On the Pell Polynomials
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1 Applied Mathematical Sciences, Vol. 5, 2011, no. 37, On the Pell Polynomials Serpil Halici Sakarya University Department of Mathematics Faculty of Arts and Sciences 54187, Sakarya, Turkey 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)
2 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 ,P 6 x) = x, P 7 x) =64x 6 +80x 4 +24x ) 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 ) 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)
3 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, ) 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;
4 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 = ) it can be seen that P m Pm+1 x) P = m x) P m x) P m 1 x) ), 29)
5 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)
6 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), [2] F. E. Hohn, Elementary Matrix Algebra, Macmillan Company, New York, [3] T. Koshy, Fibonacci and Lucas Numbers with Applications, A Wiley Pub., Received: November, 2010
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