The Hankel Transform of the Sum of Consecutive Generalized ...
The Hankel Transform of the Sum of Consecutive Generalized ... The Hankel Transform of the Sum of Consecutive Generalized ...
6It is known (for example, see Krattenthaler [?]) that the Hankel determinant h nof order n of the sequence {a n } n≥0 equalswhere {β n } n≥1 is the sequence given by:h n = a n 0β n−11 β n−22 · · · β 2 n−2β n−1 , (19)∞∑G(x) = a n x n a 0=βn=01 x 21 + α 0 x −β1 + α 1 x − 2 x 21 + α 2 x − · · ·(20)The sequences {α n } n≥0 and {β n } n≥1 are the coefficients in the recurrence relationQ n+1 (x) = (x − α n )Q n (x) − β n Q n−1 (x), (21)where {Q n (x)} n≥0 is the monic polynomial sequence orthogonal with respect to thefunctional L determined byL[x n ] = a n (n = 0, 1, 2, . . .). (22)In the next section this functional will be constructed for the sum of consecutivegeneralized Catalan numbers.3. The weight function corresponding to the functionalWe would like to express L[f] in the form:∫L[f(x)] = f(x)dψ(x),Rwhere ψ(x) is a distribution, or, even more, to find the weight function w(x) suchthat w(x) = ψ ′ (x).Denote by F (z; L) the functionF (z; L) =∞∑a k z −k−1 .k=0From the generating function (??), we have:Example 3.1. From (??), we have:F (z; 1) = z −1 G ( z −1 ; 1) = 1 2F (z; L) = z −1 G ( z −1 ; L). (23){z − 1 − (z + 1)√1 − 4 z}.
7Example 3.2. From (??), we have:{ ()}F (z; 2) = −1 1 + z 2 − z + (z + 1)√1 − 6 2zz + 1 .z 2∫F (z; 2)dz = z + 1 z(z − 1)ρ(1/z, 2) + log(z)4− 1 2 log (1 + z ( ρ(1/z, 2) − 3 )) − 7 2 log( z − 3 + zρ(1/z, 2) ) .When we replace relation (8) in previous relation and after some simplificationswe obtain that2(z + 1)F (z; L) = −1 +L − 1 + z + √ L 2 + (z − 1) 2 − 2L(z + 1)2(z + 1)= −1 +L − 1 + z(1 + zρ( 1, L)) zDenote R(z; L) = zρ( 1, L) = √ Lz 2 + (z − 1) 2 − 2L(z + 1). From the theory ofdistribution functions (see Chihara [?]), we have Stieltjes inversion formulaψ(t) − ψ(0) = − 1 π limy→0 + ∫ t0IF (x + iy; L)dx. (24)using which we can calculate the distribution function ψ(x) and weight functionw(x) = ψ ′ (x). It can be shown that holds:∫F (z; L)dz = 1 []z 2 − 2Lz − (z − L + 1)R(z; L) − l 1 (z) + l 2 (z) (25)4where we denoted:[]l 1 (z) = 2(3L + 1) log z − (L + 1) + R(z; L)l 2 (z) =[ −(L − 1)R(z; L) − (L − 1) 2 + z(L + 1)]2(L − 1) logz 2 (L − 1) 3Denote with F(z; L) = ∫ F (z; L)dz. Now we will replace z = x + iy and find a limitlim y→0 F(x + iy; L). First rewrite function R(z; L) in the following form:R(z; L) = √ (z − L − 1) 2 − 4LNow replace z = x + iy and let y tends to 0 + . Then we have:where{ √R(x; L) = lim R(x + iy; L) = √ i 4L − (x − L − 1)2 , x ∈ (a, b);y→0 + (x − L − 1)2 − 4L, otherwise,a = ( √ L − 1) 2 , b = ( √ L + 1) 2 . (26)
- Page 1 and 2: The Hankel Transform of the Sum ofC
- Page 3 and 4: 3is given by−n)/2h n =L(n2 2 n+1
- Page 5: 5Now,Also,∞∑T (2n, n; L) t n =n
- Page 9 and 10: 9Example 4.1. For L = 4 we getwhere
- Page 11 and 12: 11Example 4.2. For L = 4 we getwher
- Page 13: We will apply the mathematical indu
7Example 3.2. From (??), we have:{ ()}F (z; 2) = −1 1 + z 2 − z + (z + 1)√1 − 6 2zz + 1 .z 2∫F (z; 2)dz = z + 1 z(z − 1)ρ(1/z, 2) + log(z)4− 1 2 log (1 + z ( ρ(1/z, 2) − 3 )) − 7 2 log( z − 3 + zρ(1/z, 2) ) .When we replace relation (8) in previous relation and after some simplificationswe obtain that2(z + 1)F (z; L) = −1 +L − 1 + z + √ L 2 + (z − 1) 2 − 2L(z + 1)2(z + 1)= −1 +L − 1 + z(1 + zρ( 1, L)) zDenote R(z; L) = zρ( 1, L) = √ Lz 2 + (z − 1) 2 − 2L(z + 1). From <strong>the</strong> <strong>the</strong>ory <strong>of</strong>distribution functions (see Chihara [?]), we have Stieltjes inversion formulaψ(t) − ψ(0) = − 1 π limy→0 + ∫ t0IF (x + iy; L)dx. (24)using which we can calculate <strong>the</strong> distribution function ψ(x) and weight functionw(x) = ψ ′ (x). It can be shown that holds:∫F (z; L)dz = 1 []z 2 − 2Lz − (z − L + 1)R(z; L) − l 1 (z) + l 2 (z) (25)4where we denoted:[]l 1 (z) = 2(3L + 1) log z − (L + 1) + R(z; L)l 2 (z) =[ −(L − 1)R(z; L) − (L − 1) 2 + z(L + 1)]2(L − 1) logz 2 (L − 1) 3Denote with F(z; L) = ∫ F (z; L)dz. Now we will replace z = x + iy and find a limitlim y→0 F(x + iy; L). First rewrite function R(z; L) in <strong>the</strong> following form:R(z; L) = √ (z − L − 1) 2 − 4LNow replace z = x + iy and let y tends to 0 + . <strong>The</strong>n we have:where{ √R(x; L) = lim R(x + iy; L) = √ i 4L − (x − L − 1)2 , x ∈ (a, b);y→0 + (x − L − 1)2 − 4L, o<strong>the</strong>rwise,a = ( √ L − 1) 2 , b = ( √ L + 1) 2 . (26)
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