aa r X i v : . [ m a t h . N T ] F e b FAST-GROWING SERIES ARE TRANSCENDENTAL
ROBERT J. MACG. DAWSON AND GRANT MOLNAR
Abstract.
Let K be a field endowed with an absolute value, let X ∈ K [[ z ]], and let R be asubring of K [[ z ]]. If the coefficients of X grow sufficiently rapidly relative to the coefficientsof the series in R , then X is transcendental over R . We prove this result by establishing arelationship between the coefficients of A ( X ) and A ′ ( X ), where A ( t ) is a polynomial over K [[ z ]]. Introduction
Fix a field K , and consider the ring of formal power series K [[ z ]] over K . For C ( z ) = C a power series in K [[ z ]], we write ( C ) n for the n th coefficient of C , or C n if no confusionarises. Throughout this paper, indices of series and sequences are always nonnegative.Suppose we are given a ring R ⊆ K [[ z ]] and a power series X ∈ K [[ z ]]. It is natural toask what relationship R bears to X . For instance, we may ask whether X is algebraic ortranscendental over R . (Recall that X is algebraic over R if there is a nonzero polynomial F ( t ) ∈ R [ t ] such that F ( X ) = 0, and X is transcendental over R otherwise.)Transcendental elements are useful building blocks in the theory of commutative rings.Indeed, X is transcendental over R precisely if the ring R [ X ] satisfies the following universalproperty: for every R -algebra S , and every element x ∈ S , there is a unique R -algebramorphism f : R [ X ] → S such that f ( X ) = x . But if X is not transcendental over R ,our options for f ( X ) are sharply curtailed. For instance, if X R but X ∈ R , then an R -algebra morphism f : R [ X ] → S must map X to a square root of X in S ; any otherseries algebraic over R is subject to similar restrictions.However, transcendental elements are slippery. Indeed, many decades elapsed betweenEuler’s formulation of the concept for real numbers and Liouville’s 1844 construction of atranscendental number [4]; Hermite did not show e to be transcendental until 1873 [3]. (Foran historical overview of the history of transcendental numbers, see the first chapter of Baker[1] or Section 22.2.3 of Suzuki [5].)For certain rings of series there are easy and rather obvious constructions and existenceproofs. The first of these follows Cantor’s 1874 proof [2] that almost all real numbers aretranscendental. Example 1.1.
Let K = Q and R = Q [ z ] . Then Q , the ring Q [ z ] of rational polynomials,the ring Q [ z ][ t ] of series polynomials with coefficients in Q [ z ] , and the set of series occurringas the root of a nonzero polynomial A ( t ) ∈ Q [ z ][ t ] are all countable. But the ring Q [[ z ]] ofseries with rational coefficients is uncountable, and thus almost all series are transcendentalover Q [ z ] . Even the ring of absolutely convergent series is uncountable, and so almost allabsolutely convergent series are transcendental over Q [ z ] . The next example follows Liouville’s construction of a decimal expansion that must rep-resent a transcendental number.
Example 1.2.
Let K = C and R = C [ z ] . Let L ( z ) = P n ≥ z n ; we note that the set ofnonzero coefficients of L p − is a proper subset of the set of nonzero coefficients of L p . For p < q , define c ( p, q ) := 2 q (1 − − p ) ; then we have ( L p ) c ( p,q ) = q ! , ( L j ) c ( p,q ) = 0 for all j < p ,and ( L j ) n = 0 for all j ≤ p and n with < | c ( p, q ) − n | < q − p .Suppose now that A ( t ) := P j ≤ m A j t j ∈ C [ z ][ t ] is a degree m polynomial. We select n with ( A m ) n = 0 , and let d := max { deg( A j ) : j ≤ m } , where we adopt the convention deg(0) := −∞ . Then for any q large enough that q − p > d , we have ( A ( L )) c ( p,q )+ n = ( L m ) c ( p,q ) · ( A m ) n = 0 . Thus L is transcendental over C [ z ] . Suppose K is endowed with an absolute value |·| , and X ∈ K [[ z ]]. If for some r > | X n | = O ( r n ) as n → ∞ , we say X exhibits exponential growth . Otherwise, X exhibits superexponential growth . Remark 1.3. If K is complete with respect to |·| , then a power series C ( z ) ∈ K [[ z ]] exhibitsexponential growth precisely if C ( z ) converges for z in a neighborhood of the origin. Inparticular, if K = C then C ( z ) exhibits exponential growth precisely if C ( z ) defines ananalytic function in a neighborhood of the origin.It is possible that the technique embodied in Example 1.2 can be extended to constructseries transcendental over (for instance) the ring of series exhibiting exponential growth. Inthis paper we have chosen a different approach, in which the rate of growth of the coefficientsthemselves, rather than of the gaps between them, is incompatible with the existence of apolynomial that takes the series to 0. To illustrate the idea, we give a simple result in whicha fast-growing series is shown to be irrational over a ring of series exhibiting exponentialgrowth. Proposition 1.4.
Suppose K is endowed with an absolute value |·| , and let C, D, X ∈ K [[ z ]] such that CX = D . If C = 1 and there exist c, d > such that for all n we have | C n | < c and | D n | < d , then for all n we have | X n | ≤ d (1 + c ) n . Proof.
We use strong induction. Clearly X = D , so | X | < d (1 + c ) . Suppose the claimholds for j < n ; then X n = D n − n − X j =0 C n − j X j | X n | ≤ d + n − X j =0 cd (1 + c ) j by hypothesis;= d + cd (1 + c ) n − c ) − d (1 + c ) n . (cid:3) No generality is lost in assuming that C = 1. Indeed, if CX = D then by cancelingleading zeros in C and D we can assume C is nonzero, and by dividing C and D by thesame constant we can assume C = 1. A simple scaling argument extends Proposition 1.4to all series exhibiting exponential growth. Corollary 1.5.
Suppose K is endowed with an absolute value |·| , and let C, D, X ∈ K [[ z ]] such that CX = D . If C = 1 and there exist c, d, r > such that for all n we have | C n | < cr n AST-GROWING SERIES ARE TRANSCENDENTAL 3 and | D n | < dr n , then | X n | ≤ d (1 + c ) n r n . Corollary 1.6.
Suppose K is endowed with an absolute value |·| . If X exhibits superexpo-nential growth, then X is irrational over the ring of series exhibiting exponential growth. Example 1.7.
The power series P n ≥ n ! z n ∈ C [[ z ]] is irrational over the ring of seriesexhibiting exponential growth. It is a fortiori irrational over the ring of Abel-summableseries, the ring generated by the convergent series, and the ring of absolutely convergentseries, since these are all subrings of the ring of series exhibiting exponential growth. The object of this note is to show that if K is a characteristic 0 field equipped with anabsolute value |·| and every series in R has coefficients exhibiting at most modest growth,then any series X ∈ K [[ z ]] with sufficiently fast-growing coefficients is transcendental over R . We state our main theorem (Theorem 3.3) here. Theorem.
Suppose K is endowed with an absolute value |·| , and suppose moreover that K is of characteristic . Fix R a subring of K [[ z ]] , and suppose we have a monotone increasingfunction ρ : N → R > such that for every C ∈ R , we have C n = O ( ρ ( n )) as n → ∞ .Suppose that | X | ≥ , and that for every fixed λ and m , the power series X satisfies thefollowing conditions as n → ∞ : ρ ( n ) X ℓ ≤ n | X ℓ | m = o ( | X n − λ | ); ρ ( n ) | X n − λ − | X ℓ< n | X ℓ | m = o ( | X n − λ | ) . Then X is transcendental over R . In particular, Theorem 3.3 applies to the ring of power series exhibiting exponentialgrowth, for instance when we take ρ ( n ) := n ! (see Example 3.5 below).The idea behind the proof of Theorem 3.3 is fairly straightforward: we suppose that wehave a polynomial A ( t ) := P j ≤ m A j t j ∈ R [ t ] such that A ( X ) = 0, and deduce that A ( t ) = 0.To do so, we make a careful examination of A ( X ) n for n large; if the coefficients of X growsufficiently rapidly, then the behavior of X n will dominate A ( X ) n unless the coefficient of X n in A ( X ) n is zero. But as A ( X ) = 0, the coefficient of X n must be zero, and so thebehavior of X n − will dominate A ( X ) n unless this coefficient is zero; proceeding inductively,for any ℓ small, we conclude the coefficient of X n − ℓ is zero. It turns out that for ℓ smallrelative to n , the coefficient of X n − ℓ in A ( X ) n is independent of n : in fact, the coefficientof X n − ℓ is precisely A ′ ( X ) ℓ , where A ′ ( X ) := P j ≤ m jA j X j − is the formal derivative of A ( t )evaluated at X (see Theorem 2.2 below). Consequently, if A ( X ) = 0 we would expect tohave A ′ ( X ) = 0, and this is indeed the case; now an easy bit of algebra tells us that A ( t ) = 0as desired. In the remainder of this paper, we formalize the intuition outlined above andfurnish some related results. 2. Algebraic preliminaries
In this section, we give a relationship between the coefficients of A ( X ) and A ′ ( X ) whichemphasizes the high-index coefficients of X . ROBERT J. MACG. DAWSON AND GRANT MOLNAR
Lemma 2.1.
Fix a power series X ∈ K [[ z ]] . For any m , we have X m = X [ m ] + mX h m i , (1) where the power series X [ m ] and X h m i are defined by (cid:0) X [ m ] (cid:1) n := X k + ... + k m = nk ,...,k m ≤ n X k . . . X k m and (cid:0) X h m i (cid:1) n := X ℓ< n (cid:0) X m − (cid:1) ℓ X n − ℓ . Proof.
By definition, we have( X m ) n = X k + ... + k m = n X k . . . X k m = X k + ... + k m = nk ,...,k m ≤ n X k . . . X k m + m X j =1 X k + ... + k m = nk j > n X k . . . X k m . If j = 1 and ℓ := n − k , then ℓ < n/
2. By symmetry, the last sum is independent of j , so( X m ) n = (cid:0) X [ m ] (cid:1) n + m X ℓ 2, and m sets in which each summand has a (necessarilysingle) factor X k z k with k > n/ 2. We illustrate this with the summands of ( X ) , thecoefficient of z in X = ( X + X z + X z + . . . ) , arranged as X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X . The terms in any of the three boldfaced triangles sum to ( X h i ) , while ( X [3] ) is the sumover the central inverted triangle.These definitions extend to full power series. For instance, we have X [3] ( z ) = X + 3 X X z + 3 X X z + X z + (3 X X + 3 X X ) z + · · · and X h i ( z ) = X X z + ( X X + 2 X X X ) z + ( X X + 2 X X X ) z + · · · . But observe that for m ∈ { , , } and n arbitrary, the set { ( k , . . . , k m ) ∈ Z m : k + . . . + k m = n and 0 ≤ k , . . . , k m ≤ n/ } may be empty or singleton, so there are a few irregularities for low powers: we have X [0] = 1, X [1] = X , and X [2] = 1 + X z + X z + · · · . AST-GROWING SERIES ARE TRANSCENDENTAL 5 Let δ ( A, X ) := X j ≤ m A j X [ j ] , and write δ n ( A, X ) := δ ( A, X ) n . We may think of δ ( A, X ) as a “pseudopolynomial” versionof A ( X ) using only the cores of the powers X m . Let also ǫ n ( A, X ) := X j ≤ m X k + p + q = nq
2, let γ n,λ ( A, X ) := X λ ≤ ℓ< n A ′ ( X ) ℓ X n − ℓ . While innocuous on its face, Lemma 2.1 is instrumental in the proof of the following algebraicidentity. Theorem 2.2. Fix a power series X ∈ K [[ z ]] and a polynomial A ( t ) ∈ K [[ z ]][ t ] . For any n and any λ < n/ , we have (2) A ( X ) n = X ℓ<λ A ′ ( X ) ℓ X n − ℓ + γ n,λ ( A, X ) + δ n ( A, X ) + ǫ n ( A, X ) , Proof. Write A ( t ) = P j ≤ m A j t j , where m is the degree of A . By definition,(3) A ( X ) n = X j ≤ m ( A j X j ) n = X j ≤ m X k ≤ n ( A j ) k (cid:0) X j (cid:1) n − k . Substituting (1) into (3), and recalling the definition of X [ j ] , we obtain A ( X ) n = X j ≤ m X k ≤ n ( A j ) k (cid:0) ( X [ j ] ) n − k + j ( X h j i ) n − k (cid:1) = X j ≤ m ( A j X [ j ] ) n + j X k ≤ n ( A j ) k X q< n − k (cid:0) X j − (cid:1) q X n − k − q = δ n ( A, X ) + X j ≤ m j X k ≤ n ( A j ) k X q< n − k (cid:0) X j − (cid:1) q X n − k − q Let p := n − k − q , so q < ( n − k ) / q < p . Thus A ( X ) n − δ n ( A, X ) = X k + p + q = nq n/ X j ≤ m X k + p + q = nq
Components of A ( X ) n in barycentric coordinatesIn the second case, k and q have no restriction other than that k + p + q = n , and X k + p + q = n n
2, the summand is part of the core. Otherwise, let p besome maximal ℓ i , and let q the sum of the others. For fixed n , the triplets ( k, p, q ) may beconsidered as barycentric coordinates.Triplets with p ≤ q , shown in white, are the summands of the core, δ n ( A, X ). (Due tothe definition of p , we cannot have p = 0 and q > q < p ≤ n/ ǫ n ( A, X ). Finally, coordinates with p > n/ X p in A ( X ) n is ( A ′ ( X )) n − p . We subdivide this into the summands of γ n,λ ( X, A ) (darkgrey) and the set of triplets (shown in black) for which l < λ . Unlike the others, this lastset of triplets does not become more numerous with increasing n . We shall see that if thecoefficients of X grow fast enough, each of γ n,λ ( X, A ), δ n ( X, A ), and ǫ n ( X, A ) is insignificantcompared to the terms with l < λ .3. Criteria and constructions for transcendental power series In this section, we prove our main theorem, furnish some related results, and suggest adirection for future research. Lemma 3.1. Suppose K is endowed with an absolute value |·| . Fix a power series X ∈ K [[ z ]] and a polynomial A ( t ) ∈ K [[ z ]][ t ] , and suppose that A ( X ) = 0 . Suppose moreover that foreach λ , we have, as n → ∞ : | γ n,λ ( A, X ) | = o ( | X n − λ | );(4) | δ n ( A, X ) | = o ( | X n − λ | );(5) AST-GROWING SERIES ARE TRANSCENDENTAL 7 | ǫ n ( A, X ) | = o ( | X n − λ | ) . (6) Then A ′ ( X ) = 0 .Proof. Let n ≥ A ( X ) n = X ℓ<λ A ′ ( X ) ℓ X n − ℓ + γ n,λ ( A, X ) + δ n ( A, X ) + ǫ n ( A, X ) , We claim A ′ ( X ) ℓ = 0 for each ℓ . If not, then let λ be minimal such that A ′ ( X ) λ = 0. Then A ( X ) n = A ′ ( X ) λ X n − λ + γ n,λ ( A, X ) + δ n ( A, X ) + ǫ n ( A, X ) , By conditions 4, 5, and 6, we observe | γ n,λ ( A, X ) | , | δ n ( A, X ) | , | ǫ n ( A, X ) | < | A ′ ( X ) λ X n − λ | n sufficiently large. Thus | A ( X ) n | ≥ | A ′ ( X ) λ X n − λ | − | δ n ( A, X ) | − | ǫ n ( A, X ) | − | γ n,λ ( A, X ) | > | A ′ ( X ) λ X n − λ | − | A ′ ( X ) λ X n − λ | 3= 0 . On the other hand, | A ( X ) n | = 0 by assumption, and we have obtained a contradiction. Then A ′ ( X ) ℓ = 0 for all ℓ , and thus A ′ ( X ) = 0 as desired. (cid:3) Lemma 3.2. Suppose K is endowed with an absolute value |·| , and suppose moreover that K is of characteristic 0. Let R be a subring of K [[ z ]] and X ∈ K [[ z ]] . Suppose that conditions4 through 6 of Lemma 3.1 hold for all series polynomials A ( t ) ∈ R [ t ] such that A ( X ) = 0 .Then X is transcendental over K .Proof. Suppose A ( t ) ∈ R [ t ] is chosen with A ( X ) = 0. Repeated applications of Lemma 3.1show that every derivative of A ( X ) vanishes. Then as K is of characteristic 0, A ( t ) must beconstant, and so A ( t ) = A ( X ) = 0. Thus X is transcendental over R as desired. (cid:3) At this point, we are ready to prove our main theorem. Theorem 3.3. Suppose K is endowed with an absolute value |·| , and suppose moreoverthat K is of characteristic . Fix R a subring of K [[ z ]] , and suppose we have a monotoneincreasing function ρ : N → R > such that for every C ∈ R , we have C n = O ( ρ ( n )) as n → ∞ .Suppose that | X | ≥ , and that for every fixed λ and m , the power series X satisfies thefollowing conditions as n → ∞ : ρ ( n ) X ℓ ≤ n | X ℓ | m = o ( | X n − λ | ); ρ ( n ) | X n − λ − | X ℓ< n | X ℓ | m = o ( | X n − λ | ) . Then X is transcendental over R .Proof. Note that the last condition of Theorem 3.3 gives us | X n | = o ( | X n +1 | ) as n → ∞ ,and a fortiori | X n | is eventually increasing. We will prove that the conditions of Lemma 3.1hold for every polynomial A ( t ) ∈ R [ t ] with A ( X ) = 0, so Lemma 3.2 will give us our desiredresult. ROBERT J. MACG. DAWSON AND GRANT MOLNAR Fix a polynomial A ( t ) = P j ≤ m A j t j ∈ R [ t ] for which A ( X ) = 0; and fix λ ≥ 0. Wecompute (as n → ∞ ): | γ n,λ ( A, X ) | = O X λ<ℓ< n | A ′ ( X ) ℓ X n − ℓ | = O X λ<ℓ< n X j ≤ m X k ≤ ℓ (cid:12)(cid:12)(cid:12) j ( A j ) k (cid:0) X j − (cid:1) ℓ − k X n − ℓ (cid:12)(cid:12)(cid:12) = O ρ ( n ) X λ<ℓ< n X j ≤ m X k ≤ ℓ (cid:12)(cid:12)(cid:12)(cid:0) X j − (cid:1) ℓ − k X n − ℓ (cid:12)(cid:12)(cid:12) = O ρ ( n ) X λ<ℓ< n X k ≤ ℓ (cid:12)(cid:12)(cid:12)(cid:0) X m − (cid:1) ℓ − k X n − ℓ (cid:12)(cid:12)(cid:12) = O ρ ( n ) | X n − λ − | X λ<ℓ< n X k ≤ ℓ (cid:12)(cid:12)(cid:12)(cid:0) X m − (cid:1) ℓ − k (cid:12)(cid:12)(cid:12) = O ρ ( n ) | X n − λ − | X λ<ℓ< n X k + ··· + k m − ≤ ℓ (cid:12)(cid:12) X k . . . X k m − (cid:12)(cid:12) = O nρ ( n ) | X n − λ − | X k + ··· + k m − < n (cid:12)(cid:12) X k . . . X k m − (cid:12)(cid:12) = O nρ ( n ) | X n − λ − | X ℓ< n | X ℓ | m − = O ρ ( n ) | X n − λ − | X ℓ< n | X ℓ | m = o ( | X n − λ | ) , where the second-to-last asymptotic holds because n = O (cid:16)P ℓ< n | X ℓ | (cid:17) . Thus condition 4holds.Next we compute | δ n ( A, X ) | = O X j ≤ m X k ≤ n (cid:12)(cid:12) ( A j ) k ( X [ j ] ) n − k (cid:12)(cid:12)! = O ρ ( n ) X j ≤ m X k ≤ n (cid:12)(cid:12) ( X [ j ] ) n − k (cid:12)(cid:12)! = O ρ ( n ) X k ≤ n (cid:12)(cid:12) ( X [ m ] ) n − k (cid:12)(cid:12)! AST-GROWING SERIES ARE TRANSCENDENTAL 9 = O ρ ( n ) X k ≤ n X k + ... + k m = n − kk ,...,k m ≤ n − k | X k . . . X k m | = O ρ ( n ) X k ≤ n X k ,...,k m ≤ n − k | X k . . . X k m | = O ρ ( n ) X k ≤ n X ℓ ≤ k | X ℓ | m = O nρ ( n ) X ℓ ≤ n | X ℓ | m = O ρ ( n ) X ℓ ≤ n | X ℓ | m +1 = o ( | X n − λ | ) , where again the second-to-last asymptotic holds because n = O (cid:16)P ℓ< n | X ℓ | (cid:17) . Thus condi-tion 5 holds.Finally, we compute | ǫ n ( A, X ) | = O X j ≤ m X k + p + q = nq
Assume the notation of Theorem 3.3. Suppose that | X | ≥ , and that forevery fixed λ and m , the power series X satisfies the following condition as n → ∞ : ρ ( n ) | X n − λ − | X ℓ ≤ n | X ℓ | m = o ( | X n − λ | ) . Then X is transcendental over R . Example 3.5. Let R ⊆ C [[ z ]] comprise the power series which exhibit exponential growth.Then we may take ρ ( n ) = n ! , since n ! exhibits superexponential growth. Let X = P n ≥ n ! z n ,so X n = 2 n ! . Then | X | = 2 > . Moreover, for every fixed λ and m , we have (as n → ∞ ): ρ ( n ) X ℓ ≤ n | X ℓ | m = n ! X ℓ ≤ n ℓ ! m = O n ! X ℓ ≤ ⌊ n ⌋ ! ℓ m = O (cid:16) n !2 m · ⌊ n ⌋ ! (cid:17) = O (cid:16) n !2 ⌊ n +22 ⌋ ! (cid:17) = O (cid:16) ⌊ n +42 ⌋ ! (cid:17) = o (cid:0) ( n − λ )! (cid:1) , so X is transcendental over R by Corollary 3.4. The series X is a fortiori transcendentalover the ring of Abel-summable series, the ring generated by the convergent series, and thering of absolutely convergent series, since these are all subrings of R . Example 3.5 applies without modification if we replace C with R , and can easily be adaptedif we replace C with Q p or another field of characteristic 0 equipped with a nontrivial absolutevalue.If the absolute value |·| on K is nonarchimedean, the proof of Theorem 3.3 can be adaptedto give an even stronger result. Theorem 3.6. Suppose K is endowed with a nonarchimedean absolute value |·| , and supposemoreover that K is of characteristic . Fix R a subring of K [[ z ]] , and suppose we have amonotone increasing function ρ : N → R > such that for every C ∈ R , we have C n = O ( ρ ( n )) as n → ∞ . AST-GROWING SERIES ARE TRANSCENDENTAL 11 Suppose that | X | ≥ , and that for every fixed λ and m , the power series X satisfies thefollowing conditions as n → ∞ : ρ ( n ) (cid:12)(cid:12)(cid:12) X ⌊ n ⌋ (cid:12)(cid:12)(cid:12) m = o ( | X n − λ | ); ρ ( n ) | X n − λ − | (cid:12)(cid:12)(cid:12) X ⌊ n − ⌋ (cid:12)(cid:12)(cid:12) m = o ( | X n − λ | ) . Then X is transcendental over R . Naturally, we also have the following corollary. Corollary 3.7. Assume the notation of Theorem 3.3. Suppose that | X | ≥ , and that forevery fixed λ and m , the power series X satisfies the following condition as n → ∞ : ρ ( n ) | X n − λ − | (cid:12)(cid:12)(cid:12) X ⌊ n ⌋ (cid:12)(cid:12)(cid:12) m = o ( | X n − λ | ) . Then X is transcendental over R . Our theorems provide a widely-applicable bound for growth rates of algebraic series. How-ever, in familiar cases, it is possible that the situation may be much simpler. Question 3.8. If a power series X ∈ C [[ z ]] exhibits superexponential growth, is X tran-scendental over the ring of power series exhibiting exponential growth?If a power series X ∈ C [[ z ]] exhibits superexponential growth, then any nonzero polynomial A ( t ) ∈ C [[ z ]][ t ] with coefficients analytic in a neighborhood of the origin must be of aparticular sort. Indeed, by the Weierstrass Preparation Theorem [6], if X = 0 and A (0)vanishes at the origin, then we may assume A ( t ) has the form A ( t ) = t m + z X j Transcendental Number Theory , Cambridge, C.U.P., 1975[2] Cantor, G. ¨Uber eine Eigenschaft des Inbegriffs aller reellen algebraischem Zahlen, Jour. Reine Angew.Math. 77 (1874), 258-62[3] Hermite, C., Sur la fonction exponentielle, Comptes Rendus 77 (1873), 18-24, 74-9, 226-33, 285-93;Oeuvres III, 150-81[4] Liouville, J., Sur des classes tr`es-´etendues de quantit´es dont la valeur n’est ni alg´ebrique, ni mˆemereductible `a des irrationnelles alg´ebriques, Comtes Rendus 18 (1844), 883-5, 910–11; J. Math. pures appl.16 (1851), 133-42.[5] Suzuki, J, A History of Mathematics , Prentice-Hall, Upper Saddle River, (2002)[6] Weierstrass, K. Mathematische Werke. II. Abhandlungen 2 , Georg Olms Verlagsbuchhandlung,Hildesheim; Johnson Reprint Corp., New York (1967), 135-142 Email address : [email protected] Dept. of Mathematics and Computing Science, Saint Mary’s University, Halifax, NS,Canada, B3H 3C3