EXPLICIT INVERSE OF THE PASCAL MATRIX PLUS ONE
SHENG-LIANG YANG AND ZHONG-KUI LIU
Received 5 June 2005; Revised 21 September 2005; Accepted 5 December 2005
This paper presents a simple approach to invert the matrix Pn + In by applying the Euler polynomials and Bernoulli numbers, where Pn is the Pascal matrix.
Copyright © 2006 Hindawi Publishing Corporation. All rights reserved.
1. Introduction The Pascal matrix has been known since ancient times, and it arises in many different areas of mathematics. However, it has been studied carefully only recently, see [1, 3–5]. For any integer n>0, the n × n Pascal matrix Pn is defined with the binomial coefficients by ⎧⎛ ⎞ ⎪ ⎪ i − 1 ⎨⎝ ⎠ if i ≥ j ≥ 1, P i j = n( , ) ⎪ j − 1 (1.1) ⎪ ⎩0 otherwise.
−1 It is known that the n × n inverse matrix Pn is given by ⎧ ⎛ ⎞ ⎪ ⎪ i − 1 ⎨(−1)i− j ⎝ ⎠ if i ≥ j ≥ 1, P i j = n( , ) ⎪ j − 1 (1.2) ⎪ ⎩0 otherwise.
The Hadamard product A ◦ B of two matrices is the matrix obtained by coordinate- wise multiplication: (A ◦ B)(i, j) = A(i, j)B(i, j). Let Γn be the n × n lower triangular ma- trices defined by ⎧ ⎨(−1)i− j if i ≥ j ≥ 1, Γn(i, j) = (1.3) ⎩0 otherwise,
Hindawi Publishing Corporation International Journal of Mathematics and Mathematical Sciences Volume 2006, Article ID 90901, Pages 1–7 DOI 10.1155/IJMMS/2006/90901 2 Explicit inverse of the Pascal matrix plus one
−1 then the inverse of the Pascal matrix can be represented as the Hadamard product Pn = Pn ◦ Γn.Forexample,ifn = 5, then
⎛ ⎞ 10000 ⎜ ⎟ ⎜ ⎟ ⎜11000⎟ ⎜ ⎟ ⎜ ⎟ P = ⎜12100⎟, 5 ⎜ ⎟ ⎜ ⎟ ⎝13310⎠ 14641 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ 10000 10000 10000 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜−11000⎟ ⎜11000⎟ ⎜−11 0 00⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ − ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ P 1 = ⎜ 1 −21 00⎟ = ⎜12100⎟ ◦ ⎜ 1 −11 00⎟. 5 ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎝−13−310⎠ ⎝13310⎠ ⎝−11−110⎠ 1 −46−41 14641 1 −11−11 (1.4)
Now we consider the sum of the Pascal matrix and the identity matrix Pn + In,where In is the n × n identity matrix. We call Pn + In the Pascal matrix plus one simply. An interesting fact is that the inverse of Pn + In is related to Pn closely. For instance,
⎛ ⎞ 20 0 0 00 ⎜ ⎟ ⎜ ⎟ ⎜12 0 0 00⎟ ⎜ ⎟ ⎜ ⎟ ⎜12 2 0 00⎟ P I = ⎜ ⎟ 6 + 6 ⎜ ⎟, ⎜13 3 2 00⎟ ⎜ ⎟ ⎜ ⎟ ⎝14 6 4 20⎠ 1 5 10 10 5 2
⎛ ⎞ 1 ⎜ 00000⎟ ⎜ 2 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜− 1 1 ⎟ ⎜ 0000⎟ ⎜ 4 2 ⎟ ⎜ ⎟ ⎜ 2 1 ⎟ ⎜ 0 − 000⎟ − ⎜ ⎟ P I 1 = ⎜ 4 2 ⎟ 6 + 6 ⎜ ⎟ ⎜ 1 3 1 ⎟ ⎜ 0 − 00⎟ ⎜ ⎟ ⎜ 8 4 2 ⎟ ⎜ ⎟ ⎜ 4 4 1 ⎟ ⎜ 0 0 − 0⎟ ⎜ 8 4 2 ⎟ ⎝ 1 10 5 1⎠ − 0 0 − 4 8 4 2 S.-L. Yang and Z.-K. Liu 3 ⎛ ⎞ 1 ⎜ 00000⎟ ⎜ 2 ⎟ ⎜ ⎟ ⎜ ⎟ ⎛ ⎞ ⎜− 1 1 ⎟ ⎜ 0000⎟ 10 0 0 00 ⎜ 4 2 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜11 0 0 00⎟ ⎜ 1 1 ⎟ ⎜ ⎟ ⎜ − ⎟ ⎜ ⎟ ⎜ 0 000⎟ = ⎜12 1 0 00⎟ ◦ ⎜ 4 2 ⎟. ⎜ ⎟ ⎜ ⎟ ⎜13 3 1 00⎟ ⎜ 1 1 1 ⎟ ⎜ ⎟ ⎜ 0 − 00⎟ ⎝14 6 4 10⎠ ⎜ ⎟ ⎜ 8 4 2 ⎟ ⎜ ⎟ 1 5 10 10 5 1 ⎜ 1 1 1 ⎟ ⎜ 0 0 − 0 ⎟ ⎜ 8 4 2 ⎟ ⎝ 1 1 1 1 ⎠ − 0 0 − 4 8 4 2 (1.5) {a }∞ P This suggests that there may exist a sequence of constants n n=0 such that ( n + − In) 1 = Pn ◦ Δn,wherethematrixΔn is a lower triangular matrix with generic element Δn(i, j) = ai− j when i ≥ j. Aggarwala and Lamoureux [2] have showed that these con- stants are values of the Dirichlet eta function evaluated at negative integers, or more gen- erally, certain polylogarithm functions evaluated at the number −1. In this note, we will give a new simple approach to invert the matrix Pn + In by applying the Euler polyno- mials. As a result, we will show that these constants are values of the Euler polynomials evaluated at the number 0. The Euler polynomials En(x) are defined by means of the following generating func- tion (see [7]): ∞ tn etx E x = 2 n( )n et , (1.6) n=0 ! +1 ∞ E x E x tn/n = ∞ E x tn/n ∞ E x tn/n = since n=0( n( +1)+ n( ))( !) n=0 n( +1)( !) + n=0 n( )( !) et(x+1)/ et etx/ et = etx = ∞ xn tn/n ffi 2 ( +1)+2 ( +1) 2 n=0 2 ( !). Comparing the coe cients of tn/n! in this equation, we obtain
n En(x +1)+En(x) = 2x , n ≥ 0. (1.7)
The following lemmas are well known and can be found in [9], we give a short proof for the sake of completeness. Lemma 1.1. For all n ≥ 0, n n E x y = E x yn−k n( + ) k k( ) , (1.8) k= 0 n n E x = E x . n( +1) k k( ) (1.9) k=0 ∞ E x y tn/n = et(x+y)/ et = etx/ et ety = ∞ E x tn/ Proof. n=0 n( + )( !) 2 ( +1) (2 ( +1)) ( n=0 n( )( n ∞ yn tn/n = ∞ n n E x yn−k tn/n ffi !))( n=0 ( !)) n=0( k=0 k k( ) )( !). Comparing the coe cients of tn/n! in this equation, we obtain (1.8). In particular, when y = 1, we get (1.9). 4 Explicit inverse of the Pascal matrix plus one
From (1.7)and(1.9), we obtain n n 1 E x 1E x = xn n ≥ . k k( )+ n( ) , 0 (1.10) 2 k=0 2 x = E y = n n E yk If we set 0in(1.8), we get n( ) k=0 k n−k(0) , that is, n n E x = E xk n ≥ . n( ) k n−k(0) , 0 (1.11) k=0
T Let E(x)andX(x)bethen × 1matricesdefinedbyE(x) = [E0(x),E1(x),...,En−1(x)] , n− T X(x) = [1,x,...,x 1] ,andletE¯n be n × n lower triangular matrices defined by ⎧⎛ ⎞ ⎪ i − ⎪⎝ 1⎠ ⎨ Ei− j (0) if i ≥ j ≥ 1, j − E¯n(i, j) = 1 (1.12) ⎪ ⎩⎪ 0 otherwise.
Then (1.10), (1.11) can be represented as matrix equations, respectively,
1 Pn + In E(x) = X(x), 2 (1.13) E(x) = E¯nX(x).
Thus, we have