Tom Copeland, Nov. 16, e trx 1 = A(t) x t = e tψ.(x) = n 0 n!

Transcription

1 The Creation / Raising Operators for Appell Sequences Tom Copelan, tcjpn@msn.com Nov. 6, 5 Part I Raising ops for logarithmic Appell sequences Dene the raising op for the logarithmic Appell sequence R x by e trx = At x t = e tψ.x = n ψ n x tn with A =. Then, with n t n t= At = D n t=at = a n, i.e., At = e a.t, D n t= e trx = R n x = ψ n x = lnx + a. n = = lnx n a, an clearly R x ψ n x = ψ n+ x. Acting on the top equation with the shift op e sdt ft = fs + t gives e sdt e trx = e s+trx = e sdt At x t = As + t x s+t = e s+tψ.x. implying the evolution equation evaluate D s at s = D t Atx t = R x e trx = R x Atx t an so e srx e trx = e srx At x t = As + t x s+t, e trx x s t As + t = x x s = x t AxD x + t As AxD x x s.

2 This can be rewritten in terms of the reciprocal Âs= As as a generalize fractional integro-erivative, or shift, operator acting on a particular basis : e trx Extracting the ops gives xs e trx = x t AxD x + t AxD x A,x Âs = D t x s Âs = xs+t Âs + t. = x t Aφ.: xd x : +t Aφ.: xd x : = x t e c.:xdx: = D t A,x = xt ÂxD x ÂxD x + t = xt Âφ: xd x : Âφ: xd x : + t = xt e ĉ.:xdx: with c n = An+t An, ĉ = Ân, φx equal to the Bell / Touchar / exponential Ân+t polynomials Stirling polynomials of the secon in, OEIS A877, an by enition : xd x : n = x n Dx. n The nite ierence expressions incorporating b. n = follow from the umbral relations AxD x + t AxD x fx = n An + t An b = n b = f n x n = e c.:xdx: fx = f[x c.x] = fc.x. Careful! From the binomial expansion b. = b, which nee not be equal in general to unity; however, it is unity for Appell sequences. Evaluating the erivative w.r.t. t at the origin of the exponentiate raising op gives xd R x = lnx + A x AxD x = lnx + xd x ln[axd x]. The associate lowering operator is ene as Lψ n x = n ψ n x = n lnx + a. n = xd x lnx + a. n = xd x ψ n x, so the commutator for the operators acting on any ψ n x is [L x, R x ] = L x R x R x L x = Ientity. An, from the properties of the Pincherle erivative, Note also that lnx + [ln AL x, R x ] = lnx + e trx e srx = e Rx L x ln AL x = R x.

3 implies x t At + xd x x s As + xd x = x At + s + xd x AxD x AxD x AxD x = x s As + xd x x t At + xd x AxD x AxD x an the same relations for the operator x t ÂxD x At + xd x = x t At+xD x AxD. x For example, with Ax = x!, this implies x t t + xdx t x s s + xdx s t + s + = x xdx. t + s Part II Recursion relations an integral reps for the raising op The raising op can be expresse several ways as in the entry Bernoulli Appells here. The most convenient for binomial relations is, with a change variables to lnx = z, R z = z + D z ln [MD z ] = z + D z ln [ e m.dz] = z + D z e c.dz = z + c. e c.dz with c n regare as the formal cumulants an m n as the formal moments of OEIS A767 ene by with M =. Then e c.t = ln [ e m.t] = ln[mt] R n z = z + m. n = p n z, an a recursion relation follows from the Appell binomial property p n z + h = z +m.+h n = p.z+h n as in the Appell Polynomials, Cumulants,... entry : R z p n z = p n+ z = z + c. e c.dz p n z = z p n z + c. p n c. + z = z p n z + c. c. + p.z n = z p n z + = z + c p n z + = = n c + p n z. These polynomials are precisely the general Bell polynomials c n+ p z p n z = B n z + c, c,..., c n = B n p z, c,..., c n = B n x[],..., x[n], 3

4 or partition polynomials for the rene Stirling numbers of the secon in, of OEIS A364, which are an Appell sequence in the istinguishe ineterminate c, i.e., c B n c,..., c n = n B n c,..., c n. They can also be expresse as the cycle inex polynomials of the symmetric groups, or the partition polynomials for the rene Stirling numbers of the rst in, p n z = CIP n z + b, b,.., b n = CIP n p z, b,.., b n = CIP n x[],..., x[n], of A3639 with x[] = b = p z = z + c an x[n] = b n = c n /n! for b. n > with raising op R = = b + b n+ Db n b.d, which are an Appell b n sequence in the istinguishe ineterminate b, i.e., b CIP n b,..., b n = n CIP n b,..., b n. The recursion relation leas to an integral representation for the raising op. Compare p n z p.z ω n = p n z p n z ω = + ω p n z = with the recursion relation. If we can n a istribution µω such that its moments are the cumulants c + for >, we have our integral rep ˆ ˆ c + = + ω µω ω = α e ωα µω ω α=, or = ˆ c. e c.α = e ωα µω ω. If this is essentially a Laplace transform, then the inverse Lapace transform in an appropriate region of evaluation gives µω. Uner suitable conitions, the coecients can be regare as iscrete samples c + = +! C + of a Mellin transform of the istribution: Cs = ˆ µω ω s s! ω. The istribution may nee to be moie the integral regularize over strips of s to obtain an invariant Cs over the Mellin ual space, just as for Cs =. 4

5 Since p n z p.z ω n = p n z p n z ω, the recursion relation can be expresse as R z p n z = p n+ z = z + c p n z + = z + c p n z + ˆ l Or, with the change of variables ω = z t, Now let then R z p n z = p n+ z = z + c p n z + l = c + p n z. [p n z p n z ω] µω ω. ˆ z l z l [p n z p n t] µz t t. z = lnx, t = lnv, an p n z = p n lnx = ψ n x, R x ψ n x = ψ n+ x = lnx + c ψ n x + So, for functions analytic about the origin, e l x e l [ψ n x ψ n v] µ[lnx/v] v v. an R z fz = z + c fz + ˆ z l z l [fz ft] µz t t, e l R x gx = lnx + c gx + [gx gt] µ[lnx/t] t t. x e l EXAMPLES: As gleane from the Riemann zeta Appell sequence of the MathOverow question Riemann zeta function at positive integers an an Appell sequence of polynomials relate to fractional calculus an the entry here On the Mellin interpolation of ierential ops an associate innigens an Appell polynomials..., the istributions for the Riemann zeta Appell sequences are signe an unsigne µω = e ω, an the cumulants within an overall sign for n > are ˆ c n = n n! ζn = n ω n e ω ω 5

6 with c. e c.α = n n+ ζn + α n = Ψ + α + γ = α ec.α c = α ln [em.α ] c = α ln[α!] + γ for α <, where Ψ is the igamma, or Psi, function with c = α lnα! α= = γ = Ψ = , the negate Euler-Mascheroni constant. Also, l =, l =, an ˆ z R z fz = z γ fz + [fz ft] e z t t, an R x gx = lnx γ gx + = lnx gx + πi = lnx gx + π ˆ π π z x =x = lnx γ gx π gx gt x t lnz x γ z x t gz z iθ lnx γ gx + e iϑ θ. ˆ π π iθ gx + e iϑ θ. Specically, for this choice of overall sign for the cumulants, we have obtaine the raising operator or innigen associate to the shifte Laguerre operator xd x x, iscusse in the earlier entry: R x x s = lnx γ x s x s t s ˆ + t = R x x s = [lnx γ+ x t = [lnx + Ψ s] x s = [lnx + Ψ + xd x ] x s = lnxd x x x s = [ lnx + lnd x ] x s. Then acting on functions analytic at the origin, R x = lnx + Ψ + xd x = lnx γ + n ˆ = lnx γ + = lnx γ + n n + n + + xd x t xdx t t H n x n D n x= t s t t] x s 6

7 = lnx γ + n H. n xn Dx n n = lnx γ + n x n Dx n n n = lnx γ + n xdx n n n = lnx γ + b. xdx, where b = H = note that b. = b =, in this case, an b n = /n otherwise an the harmonic numbers H n are ene for n > by so H n = = b. n = = = ˆ = n b = u n u = γ + Ψn + u n n =, R x fx = [lnx γ + b. xdx ] fx = lnx γ fx + f [ b. x] = lnx γ fx + n H n f n xn = lnx γ fx + fx ft x t The operator can also be expresse in terms of the entire function exponential integral ˆ z e t Einz = t = n z n t n, n with : xd x : n = x n D n x by enition, as R x = lnx γ + Ein: xd x :. t. For the Bernoulli polynomials iscusse in the Bernoulli Appells entry, the raisng op for the logarithmic Appell sequence is R x = lnx + n B n+ n + n xd x n = lnx + n ζn π n xd x n n = lnx + xd x e xdx 7

8 = lnx = lnx + xd x coth xdx + i xd x Ψ π = lnx + i [ Ψ π i xd x π = lnx + πin xd x πin + xd x n [ˆ ] = lnx + i ω i xdx xdx π i ω π ω π ω = lnx ˆ ωxdx sin π π e ω ω = lnx ˆ lnu xdx sin π π u u [ sinh xd x xd x ln ] [ xd x e xdx = lnx + xd x ln [ = lnx sini xd x xd x ln e xdx i xdx = lnx + {ˆ π π Im { = lnx + π Im πi We then obtain, for t about the origin, e trx x s = = x t e t As + t As x s+t = s + t s ] sinh xdx xdx x sinh s = x t e t xdx+t xdx+t i t+xd x = x t t e t = x t e t π i t π x π i xd x ] = lnx+ [ xd x ln i xd x } iθ + γ + e iθ i xdx π θ π z = lnz + γ z e s e s+t xs+t = e t[lnx + i i t sin i t x s = x t e t : Dx x : i t π i t+xd x π i t π z i xdx π z }. π [Ψ+ i xdx π xd x e xdx ]! i xd x i t+xd x π i t x s π i t+xd x π! i s! π x s x i π xd x x s. ]! e xdx Ψ i xdx π ]] x s In the last equality, the last factor from the left in the moulus transforms x s to x i s π an the rst factor reverses the transformation as x q xd x x s = x s x q s = x qs = n x q n : xd x : n x s. 8 n

9 Recall that by enition : AB : n = A n B n so that : xd x : n = x n D n an : D x x : n = Dxx n n. The mile factor is a generalize Laguerre operator giving : D x x : i t n x i s i π = D t π x x i t π x i s π = i t + s! D i t x i π π i x π! π i = π! i s x i s π π! with one convolution rep of the fractional integro-erivative being the Haamar nite part of D i t π x x i π ˆ = x i s π = x u i t π i t π u i u i t π π u =! i t π! u i t π i t π ui π u = x i s π! = i t! π i π i t π i t! π n x i s π, i π n x ui π u sinπn i t π πn i t π with the summation vali for Real i π >. Other reps are given in other notes at this site an are reecte in the integral reps above. The last expression of the exponentiate raising op may also be expane out an the mile two factors reuce as e trx x s = x t e t x π i xd x : Dx x : i t π x xd x : Dx x : i t π x i π xd x. Part III Convolution rep for D t A,x = exptr x from the Mellin transform Acting on a function representable as an inverse Mellin transform, e trx fx = πi ˆ σ+i σ i f M se trx x s s = πi ˆ σ+i σ i f M s x t AxD x + t x s s AxD x with = ˆ σ+i f M s πi σ i A s + t x s+t s A s ˆ = x t σ+i f M s πi K M s; t x s s σ i K M s; t = As + t As = Âs Âs + t. 9

10 Then from the Mellin convolution theorem, ˆ ˆ e trx fx = x t Ku; t fxuu = x t x Ku ; t fuu x = x t n An + t x n f n An = xt e c.:xdx=: fx = x t e c.:xdx: fx where Kx; t is the inverse Mellin transform of KM s; t an f n = D n x= fx. For example, let As = /s!, corresponing to the innigen R x = lnx Ψ + xd x = lnd x for the fractional integro-erivatives Dx t. Then for σ > an t non-integral, Kx; t = ˆ σ+i K M s; t x s s = ˆ σ+i s! πi σ i πi σ i s + t! x s s = ˆ σ+i πi σ i π sinπs s + t! x s s! s = H x n t n! xn = H x n where Hx is the Heavisie step function, so xα βrx e α! = x α D β A,x α! = xα+β α + β! ˆ = x β x Ku uα ; β u = xβ x α! = x u β β! ˆ u x β x β! u α α! u. xt t! u α α! u As another illustration note that if Ax =, then K M s; t =, an the inverse Mellin transform gives Kx; t = δ x; therefore, D t A,x = xt, consistent with R x = lnx an e trx = x t.,