
We present a new method for algebraic independence results in thecontext of Mahler’s method. In particular, our method uses the asymptoticbehaviour of a Mahler function f ( z ) as z goes radially to a root of unity todeduce algebraic independence results about the values of f ( z ) at algebraicnumbers. We apply our method to the canonical example of a degree twoMahler function; that is, we apply it to F ( z ), the power series solution to thefunctional equation F ( z ) − (1 + z + z ) F ( z ) + z F ( z ) = 0. Specifically,we prove that the functions F ( z ), F ( z ), F (cid:48) ( z ), and F (cid:48) ( z ) are algebraicallyindependent over C ( z ). An application of a celebrated result of Ku. Nishiokathen allows one to replace C ( z ) by Q when evaluating these functions at anonzero algebraic number α in the unit disc. Introduction
We say a function f ( z ) ∈ C [[ z ]] is a Mahler function provided there are integers k (cid:62) d (cid:62) a ( z ) , a ( z ) , . . . , a d ( z ) ∈ C [ z ] with a ( z ) a d ( z ) (cid:54) = 0such that(1) a ( z ) + a ( z ) f ( z ) + a ( z ) f ( z k ) + · · · + a d ( z ) f ( z k d ) = 0 . We call the (minimal) integer d the degree of the Mahler function f . In the lastfew decades the study of Mahler functions has been given renewed importancebecause of their relationships to theoretical computer science and linguistics [3]. Inparticular, the generating function of an automatic sequence is a Mahler function.While transcendence questions concerning Mahler functions have more or lessbeen answered, much less is known about the deeper area of algebraic indepen-dence of the functions and their derivatives. All of the current results in the latterdirection, and certainly the most practical examples, concern only Mahler functionsof degree one [4, 5, 6]. Until now, these results relied on a hypertranscendence cri-terion due to Ke. Nishioka [21]. Recall that a function is called hypertranscendental provided it does not satisfy an algebraic differential equation; in other words, thefunction and all its derivatives are algebraically independent over the field of ratio-nal functions.In this paper, we introduce a new method for proving algebraic independenceresults for Mahler functions and their derivatives. We apply this method to adegree two Mahler function introduced by Dilcher and Stolarsky [12], which hasquite recently become the canonical example of a degree two Mahler function.Specifically, we consider the function F ( z ) ∈ Z [[ z ]] satisfying the functional equation(2) F ( z ) = (1 + z + z ) F ( z ) − z F ( z ) , Date : 26 December 2014.
Revised : 4 April 2015.2010
Mathematics Subject Classification.
Primary 11J91; Secondary 11J81, 12H10, 30B30,33F05, 39A45, 65D20.The research of R. P. Brent was support by ARC grant DP140101417, the research of M. Coonswas supported by ARC grant DE140100223, and the research of W. Zudilin was supported by ARCgrant DP140101186. a r X i v : . [ m a t h . N T ] A p r RICHARD P. BRENT, MICHAEL COONS, AND WADIM ZUDILIN which starts F ( z ) = 1 + z + z + z + z + z + z + z + · · · . Among various combinatorial properties, Dilcher and Stolarsky [12] showed that allthe coefficients of F ( z ) are in { , } . Coons [10] proved that F ( z ) is transcendentaland Adamczewski [1] gave the transcendence of the values F ( α ) for any nonzeroalgebraic number α inside the unit disc. Recently, Bundschuh and V¨a¨an¨anen [7]proved that F ( z ) and F ( z ) are algebraically independent over C ( z ), and veryrecently [8] they showed that the functions F ( z ), F ( z ), and F ( z ) are algebraicallyindependent over C ( z ).Our central result is the following theorem. Theorem 1.
The functions F ( z ) , F ( z ) , F (cid:48) ( z ) , and F (cid:48) ( z ) are algebraically inde-pendent over C ( z ) . An application of Theorem 1 along with Mahler’s powerful method implies thealgebraic independence result for the values of the functions.
Theorem 2.
For each non-zero algebraic number α inside the unit disc, the num-bers F ( α ) , F ( α ) , F (cid:48) ( α ) , and F (cid:48) ( α ) are algebraically independent over Q . Indeed, one expects the stronger version of algebraic independence of the functions F ( z ) and F ( z ) along with all of their derivatives, though the present methodsseem inadequate for a result of this generality.As alluded to in the above paragraphs, the novelty of our approach is the avoid-ance of the hypertranscendence criterion of Ke. Nishioka [21]. Ke. Nishioka’s cri-terion is very specialised and only applicable to Mahler functions of degree one.In contrast, our method partly relies on understanding the radial asymptotics ofMahler functions and can be applied to Mahler functions of any degree. The analyt-ical problem of determining this type of asymptotic behaviour for Mahler functionsis very classical, even for degree one Mahler functions; e.g., see Mahler [20], deBruijn [11], Dumas [14], and Dumas and Flajolet [15]. The importance of suchasymptotics also appear (though in a weaker form) in recent work of Adamczewskiand Bell [2].In the case of F ( z ), we prove the following result. Theorem 3. As z → − , we have F ( z ) = C ( z )(1 − z ) lg ρ · (1 + O (1 − z )) , where lg denotes the base- logarithm, ρ := (1 + √ / denotes the golden ratio,and C ( z ) is a positive oscillatory term, which in the interval (0 , is bounded awayfrom and ∞ , real analytic, and satisfies C ( z ) = C ( z ) . This paper is organised as follows. In Section 2 we prove Theorem 3 by a carefulstudy of the continued fraction for F ( z ) /F ( z ). In Section 3 we use this knowledgeto establish Theorem 1: assuming a polynomial relation in F ( z ), F ( z ), F (cid:48) ( z ),and F (cid:48) ( z ) the asymptotic behaviour of F ( z ) as z → − allows us to significantlyshorten it; then using a linear algebra argument, we show that this reduced algebraicrelation is not possible. The related algebraic statement, Theorem 4, is proved ingenerality in Section 4. The derivation of Theorem 2 from Theorem 1 is performedat the end of Section 3. Finally, in Section 5 we discuss an alternative proof ofTheorem 3 that can be used in the asymptotical study of general Mahler functionsat arbitrary roots of unity.We would like to point out that the methods of the paper apply with no difficultyto the ‘satellite’ function G ( z ) (for definitions and related results see Dilcher and LGEBRAIC INDEPENDENCE OF MAHLER FUNCTIONS 3
Stolarsky [12], Adamczewski [1], and Bundschuh and V¨a¨an¨anen [7, 8]), so that allthree theorems above remain true when we replace F ( z ) in their statements with G ( z ). 2. A continued fraction related to F ( z ) and asymptotics In this section, we prove Theorem 3 as stated in the introduction. In order tocarry out our method, it is useful to define the auxiliary function µ : [0 , → R given by(3) µ ( z ) := F ( z ) F ( z ) . From (2) and (3), µ ( z ) satisfies the recurrence(4) µ ( z ) = 1 + z + z − z µ ( z ) . Our strategy is to analyse the asymptotic behaviour of µ ( z ) and then deduce thecorresponding behaviour of F ( z ).Note that µ ( z ) may be written as a continued fraction µ ( z ) = 1 + z + z − z z + z · − z z + z · − ... . Also, from (3), F ( z ) is given by the infinite product F ( z ) = ∞ (cid:89) k =0 µ ( z k ) . In this sense we have an ‘explicit solution’ for F ( z ) as an infinite product of con-tinued fractions.Before continuing, we make some remarks on notation. Since logarithms todifferent bases occur naturally in the analysis, we write ln x for the natural logarithmand lg x for the logarithm to the base 2. As in the statement of Theorem 3, wedefine ρ := (1 + √ / ≈ .
618 to be the golden ratio, and note that ρ = ρ + 1.The following few lemmas provide the needed background for the proof of The-orem 3 concerning the asymptotics of F ( z ) as z → − . Lemma 1.
The power series F ( z ) = ∞ (cid:88) n =0 c n z n has coefficients c n ∈ { , } . Also, F ( z ) is strictly monotone increasing and un-bounded for z ∈ [0 , , and cannot be analytically continued past the unit circle.Proof. Since the coefficients c n are in { , } (see [7, 12]) and infinitely many arenonzero, the strict monotonicity and unboundedness of F ( z ) follow easily. Thus, F ( z ) has a singularity at z = 1.From the functional equation (2) it follows that F ( z ) has a singularity at z = e πi/ k for all nonnegative integers k . Thus, there is a dense set of singularities onthe unit circle, so the unit circle is a natural boundary. See also Bundschuh andV¨a¨an¨anen [7, Theorem 1.1] for a proof of the last part of the lemma. (cid:3) RICHARD P. BRENT, MICHAEL COONS, AND WADIM ZUDILIN
Lemma 2. If z ∈ [0 , , then µ ( z ) (cid:62) . Moreover, if µ := lim z → − µ ( z ) and µ (cid:48) := lim z → − µ (cid:48) ( z ) , then µ = 3 + √
52 = ρ ≈ . and µ (cid:48) = 21 + 8 √ ≈ . . Proof.
Suppose that z ∈ [0 , F ( z ) is monotonic increasing on [0 , F ( z ) (cid:62) F ( z ) (cid:62)
1, so µ ( z ) (cid:62) Q ( z ) := 1 + z + z + (cid:112) (1 + z + z ) − z Q ( z ) = 1 + z + z − z Q ( z ) . Observe that Q ( z ) is a continuous monotone increasing function on [0 , Q (1) = (3 + √ / z ∈ (0 , z k := z / k , so z k − = z k for k (cid:62) k →∞ z k = 1. For notational convenience, we also define y k := µ ( z k ) and Q k := Q ( z k ); in particular, Q k = 1 + z k + z k − z k /Q k and y k = 1 + z k + z k − z k /y k − from the functional equation (4). Since lim k →∞ Q k = Q (1) >
2, we can assumethat k (cid:62) Q k (cid:62) k (cid:62) k . Thus | Q k − y k | = | z k ( y − k − − Q − k ) | = (cid:12)(cid:12)(cid:12)(cid:12) z k ( Q k − y k − ) Q k y k − (cid:12)(cid:12)(cid:12)(cid:12) (cid:54) | Q k − y k − | (cid:54) | Q k − − y k − | | Q k − Q k − | , using | z k | (cid:54) y k − (cid:62) | Q k | (cid:62)
2, and the triangle inequality. It follows fromlim k →∞ ( Q k − Q k − ) = 0 that lim k →∞ ( Q k − y k ) = 0. Thus lim k →∞ y k = Q (1),which completes the proof of µ = ρ .Differentiating each side of the recurrence (4) gives(5) µ (cid:48) ( z ) = 1 + 2 z − z µ ( z ) + 4 z µ (cid:48) ( z ) µ ( z ) . As z → − , 1 + 2 z − z µ ( z ) → − µ and 4 z µ ( z ) → µ , so (5) may be written as µ (cid:48) ( z ) = 3 − /µ + o (1) + (4 /µ + o (1)) µ (cid:48) ( z ) . Using the latter expression, it can be shown that µ (cid:48) ( z ) → µ (cid:48) , where µ (cid:48) satisfies µ (cid:48) = 3 − /µ + (4 /µ ) µ (cid:48) , so µ (cid:48) = 3 − /µ − /µ = 21 + 8 √ . We omit the details, but note that | /µ | <
1, so the iteration m k = 3 − /µ + (4 /µ ) m k − converges, and lim k →∞ m k = µ (cid:48) . (cid:3) LGEBRAIC INDEPENDENCE OF MAHLER FUNCTIONS 5
In view of Lemma 2, we define by continuity µ (1) := µ and µ (cid:48) (1) := µ (cid:48) . Since µ (cid:48)(cid:48) ( z ) is unbounded as z → − , this process cannot be continued; see Figure 1 fora graph of µ ( z ) and µ (cid:48) ( z ) for z ∈ [0 , Figure 1.
The functions µ ( z ) (dashed) and µ (cid:48) ( z ) (solid) for z ∈ [0 , Lemma 3.
Let α be any constant satisfying α < ρ ≈ . . Then, for t ∈ (0 , ∞ ) , we have (6) µ (cid:48)(cid:48) ( e − t ) = O ( t α − ) and (7) µ ( e − t ) = µ − tµ (cid:48) + O ( t α ) . Proof.
Let z = e − t ∈ (0 , z gives(8) µ (cid:48)(cid:48) ( e − t ) = A ( t ) + B ( t ) µ (cid:48)(cid:48) ( e − t ) , where A ( t ) is uniformly bounded, say | A ( t ) | (cid:54) A , and(9) B ( t ) = 16 e − t µ ( e − t ) = 16 µ + O ( t )as t → + .We now prove by induction on k (cid:62) t is sufficiently small, t k := t / k ,and C is sufficiently large, then(10) µ (cid:48)(cid:48) ( e − t k ) < Ct α − k holds for all k (cid:62) α , we have δ := 1 − µ · α − > . By (9), there exists ε > t ∈ (0 , ε ), we have B ( t ) < (16 /µ )(1+ δ ).Thus, for all t ∈ (0 , ε ), 4 α − B ( t ) < (1 − δ )(1 + δ ) = 1 − δ . For an arbitrary t ∈ (0 , ε ), choose(11) C > max { µ (cid:48)(cid:48) ( e − t ) /t α − , A/δ } . Thus µ (cid:48)(cid:48) ( e − t ) < Ct α − , so the inductive hypothesis (10) holds for k = 0. Supposethat it holds for some k (cid:62)
0. Then from (8), µ (cid:48)(cid:48) ( e − t k +1 ) = µ (cid:48)(cid:48) ( e − t k / ) (cid:54) A ( t k /
4) + B ( t k / µ (cid:48)(cid:48) ( e − t k ) < A + 4 − α (1 − δ ) Ct α − k = A + C (1 − δ ) t α − k +1 = ( A − Cδ t α − k +1 ) + Ct α − k +1 < Ct α − k +1 , RICHARD P. BRENT, MICHAEL COONS, AND WADIM ZUDILIN where on the final step we used
A < Cδ < Cδ t α − k +1 , by the choice (11) of C andalso since t k +1 ∈ (0 ,
1) and α <
2. Thus, (10) holds for all k (cid:62)
0, by induction.This proves (6). To prove (7) we integrate twice over the interval [0 , t ]. (cid:3) With a similar (but more precise) proof, we can show that the bounds (6) and(7) of Lemma 3 hold for α = 2 lg ρ . We omit the details since this result is notnecessary in what follows. Numerical experiments indicate that the constant 2 lg ρ is best possible – see Table 1, where the last column gives ( µ ( e − t ) − ( µ − tµ (cid:48) )) /t ρ .Observe the small oscillations in the last column (these are discussed at the end ofthis section). Table 1.
Approximation of µ ( e − t ) for t = 2 − k , 20 (cid:54) k (cid:54) e ( t ) := µ ( e − t ) − ( µ − tµ (cid:48) ) k t = 2 − k µ ( e − t ) e ( t ) e ( t ) /t ρ
20 9 . · − . · − . · − . · − . · − . · − . · − . · − . · − . · − µ ( z ). Lemma 4.
The function µ ( z ) is strictly monotone increasing for z ∈ [0 , .Sketch of proof. Suppose that z ∈ [0 ,
1) and N (cid:62)
1. Since F ( z ) = (cid:80) n (cid:62) c n z n ,where the c n ∈ { , } , we can bound the ‘tails’ (cid:88) n (cid:62) N c n z n (cid:54) z N − z and (cid:88) n (cid:62) N nc n z n − (cid:54) N z N − − z + z N (1 − z ) . Thus, given z < ε >
0, we can easily find N = N ( ε ) such that, for all z ∈ [0 , z ], 0 (cid:54) F ( z ) − F N ( z ) (cid:54) ε and 0 (cid:54) F (cid:48) ( z ) − F (cid:48) N ( z ) (cid:54) ε, where F N ( z ) := (cid:80) N − n =0 c n z n is the truncated power series approximating F ( z ).From (3) we have µ ( z ) (cid:62) µ (cid:48) ( z ) µ ( z ) = F (cid:48) ( z ) F ( z ) − z F (cid:48) ( z ) F ( z ) . Take z = 3 /
4. Using the above and a rigorous numerical computation, we can showthat µ (cid:48) ( z ) > z ∈ [0 , z ], and also that µ ( z ) > /
3. Thus, for z ∈ [ z , z / ], wehave µ ( z ) > /
3. In particular, µ ( z ) > / z k := (3 / / k , so z k +1 = z / k . We prove, by induction on k (cid:62)
0, that µ (cid:48) ( z ) > z ∈ [0 , z k ]. The base case ( k = 0) has been established. Assume thatthe result holds for k (cid:54) K ; hence, for z ∈ [ z K , z K +1 ], we have µ ( z ) > /
3. Now,from (5), µ (cid:48) ( z ) (cid:62) z − z µ ( z ) (cid:62) z − z = (1 − z )(1 + 3 z + 3 z ) > . In other words, the result holds for k = K + 1, thus it holds for all k (cid:62)
0, byinduction. Since lim k →∞ z k = 1, this completes the proof. (cid:3) LGEBRAIC INDEPENDENCE OF MAHLER FUNCTIONS 7
We are now in a position to treat the asymptotics of F ( z ) as z → − . To thisend, we define the Mellin transforms(12) F ( s ) := (cid:90) ∞ ln F ( e − t ) t s − d t and(13) M ( s ) := (cid:90) ∞ ln µ ( e − t ) t s − d t where the integrals converge for (cid:60) ( s ) >
0, and by analytic continuation elsewhere.From (3) and well-known properties of Mellin transforms (see, for example, [17,Appendix B.7]), we have(14) (1 − − s ) F ( s ) = M ( s ) . We deduce the asymptotic behaviour of F ( e − t ) for small positive t from knowl-edge of the singularities of F ( s ). Before doing this, we use analytic continuation toextend the definitions (12) and (13) into the left half-plane.Define (cid:101) µ ( t ) := ln µ ( e − t ) − ln( µ ) e − λt , where λ := µ (cid:48) µ ln µ ≈ . λ will soon be clear.Clearly (cid:101) µ ( t ) = O ( e − t ) as t → + ∞ . Also, from Lemma 3, as t → + we have forany constant α < ρ ≈ . (cid:101) µ ( t ) = ( λ ln µ − µ (cid:48) /µ ) t + O ( t α ) = O ( t α ) , by our choice of λ .From (13) and the definition of (cid:101) µ ( t ), we have(15) M ( s ) = (cid:102) M ( s ) + ln( µ ) λ − s Γ( s ) , where(16) (cid:102) M ( s ) := (cid:90) ∞ (cid:101) µ ( t ) t s − d t. However, the integral in (16) converges for (cid:60) ( s ) > − α . Since α may be chosen arbi-trarily close to 2 lg ρ , this implies that (15) and (16) give the analytic continuation of M ( s ) into a meromorphic function in the half-plane H := { s ∈ C : (cid:60) ( s ) > − ρ } .Since (cid:102) M ( s ) has no singularities in H , it follows from (15) that the singularities of M ( s ) in H are precisely those of ln( µ ) λ − s Γ( s ). Also, from (14), the singularitiesof F ( s ) in H are precisely those of M ( s ) / (1 − − s ). We conclude that the Mellintransform F ( s ) has three types of singularities in H , as follows:(a) a double pole at s = 0, since Γ( s ) has a pole there, and the denominator1 − − s vanishes at s = 0;(b) simple poles at s = ikπ/ ln 2 for k ∈ Z \ { } , since the denominator 1 − − s vanishes at these points;and(c) a simple pole at s = −
1, since Γ( s ) has a pole there.We are now ready to prove the following result, which gives the asymptotic be-haviour of F ( z ) as z → − . It is convenient to express the result in terms ofln F ( e − t ). Indeed, Theorem 3 is a weaker result, written in terms of F ( z ), of thefollowing statement. RICHARD P. BRENT, MICHAEL COONS, AND WADIM ZUDILIN
Proposition 1.
For small positive t , (17) ln F ( e − t ) = − lg ρ · ln t + c + ∞ (cid:88) k =1 a k ( t ) + c t + O ( t α ) , where c is given by (19) , c is given by (20) , α < ρ ≈ . , and a k ( t ) = 1ln 2 (cid:60) (cid:18) M (cid:18) ikπ ln 2 (cid:19) exp( − ikπ lg t ) (cid:19) . Proof.
We consider the three types of singularities of F ( s ) in H . For case (a), thedouble pole at s = 0, we need the first two terms in the Laurent expansion of F ( s ).It is convenient to define L ( s ) := M ( s )Γ( s ) , so, from (15),(18) L ( s ) = (cid:102) M ( s )Γ( s ) + ln( µ ) λ − s . Taking the limit as s → L (0) = ln µ = 2 ln ρ ≈ . . Differentiating both sides of (18) and then taking the limit as s → L (cid:48) (0) = (cid:102) M (0) − λ · ln ρ ≈ . . Near s = 0 we have L ( s ) = L (0) (cid:18) L (cid:48) (0) L (0) s + O ( s ) (cid:19) , Γ( s ) = 1 s (1 − γs + O ( s )) , and (1 − − s ) − = 12 s ln 2 (1 + s ln 2 + O ( s )) , so F ( s ) = L (0)2 ln 2 · s + c s + O (1) , where L (0) / ln 4 = lg ρ and(19) c = (ln 2 − γ ) L (0) + L (cid:48) (0)2 ln 2 ≈ . . Now, the ‘Mellin dictionary’ of [17, pg. 765] shows that the double pole at s = 0contributes the two leading terms − lg ρ · ln t + c of (17).For case (b), the poles at s = ikπ/ ln 2 for k ∈ Z \{ } are simple and have residue M ( ikπ/ ln 2) / ln 4. Thus from the simple pole at ikπ/ ln 2 we get a term T k ( t ) := 1ln 4 M (cid:18) ikπ ln 2 (cid:19) exp( − ikπ lg t ) . Combining the terms T k ( t ) and T − k ( t ) for k (cid:62)
1, the imaginary parts cancel andwe are left with the oscillatory term a k ( t ) in (17). Of course, in order to write theinfinite sum over the a k ( t ) as stated in the proposition, we must show that this sum The reader may think of L ( s ) as the Dirichlet series (cid:80) ∞ n =1 b n n − s , where the b n are definedto be the coefficients in the power series ln µ ( z ) = (cid:80) ∞ n =1 b n z n . Be warned that µ ( z ) has a zero at z ≈ − . . i , so the power series has radius of convergence R = | z | ≈ . <
1. Thus,the b n have faster than polynomial growth, and the Dirichlet series does not converge anywhere. LGEBRAIC INDEPENDENCE OF MAHLER FUNCTIONS 9 converges. Note that (cid:101) µ (cid:48)(cid:48) ( t ) = O ( t α − ) as t → + , and (cid:101) µ (cid:48)(cid:48) ( t ) decreases exponentiallyas t → + ∞ . Suppose y ∈ R \{ } . Then, using integration by parts once, we have (cid:102) M ( iy ) = (cid:90) ∞ (cid:101) µ ( t ) t iy − d t = (cid:20)(cid:101) µ ( t ) t iy iy − (cid:90) (cid:101) µ (cid:48) ( t ) t iy iy d t (cid:21) ∞ = − iy (cid:90) ∞ (cid:101) µ (cid:48) ( t ) t iy d t, and twice, we have (cid:102) M ( iy ) = 1 iy ( iy + 1) (cid:90) ∞ (cid:101) µ (cid:48)(cid:48) ( t ) t iy +1 d t [2 pt ] = I ( y ) + I ( y ) iy ( iy + 1) , where I ( y ) = (cid:90) (cid:101) µ (cid:48)(cid:48) ( t ) t iy +1 d t and I ( y ) = (cid:90) ∞ (cid:101) µ (cid:48)(cid:48) ( t ) t iy +1 d t. Now, using the asymptotic bounds on (cid:101) µ (cid:48)(cid:48) ( t ) for small and large t , respectively, wehave | I ( y ) | (cid:28) (cid:90) t α − d t (cid:28) | I ( y ) | (cid:28) (cid:90) ∞ te − t d t (cid:28) , so as y → ∞ , (cid:102) M ( iy ) (cid:28) | y | − . Also, it follows from the complex version of Stirling’s formula that Γ( iy ) (cid:28) e − πy/ ,so M ( iy ) (cid:28) | y | − as y → ∞ . Thus, ∞ (cid:88) k =1 |M ( ikπ/ ln(2)) | (cid:28) ∞ (cid:88) k =1 k − < ∞ , and the series (cid:80) ∞ k =1 a k ( t ) is uniformly and absolutely convergent.For case (c), the factor (1 − − s ) − is − / s = −
1, so F ( s ) has a pole withresidue(20) c = λ ln µ µ (cid:48) µ = 23 + 3 √ ≈ . s = −
1. This accounts for the term c t in (17).Finally, the error term O ( t α ) in (17) follows from the fact that we have onlyconsidered the singularities of F ( s ) in H . (cid:3) We may write a k ( t ) as a k ( t ) = A k cos( kπ lg t ) + B k sin( kπ lg t ) . Define C k := (cid:112) A k + B k = max t> | a k ( t ) | . The constants A k , B k and C k for k (cid:54) Table 2.
The constants A k , B k and C k related to a k ( t ) for k (cid:54) k A k B k C k . · − − . · − . · − − . · − +1 . · − . · − . · − − . · − . · − . · − +3 . · − . · − Proof of Theorem . If we define C ( z ) := (1 − z ) lg ρ F ( z ), then clearly C ( z ) is pos-itive for z ∈ [0 , C (0) = 1. Also, for small positive t , Proposition 1 gives C ( e − t ) = D ( t ) e O ( t ) , where D ( t ) = exp (cid:18) c + ∞ (cid:88) k =1 a k ( t ) (cid:19) is a continuous function, which is periodic in the variable lg t . Since F ( e − t ) > t ∈ (0 , ∞ ), we must have 0 < inf t> D ( t ) (cid:54) sup t> D ( t ) < ∞ . Thus 0 < inf 14 for all z ∈ (0 , . < C ( z ) < . 14 for all z ∈ [1 / , M ( πik/ ln 2) for k ∈ Z \ { } , by (15) it suffices to evaluate (cid:102) M ( πik/ ln 2), since the term involving the Γ-function can be evaluated by standardmethods. For purposes of numerical computation, we transform the integral (16)as follows.Changing variables t = e u , we have (cid:102) M (cid:18) πik ln 2 (cid:19) = (cid:90) + ∞−∞ (cid:101) µ ( e u ) e πiku/ ln 2 d u. Now let v := ku/ (2 ln 2), so (cid:102) M (cid:18) πik ln 2 (cid:19) = 2 ln 2 k (cid:90) + ∞−∞ (cid:101) µ ( e v/k ) e πiv d v. Using the 1-periodicity of e πiv , we obtain(21) (cid:102) M (cid:18) πik ln 2 (cid:19) = 2 ln 2 k (cid:90) f k ( v ) e πiv d v, where f k ( v ) is a 1-periodic function defined by a rapidly convergent series; f k ( v ) := (cid:88) j ∈ Z (cid:101) µ ( e v + j ) /k ) . The integral in (21) can be evaluated by any method which is suitable for periodicintegrands (a simple and good choice is the trapezoidal rule [24]).Table 3 shows the results of a numerical computation using Proposition 1. Weused 8 terms in the sum over a k ( t ). In the table, e ( t ) is defined as the approx-imation given by (17) minus the exact value ln F ( e − t ). It appears from the lastcolumn of the table that the error is of order t ρ . Also, the last column does notappear to tend to a limit as t → + ; instead it fluctuates in a small interval. Thesame phenomenon may be observed in the last column of Table 1. This suggeststhat M ( s ) and F ( s ) have poles at s = − ρ + ikπ/ ln 2 for k ∈ Z , as expectedfrom the form of (8). LGEBRAIC INDEPENDENCE OF MAHLER FUNCTIONS 11 Table 3. Approximation of ln( F ( e − t )) using Proposition 1. Here e ( t ) is defined as the approximation given by (17) minus the exactvalue ln F ( e − t ). t ln( F ( e − t )) e ( t ) e ( t ) /t ρ . · − . · − . · − . · − . · − . · − . · − . · − . · − . · − . · − . · − . · − . · − . · − . · − . · − . · − . · − . · − Algebraic independence of F ( z ) , F ( z ) , F (cid:48) ( z ) , and F (cid:48) ( z )In this section, we prove Theorem 1 up to a certain algebraic statement con-cerning the nonexistence of polynomials satisfying a certain functional equation.Because of possible independent interest, we provide a much more generalised ver-sion of the statement than immediately needed for our current purpose. It is asfollows. Theorem 4. There are no polynomials p m ,...,m s ( z ) ∈ C [ z ] ( besides all being triv-ial ) such that (22) λ ( z ) (cid:88) (cid:54) m j (cid:54) M j j =0 , ,...,s p m ,...,m s ( z ) y m · · · y m s s = (cid:88) (cid:54) m j (cid:54) M j j =0 , ,...,s p m ,...,m s ( z ) s (cid:89) i =0 (1 + z + z − zy i ) M i − m i for some rational function λ ( z ) . This nonexistence result is proved in the next section.To start our proof of Theorem 1, we show that Theorem 3 gives a recipe forcomputing the radial asymptotics of F ( ξz ) as z → − for any root of unity ξ of degree 4 n . Consider, for example, ξ ∈ {± i, − } and substitute z = ξ z intoequation (2). Using the asymptotics provided in Theorem 3, then as z → − wehave F ( ξ z ) = (1 + ξ z + ξ z ) F ( z ) − z F ( z )= (1 + ξ + ξ ) C ( z )(1 − z ) lg ρ (1 + O (1 − z )) − C ( z )(1 − z ) lg ρ (1 + O (1 − z ))= (1 + ξ + ξ ) C ( z )(4(1 − z )) lg ρ (1 + O (1 − z )) − C ( z )(16(1 − z )) lg ρ (1 + O (1 − z ))= (cid:18) (1 + ξ + ξ ) 3 − √ − − √ (cid:19) C ( z )(1 − z ) lg ρ (1 + O (1 − z ))= Ω( ξ ) C ( z )(1 − z ) lg ρ (1 + O (1 − z )) , because 4 − lg ρ = (3 − √ / 2. Similarly, if ξ = ξ then as z → − we have F ( ξ z ) = (1 + ξ z + ξ z ) F ( ξ z ) − ξ z F ( z ) = Ω( ξ ) C ( z )(1 − z ) lg ρ (1 + O (1 − z )) , where Ω( ξ ) = (1 + ξ + ξ )Ω( ξ ) 3 − √ − ξ − √ , and in general this iteration defines the function Ω( ξ ) at any root of unity ξ ofdegree 4 n for n (cid:62) ξ ) = (1 + ξ + ξ )Ω( ξ ) 3 − √ − ξ Ω( ξ ) 7 − √ , and Ω(1) = 1, together with the related radial asymptotics of F ( ξz ). Note that(3 − √ / ρ − and (7 − √ / ρ − . Lemma 5. Let ξ be a root of unity of degree n for some n (cid:62) . Then as z → − ,we have F ( ξz ) = Ω( ξ ) C ( z )(1 − z ) lg ρ (1 + O (1 − z )) , where the function Ω( z ) satisfies Ω(1) = 1 and (23) Ω( z ) = (1 + z + z ) ρ − Ω( z ) − z ρ − Ω( z ) . We stress that the function Ω( z ) and so its relative(24) ω ( z ) := ρ Ω( z )Ω( z )are only defined on the set of roots of unity of degree 4 n where n = 0 , , , . . . . Lemma 6. The function (24) is transcendental over the field of rational functions.Proof. We start observing that the functional equation (23) translates into(25) ω ( z ) = 1 + z + z − z ω ( z )for the function (24). Assume, on the contrary, that the function ω ( z ) is algebraic,hence satisfies, on the set of the roots of unity, a non-trivial relation M (cid:88) m =0 p m ( z )( ω ( z ) /z ) m = M (cid:88) m =0 p m ( z ) y m (cid:12)(cid:12) y = ω ( z ) /z = 0 , which we suppose to have the least possible M . Multiply the relation by ( z/ω ( z )) M ,substitute z for z in the relation, and apply (25) to arrive at M (cid:88) m =0 p m ( z )(1 + z + z − zy ) M − m (cid:12)(cid:12) y = ω ( z ) /z = M (cid:88) m =0 p m ( z )(1 + z + z − ω ( z )) M − m = 0 . If the two algebraic relations are not proportional, that is, if (cid:80) Mm =0 p m ( z )(1 + z + z − zy ) M − m is not λ ( z ) (cid:80) Mm =0 p m ( z ) y m for some λ ( z ) ∈ C ( z ), then a suitablelinear combination of the two will eliminate the term y M and result in a non-trivial algebraic relation for ω ( z ) of degree smaller than M , a contradiction. Theproportionality, on the other hand, is not possible in view of Theorem 4 applied inthe case s = 0. Thus, ω ( z ) cannot satisfy an algebraic relation over C ( z ). (cid:3) LGEBRAIC INDEPENDENCE OF MAHLER FUNCTIONS 13 Recall the function µ ( z ) defined in (3). It follows from (4) and (25) and from µ (1) = ρ = ω (1) that the functions µ ( z ) and ω ( z ) coincide on the set of roots ofunity of degree 4 n . Thus, Lemma 6 implies that µ ( z ) is a transcendental function —the fact which is already a consequence of the algebraic independence of F ( z ) and F ( z ). Note, however, that in the opposite direction the transcendence of µ ( z ) doesnot directly imply Lemma 6, because the function ω ( z ) is defined on a smaller setof certain roots of unity and is not even known to be analytic.Another immediate consequence of the transcendence of ω ( z ) is the followingresult. Lemma 7. Assume that with polynomials p ( z ) , . . . , p M ( z ) ∈ C [ z ] we have M (cid:88) m =0 p m ( ξ )Ω( ξ ) m Ω( ξ ) M − m = 0 for any root of unity of degree n , where n = 0 , , , . . . . Then p m ( z ) = 0 for each m = 0 , , . . . , M .Proof. Indeed, the equation from the hypothesis of the lemma is equivalent to theidentity M (cid:88) m =0 p m ( z ) ρ − m ω ( z ) m = 0on the set of roots of unity of degree 4 n where n = 0 , , , . . . . This contradicts thetranscendence of ω ( z ) established in Lemma 6. (cid:3) Denoting F k ( z ) := (cid:18) z dd z (cid:19) k F ( z )and using the fact that C ( z ) is real analytic, we have, as z → − , that F k ( ξz ) = k − (cid:89) j =0 (lg ρ + j ) · Ω( ξ ) C ( z )(1 − z ) lg ρ + k (1 + O (1 − z )) , since this is true for k = 0 and we simply differentiate it as many times as needed.From now on, we can consider the limit as z → − along the sequence exp( − t − n )for integers n (cid:62) 1, for some fixed t , so that C ( z ) is constant along the sequence.Finally, we can write the functional equation for the derivatives in the form(26) F k ( z ) = (cid:0) k (1 + z + z ) F k ( z ) − k z F k ( z ) (cid:1) · (1 + o (1))as z approaches any root of unity of degree 4 n , because the terms in o (1) involvethe derivatives of F of order smaller than k . (In fact, we will only use (26) for k = 1.)We are now in a position to present the proof of Theorem 1. Proof of Theorem . For the sake of a contradiction, assume that the theorem isfalse and that we have an algebraic relation(27) (cid:88) m =( m ,m ,m ,m ) ∈ M p m ( z ) F ( z ) m F ( z ) m F ( z ) m F ( z ) m = 0 , where the set M of multi-indices m ∈ Z (cid:62) is finite and none of the polynomials p m ( z ) in the sum is identically zero. Without loss of generality, we can assumethat the polynomial (cid:80) m p m ( z ) y m y m y m y m in five variables is irreducible.In the first part of our proof, we discuss the algebraic independence of F ( z ) , F ( z ),and F ( z ) only (so that the dependence on y is suppressed); this scheme is generalfor this particular case as well as for the one with F ( z ). Let ξ be any root of unity of degree 4 n for some n (cid:62) 0. Note that as z → − ,we have F ( ξz ) m F ( ξz ) m F (cid:0) ( ξz ) (cid:1) m F (cid:0) ( ξz ) (cid:1) m = C m · Ω( ξ ) m + m Ω( ξ ) m + m (1 − z ) (lg ρ ) | m | +( m + m ) (1+ o (1))where | m | := m + m + m + m , C m := C | m | m (cid:18) − √ (cid:19) m + m (lg ρ ) m + m , and C = C ( e − t / ) does not depend on ξ or z , the latter chosen along the sequence.Denote by M (cid:48) the subset of all multi-indices of M for which the quantity β := (lg ρ ) | m | + m + m is maximal; in particular, | m | and m + m are the same for all m ∈ M (cid:48) .Substituting ξz for z in (27), multiplying all the terms in the resulted sum by(1 − z ) β , and letting z → − , we deduce that(28) (cid:88) m ∈ M (cid:48) C m · p m ( ξ ) · Ω( ξ ) m + m Ω( ξ ) m + m = 0for any root of unity ξ under consideration. If there is no dependence on F ( z )in (27) (hence in (28)) then the summation in m is suppressed; in this case M := | m | = m + m + m and M (cid:48) := m are constant for all indices m ∈ M (cid:48) , so thatequation (28) becomes (cid:88) m =( m ,M (cid:48) ,M − M (cid:48) − m , ∈ M (cid:48) C m · p m ( ξ ) · Ω( ξ ) m Ω( ξ ) M − m = 0for any root of unity ξ of degree 4 n . By Lemma 7, this is only possible when p m ( z ) = 0 identically, a contradiction to our choice of M . This means that thefunctions F ( z ), F ( z ) and F ( z ) are algebraically independent.The same argument in the case of general (27) implies that (cid:88) m =( m ,N − m ,M − m ,M − ( N − m )) ∈ M (cid:48) C m · p m ( z ) = 0for any N , where M := m + m and M := m + m are constant on M (cid:48) .We next iterate relation (27) and compute, again, the asymptotics of the leadingterm as z tends radially to a root of unity ξ of degree 4 n . For this, we substitute z for z in the relation (27), multiply the result by 16 m + m z | m | and use theexpressions for F ( z ) and F ( z ) given by the Mahler functional equations in F ( z ), F ( z ), F ( z ), and F ( z ) (see also the proof of Theorem 2 below) to arriveat (cid:88) n ∈ N q n ( z ) F ( z ) n F ( z ) n F ( z ) n F ( z ) n = 0 , where N is defined analogous to M . The terms contributing the leading asymptotics(which, of course, remains attached to the same β as before) correspond to themulti-indices n ∈ N (cid:48) with the property β = (lg ρ ) | n | + n + n . Because of (26),controlling the coefficients q n ( z ) for n ∈ N (cid:48) is much easier than for general n ∈ N .Note that N (cid:48) is characterised by constant M = n + n and M = n + n aswas before M (cid:48) . The above transformation for the leading asymptotics terms in (27) LGEBRAIC INDEPENDENCE OF MAHLER FUNCTIONS 15 assumes the form (cid:88) m ∈ M (cid:48) m z m + m ) p m ( z ) F ( z ) m (cid:0) (1 + z + z ) F ( z ) − F ( z ) (cid:1) m × F ( z ) m (cid:0) z + z ) F ( z ) − F ( z ) (cid:1) m = (cid:88) m ∈ M (cid:48) m z m + m ) p m ( z ) × m (cid:88) n =0 ( − n (cid:18) m n (cid:19) (1 + z + z ) m − n F ( z ) n F ( z ) m + m − n × m (cid:88) n =0 ( − n (cid:18) m n (cid:19) m − n (1 + z + z ) m − n F ( z ) n F ( z ) m + m − n implying q n ( z ) = ( − n + n (cid:88) n (cid:54) m (cid:54) n + n n (cid:54) m (cid:54) n + n n + n − m m − n (cid:18) m n (cid:19)(cid:18) m n (cid:19) × z | n |− m − m ) (1 + z + z ) m + m − n − n p ( n + n − m ,n + n − m ,m ,m ) ( z )= ( − n + n (cid:88) n (cid:54) m (cid:54) M n (cid:54) m (cid:54) M M − m − n (cid:18) m n (cid:19)(cid:18) m n (cid:19) × z M − m + M − m ) (1 + z + z ) m + m − n − n p ( M − m ,M − m ,m ,m ) ( z )= ( − n + n M (cid:88) (cid:54) m (cid:54) M − n (cid:54) m (cid:54) M − n m − n − M (cid:18) M − m n (cid:19)(cid:18) M − m n (cid:19) × z m + m ) (1 + z + z ) M + M − m − m − n − n p ( m ,m ,M − m ,M − m ) ( z )for all n ∈ N (cid:48) , where for the last equality we switched to summation over m = M − m and m = M − m .In view of our assumption of the irreducibility of the original algebraic rela-tion (27), the newer relation must be proportional to it; that is, the polynomial (cid:80) m ∈ M p m ( z ) y m y m y m y m divides (cid:80) n ∈ N q n ( z ) y n y n y n y n in the polynomialring C [ z, y , y , y , y ]. In particular, the leading asymptotic parts of these polyno-mials, (cid:88) n ∈ N (cid:48) q n ( z ) y n y n y n y n = y M y M (cid:88) n ∈ N (cid:48) q n ( z ) (cid:18) y y (cid:19) n (cid:18) y y (cid:19) n and (cid:88) m ∈ M (cid:48) p m ( z ) y m y m y m y m = y M y M (cid:88) m ∈ M (cid:48) p m ( z ) (cid:18) y y (cid:19) m (cid:18) y y (cid:19) m , must be proportional, hence their quotient must be a polynomial in z . In otherwords, the sets N (cid:48) and M (cid:48) coincide (unless the former is empty, meaning that q m ( z ) = 0 identically for all m ∈ M (cid:48) ) and q m ( z ) = q ( z ) p m ( z ) for all m ∈ M (cid:48) forsome q ( z ) ∈ C [ z ]. We define (cid:98) p m ,m ( z ) := 4 m p m ,m ,M − m ,M − m ( z ) for 0 (cid:54) m (cid:54) M , (cid:54) m (cid:54) M , so that M (cid:88) m =0 M (cid:88) m =0 p ( m ,m ,M − m ,M − m ) ( z ) y m y m = M (cid:88) m =0 M (cid:88) m =0 (cid:98) p m ,m ( z ) y m (cid:18) y (cid:19) m , and the above proportionality relation reads q ( z ) M (cid:88) m =0 M (cid:88) m =0 (cid:98) p m ,m ( z ) y m y m = 4 M M (cid:88) m =0 M (cid:88) m =0 (cid:98) p m ,m ( z )(1 + z + z − zy ) M − m (1 + z + z − zy ) M − m . However, it follows from the case s = 1 of Theorem 4 that this is not possible; thatis, the polynomials must all be identically zero. (cid:3) In this final part of the section, we prove Theorem 2, by applying Theorem 1along with a general result in Mahler’s method due to Ku. Nishioka [22]; see alsoher monograph [23], in particular, Theorem 4.2.1 there. Proposition 2 (Ku. Nishioka [22]) . Let K be an algebraic number field and let k (cid:62) be a positive integer. Let f ( z ) , . . . , f d ( z ) ∈ K [[ z ]] and write f ( z ) for thecolumn-vector ( f ( z ) , . . . , f d ( z )) T . If f ( z k ) = B ( z ) f ( z ) for some matrix B ( z ) ∈ K ( z ) d × d and α is a nonzero algebraic number in the radiusof convergence of f ( z ) such that α k j is not a pole of B ( z ) for any j (cid:62) , then tr deg Q Q ( f ( α ) , . . . , f d ( α )) (cid:62) tr deg K ( z ) K ( z )( f ( z ) , . . . , f d ( z )) . Proof of Theorem . Apply Theorem 1 and Proposition 2 with K = Q , k = d = 4, f ( z ) = ( F ( z ) , F ( z ) , F (cid:48) ( z ) , F (cid:48) ( z )) T , and B ( z ) = − z z + z z z − z +2 z z − z z + z z . (cid:3) Linear algebra and Fibonacci numbers This section is entirely devoted to the proof of Theorem 4.Assume, on the contrary, that a non-trivial collection of polynomials p m ,...,m s ( z )satisfying (22) exists. If the greatest common divisor of the polynomials is p ( z ) thendividing them all by p ( z ) we arrive at the relation (22) for the newer normalisedpolynomials but with λ ( z ) replaced by λ ( z ) p ( z ) /p ( z ). Therefore, we can assumewithout loss of generality that the polynomials p m ,...,m s ( z ) in (22) are relativelyprime. Furthermore, without loss of generality we can assume that the existingpolynomials all have rational coefficients as the identity (22) itself happens to beover the field of rationals, so that p m ,...,m s ( z ) ∈ Q [ z ] and λ ( z ) ∈ Q ( z ).We first analyse the s = 0 case of relation (22). To this end, suppose there exist p ( z ) , . . . , p M ( z ) ∈ Q [ z ] with gcd( p ( z ) , . . . , p M ( z )) = 1 such that(29) λ ( z ) M (cid:88) m =0 p m ( z ) y m = M (cid:88) m =0 p m ( z )(1 + z + z − zy ) M − m for some rational function λ ( z ). Assuming λ ( z ) is nonzero, write λ ( z ) = a ( z ) /b ( z ),where gcd( a ( z ) , b ( z )) = 1, so that (29) becomes(30) a ( z ) M (cid:88) m =0 p m ( z ) y m = b ( z ) M (cid:88) m =0 p m ( z )(1 + z + z − zy ) M − m . LGEBRAIC INDEPENDENCE OF MAHLER FUNCTIONS 17 It follows immediately that any polynomial p m ( z ) on the left-hand side of (30) isdivisible by b ( z ), hence b := b ( z ) is a constant. By substituting x = 1 + z + z − zy ,we write (30) as a ( z ) M (cid:88) m =0 p m ( z ) z M − m (1 + z + z − x ) m = bz M M (cid:88) m =0 p m ( z ) x M − m , from which we conclude as before that each p m ( z ) is divisible by a ( z ) /z N where z N is the highest power of z dividing a ( z ). As gcd( p ( z ) , . . . , p M ( z )) = 1 we findout that a := a ( z ) /z N is a constant. In summary, λ ( z ) = λz N for some λ ∈ Q and N ∈ Z (cid:62) ; that is,(31) λz N M (cid:88) m =0 p m ( z ) y m = M (cid:88) m =0 p m ( z )(1 + z + z − zy ) M − m . Note that the rational constant λ must be nonzero as otherwise, by substituting z = 1 into (31), all p m ( z ) would have the common divisor z − Lemma 8. Assuming relation (22) holds, we have λ ( z ) = λz N for some λ ∈ Q \{ } and N ∈ Z (cid:62) . In our further investigation we will be interested in specialising identity (22) bychoosing z to be an appropriate root of unity. Any such specialisation leads to alinear relation on the space of polynomials in y , y , . . . , y s of degree at most M j in y j for each j = 0 , , . . . , s .Take a matrix γ := (cid:18) a bc d (cid:19) ∈ SL ( C ) , so that its determinant ad − bc = 1, and assume that its eigenvalues µ and µ − aredistinct. Consider the linear operator U M ( γ ) : y m (cid:55)→ ( ay + b ) m ( cy + d ) M − m on the linear space P M [ y ] of polynomials of degree at most M . Let the two rowvectors ( α , β ) and ( α , β ) be the eigenvectors of γ ; that is,( α , β ) γ = µ ( α , β ) and ( α , β ) γ = µ − ( α , β ) . Lemma 9. The spectrum of U M ( γ ) is the set { µ k : − M (cid:54) k (cid:54) M, k ≡ M (mod 2) } , with the corresponding eigenpolynomials r k ( y ) = r k ( γ ; y ) = ( α y + β ) ( M + k ) / ( α y + β ) ( M − k ) / , − M (cid:54) k (cid:54) M, k ≡ M (mod 2) . Proof. This follows immediately from the fact that the operator U ( γ ) maps α y + β onto µ ( α y + β ) and α y + β onto µ − ( α y + β ). (cid:3) Lemma 9 allows us to describe the spectrum of the (tensor-product) operator U = U M ( γ ) ⊗ · · · ⊗ U M s ( γ ) : s (cid:89) j =0 y m j j (cid:55)→ s (cid:89) j =0 ( ay j + b ) m j ( cy j + d ) M j − m j that acts on the space of polynomials in y , y , . . . , y s of degree at most M j in y j for j = 0 , , . . . , s , as well as to explicitly produce the corresponding eigenpolynomials. Lemma 10. The spectrum of U is contained in µ Z , with the ( linearly independent ) eigenpolynomials s (cid:89) j =0 r k j ( γ ; y j ) , − M j (cid:54) k j (cid:54) M j , k j ≡ M j (mod 2) for j = 0 , , . . . , s, corresponding to the eigenvalues µ k + k + ··· + k s , respectively. Substitution z = 1 into (22), where λ ( z ) = λz N , brings our case to λ (cid:88) (cid:54) m j (cid:54) M j j =0 , ,...,s p m ,...,m s (1) y m · · · y m s s = (cid:88) (cid:54) m j (cid:54) M j j =0 , ,...,s p m ,...,m s (1)(3 − y ) M − m · · · (3 − y s ) M s − m s . This corresponds to an eigenvector r ( y , . . . , y s ) of the operator U when γ = (cid:0) − (cid:1) ∈ SL ( C ). We get µ = (3 + √ / ρ , α y + β = µ − y − α y + β = µy − . (Of course, we exclude the trivial case r ( y , . . . , y s ) = 0 as it would imply thatall p m ,...,m s ( z ) are divisible by z − p m ,...,m s ( z ) ∈ Q [ z ], we have r ( y , . . . , y s ) ∈ Q [ y , . . . , y s ], so that none of the irrational values in µ Z can show upas λ . In other words, λ = 1 and the structure of the tensor product above dictates M + · · · + M s to be even and further produces(32) r ( y , . . . , y s ) = (cid:88) | k j | (cid:54) M j , k j ≡ M j (mod 2) k + ··· + k s =0 C k ,...,k s · r k ( y ) · · · r k s ( y s )where C k ,...,k s ∈ Q [ µ ]. We do not require the form (32), but only the fact λ = 1. Lemma 11. Assuming relation (22) holds, we have λ ( z ) = z N for some N ∈ Z (cid:62) :(33) z N (cid:88) (cid:54) m j (cid:54) M j j =0 , ,...,s p m ,...,m s ( z ) y m · · · y m s s = (cid:88) (cid:54) m j (cid:54) M j j =0 , ,...,s p m ,...,m s ( z ) s (cid:89) i =0 (1 + z + z − zy i ) M i − m i . We now take any prime p > 3, a root of unity ζ p of degree p , and the matrix γ p := g ( ζ ( p − / − p ) g ( ζ ( p − / − p ) · · · g ( ζ p ) g ( ζ p ) , (34) where g ( z ) = (cid:18) − z z + z + 1 (cid:19) . If we write γ p = (cid:0) a bc d (cid:1) , iterate the right-hand side of (33), and substitute z = ζ p ,then we obtain(35) ζ Np (cid:88) (cid:54) m j (cid:54) M j j =0 , ,...,s p m ,...,m s ( ζ p ) y m · · · y m s s = (cid:88) (cid:54) m j (cid:54) M j j =0 , ,...,s p m ,...,m s ( ζ p ) s (cid:89) i =0 ( ay i + b ) m i ( cy i + d ) M i − m i , because ζ ( p − / p = ζ p − p = ζ p . Note that det g ( z ) = z , so thatdet γ p = ( p − / − (cid:89) j =0 ζ j p = ζ (2 p − − / p = 1 LGEBRAIC INDEPENDENCE OF MAHLER FUNCTIONS 19 for primes p > 3, thus establishing that γ p ∈ SL ( C ). Lemma 12. Let µ and µ − be the eigenvalues of γ p . If µ p (cid:54) = 1 then N ≡ p ) in (33) .Proof. Comparing relation (35) with the result of Lemma 10 we conclude that ζ Np ∈ µ Z , and the latter is only possible when ζ Np = 1. (cid:3) Lemma 13. There are infinitely many primes p for which the eigenvalues µ of thecorresponding γ p are not p -th roots of unity. We have checked by direct computation that the only primes p in the range3 < p < µ p (cid:54) = 1 is violated, are p = 5 and p = 11 (and µ = 1 in the two cases). It is therefore natural to expect that we always have thecondition of Lemma 12 satisfied for primes p > Proof of Lemma . For positive exponents e , . . . , e s consider g ( z e ) g ( z e ) · · · g ( z e s ) = (cid:18) − a ( z ) b ( z ) − c ( z ) d ( z ) (cid:19) . Using induction on s , the coefficients of polynomials a ( z ) , b ( z ) , c ( z ) , d ( z ) are non-negative integers; furthermore, (cid:18) − a (1) b (1) − c (1) d (1) (cid:19) = g (1) s = (cid:18) − F F − F F (cid:19) s = (cid:18) − F s − F s − F s F s +2 (cid:19) , where F = 0, F = 1, and F n = F n − + F n − is the Fibonacci sequence. It is nothard to verify that(36) F n ≡ n ≡ , n ≡ , n ≡ , n = 0 , , , . . . . Consider now any prime p ≡ 15 (mod 28); this means that p − s = ( p − / 2, the number of the matrices g ( · ) in the product for γ p , isodd. The trace of the matrix γ p is a polynomial in ζ p containing F s +2 monomials ζ mp minus F s − monomials ζ mp . Note the the sums of the form (cid:80) p − j =1 ζ jmp = 0, eachinvolving p − j (cid:54)≡ p ), and such sums only, can be cancelled fromconsideration, thus leaving us with F s +1) − F s − − ( p − N ≡ ± γ p in its irreducible form isthe sum of at least three monomials ζ mp , corresponding to not necessarily different m ∈ { , , . . . , p − } ; in particular, the trace cannot be written in the form ζ (cid:96)p + ζ − (cid:96)p for some (cid:96) . Therefore, the eigenvalues µ, µ − of γ p are not of the form ζ (cid:96)p , ζ − (cid:96)p . Thiscompletes the proof of Lemma 13. (cid:3) Proof of Theorem . It follows from Lemmas 12 and 13 that N ≡ p ) in (33)for infinitely many primes p . This means that N = 0 and equation (33) assumesthe form(37) (cid:88) (cid:54) m j (cid:54) M j j =0 , ,...,s p m ,...,m s ( z ) y m · · · y m s s = (cid:88) (cid:54) m j (cid:54) M j j =0 , ,...,s p m ,...,m s ( z ) s (cid:89) i =0 (1 + z + z − zy i ) M i − m i . As y , . . . , y s are replaced with c y , . . . , c s y s , where ( c , . . . , c s ) varies over R s +1 ,the generic degree in z of the right-hand side in (37) is bounded from below by d , where d denotes the maximal degree of the polynomials p m ,...,m s ( z ), while thedegree in z of the left-hand side in (37) is at most d . This can only happen whenthe polynomials are constant; however in the latter circumstances we will still havea positive degree in z for the right-hand side in (37), a contradiction completingthe proof of Theorem 4. (cid:3) Mahler functions at roots of unity In this section, we discuss the structure of general Mahler functions at roots ofunity and provide an alternative approach to the proof of Theorem 3, which can beused in the asymptotical study of Mahler functions of any degree.The simplest possible Mahler functions are given as infinite products, so that itis natural to investigate the asymptotics of P ( z ) := ∞ (cid:89) j =0 − αz k j )as z → − , where α ∈ C , | α | (cid:54) 1. The recent paper [2] provides crude estimates forthe asymptotics of such products, though earlier works [11, 15, 20] already discussthe asymptotics in the ‘most natural’ case | α | = 1; see also the paper [13].As in Section 2, we make use of the Mellin transform, and so we define P ( s ) := (cid:90) ∞ ln P ( e − t ) t s − d t, which maps e − λt to Γ( s ) λ − s . Sinceln P ( e − t ) = ∞ (cid:88) j =0 ∞ (cid:88) l =1 l α l e − lk j t , we have(38) P ( s ) = Γ( s ) ∞ (cid:88) j =0 ∞ (cid:88) l =1 l α l ( lk j ) s = Γ( s )1 − k − s ∞ (cid:88) l =1 α l l s +1 . Thus the asymptotics of ln P ( e − t ) as t → + is related to the values of the mero-morphic continuation of the Dirichlet series P ( s )Γ( s ) = 11 − k − s ∞ (cid:88) l =1 α l l s +1 at negative integers [26, Proposition 2]. Without reproducing the standard analyt-ical argument in this situation (see [11, 15] for details) one gets, as t → + , P ( e − t ) = C ( t ) t (ln(1 − α )) / (ln k ) (1 + O ( t ))if α (cid:54) = 1, and P ( e − t ) = C ( t ) t − / e (ln t ) / (2 ln k ) (1 + O ( t ))if α = 1, for some positive and (2 πim/ ln k )-periodic function C ( t ) of t . Clearly,the asymptotics so obtained allow one to write out the asymptotic behaviour of anysolution of the Mahler equation f ( z ) = m (cid:89) j =1 (1 − ξ j z ) · f ( z k )along any radial limit as z → ξ , where ξ , . . . , ξ m and ξ are roots of unity.This analysis, Theorem 3 and the approach we discuss below allow us to expectsimilar asymptotic behaviour for other Mahler functions f ( z ) satisfying functionalequations of the form (1). That is, under some natural conditions imposed on the LGEBRAIC INDEPENDENCE OF MAHLER FUNCTIONS 21 polynomials a ( z ) , a ( z ) , . . . , a d ( z ), the asymptotics of f ( z ) as z → − is either ofthe form C ( z )(1 − z ) c (1 + O (1 − z ))or C ( z )(1 − z ) c e c ln (1 − z ) (1 + O (1 − z ))for some rational c or c of the form ‘ln(an algebraic integer) / ln k ’ and c of theform ‘rational / ln k ’, where the function C ( z ) is assumed to have some oscillatorybehaviour.For our alternative method to prove Theorem 3, we consider the function F ( z )as defined in the introduction. If we let z = e − t where t = 4 − x and denote f ( x ) = F ( z ) = F ( e − t ), then the functional equation (2) assumes the form f ( x + 2) − (1 + e − t + e − t ) f ( x + 1) + e − t f ( x ) = 0 . Using | − e − t | < t for t > 0, we can then recast this equation in the form f ( x + 2) − (3 + a ( x )) f ( x + 1) + (1 + a ( x )) f ( x ) = 0 , where | a ( x ) | , | a ( x ) | (cid:54) · − x . Denoting the zeroes of the characteristic polynomial λ − λ + 1, by λ := (3 − √ / λ := (3 + √ / ρ , and applying thequantitative version of Perron’s theorem due to Coffman [9] (see [25, Theorem 2]and comments to it within for the explicit statement, as well as [16] and [19] forthe predecessors), we deduce that f ( x ) = ˜ Cλ xj (1 + O (4 − x ))as x → + ∞ along x ≡ x (mod Z ), for some ˜ C = ˜ C ( x ) > j ∈ { , } . Asimple analysis then shows that j = 2.Note that ˜ C ( x ) is a 1-periodic real-analytic function in an interval x > σ becauseof the analytic dependence of the solution of the difference equation on the initialdata. This implies that as t → + F ( e − t ) = ˆ C ( t ) t lg ρ (1 + O ( t )) , and further F ( z ) = C ( z )(1 − z ) ln ρ (1 + O (1 − z )) as z → − , where C ( z ) is real-analytic and satisfies C ( z ) = C ( z ) for z ∈ (0 , F ( z ).Denote ζ n a primitive root of unity of (odd) degree n . For ζ we clearly have ζ = ζ , therefore the defining equation (2) for F ( z ) transforms to the equation˜ F ( z ) = (1 + ζ z + ζ z ) ˜ F ( z ) − ζ z ˜ F ( z )for the function ˜ F ( z ) := F ( ζ z ). The characteristic polynomial of this recursion as z → − is λ + ζ , with the absolute values of both roots equal to 1, so that thetheorem of Poincar´e [18] applies to imply that ˜ F ( z ) has oscillatory behaviour as z → − .Similarly, the difference equation for F ( z ) gives rise to a difference equationfor ˜ F ( z ) = F ( ζ n z ), because the former is equivalent to a relation between F ( z ), F ( z k ) and F ( z k ) for any k (cid:62) 1. In particular, one can take k = ( p − / n = p is a prime (compare with the construction of γ p in Section 4). Taking n = 5and choosing k = 2, so that ζ k n = ζ n , we find the corresponding characteristic polynomial λ − λ + 1. Again, the double zero λ = 1 (of absolute value 1) of thepolynomial leads to the oscillatory behaviour of ˜ F ( z ) as z → − ; the same storyhappens for the choice n = 11 and k = 6. The first ‘interesting’ situation originatesat n = 7. Here k = 3 and the characteristic polynomial of the difference equationrelating F ( ζ z ), F ( ζ z ) and F ( ζ z ) is λ + ( ζ + ζ + ζ ) λ + 1 = λ + − ± √− λ + 1 . Its roots are 1 ± √− ± (cid:112) − ± √− , whose absolute values are approximately 0 . . F ( ζ z ) ∼ C ( z )(1 − z ) ln(0 . / (3 ln 4) as z → − , where C ( z ) oscillates.The characteristic polynomial for n = p is exactly the characteristic polynomialof the matrix γ p in (34). It may be interesting to look for the known L -functionsas potential Mellin transforms for such a sophisticated behaviour at different rootsof unity: recall that the Mellin transform (38) used at the beginning of this sectionwas, up to the unwanted factor (1 − k − s ) − , a Dirichlet L -function. 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