An 4n-point Interpolation Formula for Certain Polynomials
aa r X i v : . [ m a t h . C O ] S e p An n -point Interpolation Formula forCertain Polynomials Sandy H.L. Chen , Amy M. Fu Center for Combinatorics, LPMC-TJKLCNankai University, Tianjin 300071, P.R. China [email protected], [email protected] Abstract
By using some techniques of the divided difference operators, we establish an4 n -point interpolation formula. Certain polynomials, such as Jackson’s φ terminatingsummation formula, are special cases of this formula. Based on Krattenthaler’s identity,we also give Jackson’s formula a determinantal interpretation. Recall that the i -th divided difference operator ∂ i , acting on functions f ( x , . . . , x n ) ofseveral variables, is defined by f ( x , . . . , x i , x i +1 , . . . ) ∂ i = f ( x , . . . , x i , x i +1 , . . . ) − f ( x , . . . , x i +1 , x i , . . . )( x i − x i +1 ) . To be more general, we introduce the operator c ∂ i which we call the i -th c -divideddifference operator : f ( x , . . . , x i , x i +1 , . . . ) c ∂ i = f ( x , . . . , x i , x i +1 , . . . ) − f ( x , . . . , x i +1 , x i , . . . )( x i − x i +1 )(1 − c/x i x i +1 ) . (1.1)Note that ∂ i = ∂ i .Several properties of the c -divided difference operators will be given in next section,see Lemmas 2.1 to 2.5. Employing those lemmas, we obtain the main result of thispaper. Theorem 1.1
Given two sets of variables A = { a , c/a , . . . , a n , c/a n } , B = { b , c/b , . . . , b n , c/b n } , we have the following n -point interpolation formula for certain polynomials f ( y ) ofdegree n with symmetry y − n f ( y ) = ( c/y ) − n f ( c/y ) when c = 0 : f ( y ) = f ( b ) Q ni =1 ( b − a i )( b − c/a i ) n Y i =1 ( y − a i )( y − c/a i )+ f ( a ) Q ni =1 ( a − b i )( a − c/b i ) n Y i =1 ( y − b i )( y − c/b i )+ n − X j =1 C j · j Y i =1 ( y − b i )( y − c/b i ) n − j Y i =1 ( y − a i )( y − c/a i ) , (1.2)1 here C j = f ( b ) b − j Q n − j +1 i =1 ( b − a i )( b − c/a i ) c ∂ · · · c ∂ j ( b j +1 − a n − j +1 )(1 − c/a n − j +1 b j +1 ) . Note that Theorem 1.1 leads to the following 2 n -point interpolation formula givenin [1] if c = 0: f ( y ) = f ( b ) Q ni =1 ( b − a i ) n Y i =1 ( y − a i ) + f ( a ) Q ni =1 ( a − b i ) n Y i =1 ( y − b i )+ n − X j =1 f ( b ) Q n − j +1 i =1 ( b − a i ) ∂ · · · ∂ j ( b j +1 − a n − j +1 ) · j Y i =1 ( y − b i ) n − j Y i =1 ( y − a i ) . The symmetry y − n f ( y ) = ( c/y ) − n f ( c/y ) when c = 0 implies that f ( y ) can bewritten as a product Q ni =1 ( y − x i )( c − x i y ). Considering the case A = { a, c/a, . . . , aq − n , cq n − /a } , B = { b, c/b, . . . , bq − n , cq n − /b } and y = x n +1 , one can check that Theorem 1.1 implies the following identity by ex-panding the determinant with respect to the last row:det (cid:0) P n − j +1 ( x i , aq j − n ) P n − j +1 ( x i , c/a ) P j − ( x i , bq − n ) P j − ( x i , cq n − j +1 /b ) (cid:1) n +1 i,j =1 = Y ≤ i
Clearly, y − n p n ( y ) can be rewritten as Q ni =1 ( y − x )(1 − c/y x ). When m = 1and n = 0 or n = 1, it is easy to verify that (2.1) holds. In view of Lemma 2.1, we canprove Lemma 2.3 by induction on the length of the operators. Lemma 2.4
We have j Y k =1 ( y − b k )(1 − c/y b k ) c ∂ · · · c ∂ i (cid:12)(cid:12)(cid:12) y k = b k , ≤ k ≤ i +1 = (cid:26) , j = i ;1 , j = i. (2.2) Proof.
For j ≤ i , Lemma 2.4 is a direct consequence of Lemma 2.3. For j > i , we have j Y k =1 ( y − b k )(1 − c/y b k ) c ∂ · · · c ∂ i (cid:12)(cid:12)(cid:12) y k = b k , ≤ k ≤ i +1 = j Y k =2 ( y − b k )(1 − c/y b k ) c ∂ · · · c ∂ i (cid:12)(cid:12)(cid:12) y k = b k , ≤ k ≤ i +1 = · · · = j Y k = i +1 ( y i +1 − b k )(1 − c/y i +1 b k ) (cid:12)(cid:12)(cid:12) y k = b k ,i +1 ≤ k ≤ i +1 = 0 . We complete the proof.
Lemma 2.5
We have j Y k =1 ( y − b k )(1 − c/y b k )( y − a k )(1 − c/y a k ) i − Y k =1 ( y − a k )(1 − c/y a k ) c ∂ · · · c ∂ i (cid:12)(cid:12)(cid:12) y k = b k , ≤ k ≤ i +1 = ( , j = i ; b j +1 − a j )(1 − c/a j b j +1 ) , j = i. (2.3) Proof.
For j < i , we have j Y k =1 ( y − b k )(1 − c/y b k ) i − Y k = j +1 ( y − a k )(1 − c/y a k ) c ∂ · · · c ∂ i (cid:12)(cid:12)(cid:12) y k = b k , ≤ k ≤ i +1 = i X l =0 j Y k =1 ( y − b k )(1 − c/y b k ) c ∂ · · · c ∂ l (cid:12)(cid:12)(cid:12) y k = b k , ≤ k ≤ l +1 × i − Y k = l +1 ( y l +1 − a k )(1 − c/y l +1 a k ) c ∂ l +1 · · · c ∂ i (cid:12)(cid:12)(cid:12) y k = b k ,l +1 ≤ k ≤ i +1 . j ≥ i , we have Q jk =1 ( y − b k )(1 − c/y b k ) Q jk = i ( y − a k )(1 − c/y a k ) c ∂ · · · c ∂ i (cid:12)(cid:12)(cid:12) y k = b k , ≤ k ≤ i +1 = ( y − b )(1 − c/yb ) c ∂ (cid:12)(cid:12)(cid:12) y = b Q jk =2 ( y − b k )(1 − c/y b k ) Q jk = i ( y − a k )(1 − c/y a k ) c ∂ · · · c ∂ i (cid:12)(cid:12)(cid:12) y k = b k , ≤ k ≤ i +1 = Q jk =2 ( y − b k )(1 − c/y b k ) Q jk = i ( y − a k )(1 − c/y a k ) c ∂ · · · c ∂ i (cid:12)(cid:12)(cid:12) y k = b k , ≤ k ≤ i +1 = · · · = Q jk = i ( y i − b k )(1 − c/y i b k ) Q jk = i ( y i − a k )(1 − c/y i a k ) c ∂ i (cid:12)(cid:12)(cid:12) y k = b k ,i ≤ k ≤ i +1 = (cid:26) , j > i, / ( b j +1 − a j )(1 − c/a j b j +1 ) , j = i. We complete the proof.
Proof of Theorem 1.1
Given a polynomial f ( y ) of degree 2 n with symmetry y − n f ( y ) = ( c/y ) − n f ( c/y ), weassume that f ( y ) = n X j =0 C j j Y k =1 ( y − b k )( y − c/b k ) n Y k = j +1 ( y − a k )( y − c/a k ) . (2.4)Taking y = b in (2.4), one has f ( b ) = C n Y k =1 ( b − a k )( b − c/a k ) . Therefore, C = f ( b ) Q nk =1 ( b − a k )( b − c/a k ) . Setting y = a n in (2.4) leads to C n = f ( a n ) Q nk =1 ( a n − b k )( a n − c/b k ) . Let g ( y ) = f ( y ) / Q nk =1 ( y − a k )( y − c/a k ). Rewrite (2.4) as g ( y ) = g ( b ) + n − X j =1 C j Q jk =1 ( y − b k )(1 − c/yb k ) Q jk =1 ( y − a k )(1 − c/ya k )+ f ( a n ) Q ni =1 ( a n − b i )( a n − c/b i ) n Y i =1 ( y − b k )(1 − c/yb k )( y − a k )(1 − c/ya k ) . Q i − k =1 ( y − a k )(1 − c/ya k ), then applying the operator c ∂ · · · c ∂ i , one has g ( y ) i − Y k =1 ( y − a k )(1 − c/y a k ) c ∂ · · · c ∂ i (cid:12)(cid:12)(cid:12) y k = b k , ≤ k ≤ i +1 = i − X j =1 C j j Y k =1 ( y − b k )(1 − c/y b k ) i − Y k = j +1 ( y − a k )(1 − c/y a k ) c ∂ · · · c ∂ i (cid:12)(cid:12)(cid:12) y k = b k , ≤ k ≤ i +1 + n − X j = i C j Q jk =1 ( y − b k )(1 − c/y b k ) Q jk = i ( y − a k )(1 − c/y a k ) c ∂ · · · c ∂ i (cid:12)(cid:12)(cid:12) y k = b k , ≤ k ≤ i +1 . By Lemma 2.5, we have g ( y ) i − Y k =1 ( y − a k )(1 − c/y a k ) c ∂ · · · c ∂ i (cid:12)(cid:12)(cid:12) y k = b k , ≤ k ≤ i +1 = C i ( b i +1 − a i )(1 − c/b i +1 a i ) . Thus C i = g ( y ) i − Y k =1 ( y − a k )(1 − c/y a k ) c ∂ · · · c ∂ i (cid:12)(cid:12)(cid:12) y k = b k , ≤ k ≤ i +1 ( b i +1 − a i )(1 − c/b i +1 a i )= f ( b ) b − i Q nk = i ( b − a k )( b − c/a k ) c ∂ · · · c ∂ i ( b i +1 − a i )(1 − c/b i +1 a i ) . Replacing a i by a n − i +1 , we complete the proof. φ terminating summation formula Letting x i = uq i − for 1 ≤ i ≤ n and x n +1 = y , we shall show that (1.3) in this caseis equivalent to Corollary 1.2. In other words, we shall give Jackson’s φ terminatingsummation formula a determinantal interpretation.Our proofs in this section involve the following well-known symmetric functions.Given two sets of variables X and Y , the i -th supersymmetric complete function h i ( X − Y ) is defined by h i ( X − Y ) = [ t i ] Q y ∈ Y (1 − yt ) Q x ∈ X (1 − xt ) = n X i =0 ( − i e i ( Y ) h n − i ( X ) , (3.1)where [ t i ] f ( t ) means the coefficient of t i in f ( t ), e i ( X ) and h i ( Y ) are i -th elementarysymmetric function and i -th complete symmetric function, respectively.6xpanding the determinant of (1.3) along the last row in the case x i = uq i − for1 ≤ i ≤ n and x n +1 = y , we have n Y i =1 ( uq i − − y )( c − uq i − y ) b ( n +12 ) q − ( n +1) n ( n − / × Y ≤ i For ≤ k ≤ n + 1 , we have F n,k ( U, A, B ) = (cid:20) nk − (cid:21) q − n ( n − n − b ( n ) Y ≤ i Lemma 3.3 For ≤ k ≤ n + 1 , we have det( e n − i +1 ( Y j,k )) ni,j =1 = (cid:20) nk − (cid:21) c ( n +12 ) q ( n − k +12 ) − n ( n − n − n − Y i =1 ( b − aq i − ) n − i n − Y j =0 j = n − k +1 n Y i = j +1 i = n − k +1 (1 − cq i + j − /ab ) , (3.5) where the sets of n variables Y j,k , ≤ j ≤ n , are defined as above.Proof. Obviously, e n ( Y j,k ) = c n . We shall use induction on n . When n = 1, we havedet( e ( a, c/a )) = det( e ( b, c/b )) = c. Thus (3.5) is true for n = 1. Assume that (3.5) holds for 1 ≤ m ≤ n − 1, where n ≥ m = n .When k = 1, the substraction of two successive columns of the determinant gives e i ( Y j, ) − e i ( Y j − , ) = q j − n ( b − a )(1 − cq n − j /ab ) e i − ( Y ′ j − , ) , where Y ′ j − , = Y j, \ { bq j − n , cq n − j /b } .It is easy to verify thatdet( e n − i +1 ( Y j, )) ni,j =1 = c ( n +12 ) q ( n ) − n ( n − n − n − Y i =1 ( b − aq i − ) n − i n − Y j =0 n Y i = j +1 (1 − cq i + j − /ab ) , which is equal to the right side of (3.5). The case k = n + 1 is similar.We now consider the case 2 ≤ k ≤ n . According to e i ( Y k,k ) − e i ( Y k − ,k )= q k − n ( b − a )(1 − cq n − k /ab ) e i − ( Y ′ k − ,k − )+ q k − n − ( b − a )(1 − cq n − k +2 /ab ) e i − ( Y ′ k − ,k ) , we havedet( e n − i +1 ( Y j,k )) = c n q − ( n )( b − a ) n − n − Y i =0 (1 − cq i /ab ) × det( e n − i − ( Y ′ j,k − )) q n − k − cq n − k /ab + det( e n − i − ( Y ′ j,k )) q n − k +1 − cq n − k +2 /ab ! , (3.6)8here for i = k − i = k , we have Y ′ j,i = (cid:26) Y j,k \ { aq j − n , cq n − j /a } , ≤ j < i,Y j +1 ,k \ { bq j +1 − n , cq n − j − /b } , i ≤ j ≤ n − . Our induction hypothesis implies thatdet( e n − i − ( Y ′ j,k ) = (cid:20) n − k − (cid:21) c ( n ) q ( n − k ) − ( n − n − n − × n − Y i =1 ( bq − − aq i − ) n − i − n − Y j =0 j = n − k n − Y i = j +1 i = n − k (1 − cq i + j /ab ) , anddet( e n − i − ( Y ′ j,k − ) = (cid:20) n − k − (cid:21) c ( n ) q ( n − k +12 ) − ( n − n − n − × n − Y i =1 ( bq − − aq i − ) n − i − n − Y j =0 j = n − k +1 n − Y i = j +1 i = n − k +1 (1 − cq i + j /ab ) . Therefore,det( e n − i +1 ( Y j,k )) ni,j =1 = c ( n +12 ) q ( n − k +12 ) − n ( n − n − n − Y i =1 ( b − aq i − ) n − i n − Y j =0 j = n − k +1 n Y i = j +1 i = n − k +1 (1 − cq i + j − /ab ) × (cid:20) n − k − (cid:21) − cq n − k /ab − cq n − k +1 /ab + (cid:20) n − k − (cid:21) − cq n − k /ab − cq n − k +1 /ab q n − k +1 ! . With the aid of the following recurrence (cid:20) nk − (cid:21) = (cid:20) n − k − (cid:21) + q n − k +1 (cid:20) n − k − (cid:21) = q k − (cid:20) n − k − (cid:21) + (cid:20) n − k − (cid:21) , we complete the proof.We are now ready to complete the proof of Theorem 3.1. Proof of Theorem 3.1. View F n,k ( U, A, B ) as a polynomial in u of degree n + n with coefficients expressedin terms of the other variables. Applying Lemma 3.2, we first prove that F n,k ( U, A, B )has 2 n roots: aq − n , . . . , aq k − − n , cq − k /a, . . . , c/a, bq k − n +1 , . . . , bq − n , c/b, . . . , cq n − k /b. u = aq i − n in F n,k ( U, A, B ), where 1 ≤ i ≤ k − 1. We take D = ∅ , D = { aq } , . . . , D i − = { aq, aq , . . . , aq i − } ,D i = { aq i − n , aq, aq , . . . , aq i − } , . . . , D n − = { aq i − n , . . . , aq − , aq, aq , . . . , aq i − } , then apply Lemma 3.2.Since e k ( X ) = 0 if the cardinality of X is less than k , F n,k ( U, A, B ) can be trans-formed into a determinant whose ( i, j )-th entry is equal to 0 if( i, j ) ∈ { ( i, j ) : 1 ≤ j ≤ k − ≤ i ≤ n − k +2 , or k ≤ j ≤ n − ≤ i ≤ n − j } . Thus F n,k ( U, A, B ) | u = aq i − n = 0 for 1 ≤ i ≤ k − u = cq − i /a , where 1 ≤ i ≤ k − 1, is similar to the above if we take D = ∅ , D = { cq − /a } , . . . , D i − = { cq − i /a, . . . , cq − /a } ,D i = { cq − i /a, . . . , cq − /a, cq n − i /a } , . . . , D n − = { cq − i /a, . . . , cq − /a, cq/a, . . . , cq n − i /a } . For the cases u = bq − n − i and u = cq i − /b , where 1 ≤ i ≤ n − k + 1, we take D = ∅ , D = { bq − n } , . . . , D i − = { bq − i − n , . . . , bq − n } ,D i = { bq − i − n , . . . , bq − n , bq − i } , . . . , D n − = { bq − i − n , . . . , bq − n , bq − n , . . . , bq − i } and D = ∅ , D = { cq n /b } , . . . , D i − = { cq n /b, . . . , cq n + i − /b } ,D i = { cq i − /b, cq n /b, . . . , cq n + i − /b } , . . . , D n − = { cq i − /b, . . . , cq n − /b, cq n /b, . . . , cq n + i − /b } , respectively.In view of Lemma 3.2, F n,k ( U, A, B ) in both cases becomes a determinant whose( i, j )-th entry is equal to 0 if( i, j ) ∈ { ( i, j ) : 2 ≤ j ≤ k − ≤ i ≤ j − , or k ≤ j ≤ n and 1 ≤ i ≤ k } . Therefore F n,k ( U, A, B ) | u = bq − n − i = F n,k ( U, A, B ) | u = cq i − /b = 0 for 1 ≤ i ≤ n − k + 1.Secondly, we show that Q ≤ i 10o complete the proof, we need to determine C . Setting u = 0, then applying (3.5),we have C = ( − n − k +1 det( e n − i +1 ( − Y j,k )) ni,j =1 c ( n +12 ) q ( n − k +12 ) = (cid:20) nk − (cid:21) q − n ( n − n − × n − Y i =1 ( b − aq i − ) n − i n − Y j =0 j = n − k +1 n Y i = j +1 i = n − k +1 (1 − cq i + j − /ab ) , as desired.Putting (3.3) into (3.2), then divided both sides of (3.2) by Y ≤ i This work was supported by the PCSIRT Project of the Ministryof Education, and the National Science Foundation of China. References [1] S.H.L. Chen and A.M. Fu, A 2 n -point Interpolation Formula with Its Applicationsto q -Identities, Submitted.[2] W.Y.C. Chen, A.M. Fu and B.Y. Zhang, The homogenous qq