Applied Mathematical Sciences, Vol. 5, 2011, no. 37, 1833-1838 On the Pell Polynomials Serpil Halici Sakarya University Department of Mathematics Faculty of Arts and Sciences 54187, Sakarya, Turkey shalici@sakarya.edu.tr Abstract In this study, Pell polynomials and its some properties are investigated. The some sum formulae for these polynomials are derived by matrices. Mathematics Subject Classification: 11B37, 11B39, 11B50 Keywords: Binet Formulas, Pell Polynomials, Pell Numbers 1 Introduction The Fibonacci and Lucas polynomials are defined by and f n x) =xf n 1 x)+ f n 2 x),n 3, 1) L n x) =xl n 1 x)+l n 2 x),n 2 2) respectively. Where f 1 x) =1, f 2 x) =x, L 0 x) = 2 and L 1 x) =x.the polynomials studied by Jacobstal in [3] are defined by J n x) =J n 1 x)+xj n 2 x),n 3. 3) where J 1 x) = 1 and J 2 x) = 1. It can be seen that for n 0, L n 1) = L n, f n 1) = F n,j n 1) = F n. The some known identities related with these polynomials; f n+m x) = f n x) f m 1 x)+ f n+1 x) f m x), 4) x 2 +4 ) f 2 n x) =L n+1 x)+l n 1 x), 5)
1834 S. Halici f n+1 x) f n 1 x) f 2 n x) = 1)n. 6) Similarly, Pell, Pell-Lucas polynomials are defined as where P 0 x) = 0 and P 1 x) =1,n 2 and P n x) =P n 1 x)+p n 2 x), 7) Q n x) =Q n 1 x)+q n 2 x), 8) respectively. Where Q 0 x) = 2 and Q 1 x) =, n 2[1]. P 5 x) =16x 4 +1 2 +1,P 6 x) =3 5 +3 3 +6x, P 7 x) =64x 6 +80x 4 +24x 2 +1. 9) Thus, if P n x) isnth Pell polynomial, then for n 2, the following properties can be written. Pell polynomials have not the same degree. The leading coefficient of P n x) is2 n 1. For n odd number, the coefficients of P n x) are even numbers, except for constant term. For n N even number, we say that divides P n x) which x 0. For all n, deg [P n x)] = n 1. Notice that for n N even and n 2, the last terms of P n x) are nx. nj 0 1 2 3 0 0 1 1 2 2 3 4 1 4 8 4 5 16 12 1 6 32 32 6 7 64 80 24 1 8 128 192 80 8 10) Table 1. The Coefficients of the polynomials P n x). Let P n, j) denote the element in row n and column j, where j 0, n 1. From the Table 1, P n, 0) = 2P n 2, 0) + P n 1, 0). 11) can be written. For example, we have P 7, 0) = 2P 5, 0) + P 6, 0) = 64. It can be seen that every row sum in the array of coefficient in the Table 1 is a Pell number. Moreover, we can write n 2 P n, j) =P n. 12)
On the Pell polynomials 1835 Some relationships involving P n x) and Q n x) can be found in [1]. Some of these are; P n x)q n x) =P 2n, 13) P n+1 x)+q n 1 x) =Q n x), 14) P n+1 x)p n 1 x) Pn 2 x) = 1)n, 15) Q n+1 x)q n 1 x) Q 2 nx) = 1) n 1 4x 2 +1). 16) In [1], Horadam obtained some sum formulae for these polynomials. Some of these formulas; P 2r x) = P 2n+1x) 1, 17) Q 2r x) = Q 2n+1x), 18) P r x) = P n+1x)+p n x) 1, 19) Q r x) = Q n+1x)+q n x) 2. 20) 2 Some Properties of Pell Polynomials In this section we give some formulas for sums of the Pell Polynomials using by matrices. If P n n th Pell polynomial, then we can write x + x2 +1 ) n x x2 +1 ) n P n x) = 2. 21) x 2 +1 When n = 2, this equation is equal to P 2 x) as expected. The polynomial P n x) can be also computed using by the matrices. Let An) be a family of tridiagonal matrices, as follows. a 1,1 a 1,2 a 2,1 a 2,2 a 2,3 An) = a 3,2 a 3,3... 22)...... a n 1,n a n,n 1 a n,n It is well known that the determinants of An) matrices can be described by deta1)) = a 1,1 deta2)) = a 2,2 a 1,1 a 2,1 a 1,2 detan)) = a n,n detan 1)) a n,n 1 a n 1,n detan 2)). Thus, we can give the following theorem;
1836 S. Halici Theorem 2.1 For, n 0 if D n x) is a n n tridiagonal matrix D n x) = 1 i 0 i i... then, we have det D n x)) = P n x)....... i i,d 0 x) = 0 23) Proof. We can easily seen that det D 1 x)) = 1, det D 2 x)) =. We assume that D n 1 x) = P n 1 x) and D n 2 x) = P n 2 x). So, we can write D n x) = D n 1 x) i 2 D n 2 x) =P n 1 x)+p n 2 x) =P n x). 24) Thus, the proof is completed. Now, let us define a matrix different from matrix D n x) as follows; Dnx) = then, n 1 we have that 2 i 0 x i i......... i i,d 0 x) = 0 25) D n x) = Q n 1 x), 26) where Q n x) denotes the n th Pell-Lucas polynomials. For the Pell polynomials, P n x) = 1) n+1 P n x) can be written. That is Pell Polynomials are even functions, if n is a odd number and this polynomials are odd functions, if n is a even number. So, for the n th Pell Polynomial we get P n x) t n = 1) n+1 P n x) t n. 27) n=0 n=0 The following matrix generates Pell and Pell-Lucas polynomials [1]; ) 1 P = 1 0 28) it can be seen that P m Pm+1 x) P = m x) P m x) P m 1 x) ), 29)
On the Pell polynomials 1837 and det P n )= 1) n. Also, for all n, m Z we can use the matrix P to establish some properties of these polynomials; i.e. P n+m x) =P n+1 x) P m x)+p n x) P m 1 x), 30) 1) n P m n x) =P m x) P n+1 x) P n x) P m+1 x). 31) By using the matrix P m x) we can get the determinant of P m+n x)+ 1) n P m n x)) is P m+1 x) Q n x) P m x) Q n x)) P m+1 x) Q n x)+p m x) Q n x)) 32) Theorem 2.2 If n N and m, k Z with m is different from zero, then we have P mj+k x) = P m x) 1 P m x)) n+1 ) P k x) 1+ 1) m. 33) Q m x) Proof. It is well known that I P m x)) n+1) =I P m x))) P mj x). 34) If deti P m x)) is different from zero, then we can write I P m x))) 1 I P m x)) n+1 = P mj x), 35) and I P m x))) 1 I P mn+m+k x) ) = P mj+k x). 36) Also, from ) P m Pm+1 x) P = m x), 37) P m x) P m 1 x) n ) P mj+k P x) = mj+k+1 x) n P mj+k x) n P mj+k x) n, 38) P mj+k 1 x) can be written. If A = I P m x)), then A = and deta) =1+ 1 m ) Q m x). 39) Thus, it follows from A 1 = 1 1 Pm 1 x) P m x) 1+ 1 m ) Q m x) P m x) 1 P m+1 x) ) 40)
1838 S. Halici Using the above equations, we can write I P m x))) 1 I P mn+m+k x) ) = P mj+k x), 41) P mj+k x) = P m x) 1 P m x)) n+1 ) P k x) 1+ 1) m. 42) Q m x) Thus, the proof is completed. Now, it will be shown that the polynomials P n x) satisfy the following recurrence relation. Theorem 2.3 If t is a square matrix such that t 2 =t + I, then for all n Z we have t n = P n x) t + P n 1 x) I. 43) Where P n x) is the nth Pell polynomial and I is a unit matrix. Proof. If n = 0, then the proof is obvious. It can be shown that by induction that t n = P n x) t + P n 1 x) I. 44) For every n N we will show that t n = P n x) t + P n 1 x) I. Let y = t = t 1. Since y 2 =y + I and y 3 =4x 2 y + + y, we write y n = P n x) y + P n 1 x) I. 45) Also, from y = t we write y n = 1) n t n = 1) n P n x) y + 1) n P n 1 x) I. 46) So, we obtain that t n = P n x) t + P n 1 x) I. 47) Thus, for all n Z t n = P n x) t + P n 1 x) I, 48) which is desired. References [1] A. F. Horadam, Pell Identities, The Fibonacci Quarterly, 91971), 245-252. [2] F. E. Hohn, Elementary Matrix Algebra, Macmillan Company, New York, 1973. [3] T. Koshy, Fibonacci and Lucas Numbers with Applications, A Wiley Pub., 2001. Received: November, 2010