The Hankel Transform of the Sum of Consecutive Generalized ...

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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

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