Integral Representations and New Generating Functions of Chebyshev Polynomials

Integral Representations and New Generating Functions of Chebyshev Polynomials

Integral representations and new generating functions of Chebyshev polynomials Clemente Cesarano Faculty of Engineering, International Telematic University UNINETTUNO Corso Vittorio Emanuele II, 39 – 00186 – Roma, Italy email: [email protected] ABSTRACT – In this paper we use the two-variable Hermite polynomials and their operational rules to derive integral representations of Chebyshev polynomials. The concepts and the formalism of the Hermite polynomials Hn(x; y) are a powerful tool to obtain most of the properties of the Chebyshev polynomials. By using these results, we also show how it is possible to introduce relevant generalizations of these classes of polynomials and we derive for them new identities and integral representations. In particular we state new generating functions for the first and second kind Chebyshev polynomials. Keywords: Chebyshev polynomials, Integral representations, Generating functions. 2000 Mathematics Subject Classification: 33C45, 33D45. 1 Introduction The Hermite polynomials [1] can be introduced by using the concept and the formalism of the generating function and related operational rules. In the fol- lowing we recall the main definitions and properties. (2) Definition – The two-variable Hermite Polynomials Hm (x; y) of Kamp´ede F´erietform [2], [3] are defined by the following formula: [ m ] X2 m! H(2)(x; y) = ynxm¡2n (1.1) m n!(m ¡ 2n)! n=0 1 In the following we will indicate the two-variable Hermite polynomials of (2) Kamp´ede F´erietform by using the symbol Hm(x; y) instead than Hm (x; y). The two-variable Hermite polynomials Hm(x; y) are linked to the ordinary Hermite polynomials by the following relations: µ ¶ 1 H x; ¡ = He (x) m 2 m where: [ m ] X2 (¡1)rxn¡2r He (x) = m! m r!(n ¡ 2r)!2r r=0 and: Hm (2x; ¡1) = Hm(x) where: [ m ] X2 (¡1)r(2x)n¡2r H (x) = m! m r!(n ¡ 2r)! r=0 and it is also important to note that the Hermite polynomials Hm(x; y) satisfy the relation: m Hm(x; 0) = x : (1.2) Proposition 1 – The polynomials Hm(x; y) solve the following partial differ- ential equation: @2 @ H (x; y) = H (x; y) (1.3) @x2 m @y m Proof – By deriving, separately with respect to x and to y, in the (1.1), we obtain: @ H (x; y) = mH (x; y) @x m m¡1 @ H (x; y) = H (x; y): @y m m¡2 From the first of the above relation, by deriving again with respect to x and by noting the second relation, we end up with the (1.3). The Proposition 1 help us to derive an important operational rule for the Hermite polynomials Hm(x; y). In fact, by considering the differential equation 2 (1.3) as linear ordinary in the variable y and by remanding the (1.2) we can immediately state the following relation: 2 y @ m Hm(x; y) = e @x2 x (1.4) The generating function of the above Hermite polynomials can be state in many ways, we have in fact Proposition 2 – The polynomials Hm(x; y) satisfy the following differential difference equation: d Y (z) = a nY (z) + b n(n ¡ 1)Y (z) (1.5) dz n n¡1 n¡2 Yn(0) = ±n;0 where a and b are real numbers. Proof – By using the generating function method, by putting: X+1 tn G (z; t) = Y (z) n! n n=0 with t continuous variable,we can rewrite the (1.5) in the form: d ¡ ¢ G (z; t) = at + bt2 G (z; t) dz G (0; t) = 1 that is a linear ordinary differential equation and then its solution reads: ¡ ¢ G (z; t) = exp xt + yt2 where we have putted az = x and bz = y. Finally, by exploiting the r.h.s of the previous relation we find the thesis and also the relation linking the Hermite polynomials and their generating function: ¡ ¢ X+1 tn exp xt + yt2 = H (x; y): (1.6) n! n n=0 The use of operational identities, may significantly simplify the study of Her- mite generating functions and the discovery of new relations, hardly achievable by conventional means. By remanding that the following identity µ ¶ 2 n ¡ 1 d n d e 4 dx2 (2x) = 2x ¡ (1) (1.7) dx 3 is linked to the standard Burchnall identity, we can immediately state the fol- lowing relation. Proposition 3 – The operational definition of the polynomials Hn(x) reads 2 ¡ 1 d n e 4 dx2 (2x) = Hn(x) (1.8) Proof – By exploiting the r.h.s of the (1.7), we immediately obtain the Burchnall identity µ ¶ d n Xn 1 ds 2x ¡ = n! (¡1)s H (x) (1.9) dx (n ¡ s)!s! n¡s dxs s=0 after using the decoupling Weyl identity, since the commutator of the operators of l.h.s. is not zero. The derivative operator of the (1.9) gives a not trivial contribution only in the case s = 0 and then we can conclude with: µ ¶ d n 2x ¡ (1) = H (x) ; dx n which prove the statement. The Burchnall identity can be also inverted to give another important rela- tion for the Hermite polynomials Hn(x). We find in fact Proposition 4 – The polynomials Hn(x) satisfy the following operational iden- tity: µ ¶ µ ¶ 1 d Xn n ds H x + = (2x)n¡s (1.10) n 2 dx s dxs s=0 tn Proof – By multiplying the l.h.s. of the above relation by n! and then summing up, we obtain µ ¶ X+1 n t 1 d 2 x+ 1 d t¡t2 H x + = e ( 2 )( dx ) : n! n 2 dx n=0 By using the Weyl identity, the r.h.s. of the above equation reads 2 x+ 1 d t¡t2 2xt t d e ( 2 )( dx ) = e e dx and from which the (1.10) immediately follows, after expanding the r.h.s and by equating the like t-powers. The previous results can be used to derive some addition and multiplication relations for the Hermite polynomials. 4 Proposition 5 – The polynomials Hn(x) satisfy the following identity 8 n; m 2 N: min(n;m) µ ¶µ ¶ X n m H (x) = (¡2)s s!H (x)H (x): (1.11) n+m s s n¡s m¡s s=0 Proof – By using the Proposition 3, we can write µ ¶ µ ¶ µ ¶ d n d m d n H (x) = 2x ¡ 2x ¡ = 2x ¡ H (x) n+m dx dx dx m and by exploiting the r.h.s. of the above relation, we find µ ¶ Xn n ds H (x) = (¡1)s H (x) H (x): n+m s n¡s dxs m s=0 After noting that the following operational identity holds ds 2sm! H (x) = H (x) dxs m (m ¡ s)! m¡s we obtain immediately our statement. From the above proposition we can immediately derive as a particular case, the following identity Xn (¡1)s [H (x)]2 H (x) = (¡1)n2n(n!)2 s : (1.12) 2n 2s(s!)2(n ¡ s)! s=0 The use of the identity (1.10), stated in Proposition 4, can be exploited to obtain the inverse of relation contained in the (1.12). We have indeed 2 Proposition 6 – Given the Hermite polynomial Hn(x), the square [Hn(x)] can be written as Xn H (x) H (x)H (x) = [H (x)]2 = 2n(n!)2 2n : (1.13) n n n 2s(s!)2(n ¡ s)! s=0 Proof – We can write · µ ¶ µ ¶¸ 2 2 ¡ 1 d 1 d 1 d [H (x)] = e 4 dx2 H x + H x + : n n 2 dx n 2 dx By using the relation (1.10), we find, after manipulating the r.h.s. " # n 2n 2 X 2 ¡ 1 d n 2 (2x) [H (x)] = e 4 dx2 2 (n!) n 2s(s!)2(n ¡ s)! s=0 and then, from the Burchnall identity (1.7), the thesis. 5 A generalization of the identities stated for the one variable Hermite poly- nomials can be easily done for the polynomials Hn(x; y). We have in fact Proposition 7 – The following identity holds µ ¶ µ ¶ @ n Xn n @s x + 2y (1) = (2y)s H (x; y) (1): (1.14) @x s n @xs s=0 tn Proof – By multiplying the l.h.s. of the above equation by n! and then summing up, we find µ ¶ X+1 n n t @ t x+2y @ x + 2y = e ( @x )(1): n! @x n=0 By noting that the commutator of the two operators of the r.h.s. is · ¸ @ tx; t2y = ¡2t2y @x we obtain µ ¶ X+1 n n t @ xt+yt2 2ty @ x + 2y = e e @x (1): (1.15) n! @x n=0 After expanding and manipulating the r.h.s. of the previous relation and by equating the like t powers we find immediately the (1.14). By using the Proposition 6 and the definition of polynomials Hn(x; y), we can derive a generalization of the Burchnall-type identity: µ ¶ 2 n y @ n @ e @x2 x = x + 2y ; (1.16) @x and the related inverse: µ ¶ µ ¶ @ Xn n @s H x ¡ 2y ; y = (¡2y)s xn¡s : (1.17) n @x s @xs s=0 We can also generalize the multiplication rules obtained for the Hermite polynomials Hn(x), stated in Proposition 5. Proposition 8 – Given the Kamp´ede F´erietHermite polynomials Hn(x; y). We have min(n;m) X H (x; y)H (x; y) H (x; y) = m!n! (2y)s n¡s m¡s : (1.18) n+m (n ¡ s)!(m ¡ s)!s! s=0 6 Proof – By using the relations stated in the (1.14) and (1.16), we can write: µ ¶ @ n H (x; y) = x + 2y H (x; y) ; n+m @x m and then µ ¶ Xn n @s H (x; y) = (2y)s H (x; y) H (x; y): (1.19) n+m s n @xs m s=0 By noting that @s m! xm = xm¡2s @xs (m ¡ 2s)! we obtain @s m! H (x; y) = H (x; y) ;: @xs m (m ¡ s)! m¡s After substituting the above relation in the (1.19) and rearranging the terms we immediately obtain the thesis.

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