On Taylor series of zeros of complex-exponent polynomials
aa r X i v : . [ m a t h . C V ] J a n On Taylor series of zeros of complex-exponentpolynomials
Mario DeFrancoJanuary 7, 2021
Abstract
We prove a factorization formula for the Taylor series coefficients of a zero ofa polynomial as a function of the polynomial’s coefficients. This result extendsto more general functions which we call “complex-exponent polynomials”. Toprove this formula, we prove theorems about derivations on commutative rings.We also show that, when applied to polynomials, our formula recovers the resultsof Sturmfels obtained with GKZ systems (“Solving algebraic equations in termsof A -hypergeometric series”. Discrete Math. 210 (2000) pp. 171-181) . Given a d -tuple ⇀ a of complex numbers ⇀ a = ( a , . . . , a d ) , a polynomial p ( z ) is a function p : C → C z d X k =1 a k z k − . If a d = 0, then p ( z ) has degree d −
1. A zero (or root) of p ( z ) is a number φ ∈ C such that p ( φ ) = 0 . As φ depends on the coefficients a , . . . , a d , we may think of φ as a function of ⇀ a φ = φ ( ⇀ a ) . There has been much work to describe this dependence on ⇀ a . For degrees 1through 4 there are the linear, quadratic, cubic, and quartic formulas, respec-tively, that give all zeros of p ( z ) in C . These are known as the solutions by adicals; that is, φ ( ⇀ a ) is a function consisting of a finite number of applicationsof addition, subtraction, multiplication, division, and taking n -th roots (raisingto the exponent n − ), where n is a positive integer. Knowledge of the quadraticformula stems from antiquity; for example, Babylonian cuneiform tablets fromthe second millennium B.C. describe methods to solve a quadratic equation(see Berriman [2]). The cubic formula was discovered in the sixteenth centuryA.D. (see Boyer and Merzbach [4] and Guilbeau [9]), due to Scipione Del Ferroand also being attributed to Niccol`o Tartaglia and Gerolamo Cardano, whosebook Ars Magna [5], published in 1545, contains this formula. This book alsocontains the quartic formula discovered by Lodovico Ferrari (see O’Connor andRobertson [13]). A solution by radicals does not exist for degree 5 or greater bythe Abel-Ruffini theorem ( [1], [16]) proved completely by Niels Henrik Abel in1824. Evariste Galois also proved this fact in 1831 (see [6] and Radloff [15]).Nevertheless, the function φ ( ⇀ a ) can still be described using infinite series fora polynomial of any degree. To study these series, Birkeland [3] used Lagrangeinversion; Mayr [11] used a system of differential equations; and Herrera [10]used reversion of Taylor series. Sturmfels [17] considers the zero as a solutionto a system of A -hypergeometric differential equations introduced by Gel’fand,Kapranov, and Zelevinsky [7], [8]. McDonald [12] and Passare and Tsikh [14]also study these hypergeometric series.In this paper we calculate the Taylor series coefficients of φ ( ⇀ a ) directly andprove the main result (Theorem 1.3) that they have a factorization formula.This theorem applies not only to polynomials but also to more general functionswhich we call “complex-exponent polynomials” defined in section 1.2. We proveTheorem 1.3 in section 2; we use lemmas about Stirling numbers of the first andsecond kind from section 3, and also theorems involving sums over set partitionsand subsets from section 4. In section 5, we prove in Corollary 5.5 that thisformula, when applied to polynomials, recovers the results of Sturmfels [17].In section 4, the final step in the chain of reasoning is Theorem 4.3 aboutan arbitrary derivation on a commutative ring. We prove another such theorem(Theorem 6.4) in section 6 which is used to give an alternative proof of Theorem1.3 in the case β = 1. We define a complex-exponent polynomial using essentially the same definitionabove for a polynomial but allowing the exponents of z to be complex numbersinstead of natural numbers. This complex exponentiation is defined in theconventional way by changing the domain from C to the Riemann surface L forthe logarithm. This surface L is parametrized by L = { ( r, θ, n ) : r ∈ R + , θ ∈ ( − π, π ] , n ∈ Z } . Then for z ∈ L corresponding to ( r, θ, n ), define z γ by z γ = e γ ln( r )+ iγθ +2 πinγ ∈ C . efinition 1.1. Let ⇀ a and ⇀ γ be two d -tuples of complex numbers ⇀ a = ( a , . . . , a d ) ⇀ γ = ( γ , . . . , γ d ) . Define a complex-exponent polynomial p ( z ; ⇀ a , ⇀ γ ) to be a function of the form p : L → C z d X k =1 a k z γ k which we abbreviate as p ( z ).Now we will consider a complex-exponent polynomial f ( z ) which is a “mod-ification” of another complex-exponent polynomial g ( z ). Specifically, supposewe fix a d -tuple ⇀ γ and also non-zero complex numbers b and β . Then let g ( z )be the complex-exponent polynomial g ( z ) = 1 + bz β . We call g ( z ) the “base function”. Definition 1.2.
Define f ( z ; g, ⇀ a , ⇀ γ ) to be the complex-exponent polynomial z g ( z ) + d X i =1 a i z γ i which we abbreviate as f ( z ).We next show how to express the Taylor series of a zero of f ( z ) using a zeroof g ( z ).Let b = r e iθ for some r > θ ∈ ( − π, π ], andRe( β ) = β and Im( β ) = β . Now for each m ∈ Z , it is straightforward to check that g ( z ) has a simple zeroin L corresponding to ( r, θ, n )( r, θ, n ) = ( e β m +1) π − θ − β r | β | , β ((2 m + 1) π − θ ) + β ln( r ) | β | − πn, n )(1)where n is chosen so that θ ∈ ( − π, π ]. Let α denote one of these zeros. In whatfollows we will need not the formula (1) for α , only the fact that α exists.We thus consider how this zero α varies as we vary a i . That is, let a i be thecoordinate variables of C d , and suppose there is a neighborhood U of ⇀ C d and a function φ ( ⇀ a ) φ : U → L ifferentiable in the variables a , . . . , a d at ⇀ f ( φ ( ⇀ a ); g, ⇀ a , ⇀ γ ) = 0 for all ⇀ a ∈ U (2) φ ( ⇀ α. (3)Then the above two equations are sufficient to calculate the Taylor series coef-ficients of φ ( ⇀ a ) about ⇀ ⇀ n denote a d -tuple of non-negative integers ⇀ n = ( n , . . . , n d )and denote Σ ⇀ n = d X i =1 n i . Let ∂ ⇀ n denote the partial derivative operator ∂ ⇀ n = d Y i =1 ( ∂∂a i ) n i and for any function ψ : : U → C denote ∂ ( ψ, I ) = ∂ ⇀ n ψ ( ⇀ a ) | ⇀ . Our main result is the next theorem.
Theorem 1.3.
With the above notation and when Σ ⇀ n ≥ , ∂ ⇀ n φ ( ⇀ a ) | ⇀ = − α P di =1 n i ( γ i − g ′ ( α ) Σ ⇀ n Σ ⇀ n − Y i =1 ( − iβ − d X i =1 n i γ i ) where g ′ ( α ) = bβα β − . We re-express Theorem 1.3 as Theorem 2.3 using different notation which will beused in its proof. First we present notation for multisets and multiset partitions.For an integer M ≥
0, we let [1 , M ] denote[1 , M ] = { i ∈ Z : 1 ≤ i ≤ M } . Definition 2.1.
For a positive integer N , define an ordered multiset I of [1 , d ]to be an N -tuple of integers I = ( I (1) , . . . , I ( N )) ith 1 ≤ I ( i ) ≤ d . We say that the order | I | is N . We define the multiplicitymultiplicity( I, n ) of n in I as the number of indices i such that I ( i ) = n . LetMultiset( d ) denote the set of these ordered multisets.For a positive integer k , define a set partition s of [1 , N ] with k parts to bea k -tuple s = ( s , . . . , s k )where s i are pairwise disjoint non-empty subsets of [1 , N ], k [ i =1 s i = [1 , N ] , and min( s i ) < min( s j ) for i < j. We also write a set s i as an m -tuple s i = ( s i (1) , . . . , s i ( m ))where m = | s i | and s i ( j ) < s i ( l ) for j < l. Let S ( N, k ) denote the set of such s . If H is any finite set of integers, we similarlydenote S ( H, k ) to be the set of all set partitions of H into k non-empty parts.For a multiset I and a set partition s ∈ S ( | I | , k ), define a multiset partition J of I with k parts to be a k -tuple J = ( J , . . . , J k )where J ∈ Multiset( d ) is given by J i = ( I ( s i (1)) , . . . , I ( s i ( m )))where m = | s i | . Thus the multiset partitions of I with k parts are in bijectionwith the set partitions in S ( | I | , k ). Let Parts( I, k ) denote the set of multisetpartitions of I . We let I (ˆ h ) denote the ordered multiset obtained from I byremoving the element at the h -th index: I (ˆ h ) = ( I (1) , . . . , I ( h − , I ( h + 1) , . . . , I ( N )) . We use the notation X m ∈ I γ m = N X i =1 γ I ( i ) where N = | I | .We also use the falling factorial applied to indeterminates, where “indeter-minate” refers to an arbitrary element of some polynomial ring over Z . Definition 2.2.
For an integer k ≥ x , define the fallingfactorial ( x ) k = k Y i =1 ( x − i + 1) . heorem 2.3. With the above notation and an ordered multiset I ∈ Multiset( d ) with | I | ≥ , ∂ ( φ, I ) = − α P m ∈ I ( γ m − g ′ ( α ) | I | ( − β ) | I |− ( β − − β − X m ∈ I γ m ) | I |− where g ′ ( α ) = bβα β − . Proof.
Recall by construction0 = f ( φ ( ⇀ a ))= g ( φ ( ⇀ a )) + d X i =1 a i φ ( ⇀ a ) γ i . We apply ∂ I to both sides the equation and then set ⇀ a = ⇀ I = ( i ), thenwe obtain 0 = g ′ ( α ) ∂ ( φ, I ) + α γ i . Solving for ∂ ( φ, I ) yields ∂ ( φ, I ) = − α γ i g ′ ( α ) . This proves the theorem when | I | = 1.Now we use induction on | I | . Given an I ∈ Multiset( d ) with | I | ≥
2, assumethe theorem is true for all I ′ ∈ Multiset( d ) with | I ′ | < | I | . Given any function ψ ( ⇀ a ) ψ : U → L, it follows from the definitions that ∂ ( f ◦ ψ, I ) = | I | X h =1 | I |− X k =1 ( γ I ( h ) ) k ψ ( ⇀ γ I ( h ) − k X J ∈ Parts( I (ˆ h ) ,k ) k Y i =1 ∂ ( ψ, J i )) (4)+ | I | X k =1 g ( k ) ( ψ ( ⇀ X J ∈ Parts(
I,k ) k Y i =1 ∂ ( ψ, J i ) . (5)For example, a term on the right side of line (4) corresponds to applying thepartial derivative ∂∂a I ( h ) to the coefficient a I ( h ) of f ( z ), and then applying k other partial derivatives ∂∂a J i (1) to the power of φ ( ⇀ a ) γ I ( h ) − i +1 proceeding from i = 1 to i = k . Each of these k applications by the chain ruleresults in a factor of ∂φ ( ⇀ a ) ∂a J i (1) . very other m ∈ J i then corresponds to applying ∂∂a m to this factor. Line (5)arises similarly.Now substitute φ ( ⇀ a ) for ψ ( ⇀ a ). As above, since f ( φ ( ⇀ a )) is identically zeroby construction, so is ∂ ( f ◦ φ, I ) for any I . We obtain0 = | I | X h =1 | I |− X k =1 ( γ I ( h ) ) k α γ I ( h ) − k X J ∈ Parts( I (ˆ h ) ,k ) k Y i =1 ∂ ( φ, J i ) (6)+ | I | X k =2 g ( k ) ( α ) X J ∈ Parts(
I,k ) k Y i =1 ∂ ( φ, J i ) (7)+ g ′ ( α ) ∂ ( φ, I ) . (8)By the induction hypothesis, the right of line (6) becomes | I | X h =1 | I |− X k =1 ( γ I ( h ) ) k α γ I ( h ) − k X J ∈ Parts( I (ˆ h ) ,k ) k Y i =1 − α P m ∈ Ji ( γ m − g ′ ( α ) | J i | ( − β ) | J i |− ( β − − β − X m ∈ J i γ m ) | J i |− which we simplify to α P m ∈ I ( γ m − ( − β ) | I |− g ′ ( α ) | I |− | I | X h =1 | I |− X k =1 − ( γ I ( h ) ) k ( β − ) k − X J ∈ Parts( I (ˆ h ) ,k ) k Y i =1 ( β − − β − X m ∈ J i γ m ) | J i |− . (9)We apply Theorem 4.1 to the quantity( β − ) k − X J ∈ Parts( I (ˆ h ) ,k ) k Y i =1 ( β − − β − X m ∈ J i γ m ) | J i |− (10)with ν = β − ; N = | I (ˆ h ) | ; and x i = β − γ I (ˆ h )( i ) to see that the quantity (10) isequal to c k , where c k = 1( k − k − X r =0 ( − k − − r (cid:18) k − r (cid:19) ( β − ( r + 1) − β − X m ∈ I (ˆ h ) γ m ) | I |− . Now the sum − | I |− X k =1 ( γ I ( h ) ) k c k is equal to − γ I ( h ) | I |− X k =1 c k ( γ I ( h ) − k − . (11) oting that c k are the form of coefficients in a Newton series in γ I ( h ) , we applyLemma 2.4 with F ( x ) = ( β − x − β − X m ∈ I (ˆ h ) γ m ) | I |− and set x to be γ I ( h ) to obtain that the expression (11) is equal to − γ I ( h ) ( β − X m ∈ I γ m − | I |− . Summing over h , we get that (9) is equal to α −| I | + P m ∈ I γ m ( − β ) | I |− g ′ ( α ) | I |− ( β − X m ∈ I γ m ) | I |− . (12)Now consider the terms at line (7). Applying g ( k ) ( α ) = b ( β ) k α β − k , we use the induction hypothesis and proceed as done for line (6) to see that thesum of these terms is equal to − bβα β ( − β ) | I |− α P m ∈ I ( γ m − g ′ ( α ) | I | | I | X k =2 ( β − k − ( k − k − X r =0 ( − k − − r (cid:18) k − r (cid:19) (( r +1) β − − β − X m ∈ I γ m ) . We add and subtract the term corresponding to k = 1; this term is( β − − β − X m ∈ I γ m ) | I |− . The sum including k = 1 is now a Newton series in β . Using Lemma 2.4 it isequal to ( β − X m ∈ I γ m ) | I |− . Therefore line (7) is equal to − bβα β ( − β ) | I |− α P m ∈ I ( γ m − g ′ ( α ) | I | ( β − X m ∈ I γ m ) | I |− − ( β − − β − X m ∈ I γ m ) | I |− ! . Combining with result (12) and simplifying using g ′ ( α ) = bβα β − , yields the equation0 = α P m ∈ I ( γ m − ( − β ) | I |− g ′ ( α ) | I |− ( β − − β − X m ∈ I γ m ) | I |− + g ′ ( α ) ∂ ( φ, I ) . Solving for ∂ ( φ, I ) completes the proof. emma 2.4. Suppose F ( x ) is a polynomial of degree m . Then F ( x ) = m +1 X k =1 ( x − k − ( k − k − X r =0 ( − k − − r (cid:18) k − r (cid:19) F ( r + 1) . Proof.
The Newton series of a polynomial P ( x ) of degree m is P ( x ) = m X k =0 (cid:18) xk (cid:19) k X r =0 ( − k − r (cid:18) kr (cid:19) P ( r ) . We prove this standard formula now. Both sides are polynomials in x of degree m . Evaluating x at an integer c, ≤ c ≤ m on the right side and collecting theterms P ( r ) for a fixed r give P ( r ) c X k = r ( − k − r (cid:18) ck (cid:19)(cid:18) kr (cid:19) . Applying the identity (cid:18) ck (cid:19)(cid:18) kr (cid:19) = (cid:18) c − rk − r (cid:19)(cid:18) cr (cid:19) gives P ( r ) (cid:18) cr (cid:19) c X k = r ( − k − r (cid:18) c − rk − r (cid:19) . which is equal to 0 if c = r and P ( c ) is c = r . Thus both sides are equalat m + 1 distinct inputs, and thus are equal as polynomials. This proves theNewton series formula.To prove the lemma, we use F ( x + 1) for P ( x ) and use (cid:18) xk (cid:19) = ( x ) k k ! , then substitute x x − k k −
1. This completes the proof.
Before proving Theorem 4.1 we present notation for Stirling numbers.
Definition 3.1.
For integers
N, r ≥
0, define the unsigned Stirling number ofthe first kind (cid:20) Nr (cid:21) to be ( − N − r times the coefficient of Y r when ( Y ) N is expressed in the mono-mial basis of polynomials in the indeterminate Y . Define the Stirling numberof the second kind (cid:26) Nr (cid:27) o be the coefficient of ( Y ) r when Y N is expressed in the falling factorial basisof polynomials.Equivalently we may define the Stirling numbers by the recursive relationsfor N ≥ r by (cid:20) Nr (cid:21) + N (cid:20) Nr + 1 (cid:21) = (cid:20) N + 1 r + 1 (cid:21) (13) (cid:26) Nr (cid:27) + ( r + 1) (cid:26) Nr + 1 (cid:27) = (cid:26) N + 1 r + 1 (cid:27) (14)and the conditions (cid:26) N (cid:27) = (cid:20) NN (cid:21) = 1 for N ≥ (cid:20) (cid:21) = 1 (cid:26) Nr (cid:27) = (cid:20) Nr (cid:21) = 0 for N < r.
It is straightforward to show that these two definitions are equivalent.
Definition 3.2.
For integers
N, r ≥ X , define (cid:20) Nr (cid:21) X to be the coefficient of Y r in ( Y + X − N . If N < r ≥
0, then let (cid:20) Nr (cid:21) X denote 0. Remark . It follows from the definitions that (cid:20) Nr (cid:21) = ( − N − r (cid:20) N + 1 r + 1 (cid:21) (15) (cid:20) Nr (cid:21) = ( − N − r (cid:20) Nr (cid:21) (16)Next we prove lemmas used in Section 4. Lemma 3.4.
For integers a, n ≥ , a ! a X r =0 ( − a − r (cid:18) ar (cid:19) ( r + 1) n = (cid:26) n + 1 a + 1 (cid:27) . (17) Proof.
Denote the left side of equation (17) by F ( n + 1 , a + 1). First, we claimthat ( a + 1) F ( n + 1 , a + 1) + F ( n + 1 , a ) = F ( n + 2 , a + 1) . iven an r with 0 ≤ r ≤ a and taking the coefficients of ( r + 1) n in the aboveequation, we see that the claim is implied by the identity( a + 1) (cid:18) ar (cid:19) − a (cid:18) a + 1 r (cid:19) = (cid:18) ar (cid:19) ( r + 1) . Second, we claim that F ( n + 1 , a + 1) = 0 when n < a . We have that F ( n + 1 , a + 1) = 1 a ! ( ddt t ) n ( t − a | t =1 (18)using the binomial expansion of ( t − a . Now evaluate the the right side ofequation (18) by applying ddt to products of t and ( t − t −
1) when n < a . This proves the second claim.Third, we have for n ≥ F ( n + 1 ,
1) = 1 . Thus F ( n + 1 , a + 1) satisfies the recursive definition and initial condition of (cid:26) n + 1 a + 1 (cid:27) when 0 ≤ a ≤ n . This completes the proof. Lemma 3.5.
For integers N ≥ r ≥ and indeterminates X and Y , (cid:20) Nr (cid:21) X + Y = N − r X i =0 X i (cid:18) r + ii (cid:19)(cid:20) Nr + i (cid:21) Y Proof.
By definition (cid:20) Nr (cid:21) X + Y is equal to X w ⊂ [1 ,N ] , | w | = N − r Y i ∈ w ( X + Y − i ) . In a term corresponding to a subset w with order N − r , the coefficient of X i is X w ′ ⊂ w, | w ′ | = N − r − i Y j ∈ w ′ ( Y − j ) . Given any subset v ⊂ [1 , N ] of order N − r − i , there are (cid:18) r + ii (cid:19) subsets of[1 , N ] of order N − r that contain v . Therefore (cid:20) Nr (cid:21) X + Y = N − r X i =1 X i (cid:18) r + ii (cid:19) X v ⊂ [1 ,N ] , | v | = N − r − i Y j ∈ v ( Y − j )= N − r X i =1 X i (cid:18) r + ii (cid:19)(cid:20) Nr + i (cid:21) Y . This completes the proof. emma 3.6. For an integer n ≥ , as formal power series ( − ln(1 − t )) n n ! = ∞ X k =0 (cid:2) kn (cid:3) k ! t k . (19) Proof.
We use induction on n . The lemma is true when n = 0, for then bothsides are equal to 1. Assume it is true for some n ≥
0. Multiply both sides ofequation (19) by 11 − t = ∞ X k =0 t k and integrate. We thus have( − ln(1 − t )) n +1 ( n + 1)! = ∞ X m =0 t m +1 m + 1 m X k =0 (cid:2) kn (cid:3) k ! . (20)We claim m X k =0 (cid:2) kn (cid:3) k ! = (cid:2) m +1 n +1 (cid:3) m ! . From the definition of the Stirling numbers of the first kind (cid:2) kn (cid:3) k ! = (cid:2) k +1 n +1 (cid:3) k ! − (cid:2) kn +1 (cid:3) ( k − . Summing both sides from k = 0 to k = m and using the fact that (cid:20) n + 1 (cid:21) = 0proves the claim and the induction step. This completes the proof. Lemma 3.7.
For integers ≤ r ≤ k , k X i = r (cid:26) ir (cid:27)(cid:18) ki (cid:19) = (cid:26) k + 1 r + 1 (cid:27) . (21) Proof.
Let F ( k +1 , r +1) denote the left side of equation (21). For a fixed r ≥ k . The lemma is true when k = r . Assume the lemma istrue for some k ≥ r . Apply the identity (cid:18) k + 1 i (cid:19) = (cid:18) ki (cid:19) + (cid:18) ki − (cid:19) to sum in F ( k + 2 , r + 1) and re-arrange to obtain F ( k + 2 , r + 1) = (cid:26) r − r − (cid:27)(cid:18) kr − (cid:19) + k +1 X i = r (cid:18)(cid:26) ir (cid:27) + (cid:26) i + 1 r (cid:27)(cid:19) (cid:18) ki (cid:19) . (22) ow multiply equation (21) by ( r + 1) and subtract the result from equation(22) to obtain (cid:26) r − r − (cid:27)(cid:18) kr − (cid:19) + k +1 X i = r (cid:18)(cid:26) ir (cid:27) + (cid:26) i + 1 r (cid:27) − ( r + 1) (cid:26) ir (cid:27)(cid:19) (cid:18) ki (cid:19) . (23)From equation (14) the coefficient of (cid:18) ki (cid:19) is (cid:26) ir − (cid:27) , and by the inductionhypothesis again, equation (23) is equal to (cid:26) k + 1 r (cid:27) . We have thus shown that F ( k + 2 , r + 1) − ( r + 1) (cid:26) k + 1 r + 1 (cid:27) = (cid:26) k + 1 r (cid:27) and combining with equation (14) we have F ( k + 2 , r + 1) = (cid:26) k + 2 r + 1 (cid:27) . This completes the proof.
In this section we prove Theorem 4.1 used in the proof of Theorem 2.3, and alsoTheorems 4.2 and Theorems 4.3 which are used to prove Theorem 4.1.
Theorem 4.1.
For integers ≤ k ≤ N , an indeterminate ν , and N indetermi-nates x i , ≤ i ≤ N , k − k − X r =0 ( − k − − r (cid:18) k − r (cid:19) (( r + 1) ν − N X i =1 x i ) N − (24)= ν k − X s ∈ S ( N,k ) k Y i =1 ( ν − X m ∈ s i x m ) | s i |− . (25) Proof.
We use induction on k . The theorem is true when k = 1 for then bothsides are equal to ( ν − N X i =1 x i ) N − . Assume that the theorem is true for all values less than some k ≥
2. We willexpand both lines (24) and (25) into powers of ν and x N , and show that thecoefficients are equal.First we expand line (24) as N − X n =0 ν n (cid:20) N − n (cid:21) P Ni =1 x i k − k − X r =0 ( − k − − r (cid:18) k − r (cid:19) ( r + 1) n hich by Lemma 3.4 we may write as N − X n = k − ν n (cid:20) N − n (cid:21) P Ni =1 x i (cid:26) n + 1 k (cid:27) . (26)Next, by the induction hypothesis, line (25) is equal to ν X w ⊂ [1 ,N ] ,N ∈ w N −| w |− X n = k − ν n (cid:26) n + 1 k − (cid:27)(cid:20) N − | w | − n (cid:21) P i ∈ wc x i ( ν − X i ∈ w x i ) | w |− . We expand the second factor of this sum in powers of ν to obtain ν X w ⊂ [1 ,N ] ,N ∈ w N −| w |− X n = k − ν n (cid:26) n + 1 k − (cid:27)(cid:20) N − | w | − n (cid:21) P i ∈ wc x i | w |− X n =0 ν n (cid:20) | w | − n (cid:21) P i ∈ w x i . In the above sum, the coefficient of ν n is n − X j = k − (cid:26) j + 1 k − (cid:27) X w ⊂ [1 ,N ] ,N ∈ w (cid:20) N − | w | − j (cid:21) P i ∈ wc x i (cid:20) | w | − n − − j (cid:21) P i ∈ w x i (27)Now we consider the coefficient of x m N ν n . In a term of the inner sum above,consider the right factor. By Lemma 3.5, we have (cid:20) | w | − n − − j (cid:21) P i ∈ w x i = | w |− n + j X m =0 x m N (cid:18) n − − j + m m (cid:19)(cid:20) | w | − n − − j + m (cid:21) P i ∈ w,i = N x i . Therefore the coefficient of x m N in expression (27) is n − X j = k − (cid:26) j + 1 k − (cid:27)(cid:18) n − − j + m m (cid:19) X v ⊂ [1 ,N − (cid:20) | v c | − j (cid:21) P i ∈ vc x i (cid:20) | v | n − − j + m (cid:21) P i ∈ v x i (28)where we have reindexed using the set v = w \ { N } , and v c denotes the com-plement of v in [1 , N − x m N ν n in expression (26), which by Lemma (3.5) is (cid:26) n + 1 k (cid:27)(cid:20) N − n + m (cid:21) P N − i =1 x i (cid:18) n + m m (cid:19) . (29)Applying Theorem 4.2 to the inner sum of expression (28) with a − j, b = n − − j + m , and N − M yields n − X j = k − (cid:26) j + 1 k − (cid:27)(cid:18) n − − j + m m (cid:19)(cid:18) n + m j + 1 (cid:19)(cid:20) N − n + m (cid:21) P N − i =1 x i . (30) quating expressions (29) and (30) and then simplifying the binomial coefficientsshows that it is sufficient to prove the equation (cid:26) n + 1 k (cid:27) = n − X j = k − (cid:26) j + 1 k − (cid:27)(cid:18) n j + 1 (cid:19) . This follows from Lemma 21. This complete the proof.
Theorem 4.2.
For integers M ≥ a ≥ , and b ≥ , and indeterminates x i , ≤ i ≤ M , X w ⊂ [1 ,M ] (cid:20) | w c | − a − (cid:21) P i ∈ wc x i (cid:20) | w | b (cid:21) P i ∈ w x i = (cid:18) a + ba (cid:19)(cid:20) Ma + b (cid:21) P Mi =1 x i (31) where w c denotes the complement of w in [1 , M ] .Proof. Fix an integer l ≥ n i ≥ ≤ i ≤ l . Consider a termof the form l Y i =1 x n i i . (32)The coefficient of this term on the left side of equation (31) is X w ⊂ [1 ,M ] ,v ⊂ [1 ,l ] ,v ⊂ w,v c ⊂ w c (cid:18) ( P i ∈ v c n i )! Q i ∈ v c i ! (cid:19) (cid:18) a − P i ∈ v c n i a − (cid:19) ( − | w c |− a (cid:20) | w c | a + P i ∈ v c n i (cid:21) (33) × (cid:18) ( P i ∈ v n i )! Q i ∈ v i ! (cid:19) (cid:18) b + P i ∈ v n i b (cid:19) ( − | w |− b (cid:20) | w | + 1 b + 1 + P i ∈ v n i (cid:21) where we have used Lemma 3.5 and equation (15), and where v c denotes thecomplement of v in [1 , l ]. The number of sets w of order j containing such v is (cid:18) M − lj − | v | (cid:19) so we rewrite (33) as X v ⊂ [1 ,l ] M X j =0 (cid:18) ( P i ∈ v c n i )! Q i ∈ v c i ! (cid:19) (cid:18) a − P i ∈ v c n i a − (cid:19) ( − M − j − a (cid:20) M − ja + P i ∈ v c n i (cid:21)(cid:18) M − lj − | v | (cid:19) (34) × (cid:18) ( P i ∈ v n i )! Q i ∈ v i ! (cid:19) (cid:18) b + P i ∈ v n i b (cid:19) ( − j − b (cid:20) j + 1 b + 1 + P i ∈ v n i (cid:21) Likewise the coefficient of the term (32) on the right side of equation (31) is (cid:18) a + ba (cid:19) ( P li =1 n i )! Q li =1 i ! ! (cid:18) a + b + P li =1 n i a + b (cid:19) ( − M − a − b (cid:20) M + 1 a + b + 1 + P li =1 n i (cid:21) . (35) ow equate expression (33) and (35) and simplify to obtain X v ⊂ [1 ,l ] M X j =0 a ( a − X i ∈ v c n i )! (cid:2) M − ja + P i ∈ vc n i (cid:3) (( M − j ) − | v c | )! ( b + X i ∈ v n i )! (cid:2) j +1 b +1+ P i ∈ v n i (cid:3) ( j − | v | )! (36)= ( a + b + l X i =1 n i )! (cid:2) M +1 a + b +1+ P li =1 n i (cid:3) ( M − l )! . (37)Now multiply both sides of the above equation by t M − l and sum over M ≥ X v ⊂ [1 ,l ] a ( ddt ) | v c | ( − ln(1 − t )) a + P i ∈ vc n i a + P i ∈ v c n i ! ( ddt ) | v | +1 ( − ln(1 − t )) b +1+ P i ∈ v n i b + 1 + P i ∈ v n i ! = ( ddt ) l +1 ( − ln(1 − t )) a + b +1+ P li =1 n i a + b + 1 + P li =1 n i ! where we have used Lemma 3.6. After applying one derivative from each powerof ddt , we see that the above equation is equivalent to X v ⊂ [1 ,l ] ( ddt ) | v c |− a ( − ln(1 − t )) a − P i ∈ vc n i − t ! ( ddt ) | v | ( − ln(1 − t )) b + P i ∈ v n i − t ! = ( ddt ) l ( − ln(1 − t )) a + b + P li =1 n i − t ! where in the case v c is empty we denote( ddt ) − (cid:18) a ( − ln(1 − t )) a − − t (cid:19) = ( − log(1 − t )) a . This equation follows from Theorem 4.3 with R the ring of power series in t , δ differentiation with respect to t , and f A = ( − log(1 − t )) a f B = ( − log(1 − t )) b − tf i = ( − log(1 − t )) n i . This completes the proof.
Theorem 4.3.
Let R be a commutative ring and let δ : R → R be a derivation.For an integer M ≥ and elements f A , f B , f i ∈ R, ≤ i ≤ M , X w ⊂ [1 ,M ] δ | w c |− ( f (1) A Y i ∈ w c f i ) δ | w | ( f B Y i ∈ w f i ) = δ M ( f A f B M Y i =1 f i ) (38) where w c denotes the complement of w in [1 , M ] ; f (1) A denotes δf A ; and in thecase w c is empty, δ − f (1) A denotes f A . roof. Note that we do not require R to contain 1. For an element f ∈ R andinteger n ≥
0, we denote nf = n X i =1 f and say that n is the coefficient of f . We also denote f ( n ) = δ n f. We use induction on M . The theorem is true when M = 0, for then w isempty and we have δ − ( f (1) A ) δ f B = f A f B . For an M ≥
1, assume the theorem is true for all values less than M . Considera term of the form f ( n A ) A f ( n B ) B M Y i =1 f ( n i ) i . In order for this term to arise on the left or right side of equation (38), we musthave M = n A + n B + M X i =1 n i . (39)Suppose n i ≥ i, ≤ i ≤ M . Then n A = n B = 0 and n i = 1 for each i . It can arise from the left side only when w = [1 , M ], and the coefficient ofthis term on both sides is M !.Thus suppose at least one of the n i = 0. By symmetry, we may consider aterm of the form f ( n A ) A f ( n B ) B l Y i =1 f ( n i ) i (40)with 0 ≤ l < M ; n A , n B ≥ n i ≥
1. The coefficient of this term on the leftside of equation (38) is X v ⊂ [1 ,l ] (cid:18) ( n A − P i ∈ v c n i )!( n A − Q i ∈ v c ( n i )! (cid:19) (cid:18) ( n B + P i ∈ v n i )!( n B )! Q i ∈ v ( n i )! (cid:19) (cid:18) M − ln B − | v | + P i ∈ v n i (cid:19) (41)where v c denotes the complement of v in [1 , l ]. Here v ⊂ w and n A − X i ∈ v c n i = M − | w | − n B + X i ∈ v n i = | w | . (43)In a term of the above sum, the first two factors are the coefficients of f ( n A ) A Y i ∈ v c f ( n i ) i and f ( n B ) B Y i ∈ v f ( n i ) i n δ n A − P i ∈ vc n i ( f (1) A Y i ∈ v c f ( n i ) i ) and δ n B + P i ∈ v n i ( f B Y i ∈ v f ( n i ) i )respectively. The third factor is the number of sets w ⊂ [1 , M ] that contain v and do not contain v c . This number is (cid:18) M − l | w | − | v | (cid:19) and then apply line (43). The coefficient of term (40) on the right side ofequation (38) is ( n A + n B + P li =1 n i )! n A ! n B ! Q li =1 ( n i )! . (44)Equating expressions (41) and (44) and simplifying with equation (39), we ob-tain the equation X v ⊂ [1 ,l ] n A ( n A − X i ∈ v c n i ) | v c |− ( n B + X i ∈ v n i ) | v | = ( n A + n B + l X i =1 n i ) l where in the case v c is empty we denote n A ( n A − − = 1 . But because l < M , this equation follows from the induction hypothesis with R being the ring of polynomials in the variable t , δ being differentiation withrespect to t , and f A = t n A , f B = t n B , and f i = t n i and then setting t = 1. Thiscompletes the proof. A -hypergeometric series We now prove that Theorem 2.3 recovers the formula of Sturmfels ( [17], Section3). We follow the notation of that paper (including Definition 5.1 here) withslight alterations. Note in this section that a i and their fractional powers aretreated as indeterminates as done in that paper. We prove Lemmas 5.2, 5.3 andTheorem 5.4 to prove the the recovery in Corollary 5.5.Fix an integer n ≥ ≤ i < i ≤ n . Set d = i − i . Let A denote the 2 × ( n + 1) matrix A = (cid:18) · · · n · · · (cid:19) . Viewing A as a linear transformation A : Q n +1 → Q , let Ker( A ) denote its kernel and let L denote L = Ker( A ) ∩ Z n +1 . efinition 5.1. For an integer v and rational number u , define γ ( u, v ) = v = 0( u ) | v | if v <
00 if u ∈ Z and 0 > u ≥ − v Q vi =1 ( u + i ) otherwise . Note that if u is not a negative integer, then γ ( u, v ) = u !( u + v )! . For u i ∈ Q , ≤ i ≤ n , define the series[ a u a u . . . a u n n ] = X ( v ,...,v n ) ∈ L n Y i =0 ( γ ( u i , v i ) a u i + v i i )which in this paper we call a “bracket series”. Let ξ denote a d -th root of − X i ,i ,ξ = ξ [ a /di a − /di ] + 1 d d X k =2 ξ k [ a ( k − d ) /di a i + k − a − k/di ] + 1 d [ a i − a − i ] (45)where [ a i − a − i ] denotes 0 if i = 0.We also use the notationΣ ∗ = n X i =0 , = i ,i and Π ∗ = n Y i =0 , = i ,i Lemma 5.2.
Suppose ( v , . . . , v n ) ∈ Ker( A ) . Then v i = − d Σ ∗ ( i − i ) v i (46) v i = − d Σ ∗ ( i − i ) v i (47)= 1 d (Σ ∗ ( i − i ) v i ) − Σ ∗ v i (48) Proof.
By definition i v i + i v i = − Σ ∗ iv i v i + v i = − Σ ∗ v i , Solving this system for v i and v i yields equations (46) and (47). Adding andsubtracting i to i in the sum of (47) and simplifying yields equation (48). Thiscompletes the proof. emma 5.3.
1. With notation as above, the series X i ,i ,ξ is a power series inthe variables a i for ≤ i ≤ n, i = i , i (that is, the exponents of these variablesare non-negative integers) whose coefficients depend on a i and a i .2. The term Π ∗ a n i i . (49) can appear in at most one or two bracket series in definition (45) , depending onwhether Σ ∗ ( i − i ) n i is not or is equal to − d , respectively.Proof. Note that each a i , ≤ i ≤ n, i = i , i appears in with corresponding u i equal to 0 or 1. Negative exponents can arise in two cases a ) u i = 0 and v i < b ) u i = 1 and v i < − . In either case γ ( u i , v i ) = 0. This proves statement 1.In the definition (45), there are d + 1 bracket series. If d = 1, then there areexactly two bracket series in (45), and the theorem is true. Therefore assume d ≥
2. Consider the possible u i for i = i , i . We say that the k -th bracketseries has u i + k = 1 for k = 0 , ≤ k ≤ d , except in the first bracket series inwhich u i = 0 for all i = i , i . Let C denote the number C = Σ ∗ ( i − i ) n i . Suppose C ≡ k − d for some k . Now in each bracket series u i + v i = n i for i = i , i , and ⇀ v ∈ L . For 0 ≤ j ≤ d , we thus define the ( n + 1)-tuple ⇀ v j⇀ v j = ( v , v , . . . , v n )where for i = i , i v i = n i − i = i + j − j = 1)and v i and v i are determined by Lemma 5.2: v i = − d ( C − j + 1) v i = 1 d ( C − j + 1) + 1( j = 1) − Σ ∗ n i , If the term (49) appears in the j -th bracket series, then ⇀ v j ∈ L . The numbers v i and v i are integers exactly when j ≡ k mod d . Thus if k d theterm (49) may appear only in the k -th bracket series (assuming 1 ≤ k ≤ d − k ≡ d , it may appear in either the 0-th or d -th bracket series.This completes the proof. heorem 5.4. For integers n i ≥ , i = i , i not all 0, the coefficient of theterm Π ∗ a n i i (50) in the series X i ,i ,ξ is ξ k d ( − M ( a i a i ) ( C +1) /d ( C +1 d − P ∗ n i − a P ∗ n i i Q ∗ n i ! (51) where C = Σ ∗ ( i − i ) n i = k − M d for some integers M and ≤ k ≤ d − .Proof. We first assume d ≥
2. Consider the term (50).We first compute Π ∗ γ ( u i , v i ) . Since u i + v i = n i , we check in either case u i = 0 or u i = 1 that γ ( u i , v i ) = 1 n i ! . Next let C denote the number C = Σ ∗ ( i − i ) n i . We compute γ ( u i , v i ) γ ( u i , v i ) depending on the equivalence class of C mod d .Suppose C ≡ k − d for some k, ≤ k ≤ d − C = k − M d for some integer M . Let v i + k − = n i + k − − v i = n i for all other i = i , i . These v i determine numbers v i and v i by Lemma 5.2: v i = − d ( C − k + 1)= − Mv i = 1 d ( C − k + 1) + 1 − Σ ∗ n i = M + 1 − Σ ∗ n i . Thus v i and v i are integers and ( v , . . . v n ) ∈ L , and so the term (50) appearsin the bracket series of k -th term in the sum from definition of X i ,i ,ξ . In thatseries we have u i = kd − u i = − kd . Thus u i + v i = − d ( C + 1) u i + v i = 1 d ( C + 1) − Σ ∗ n i (52) nd γ ( u i , v i ) γ ( u i , v i ) = (cid:18) ( − k/d )!( − k/d − M )! (cid:19) (cid:18) ( k/d − k/d + M − Σ ∗ n i )! (cid:19) = ( − k/d )!( − k/d − M )! ( k/d )!( k/d + M )! dk ( k/d + M ) Σ ∗ n i = sin( π ( k/d + M ))sin( πk/d ) ( k/d + M ) Σ ∗ n i k/d + M = ( − M ( k/d + M − Σ ∗ n i − where have used the identities for x ∈ C , m ∈ Z x !( − x )! = πx sin( πx )sin( π ( x + m )) = sin( πx )( − m and the fact that Σ ∗ n i is a positive integer. Putting this together, we obtainthat the coefficient of term (50) is equal to expression (51). This proves thetheorem for this case.Now suppose C = M d for some integer M . Let v i = n i for all i = i , i .These v i determine numbers v i and v i by Lemma 5.2: v i = − Cd = − Mv i = Cd − Σ ∗ n i = M − Σ ∗ n i . Thus v i and v i are integers and ( v , . . . v n ) ∈ L , and so the term (50) appearsin the bracket series [ a /di a − /di ] in the definition of X i ,i ,ξ . In that series wehave u i = kd − u i = − kd thus equations (52) still hold. Using similarreasoning above we obtain γ ( u i , v i ) γ ( u i , v i ) = ( − M d (1 /d + M − P ∗ i n i − . This proves the theorem in this case.Now suppose C = − M d for some integer M . First let v i + d − = n i + d − − v i = n i fir all other i = i , i . These v i determine numbers v i and v i byLemma 5.2: v i = − d ( C − d + 1)= − M + 1 v i = 1 d ( C − d + 1) + 1 − Σ ∗ n i = M − Σ ∗ n i . hus v i and v i are integers and ( v , . . . v n ) ∈ L , and so the term (50) appearsin the bracket series [ a i + d − , a − i ] in the definition of X i ,i ,ξ . In this series, u i = 0 and u i = −
1, so equations (52) still hold. We check that if M ≥ γ ( u i , v i ) γ ( u i , v i ) = (cid:0) ( − M − ( M − (cid:1) (cid:18) M − Σ ∗ n i )! (cid:19) = ( − M − ( M − Σ ∗ n i − , and if M ≤ γ ( u i , v i ) γ ( u i , v i ) = 0. Note that if i = 0, then here wemust have M ≥ i ≥
1, so assume that. Let v i − = n i − − v i = n i fir all other i = i , i . These v i determine numbers v i and v i byLemma 5.2: v i = − d ( C + 1)= − Mv i = 1 d ( C + 1) + 1 − Σ ∗ n i = M + 1 − Σ ∗ n i . Thus v i and v i are integers and ( v , . . . v n ) ∈ L , and so the term (50) appearsin the bracket series [ a i − , a − i ] in the definition of X i ,i ,ξ . In this series, u i = 0 and u i = −
1, so equations (52) still hold. We check that if M ≤ γ ( u i , v i ) γ ( u i , v i ) = (cid:16) ( − | M | +Σ ∗ n i − ( | M | + Σ ∗ n i − (cid:17) (cid:18) | M | ! (cid:19) = ( − M ( M − Σ ∗ n i − , and if M ≥ γ ( u i , v i ) γ ( u i , v i ) = 0. This completes the proof for d ≥ d = 1, the definition (45) has exactly two bracket series (and only if i = 0), and the proof proceeds similarly to the above case when C = − M d .This completes the proof.
Corollary 5.5.
Suppose c i and c i are non-zero complex numbers and parametrize C n − with the coordinate variables c i , i = i , i . Let g ( z ) = 1 + c i c i z d f ( z ) = g ( z ) + Σ ∗ c i z i − i and φ ( ⇀ c ) : U → L a smooth function where U ∈ C n − is a neighborhood of ⇀ such that f ( φ ( ⇀ c )) = 0 φ ( ⇀ α here g ( α ) = 0 . Denote ⇀ cc i = ( c c i , c c i , . . . , c i − c i , c i +1 c i , . . . , c i − c i , c i +1 c i , . . . , c n c i ) ∈ C n − . Then as formal series, the Taylor series of φ ( ⇀ cc i ) about ⇀ with respect tothe variables c i , i = i , i is equal to X i ,i ,ξ , where we identify c i = a i and α = ξ a /di a /di . Proof.
This follows from comparing the coefficient of Theorem 5.4 with theformula in Theorem 2.3 with β = d and ⇀ γ the ( n − ⇀ γ = ( − i , − i , . . . , − , , . . . , d − , d + 1 , . . . , n − i ) . β = 1 Suppose in Theorem 2.3 that β = 1. From the proof of that theorem, it is suffi-cient to prove Theorem 4.1 in the case ν = 1. We therefore give an alternativeproof of that theorem assuming ν = 1. A special case of Theorem 6.4 impliesthis result. First we prove the following two lemmas. Lemma 6.1.
For indeterminates a and b and an integer n ≥ , ( a + b ) n = n X i =0 (cid:18) ni (cid:19) ( a ) i ( b ) n − i and equivalently (cid:18) a + bn (cid:19) = n X i =0 (cid:18) ai (cid:19)(cid:18) bn − i (cid:19) . Proof.
Apply the product rule n times in( ddt ) n ( t a )( t b )and evaluate at t = 1 to give the right side. Taking the n -th derivative of t a + b without using the product rule gives the left side. This completes the proof. Lemma 6.2.
For integers i, n ≥ and an indeterminate b , i X r =0 ( − i − r (cid:18) ir (cid:19)(cid:18) b + rn (cid:19) = (cid:18) bn − i (cid:19) (53) roof. For any function f ( a ), the coefficient of (cid:18) ai (cid:19) in the Newton series of f ( a ) is i X r =0 ( − i − r (cid:18) ir (cid:19) f ( r ) . Consider the expression (cid:18) a + bn (cid:19) (54)as a function of a . Therefore the left side of equation (53) is the coefficient of (cid:18) ai (cid:19) in the Newton series of the function (54). From Lemma 6.1, this coefficientis also equal to (cid:18) bn − i (cid:19) . This completes the proof.Now take the left side of Theorem 4.1 when ν = 1 and apply Lemma 6.2with b = P Ni =1 x i , n = N −
1, and i = k − (cid:18) N − k − (cid:19) ( N X i =1 x i ) N − . Rename k by N − j . We thus seek to prove Theorem 6.3.
For integers ≤ j ≤ N − and indeterminates x i , ≤ i ≤ N , (cid:18) N − j (cid:19) ( N X i =1 x i ) N − = X s ∈ S ( N,N − j ) N − j Y i =1 ( X m ∈ s i x m ) | s i |− . (55)We now present the notation and definitions to state Theorem 6.4 whichwill imply Theorem 6.3. Let N and j be integers 0 ≤ j < N . For an integer k ,0 ≤ k ≤ N − j , let ⇀ u denote a k -tuple of non-negative integers ⇀ u = ( u , . . . , u k ) . Let R be a commutative ring and δ : R → R a derivation. Let ⇀ f denote an N -tuple ⇀ f = ( f , . . . , f N )where f h ∈ R . We use the same notation f ( n ) i as in the proof of Theorem 4.3.Let τ be a set of k integers τ = ( τ (1) , . . . , τ ( k ))where 1 ≤ τ ( i ) < τ ( i + 1) ≤ N − j , and denote the set of such τ by T ( N − j, k ).We also view a τ as a strictly increasing function τ : [1 , k ] → [1 , N − j ] . e require that T ( N − j,
0) consists of one element, the empty set. Define F ( j, ⇀ f , ⇀ u ) ∈ R by F ( j, ⇀ f , ⇀ u ) = X τ ∈ T ( N − j,k ) X s ∈ S ( N,N − j ) ( k Y i =1 (cid:18) | s τ ( i ) | − u i (cid:19) δ | s τ ( i ) |− − u i Y h ∈ s τ ( i ) f h ) × ( Y i/ ∈ τ δ | s i |− Y h ∈ s i f h ) (56)where a term is 0 if | s τ ( i ) | − − u i <
0, as the binomial coefficient is equal to 0.Define the number C ( j, N, ⇀ u ) by C ( j, N, ⇀ u ) = ( N − N − j + P ki =1 u i )( N − j − k )!( j − P ki =1 u i )! Q ki =1 ( u i )! Q ki =1 ( k − i + 1 + P kg = i u g ) . Theorem 6.4.
With the above definitions F ( j, ⇀ f , ⇀ u ) = C ( j, N, ⇀ u ) δ j − P ki =1 u i N Y i =1 f i . (57) Proof.
We use induction on N . It is straightforward to prove the theorem inthe case N = 1 and k = 0 or k = 1. For an N ≥
2, assume the theorem istrue for all values less than N . We will prove the induction step by comparingcoefficients of both sides of equation (57) of a term of the form N Y i =1 f ( n i ) i . (58)For this term to appear on either side of equation (57) we must have N X i =1 n i = j − k X i =1 u i , (59)so at least one n i must be 0. By the symmetry of F ( j, ⇀ f , ⇀ u ) in f i (Lemma 6.5),we may assume that the term (58) is of the form l Y i =1 f ( n i ) i . (60)Let M ( j, N, ⇀ u ) denote the set of N -tuples ⇀ n of non-negative integers ⇀ n = ( n , . . . , n N )satisfying (59). For X ∈ [1 , N ] and ⇀ n ∈ M ( j, N, ⇀ u ), define degsum( X, ⇀ n ) to bedegsum( X, ⇀ n ) = X i ∈ X n i . or an r -tuple of integers ⇀ m = ( m i ) ri =1 denote the multinomial coefficientmult( ⇀ m ) mult( ⇀ m ) = ( P ri =1 m i )! Q ri =1 m i ! . Note that mult( ⇀ m ) is independent of the ordering of ⇀ m .Expand the left side of equation (57) to obtain X ⇀ n ∈ M ( j,N, ⇀ u ) X τ ∈ T ( N − j,k ) X s ∈ S ( N,N − j ) ( s, τ, ⇀ n, ⇀ u ) k Y i =1 (cid:18) | s τ ( i ) | − u i (cid:19)! N − j Y i =1 mult(( n x ) x ∈ s i ) ! N Y i =1 f ( n i ) i (61)where ( s, τ, ⇀ n, ⇀ u ) = 1 if the following conditions are satisfied | s τ ( i ) | − − u i = degsum( s τ ( i ) , ⇀ n ) for 1 ≤ i ≤ k | s i | − s i , ⇀ n ) for i / ∈ τ (62)and ( s, τ, ⇀ n, ⇀ u ) = 0 otherwise.For an integer l, ≤ l ≤ N − v with 0 ≤ v ≤ l , let σ ∈ S ( l, v ) σ = ( σ , . . . , σ v ) . where we recall the subsets σ i are ordered such thatmin( σ i ) < min( σ i +1 ) . For a τ ∈ T ( N − j, k ), suppose τ ( h ) ≤ vτ ( h + 1) > v. Then we write τ as ( τ , τ )where τ ∈ T ( v, h ) and τ ⊂ [ v + 1 , N ] and | τ | = k − h . For a σ ∈ S ( l, v ), let S ( N, N − j ; σ ) ⊂ S ( N, N − j ) be the subset consisting of those s such that v ≤ length( s )and σ i ⊂ s i for 1 ≤ i ≤ v. Now if s ∈ S ( N, N − j ; σ ) , ( s, τ, ⇀ n, ⇀ u ) = 0, and τ = ( τ , τ ) as above, thenthe quantity k Y i =1 (cid:18) | s τ ( i ) | − u i (cid:19) is equal to h Y i =1 (cid:18) u i + degsum( σ τ ( i ) , ⇀ n ) u i (cid:19) nd depends only on σ, τ and ⇀ n .Define M ( j, N, ⇀ u ; l ) ⊂ M ( j, N, ⇀ u ) to be the subset consisting of those ⇀ n such that n i = 0 for i > l . In expression (61) consider the sub-sum X ⇀ n ∈ M ( j,N, ⇀ u ; l ) N Y i =1 f ( n i ) i ( t ) ! l X v =1 X σ ∈ S ( l,v ) v Y i =1 mult( { n x } x ∈ σ i ) ! v X h =1 X τ ∈ T ( v,h ) h Y i =1 (cid:18) u i + degsum( σ τ ( i ) , ⇀ n ) u i (cid:19)! X τ ⊂ [ h +1 ,V ] , | τ | = k − h X s ∈ S ( N,N − j ; σ ) ( s, ( τ , τ ) , ⇀ n, ⇀ u ) . (63)We claim that the innermost sum evaluates to X τ ⊂ [ h +1 ,V ] , | τ | = k − h X s ∈ S ( N,N − j ; σ ) ( s, ( τ , τ ) , ⇀ n, ⇀ u )= (cid:18) N − lj + v + k − h − l (cid:19)(cid:18) j + v + k − h − lj + v − l − P ki = h +1 u i (cid:19) mult( ǫ ) k Y i = h +1 (cid:18) k − i + P kg = i u g u i (cid:19) where ǫ is the v -tuple ǫ = ((degsum( σ τ ( i ) ) + u i + 1 − | σ τ ( i ) | ) hi =1 , (degsum( σ i ) + 1 − | σ i | ) i/ ∈ τ ,i ≤ v ) . We prove this claim now by constructing all pairs ( s, τ ) that have a non-zerocontribution to the sum. No matter what the choices of s and τ are, theconditions (62) imply | v [ i =1 s i ! ∪ k − h [ i =1 s τ ( i ) ! | = l X i =1 n i + k X i =1 u i + v + k − h (64)= j + v + k − h. By construction the elements 1 , . . . , l are in v [ i =1 s i , so we have to choose j + v + k − h − l elements from a set of order N − l tobe the remaining elements in the union (64). Let Z denote the set of these j + v + k − h − l elements. Next, from Z we must choose j + v − l − h X i =1 u i elements to fill the s i , ≤ i ≤ v , and there are mult( ǫ ) ways to do this. Nowfrom the set Z consisting of k − h + k − h X g =1 u g + h lements not chosen yet, we construct k − h sets w i , ≤ i ≤ k − h . We place theelement min( Z ) in w and choose u h +1 other elements to be in w . Proceedingin this manner, from a set Z i of order k − h − i + 1 + k − h X g = i u g + h , we construct w i to consist of min( Z i ) and a choice of u i + h other elements from Z i . Now let s be the partition consisting of the sets s i , for 1 ≤ i ≤ vw i , for 1 ≤ i ≤ k − h and singleton sets, and let τ be determined by τ ( i ) = i ′ where s i ′ = w i . This proves the claim.Putting this claim into the expression (63) and simplifying, we get that thecoefficient of N Y i =1 f ( n i ) i ( t ) for ⇀ n ∈ M ( j, N, ⇀ u ; l ) is1 Q ki =1 u i ! Q li =1 n i ! l X v =1 X σ ∈ S ( l,v ) v X h =1 X τ ∈ T ( v,h ) (cid:18) N − lj + v + k − h − l (cid:19) × ( j + v + k − h − l )! Q k − hi =1 ( k − h − i + 1 + P kg = h + i u g ) Y ( σ, τ , ⇀ n, ⇀ u ) (65)where Y ( σ, τ , ⇀ n, ⇀ u ) = | τ | Y i =1 ( u i +degsum( σ τ ( i ) )) | σ τ i ) |− Y i/ ∈ τ , ≤ i ≤ length( σ ) (degsum( σ i )) | σ i |− . The corresponding coefficient on the right side of equation (57) is C ( j, N, ⇀ u )mult( ⇀ n )which simplifies to1 Q ki =1 u i ! Q li =1 n i ! (cid:18) N − j + k − (cid:19) ( N − j + P ki =1 u i )( j + k − Q ki =1 ( k − i + 1 + P kg = i u g ) . (66) quating expressions (65) and (66) yields the equation l X v =1 X σ ∈ S ( l,v ) v X h =1 X τ ∈ T ( v,h ) (cid:18) N − lj + v + k − h − l (cid:19) ( j + v + k − h − l )! h Y i =1 ( k − i + 1 + k X g = i u g ) × Y ( σ, τ , ⇀ n, ⇀ u )= (cid:18) N − j + k − (cid:19) ( N − j + k X i =1 u i )( j + k − . (67)We prove equation (67) by comparing coefficients in the basis of the binomialfunctions in N . To express the right side of equation (67) in that basis, we applythe following identities for any non-negative integers n and d (cid:18) xn (cid:19) ( x − n ) = (cid:18) xn + 1 (cid:19) ( n + 1)and (cid:18) xn (cid:19) = d X i =0 (cid:18) di (cid:19)(cid:18) x − dn − i (cid:19) . Using x = N − , n = j + k − , and d = l −
1, we obtain that the coefficient of (cid:18) N − lj + k − a (cid:19) (68)where 0 ≤ a ≤ l is( j + k − ( j + k ) (cid:18) l − a (cid:19) + ( k + k X i =1 u k ) (cid:18) l − a − (cid:19)! . On the left side of equation (67), we have a contribution to the coefficient of thebinomial (68) when v − h = l − a . The total coefficient is thus( j + k − a )! l X v = l − a a + v − l Y i =1 ( k − i + 1 + k X g = i u g ) X τ ∈ T ( v,a + v − l ) X σ ∈ S ( l,v ) Y ( σ, τ , ⇀ n, ⇀ u ) . (69)We equate these two coefficients, divide by ( j + k − a )! and define X by X = k + k X i =1 u i and U ( n, ⇀ u ) by U ( n, ⇀ u ) = n Y i =1 ( X − i + 1 − i − X g =1 u g ) . e now view X, u i and n i as indeterminates and see that the induction step isimplied by the following polynomial identity, for all pairs of integers 0 ≤ a ≤ l : l X v = l − a U ( v − l + a, ⇀ u ) X σ ∈ S ( l,v ) X τ ∈ T ( v,v − l + a ) Y ( σ, τ , ⇀ n, ⇀ u )= (cid:18) l − a (cid:19) ( X + l X i =1 n i ) a + X (cid:18) l − a − (cid:19) ( X − l X i =1 n i ) a − . (70)By the induction hypothesis and Lemma 6.6, we have that the left side isequal to l X v = l − a ( l − v ( l − a )!( l − v )!( v − l + a )! ( X ) v − l + a ( l X i =1 n i ) l − v . Re-index this sum by v = l − a + r to obtain a X r =0 ( l − l − a + r )( l − a )!( a − r )! r ! ( X ) r ( l X i =1 n i ) a − r , which is equal to (cid:18) l − a (cid:19) a X r =0 (cid:18) ar (cid:19) ( X ) r ( l X i =1 n i ) a − r + (cid:18) l − a − (cid:19) a X r =0 (cid:18) a − r − (cid:19) ( X ) r ( l X i =1 n i ) a − r . The second sum is equal to X a − X r ′ =0 ( X − r ′ ( l X i =1 n i ) a − − r ′ where have re-indexed r = r ′ + 1. Now apply identity (6.1) to obtain the rightside of (70). This completes the induction step and the proof.Now we can prove Theorem 6.3. Proof.
Apply Theorem 6.4 with k = 0, R the ring of polynomials in t , δ differ-entiation with respect to t , and f i = t x i where x i is any non-negative integer,and then set t = 1. This shows that both sides of equation (55) are equal when x i are non-negative integers, and since both sides are polynomials in x i , theymust be equal as polynomials. This completes the proof. Lemma 6.5.
Assume Theorem 6.4 is true for all values of N less than some N .For any N -tuple ⇀ f of elements in R , the expression F ( j, ⇀ f , ⇀ u ) is symmetricin the f i . That is, if ⇀ g is a reordering of the N -tuple ⇀ f , then F ( j, ⇀ f , ⇀ u , ) = F ( j, ⇀ g , ⇀ u , ) . roof. It is sufficient to prove that F ( j, ⇀ f , ⇀ u , ) is invariant under each transpo-sition f r ( t ) ↔ f r +1 ( t ). We fix an r and consider a set partition s in the sum(56). If r and r + 1 are in the same subset for this s , then the contribution isinvariant under r ↔ r + 1. Therefore suppose that r ∈ s a and r + 1 ∈ s b forsome a = b . If at least one the following r = min( s a ) or r + 1 = min( s b )holds, then let s ′ denote the set partition obtained from s by switching r and r + 1. Then s ′ a = { r + 1 } ∪ ( s a \ { r } ) s ′ b = { r } ∪ ( s b \ { r + 1 } ) , so the expression D ( s, τ, j, ⇀ f , ⇀ u ) + D ( s ′ , τ, j, ⇀ f , ⇀ u ) (71)is invariant under r ↔ r + 1.Now suppose that both r = min( s a ) and r + 1 = min( s b ) (72)(so b = a + 1) and that both { a, a + 1 } / ∈ τ . Then the expression (71) is againinvariant under r ↔ r + 1. If a ∈ τ and a + 1 / ∈ τ , then let τ ′ be the functionobtained from τ by making τ ′ ( τ − ( a )) = a + 1. Then D ( s, τ, j, ⇀ f , ⇀ u ) + D ( s ′ , τ ′ , j, ⇀ f , ⇀ u ) (73)invariant under r ↔ r + 1. The case a + 1 ∈ τ and a / ∈ τ is handled similarly.Thus suppose equations (72) hold and that both a and a + 1 ∈ τ . Now if u τ − ( a ) = u τ − ( a +1) , then the expression (73) is not invariant under r ↔ r +1, sowe prove invariance another way: given integers n r and n r +1 , in the expansionof F ( j, ⇀ f , ⇀ u ), we prove that the coefficient of f ( n r ) r f ( n r +1 ) r +1 is equal to the coefficient of f ( n r +1 ) r f ( n r ) r +1 . For an s satisfying condition (72), let E ( s, a ) denote the set of set partitions s ′ obtained from s such that s ′ a ∪ s ′ a +1 = s a ∪ s a +1 s ′ i = s i for i = a, a + 1 r ∈ s ′ a and r + 1 ∈ s ′ a +1 . We claim the expression X s ′ ∈ E ( s,a ) D ( s ′ , τ, j, ⇀ f , ⇀ u ) s invariant under f r ↔ f r +1 . In the expansion of this sum, consider a term ofthe form m ( ⇀ f , ⇀ n, I ) = Y i ∈ I f ( n i ) i Y i/ ∈ I f i where I ⊂ ( s a ∪ s a +1 ) such that r, r + 1 ∈ I, and ⇀ n = ( n i ) i ∈ I is some | I | -tuple of integers where n i > i = r, r + 1. Denote τ − ( a ) by y .Now D ( s ′ , τ, j, ⇀ f , ⇀ u ) has a non-zero contribution to the coefficient of m ( ⇀ f , ⇀ n, I )only if | s ′ a | − − u y = X i ∈ I ∩ s ′ a n i | s ′ a +1 | − − u y +1 = X i ∈ I ∩ s ′ a +1 n i .. (74)Adding the above two equations gives | s ′ a ∪ s ′ a +1 | − − u y − u y +1 = X i ∈ I n i . Since at least one of n r and n r +1 is greater than zero by assumption, and theremaining n i are positive by assumption, we therefore have the the coefficientof m ( ⇀ f , ⇀ n, I ) can be non-zero only if | s ′ a ∪ s ′ a +1 | − − u y − u y +1 ≥ | I | − N ≥ | s ′ a ∪ s ′ a +1 | > | I | . Define ( s ′ , τ, ⇀ u , ⇀ n, I ) = 1if the conditions (74) are satisfied and 0 otherwise. The coefficient of m ( ⇀ f , ⇀ n, I )is thus B ( s, τ, ⇀ u ) X s ′ ∈ E ( s,a ) (cid:18) | s ′ a | − u y (cid:19)(cid:18) | s ′ a +1 | − u y +1 (cid:19) mult(( n i ) i ∈ I ∩ s ′ a )mult(( n i ) i ∈ I ∩ s ′ a +1 ) ( s ′ , τ, ⇀ u , ⇀ n, I ) . (75)where B ( s, τ, ⇀ u ) is some number depending only on s, τ and u .Now let S ( I, r, r + 1) denote the set of set partitions of I into two subsetssuch that r and r + 1 are not in the same set. It follows that the sum in xpression (75) is X σ ∈ S ( I, r,r +1) (cid:18) u y + degsum( σ , ⇀ n ) u y (cid:19)(cid:18) u y +1 + degsum( σ , ⇀ n ) u y +1 (cid:19) × mult(degsum( σ , ⇀ n ) + u y + 1 − | σ | , degsum( σ , ⇀ n ) + u y +1 + 1 − | σ | ) × mult(( n i ) i ∈ σ )mult(( n i ) i ∈ σ )which simplifies to( P i ∈ I n i + u y + u y +1 + 2 − | I | )! u y ! u y +1 ! Q i ∈ I n i ! X σ ∈ S ( I, r,r +1) ( u y +degsum( σ , ⇀ n )) | σ |− ( u y +1 +degsum( σ , ⇀ n )) | σ |− . The number multiplying this sum is clearly invariant under n r ↔ n r +1 , so wemust prove that the sum is also invariant. This sum is equal to X σ ∈ S ( I, (degsum( σ , ⇀ µ )) | σ |− (degsum( σ , ⇀ µ )) | σ |− (76) − X σ ∈ S ( l, \ S ( I, r,r +1) (degsum( σ , ⇀ µ )) | σ |− ( u y +1 + degsum( σ , ⇀ µ )) | σ |− (77)where ⇀ µ is obtained from ⇀ n by setting µ r = u y + n r µ r +1 = u y +1 + n r +1 µ i = n i for all other i ∈ I. The sum at line (77) is invariant under n r ↔ n r +1 because r and r + 1 arealways in the same subset. The sum at line (76) is equal to F ( | I | − , ⇀ g , ∅ ) | t =1 (78)where R is the ring of polynomials in t , δ is differentiation with respect to t , g = t n r + u y , g = t n r +1 + u y +1 , and g i = t n i ′ where i ′ is the i -th smallest elementof I , for i >
2. Since | I | < N , we may apply the assumption on Theorem 6.4to see that expression (78) is equal to (cid:18) | I || I | − (cid:19) ( u y + u y +1 + X i ∈ I n i ) | I |− which is invariant under n r ↔ n r +1 . This completes the proof. Lemma 6.6.
Let l be a positive integer, and assume Theorem 6.4 is true for N = l . Then with notation as in the proof of that theorem, the polynomial l X v = l − a U ( v − l + a, ⇀ u ) X σ ∈ S ( l,v ) X τ ∈ T ( v,v − l + a ) Y ( σ, τ , ⇀ n, ⇀ u ) (79) is equal to l X v = l − a ( l − v ( l − a )!( l − v )!( v − l + a )! ( X ) v − l + a ( l X i =1 n i ) l − v . (80) roof. We first prove that the polynomial (79) is constant in each variable u i .Applying identity (6.1) to each factor of( u i + degsum( σ τ ( i ) ) | σ τ ( i ) |− in Y ( σ, τ, ⇀ n, ⇀ u ), we obtain X τ ∈ T ( v,v − l + a ) X σ ∈ S ( l,v ) Y ( σ, τ , ⇀ n, ⇀ u )= l − v X b =0 X ⇀ c ∈ C ( l − v − b,v − l + a ) v − l + a Y i =1 ( u i ) c i X τ ∈ T ( v,v − l + a ) X σ ∈ S ( l,v ) v − l + a Y i =1 (cid:18) | σ τ ( i ) | − c i (cid:19) (degsum( σ τ ( i ) )) | σ τ i ) |− − c i Y i/ ∈ τ , ≤ i ≤ v (degsum( σ i )) | σ i |− (81)where the sum is over the set C ( l − v − b, v − l + a ) of compositions ⇀ c of l − v − b into v − l + a non-negative parts; that is, ⇀ c = ( c , . . . , c v − l + a ) and v − l + a X i =1 c i = l − v − b with c i ≥ . But the last line of (81) is F ( l − v, ⇀ f , ⇀ c ) | t =1 where R is the ring of polynomials in t , δ is differentiation with respect to t ,and f i = t n i , ≤ i ≤ l . Evaluating this using the assumption that Theorem 6.4is true for N = l thus gives that the right side of equation (81) is equal to l − v X b =0 X ⇀ c ∈ C ( l − v − b,v − l + a ) ( l − l − b )( l − a )! b ! Q v − l + ai =1 ( v − l + a − i + 1 + P v − l + ag = i c g ) v − l + a Y i =1 (cid:18) u i c i (cid:19) ( l X i =1 n i ) b Therefore we must prove that( l − l − a )! l X v = l − a U ( v − l + a, ⇀ u ) l − v X b =0 ( l − b ) b ! ( l X i =1 n i ) b X ⇀ c ∈ C ( l − v − b,v − l + a ) Q v − l + ai =1 ( v − l + a − i + 1 + P v − l + ag = i c g ) v − l + a Y i =1 (cid:18) u i c i (cid:19) (82)is constant in each u i . Note that in the above expression the indeterminate u a − r appears only in terms with l − r ≤ v ≤ l . e prove that is constant u a − , u a − , . . . , u a − r for an 1 ≤ r ≤ a −
1. We useinduction on r with base case r = 1. Now in expression (82), u a − appears onlyin terms for v = l and b = 0 as U ( a, ⇀ u ) (cid:18) la (cid:19) and in v = l − , b = 0 as U ( a − , ⇀ u ) (cid:18) la (cid:19)(cid:18) u a − (cid:19) for the composition ⇀ c with c i = 0 , ≤ i ≤ a − c a − = 1. Then u a − cancels out in the sum of these two expressions. This proves the base case.Now assume that expression (82) is constant in u a − , . . . , u a − r for some1 ≤ r ≤ a −
2. We thus may assume that u i = 0 for a − r ≤ i ≤ a − c i = 0 for a − r ≤ i ≤ length( ⇀ c ) havea non-zero contribution. Fix a b , ≤ b ≤ a and some non-negative integers γ , . . . , γ a − r − such that r + 1 − b − a − r − X i =1 γ i ≥ d . Now consider theterms in expression (82) with b = b and a composition ⇀ c satisfying c i = γ i , for 1 ≤ i ≤ a − r − c i = 0 , for a − r ≤ i ≤ length( ⇀ c ) . In these terms we must have l − v − b = a − r − X i =1 γ i + c a − r − , so the quantity c a − r − + v is fixed. This implies v ≤ l − b − a − r − X i =1 γ i . And since the u a − r − appears only in terms with v ≥ l − r −
1, we have that v takes on the d + 1 values l − r − ≤ v ≤ l − b − a − r − X i =1 γ i . For each of these values of v , let c ( v ) denote the unique composition that satisfiesthe above requirements. We thus obtain l − r − d X v = l − r − c = c ( v ) U ( v − l + a, ⇀ u ) Q v − l + ai =1 ( v − l + a − i + 1 + P v − l + ag = i c g ) v − l + a Y i =1 (cid:18) u i c i (cid:19) = U ( a − r − , ⇀ u ) Q a − r − i =1 (cid:0) u i γ i (cid:1)Q a − r − i =1 ( a − r − − i + 1 + P a − r − g = i γ g ) 1 d ! d X i =0 (cid:18) di (cid:19) ( X − ( a − r ) + 1 − a − r − X g =1 u g ) i ( u a − r − ) d − i . ow apply identity (6.1) to the sum on the right side to see that it is constantin u a − r − . This completes the induction step.Now evaluate expression (82) with u i = 0. The only non-zero contributionscome from the terms with compositions of 0, so l − v − b = 0. This yieldsexpression (80) and completes the proof. • Study the corresponding Taylor series coefficients when g ( z ) has more thantwo terms. • Use the Taylor series to derive the solutions by radicals and see if radicalsolutions using infinite series can be obtained for higher degree. • Prove a formal factorization of polynomials using these series. • For integer β >
1, interpret these coefficients as counting some kind of treewhich generalizes trees with negative vertex degree. • Prove Theorem 6.4 using induction as used in the proof of Theorem 4.3. • See if Theorem 4.3 can be generalized using τ -sequences as in Theorem6.4 or using more elements f A , f B , f C . . . . • Find an NRS-type algorithm that evaluates the other bracket series. • See if Theorem 6.4 can be used to give a proof for any β . • In the case β = 1 find an algebraic proof of formal zeros in terms of treeswith negative vertex degree. References
8] I.M. Gel’fand, A.V. Zelevinsky, and M.M. Kapranov, “Generalized Eulerintegrals and A A -hypergeometricseries”. Discrete Math. 210 (2000) pp. 171-181.-hypergeometricseries”. Discrete Math. 210 (2000) pp. 171-181.