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

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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 matrixPn+Inby applying the Euler polynomials and Bernoulli numbers, wherePnis the Pascal matrix.

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 integern >0, then×nPascal matrixPnis defined with the binomial coeﬃcients by

Pn(i,j)=

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

⎛

⎝i−1 j−1

⎞

⎠ ifi≥j≥1,

0 otherwise.

(1.1)

It is known that then×ninverse matrixP⁻_n¹is given by

Pn(i,j)=

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩ (−1)ⁱ⁻^j

⎛

⎝i−1 j−1

⎞

⎠ ifi≥j≥1,

0 otherwise.

(1.2)

The Hadamard productA◦Bof two matrices is the matrix obtained by coordinate- wise multiplication: (A◦B)(i,j)=A(i,j)B(i,j). LetΓnbe then×nlower triangular matrices defined by

Γn(i,j)=

⎧⎨

⎩

(−1)ⁱ⁻^j ifi≥j≥1,

0 otherwise, (1.3)

Hindawi Publishing Corporation

International Journal of Mathematics and Mathematical Sciences Volume 2006, Article ID 90901, Pages1–7

DOI10.1155/IJMMS/2006/90901

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then the inverse of the Pascal matrix can be represented as the Hadamard productP⁻_n¹= Pn◦Γn. For example, ifn=5, then

P5=

⎛

⎜⎜

⎝

1 0 0 0 0

1 1 0 0 0

1 2 1 0 0

1 3 3 1 0

1 4 6 4 1

⎞

⎟⎟

⎠ ,

P⁻5¹=

⎛

⎜⎜

⎝

1 0 0 0 0

−1 1 0 0 0

1 −2 1 0 0

−1 3 −3 1 0

1 −4 6 −4 1

⎞

⎟⎟

⎠

=

⎛

⎜⎜

⎝

1 0 0 0 0

1 1 0 0 0

1 2 1 0 0

1 3 3 1 0

1 4 6 4 1

⎞

⎟⎟

⎠

◦

⎛

⎜⎜

⎝

1 0 0 0 0

−1 1 0 0 0

1 −1 1 0 0

−1 1 −1 1 0

1 −1 1 −1 1

⎞

⎟⎟

⎠ .

(1.4)

Now we consider the sum of the Pascal matrix and the identity matrixPn+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 ofPn+Inis related toPnclosely. For instance,

P6+I6=

⎛

⎜⎜

⎜⎝

2 0 0 0 0 0

1 2 0 0 0 0

1 2 2 0 0 0

1 3 3 2 0 0

1 4 6 4 2 0

1 5 10 10 5 2

⎞

⎟⎟

⎟⎠ ,

P6+I6−1

=

⎛

⎜⎜

⎜⎝ 1

2 0 0 0 0 0

−1 4

1

2 0 0 0 0

0 −2 4

1

2 0 0 0

1

8 0 −3

4 1

2 0 0

0 4

8 0 −4

4 1

2 0

−1

4 0 10

8 0 −5

4 1 2

⎞

⎟⎟

⎟⎠

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=

⎛

⎜⎜

⎝

1 0 0 0 0 0

1 1 0 0 0 0

1 2 1 0 0 0

1 3 3 1 0 0

1 4 6 4 1 0

1 5 10 10 5 1

⎞

⎟⎟

⎠

◦

⎛

⎜⎜

⎜⎝ 1

2 0 0 0 0 0

−1 4

1

2 0 0 0 0

0 −1 4

1

2 0 0 0

1

8 0 −1

4 1

2 0 0

0 1

8 0 −1

4 1

2 0

−1

4 0 1

8 0 −1

4 1 2

⎞

⎟⎟

⎟⎠ .

(1.5) This suggests that there may exist a sequence of constants {an}^∞_n₌0 such that (Pn+ In)⁻¹=Pn◦Δn, where the matrixΔnis a lower triangular matrix with generic element Δn(i,j)=ai−j when i≥j. Aggarwala and Lamoureux [2] have showed that these constants 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 matrixPn+In by applying the Euler polynomials. As a result, we will show that these constants are values of the Euler polynomials evaluated at the number 0.

The Euler polynomialsEn(x) are defined by means of the following generating function (see [7]):

∞ n=0

En(x)tⁿ n!⁼

2e^tx

e^t+ 1, (1.6)

since ^∞_n₌₀(En(x+ 1) +En(x))(tⁿ/n!) =_∞

n=0En(x + 1)(tⁿ/n!) +^∞_n₌₀En(x)(tⁿ/n!)= 2e^t⁽^x⁺¹⁾/(e^t+ 1) + 2e^tx/(e^t+ 1)=2e^tx=_∞

n=02xⁿ(tⁿ/n!). Comparing the coeﬃcients of tⁿ/n! in this equation, we obtain

En(x+ 1) +En(x)=2xⁿ, 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 alln≥0,

En(x+y)= n k=0

n k

Ek(x)yⁿ⁻^k, (1.8)

En(x+ 1)= n k=0

n k

Ek(x). (1.9)

Proof. ^∞_n₌₀En(x+ y)(tⁿ/n!)=2e^t(x+y)/(e^t + 1)=(2e^tx/(e^t + 1))e^ty =(^∞_n₌₀En(x)(tⁿ/ n!))(^∞_n₌0yⁿ(tⁿ/n!))=_∞

n=0(ⁿ_k₌0

_n

k

Ek(x)yⁿ⁻^k)(tⁿ/n!). Comparing the coeﬃcients of tⁿ/n! in this equation, we obtain (1.8). In particular, wheny=1, we get (1.9).

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From (1.7) and (1.9), we obtain 1

2 n k=0

n k

Ek(x) +1

2En(x)=xⁿ, n≥0. (1.10) If we setx=0 in (1.8), we getEn(y)=_n

k=0

_n

k

En−k(0)y^k, that is,

En(x)= n k=0

n k

En−k(0)x^k, n≥0. (1.11)

LetE(x) andX(x) be then×1 matrices defined byE(x)=[E0(x),E1(x),...,En−1(x)]^T, X(x)=[1,x,...,xⁿ⁻¹]^T, and let ¯Enben×nlower triangular matrices defined by

E¯n(i,j)=

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

⎛

⎝i−1 j−1

⎞

⎠Ei−j(0) ifi≥j≥1,

0 otherwise.

(1.12)

Then (1.10), (1.11) can be represented as matrix equations, respectively, 1

2

Pn+In

E(x)=X(x),

E(x)=E¯nX(x). (1.13)

Thus, we have Pn+In₋1

=1 2E¯n

=1 2

⎛

⎜⎜

⎝ 0

0

E0(0) 0 0 ··· 0

1 0

E1(0)

1 1

E0(0) 0 ··· 0

2 0

E2(0)

2 1

E1(0)

2 2

E0(0) ··· 0

... ... ... . .. ...

n−1 0

En−1(0)

n−1 1

En−2(0)

n−1 2

En−3(0) ···

n−1 n−1

E0(0)

⎞

⎟⎟

⎠ .

(1.14)

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The Bernoulli numbersBnare defined by (see [7]) ∞

n=0

Bntⁿ n!⁼

t

e^t−1. (1.15)

It is known (see [6,8]) that the Euler polynomials can be expressed by the Bernoulli numbers as

En(x)= 1 n+ 1

n+1

k=1

2−2^k+1 n+ 1

k

Bkxⁿ⁺¹⁻^k. (1.16)

Puttingx=0 in (1.16) gives

En(0)=21−2ⁿ⁺¹Bn+1

n+ 1 , (1.17)

for all integersn≥0. Therefore, we obtain an explicit inverse of the Pascal matrix plus one as follows.

Theorem 1.2. Forn≥1, then×ninverse matrixQn=(Pn+In)⁻¹is given by

Qn(i,j)=

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩ 1 2

⎛

⎝i−1 j−1

⎞

⎠Ei−j(0) ifi≥j≥1,

0 ifi < j;

(1.18)

or

Qn(i,j)=

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

⎛

⎝i−1 j−1

⎞

⎠

1−2ⁱ⁻^j⁺¹Bi−j+1

i−j+ 1 ifi≥j≥1,

0 ifi < j.

(1.19)

In view of the Hadamard product, the inverse matrix (Pn+In)⁻¹ is the Hadamard product of the Pascal matrixPnand the matrixΔn, whereΔnis then×nlower triangular matrices defined by

Δn(i,j)=

⎧⎪

⎨

⎪⎩ 1

2Ei−j(0) ifi≥j≥1, 0 ifi < j;

(1.20) or

Δn(i,j)=

⎧⎪

⎪⎨

⎪⎪

⎩

1−2ⁱ⁻^j⁺¹Bi−j+1

i−j+ 1 ifi≥j≥1,

0 ifi < j. (1.21)

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The two functions, Euler(n,x) and Bernoulli(n), in the combinat library of the com- puter algebra system Maple are very useful in obtaining the matrixQn. For example, for n=8, we get

Q8=(P8+I8)⁻¹=

⎛

⎜⎜

⎝ 1

2 0 0 0 0 0 0 0

−1 4

1

2 0 0 0 0 0 0

0 −1 2

1

2 0 0 0 0 0

1

8 0 −3

4 1

2 0 0 0 0

0 1

2 0 −1 1

2 0 0 0

−1

4 0 5

4 0 −5

4 1

2 0 0

0 −3

2 0 5

2 0 −3

2 1

2 0

17

16 0 −21

4 0 35

8 0 −7

4 1 2

⎞

⎟⎟

⎠

. (1.22)

Note thatQn(i,j)=0 wheneveri < jori=j+ 2,j+ 4,j+ 6,....

Acknowledgments

This work is supported by Development Program for Outstanding Young Teachers in Lanzhou University of Technology and the NSF of Gansu Province of China. The authors wish to thank the referees for many valuable comments and suggestions that led to the improvement and revision of this note.

References

[1] L. Aceto and D. Trigiante, The matrices of Pascal and other greats, The American Mathematical Monthly 108 (2001), no. 3, 232–245.

[2] R. Aggarwala and M. P. Lamoureux, Inverting the Pascal matrix plus one, The American Mathe- matical Monthly 109 (2002), no. 4, 371–377.

[3] M. Bayat and H. Teimoori, PascalK-eliminated functional matrix and its property, Linear Algebra and Its Applications 308 (2000), no. 1–3, 65–75.

[4] R. Brawer and M. Pirovino, The linear algebra of the Pascal matrix, Linear Algebra and Its Appli- cations 174 (1992), 13–23.

[5] G. S. Call and D. J. Velleman, Pascal’s matrices, The American Mathematical Monthly 100 (1993), no. 4, 372–376.

[6] G.-S. Cheon, A note on the Bernoulli and Euler polynomials, Applied Mathematics Letters 16 (2003), no. 3, 365–368.

[7] L. Comtet, Advanced Combinatorics. The Art of Finite and Infinite Expansions, D. Reidel, Dor- drecht, 1974.

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[8] K. H. Rosen, J. G. Michaels, J. L. Gross, J. W. Grossman, and D. R. Shier (eds.), Handbook of Discrete and Combinatorial Mathematics, CRC Press, Florida, 2000.

[9] H. M. Srivastava and ´A. Pint´er, Remarks on some relationships between the Bernoulli and Euler polynomials, Applied Mathematics Letters 17 (2004), no. 4, 375–380.

Sheng-liang Yang: Department of Applied Mathematics, Lanzhou University of Technology, Lanzhou, Gansu 730050, China

E-mail address:[email protected]

Zhong-kui Liu: Department of Mathematics, Northwest Normal University, Lanzhou, Gansu 730070, China

E-mail address:[email protected]

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