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HTTP/2 301 server: nginx date: Tue, 30 Dec 2025 23:36:47 GMT content-type: text/html content-length: 162 location: https://tcjpn.wordpress.com/feed/ x-ac: 4.bom _dca MISS alt-svc: h3=":443"; ma=86400 strict-transport-security: max-age=31536000 server-timing: a8c-cdn, dc;desc=bom, cache;desc=MISS;dur=235.0 HTTP/2 200 server: nginx date: Tue, 30 Dec 2025 23:36:48 GMT content-type: application/rss+xml; charset=UTF-8 vary: Accept-Encoding x-hacker: Want root? Visit join.a8c.com/hacker and mention this header. host-header: WordPress.com vary: accept, content-type last-modified: Tue, 25 Nov 2025 18:04:02 GMT x-nc: HIT dca 158 content-encoding: gzip x-ac: 3.bom _dca BYPASS alt-svc: h3=":443"; ma=86400 strict-transport-security: max-age=31536000 server-timing: a8c-cdn, dc;desc=bom, cache;desc=BYPASS;dur=237.0 Shadows of Simplicity https://tcjpn.wordpress.com Just some math notes. Thu, 20 Feb 2025 15:40:30 +0000 en hourly 1 https://wordpress.com/ https://secure.gravatar.com/blavatar/f331fe985fe3dedf50522a675ee8fb05f44cdf9eb7e72ba4b6082a6b06184a40?s=96&d=https%3A%2F%2Fs0.wp.com%2Fi%2Fbuttonw-com.png Shadows of Simplicity https://tcjpn.wordpress.com Resonance https://tcjpn.wordpress.com/2025/02/20/resonance/ https://tcjpn.wordpress.com/2025/02/20/resonance/#respond Thu, 20 Feb 2025 15:40:30 +0000 https://tcjpn.wordpress.com/?p=10422 Continue reading →]]> I have several hundreds of pages of math notes I need to organize and publish here, but given current events and the plights of so many millions affected by the current callous regime in the U.S., I’m compelled by my conscience to direct people to a poignant speech that reaches out to people of kindred spirits from almost a century ago:

The Great Dictator Speech – Charlie Chaplin + Time – Hans Zimmer (INCEPTION Theme)

(https://youtu.be/w8HdOHrc3OQ?si=7FITT6LCfKdMRy9N)

]]> https://tcjpn.wordpress.com/2025/02/20/resonance/feed/ 0 tcjpn Boil and bubble: Conjuring the digamma and log(D) by umbrally mixing the logarithm with the Bernoulli function https://tcjpn.wordpress.com/2024/12/31/boil-and-bubble-conjuring-the-digamma-and-logd-by-umbrally-mixing-the-logarithm-with-the-bernoulli-function/ https://tcjpn.wordpress.com/2024/12/31/boil-and-bubble-conjuring-the-digamma-and-logd-by-umbrally-mixing-the-logarithm-with-the-bernoulli-function/#respond Wed, 01 Jan 2025 06:25:47 +0000 https://tcjpn.wordpress.com/?p=10214 Continue reading →]]> Define the logarithm of the derivative operator, $\log(D)$ , or more precisely $\ln(D)$ , as the infinitesimal generator (IG) for the fractional integroderivative op $D^t$ of the Heaviside-Pincherle family of the fractional calculus such that, for $x,s,t$ real,

$\displaystyle e^{t \; \ln(D_x)} H(x) \frac{x^s}{s!} = D^t H(x) \frac{x^s}{s!} = H(x) \frac{x^{s-t}}{(s-t)!},$

where $H(x)$ is the Heaviside step function and

$\displaystyle H(x) \frac{x^{-n}}{(-n)!} = \delta^{(n-1)}(x) = D^{n}H(x)$

with $\delta(x)$ as the Heaviside-Dirac delta function / operator. Alternatively, to circumvent issues with prematurely evaluating $\frac{x^{-n-1}}{(-n-1)!}$ as $0$ use the limit

$\displaystyle H(x) \frac{x^{-n}}{(-n)!} := \lim_{a \to 0} \frac{1}{2} H(x) \left[\frac{x^{-n+a}}{(-n+a)!} + \frac{x^{-n-a}}{(-n-a)!}\right]$

and

$\displaystyle D^t \frac{x^{s}}{s!} = \lim_{a \to 0} D^t \frac{1}{2} H(x) \left[\frac{x^{s+a}}{(s+a)!} + \frac{x^{s-a}}{(s-a)!}\right] = \lim_{a \to 0} e^{t \ln(D)} \frac{1}{2} H(x) \left[\frac{x^{s+a}}{(s+a)!} + \frac{x^{s-a}}{(s-a)!}\right].$

There are other methods to get around this apparent pitfall, such as to parse the computations as

$D^t H(x) \frac{x^{-n}}{(-n)!}= D^t D^n H(x) = e^{t\;\ln(D)}e^{n\;\ln(D)} H(x)$

$= H(x) e^{(t+n)\;\ln(D)}1= H(x) \frac{x^{-t-n}}{(-t-n)!}$

$= e^{n\;\ln(D)}e^{t\;\ln(D)} H(x) = D^n D^t H(x) = D^n H(x) \frac{x^{-t}}{(-t)!} = e^{n\ln(D)}H(x) \frac{x^{-t}}{(-t)!}.$

(See the links in my answer to and in the sidebar of the MO-Q “What’s the matrix of logarithm of derivative operator ln(D)? What is the role of this operator in various math fields?” for links to several of my older deeper notes on this family of fractional integroderivatives and the associated infinitesimal generator $\ln(D)$ . I also have an upcoming blog post updating this topic.)

Ultimately the differential operator rep for the IG can then be expressed from the calculus of the lowering and raising operators of Appell Sheffer polynomial sequences (and other methods) as

$\displaystyle \ln(D) = -\ln(x) + \psi(1+xD) = -\ln(x) + \ln(B.(1+xD)),$

where

$\psi(z) = \partial_z \ln((z-1)!)$

is the digamma function and

$B_s(z) =-s \zeta(1-s,z) = -\partial_z \zeta(1-s,z)$

is the Bernoulli function defined in terms of the Hurwitz zeta function.

The following analysis unpacks and establishes the umbral compositional / umbral translation identity

$\psi(z) = T_{z \to (z+b.) = (B.(z)}\ln(z) = \ln(B.(z)) = \ln(z+b.)$ ,

giving the digamma function as the umbral polynomial composition of the natural logarithm with the Bernoulli function $(z+b.)^s := (B.(z))^s := B_s(z)$ , or, equivalently, as the umbral number translation of the logarithm with the Bernoulli numbers $(b.)^n = b_n$ .

For $z,s$ real or complex, define the finite difference op by the action

$\displaystyle \delta_z z^s = (z+1)^s - z^s$ ,

or more generally for any function by

$\displaystyle \delta_z f(x) = f(z+1) - f(z).$

The differential op rep for action on suitable functions under analytic continuation (the equalities below assume these two conditions) is given by Taylor series arguments as

$\displaystyle \delta_z = e^{\partial_z}-1.$

Since the compositional inverse about the origin of $f(t) = e^t-1$ is $f^{<-1>}(t) = \ln(1+t)$ , we have formally (this can be made more rigorous using the Sheffer hybrid umbral differential operator calculus), the inverse relation–the derivative in terms of finite differences–as

$\displaystyle \partial_z = \ln(1 + \delta_z)$

and its action

$\displaystyle \partial_z f(z) = \ln(1 + \delta_z) f(z) = \frac{\ln(1 + \delta_z) }{\delta_z}\delta_z f(z) = e^{q.\delta_z} \delta_z f(z)$

$\displaystyle = \sum_{n \geq 0} (-1)^n \frac{1}{n+1} \delta_z^{n} \delta_z f(z)$

$\displaystyle = \sum_{n \geq 0} (-1)^n \frac{1}{n+1} \sum_{k \geq 0}(e^{\partial_z}-1)^k \delta_z f(z)$

$\displaystyle = \sum_{n \geq 0}\frac{1}{n+1} \sum_{k \geq 0} (-1)^k \binom{n}{k} e^{k\partial_z} \delta_z f(z)$

$\displaystyle = \sum_{n \geq 0}\frac{1}{n+1} \sum_{k \geq 0} (-1)^k \binom{n}{k} \delta_z f(z+k)$

$\displaystyle =\frac{\partial_z}{\delta_z} \delta_z f(z) = \frac{\partial_z}{e^{\partial_z}-1} \delta_z f(z)$

$\displaystyle = e^{b. \partial_z} (f(z+1) - f(z))$

$\displaystyle = (f(z+1+b.) - f(z+b.)) = f(B.(z+1))- f(B.(z)),$

where $q_n = (-1)^n \frac{n!}{n+1}$ and $(b.)^n = b_n$ are the Bernoulli numbers and

$\displaystyle (B.(z))^n = B_n(z) = e^{b.\partial_z}z^n = \sum_{k \geq 0} \binom{n}{k} (b.)^k z^{n-k} = (z + b.)^n$

are the Bernoulli polynomials. (The $q_n$ and associated polynomials $Q_n(z) = (z+q.)^n$ play central roles in elucidating associations of this analysis with the falling factorials, a.k.a. the Stirling polynomials of the first kind, for which $\delta_z$ is the lowering op, and Newton series.)

When $f(z)$ is one of the power monomials $z^n$ for $n =0,1,2,...$ , equivalency holds for all the reps above of the derivative. In particular,

$\displaystyle \partial_z z^n = n \; z^{n-1} = e^{b.\partial_z} \delta_z z^n = (b.+z+1)^n - (b.+z)^n = B_n(z+1) - B_n(z).$

Then for any formal Taylor series $f(z) = \sum_{n \geq 0}c_n \frac{z^n}{n!}$ about the origin, term by term umbral evaluation gives the formal derivative

$\displaystyle f(z+1+b.) - f(z+.b) = f(B.(z+1)) - f(B.(z))$

$\displaystyle = \sum_{n \geq 0}c_n \frac{B_n(z+1)-B_n(z)}{n!} = \sum_{n \geq 0}c_n \frac{z^{n-1}}{(n-1)!} = \partial_z f(z).$

For other functions, the finite difference rep gives correct convergent results for a broader domain of functions that does the differential op rep incorporating the Bernoulli numbers.

The Hurwitz zeta function $\zeta(s,z)$ is conjured up with action on $f(z) = z^s$ for $s$ real or complex for

$\displaystyle -s \zeta(1-s,z) = \partial_z (-\zeta(-s,z) )= \frac{\partial_z}{\delta_z} \delta_z (-\zeta(-s,z))$

$\displaystyle = \frac{\partial_z}{\delta_z}z^s = \frac{\ln(1+\delta_z)}{\delta_z} z^s = \sum_{n \geq 0} \frac{1}{n+1} \sum_{k=0}^n (-1)^k \binom{n}{k} (z+k)^s .$

The last expression is essentiallty the Helmut Hasse formula for the Hurwitz zeta function.

Although the Bernoulli differential op rep gives a divergent asymtotic formula, it is in the sense above, as well as through the extension of the Bernouilli polynomials via Ramanujan’s master formula, a.k.a Mellin transform interpolation and extrapolation (see below and much earlier posts), that we can viably define the Bernoulli function as an extension of the Bernoulli polynomials via

$\displaystyle e^{b. \partial_z} z^s := (x+b.)^s := B.(z)^s := B_s(z) = -s \zeta(1-s,z) = -\partial_z \zeta(1-s,z)$

$\displaystyle = \frac{\partial_z}{\delta_z} z^s = \frac{\ln(1+\delta_z)}{\delta_z}z^s = \sum_{n \geq 0} \frac{1}{n+1} \sum_{k=0}^n (-1)^k \binom{n}{k} (z+k)^s.$

The Bernoulli function then lives both in the function space spanned by the Hurwitz zeta and in the tangent space w.r.t. $z$ of the Hurwitz zeta function, and the Bernoulli op / Bernoulli number umbral translation op / Bernoulli function umbral subsitution op (a variant of the Hirzebruch-Todd differential op)

$\displaystyle T_{z \to (z+b.)=B.(z)} = \frac{\partial_z}{\delta_z}$

morphs the finite difference of a function into the derivative of the function. (Shortly, we’ll find the digamma function manifesting in the tangent space w.r.t. $s$ of $B_s(z)$ .)

The inverse transformation is effected by an op that transforms a derivative into a finite difference with the various reps

$\displaystyle f(z+1)-f(z) = \frac{\delta_z}{\partial_z} \partial_z f(z) = \int_{0}^1 \partial_t f(z+t)dt = \int_{z}^{z+1} \partial_t f(t) dt$

$\displaystyle = \frac{e^{\partial_z}-1}{\partial_z}\partial_zf(z) = e^{r.\partial_z}\partial_zf(z) = \frac{\delta_z}{\ln(1+\delta_z)} \partial_zf(z) = e^{c.\delta_z}\partial_zf(z),$

where $c_n$ are the Cauchy numbers of the first kind and $r_n = \frac{1}{n+1}$ , the reciprocals of the natural numbers, with associated Appell Sheffer Reciprocal polynomials

$\displaystyle R_n(z) = e^{r. \partial_z}z^n = \frac{e^{\partial_z}-1}{\partial_z}z^n = \int_z^{z+1} t^n dt = \frac{(z+1)^{n+1}-z^{n+1}}{n+1}$ ,

which extend to the Reciprocal function

$\displaystyle R_s(z) = \int_z^{z+1} t^s dt = \frac{(z+1)^{s+1}-z^{s+1}}{s+1}$

with the limiting case

$\displaystyle R_{-1}(z) = \ln(1+z) - \ln(z)$ .

The Reciprocal op with the reps

$\displaystyle T_{z \to R.(z)} = \frac{\delta_z}{\partial_z} = \frac{e^{\partial_z}-1}{\partial_z} = e^{r.\partial_z} = \frac{\delta_z}{\ln(1+\delta_z)} = e^{c.\delta_z} = \int_{z}^{z+1} ... \; dt$

morphs a partial derivative into a finite difference into and then is inverse to the Bernoulli op. So appears the generalized umbral compositional inversion relation for the Bernoulll function and Reciprocal function pair

$\displaystyle T_{z \to B.(z)} T_{z \to R.(z)} z^s=T_{z \to B.(z)} R_s(z)$

$\displaystyle = R_s(B.(z)) = z^s = B_s(R.(z))$

$\displaystyle =T_{z \to R.(z)} B_s(z) = T_{z \to R.(z)}T_{z \to B.(z)} z^s.$

Then to connect to the digamma function

$\displaystyle \psi(z) = \partial_z \ln((z-1)!)$ ,

note

$\displaystyle T_{z \to B.(z)} R_{-1}(z) = \ln(1+z) - \ln(z) |_{z \to B.(z)} = \ln(B.(1+z)) - \ln(B.(z))$

$\displaystyle = \partial_z \ln(z) = \frac{1}{z} = \psi(z+1) - \psi(z) = \delta_z \psi(z)$

$\displaystyle = T_{z \to B.(z)} \delta_z \ln(z) = \delta_z T_{z \to B.(z)}\ln(z)$

$\displaystyle = \delta_z \ln(B.(z)) = \delta_z \sum_{n \geq 0} \frac{1}{n+1} \sum_{k=0}^n (-1)^k \binom{n}{k} \ln(z+k)$

with the extension of the Appell umbral property

$(b + z+1)^n = (B.(z)+1)^n := (B.(z+1))^n = B_n(z+1)$ ,

$\displaystyle \psi(z) = \partial_z \ln((z-1)!) = \ln(B.(z)) = \sum_{n \geq 0} \frac{1}{n+1} \sum_{k=0}^n (-1)^k \binom{n}{k} \ln(z+k)$

$\displaystyle = \partial_{s=0}\sum_{n \geq 0} \frac{1}{n+1} \sum_{k=0}^n (-1)^k \binom{n}{k} (z+k)^s = \partial_{s=0} B_s(z) = \partial_{s=0} (-s\zeta(1-s,z))$ ,

for which the first series appears from numerical computations to be convergent for $Re(z) >0$ . The series for $B_s(z)$ is from Helmut Hasse’s analysis convergent for $z > 0$ (real) and $s =1$ , and $\ln(z+k)$ has weaker growth than $(z+k)$ , so the series suggested for $\psi(z)$ is convergent for $Re(z) > 0$ . (A singularity in the digamma function appears at $z=0$ but nowhere else to the RHS of the origin of the complex plane.) Finally, Wolfram Alpha gives the result that $\partial_{s=0} (-s\zeta(1-s,z)) = \psi(z)$ , so that if we are justified in interchanging the derivative and the summations, the convergence of the series to $\psi(z)$ is established. Numerical checks corroborate this.

Another consistency check:

The Newton series for the digamma function for $Re(z) > 0$ is

$\displaystyle \psi(z) = -\gamma - \sum_{n \geq 1} (-1)^n \frac{1}{n} \binom{z-1}{n}$ ,

so this implies

$\displaystyle \partial_z \psi(z) = - \sum_{n \geq 1} (-1)^n \frac{1}{n} \partial_z\binom{z-1}{n}$

$\displaystyle = - \sum_{n \geq 1} (-1)^n \frac{1}{n} \frac{\partial_z}{\delta_z} \delta_z\binom{z-1}{n}$

$\displaystyle = - \sum_{n \geq 1} (-1)^n \frac{1}{n} e^{b.\partial_z} \delta_z\binom{z-1}{n}$

$\displaystyle = - \sum_{n \geq 1} (-1)^n \frac{1}{n} e^{b.\partial_z} \binom{z-1}{n-1}$

$\displaystyle = - \sum_{n \geq 1} (-1)^n \frac{1}{n} \binom{B.(z-1)}{n-1}$

$\displaystyle = \sum_{n \geq 0} (-1)^{n+1} \frac{1}{n+1} \binom{B.(z-1)}{n}$

$\displaystyle = \sum_{n \geq 0} (-1)^n \binom{B.(z-1)}{n} (-\frac{1}{n+1})$

$\displaystyle =e^{b.\partial_z } \sum_{n \geq 0} (-1)^n \binom{z-1}{n}\sum_{k=0}^n (-1)^k \binom{n}{k} \frac{-1}{k+1}$

$\displaystyle = T_{z \to B.(z)} \frac{-1}{z} = -B_{-1}(z) = \zeta(2,z)$

and

$\partial_z^m \psi(z) = (-1)^{m+1} m! \zeta(1+m,z) = (-1)^{m+1} B_{-m}(z)$

(see the polygamma function in Wikipedia or my previous post “The Joy of Polygammary”), which follows from taking derivatives of the Hasse finite diff series rep of $\psi(z)$ .

The infinitesimal generator can then be expressed as

$\displaystyle \ln(D_z) = -\ln(z) + \psi(1+zD_z) = -\ln(z) + \ln(B.(1+zD_z))$

$\displaystyle = -\ln(z) + T_{zD \to B.(zD)} \ln(1+zD) = -\ln(z) + \partial_{s=0}B_s(1+zD).$

____________

Now let’s see how far we can carry the machinations of direct umbral substitution without applying any of the derivative or finite difference reps above.

The Taylor series expansion good for $Re(z) <1$

$\displaystyle \ln(1+z) = \sum_{n \geq 1} (-1)^{n+1} \frac{x^n}{n},$

suggests exploring the umbral substitution

$\displaystyle \psi(1+z) = \ln(B.(z+1)) = \ln(1+B.(z)) = \sum_{n\geq 1} (-1)^{n+1} \frac{1}{n}B_n(z)$

but a quick numerical check shows this is not convergent as we would expect, but there is also for $Re(z) >1$ , the Laurent expansion

$\displaystyle \ln(1+z) = \ln(z(1+1/z))= \ln(z) + \ln(1+1/z) = \ln(z) + \sum_{n\geq 1} (-1)^{n+1} \frac{x^{-n}}{n} .$

This gives

$\displaystyle \ln(1+z) - \ln(z) = R_{-1}(z) = \sum_{n\geq 1} (-1)^{n+1} \frac{x^{-n}}{n}$

suggesting

$\displaystyle \psi(1+z) - \psi(z) = R_{-1}(B(z)) = \frac{1}{z} = \ln(B.(1+z)) - \ln(B.(z)) = \sum_{n\geq 1} (-1)^{n+1} \frac{B_{-n}(z)}{n}$

$\displaystyle = \sum_{n\geq 1} (-1)^{n+1} \zeta(1+n,z) = \sum_{n\geq 2} (-1)^n \zeta(n,z) ,$

which is valid for $Re(z) \geq 1$ .

Keiper in his thesis mentioned below gives the identity

$\displaystyle \psi(z) = \ln(z) + \sum_{n \geq 2}(-1)^{n+1} \frac{1}{n} \zeta(n,z) ,$

implying, for $z > 0$ .

$\displaystyle \psi(z) = \ln(z) + \sum_{n \geq 1}(-1)^{n+1} \frac{1}{n+1} \frac{B_{-n}(z)}{-n}$ .

Taking the derivative of this identity gives

$\displaystyle \psi'(z) = \frac{1}{z} + \sum_{n \geq 2}(-1)^{n+1} \frac{1}{n} (-n\zeta(1+n,z))$

$\displaystyle = \frac{1}{z} + \sum_{n \geq 2}(-1)^n \zeta(1+n,z) ,$

$\displaystyle \frac{1}{z} = \psi'(z) + \sum_{n \geq 2}(-1)^n \frac{1}{n} (-n\zeta(1+n,z))$

Noting from above

$\displaystyle \psi'(z) = - B_{-1}(z) = \zeta(2,z)$

and, more generally,

$\displaystyle \partial_z^m \psi(z) = (-1)^{m+1} m! \zeta(1+m,z) = (-1)^{m+1} B_{-m}(z) ,$

then consistently

$\displaystyle \frac{1}{z} = \zeta(2,z) - \sum_{n \geq 2}(-1)^n \zeta(1+n,z) = \sum_{n\ge 2 } (-1)^{n} \zeta(n,z)$

$\displaystyle = \sum_{n\ge 1 } (-1)^{n} \frac{B_{-n}(z)}{-n} = \sum_{n\geq 1 } \frac{\partial_z^{n}}{n!} \psi(z)$

$\displaystyle = (e^{\partial_z}-1) \psi(z) = \delta_z \psi(z) = \psi(z+1) - \psi(z),$

for which the series is correctly convergent for $Re(z) \geq 1$ .

Finally, we have other reps for the infinitesimal generator

$\displaystyle \ln(D) =-\ln(x) + \psi(1+xD) = -\ln(x) + \ln(B.(1+xD))$

$\displaystyle = -\ln(x) + \ln(1+xD) + \left[ \ln( 1+ 1/x)|_{x \to (x+b.)=B.(x)} \right]$

$\displaystyle = -\ln(x) + \ln(1+xD) + \sum_{n \geq 2}(-1)^{n+1} \frac{1}{n} \zeta(n,1+xD)$

$\displaystyle = -\ln(x) + \ln(1+xD) + \sum_{n \geq 1}(-1)^{n+1} \frac{1}{n+1}\frac{B_{-n}(1+xD)}{-n}$

$\displaystyle = -\ln(x) + \sum_{n \geq 1}(-1)^{n+1} \left[\frac{1}{n+1} \frac{B_{-n}(1+xD)}{-n} + \frac{(xD)^n}{n} \right]$

$\displaystyle = -\ln(x) + \sum_{n \geq 1}(-1)^{n+1} \left[\frac{1}{n+1} \frac{B_{-n}(1+ST1.(:xD:))}{-n} + \frac{ST1_n(:xD:)}{n} \right].$

Acting on $x^s$ , the last two reps converge only for $0 \leq s < 1$ whereas the two above, for $s > -1$ , and $\psi(1+s)$ is nonsingular for $s$ any real except $s = -1,-2,-3,\cdots$ . This is not an issue in using the operators to generate the Taylor series in $t$

$\displaystyle D^{-t}1 = e^{-t \ln(D_x)} 1 =\sum_{n \geq 0} \frac{t^n}{n!}R^n 1 = \sum_{n \geq 0} f_n(x) \frac{t^n}{n!}= \sum_{n \geq 0}( \partial_{t=0}^n\frac{x^{t}}{t!}) \frac{t^n}{n!} = \frac{x^{t}}{t!}$

convergent for $x >0$ and all real $t$ as an entire function in $t$ , where here $R = -\ln(D_x) = \ln(x) -\psi(1+xD_x)$ is the raising op for $f_n(z)$ with $f_0(x) =1$ ; that is, $R \; f_n(x) = f_{n+1}(x)$ .

____________

Another consistency check is provided by another rep of the Bernoulli number umbral translation op / Bernoulli polynomial umbral subsitution op $T_{z \to (z+b.)=B.(z)}$ for action on $z^{-s}$ for $Re(s) > 0$ and for $s = 0,-1,-2, ...$ with $H(t) \frac{t^{-n-1}}{(-n-1)!} = \delta^{(n)}(t)$ is provided by the Ramanujan master heuristic / Mellin transform analytic continuation of the Bernoulli numbers and polynomials to the Hurwitz zeta function:

From the mathmage Euler, we have the hybrid Laplace-Mellin transform

$\displaystyle \int_0^{\infty} e^{-zt} \frac{t^{s-1}}{(s-1)!}dt = \frac{1}{z^s} = z^{-s},$

$\displaystyle T_{z \to (z+b.) = B.(z)} \int_0^{\infty} e^{-zt} \frac{t^{s-1}}{(s-1)!}dt$

$\displaystyle := \int_0^{\infty} T_{z \to (z+b.) = B.(z)} e^{-zt} \frac{t^{s-1}}{(s-1)!}dt$

$\displaystyle = \int_0^{\infty} e^{-B.(z)t} \frac{t^{s-1}}{(s-1)!}dt$

$\displaystyle = \int_0^{\infty} e^{-b.t}e^{-zt} \frac{t^{s-1}}{(s-1)!}dt$

$\displaystyle = \int_0^{\infty} \frac{-t}{e^{-t}-1}e^{-zt} \frac{t^{s-1}}{(s-1)!}dt$

$\displaystyle = T_{z \to (z+b.)=B.(z)} \frac{1}{z^s}=: \frac{1}{(b.+z)^s} := (B.(z))^{-s} = B_{-s}(z) = s\zeta(1+s,z).$

A Hankel complex-contour formulation of Euler’s iconic integral extends the analysis to $z^{s}$ for all complex $s$ and therefore the rep for $T_{z \to (z+b.)=B.(z)}$ and for $B_s(z)$ for all complex $s$ .

A similar integral rep applies for the inverse op $T_{z \to (z+r.)=R.(z)}$ , so the inverse nature of the relationship between the two ops is manifest in the eigenvalue identity

$\displaystyle \frac{-t}{e^{-t}-1} \frac{e^{-t}-1}{-t} = 1$

ensconced in the action

$\displaystyle T_{z \to (z+b.)}T_{z \to (z+r.)}e^{-zt} = \frac{\partial_z}{e^{\partial_z}-1} \frac{e^{\partial_z}-1}{\partial_z} e^{-zt} = \frac{-t}{e^{-t}-1} \frac{e^{-t}-1}{-t}e^{-zt} = e^{-zt}$ .

A consistency / coherency check:

From the Mellin transform formulation,

$\displaystyle B_{-n}(z) = \int_0^{\infty} \frac{-t}{e^{-t}-1} e^{-zt} \frac{t^{n-1}}{(n-1)!}dt$

$\displaystyle \sum_{n \geq 1}(-1)^n \frac{B_{-n}(z)}{-n} = \int_0^{\infty} \frac{-t}{e^{-t}-1} e^{-zt}\sum_{n \geq 1}(-1)^{n+1} \frac{t^{n-1}}{n!}dt = \int_0^{\infty} \frac{-t}{e^{-t}-1} e^{-zt} \frac{-(e^{-t}-1)}{t}dt$

$\displaystyle = \int_0^{\infty} e^{-zt}dt = \frac{1}{z}$ ,

agreeing with the other formulations.

____________

Using the Mellin transform gives the integral reps

$\displaystyle \sum_{n \geq 1}(-1)^{n+1} \frac{1}{n+1} \frac{B_{-n}(z)}{-n} = \int_0^{\infty} \frac{-t}{e^{-t}-1} e^{-zt}\sum_{n \geq 1}(-1)^n \frac{t^{n-1}}{(n+1)!}dt$

$\displaystyle = \int_0^{\infty} \frac{-t}{e^{-t}-1} e^{-zt} \frac{1}{t^2}\sum_{n \geq 1}(-1)^n \frac{t^{n+1}}{(n+1)!}dt$

$\displaystyle = \int_0^{\infty} \frac{-t}{e^{-t}-1} e^{-zt} \frac{-1}{t^2} \sum_{n \geq 1}(-1)^{n+1} \frac{t^{n+1}}{(n+1)!}dt$

$\displaystyle = \int_0^{\infty} \frac{-t}{e^{-t}-1} e^{-zt} \frac{-1}{t^2} (e^{-t}+t-1) dt = \int_0^{\infty} e^{-zt} \frac{(e^{-t}-1+t)}{t(e^{-t}-1)} dt$

$\displaystyle = \int_0^{\infty} e^{-zt} \left(\frac{1}{t}+ \frac{1}{e^{-t}-1}\right) dt$

$\displaystyle = \int_0^{\infty} e^{-zt} \frac{-1}{t}\left(-1+ \frac{-t}{e^{-t}-1}\right) dt.$

Then

$\displaystyle \psi(z) = \ln(z) + \int_0^{\infty} e^{-zt} \left(\frac{1}{t}+ \frac{1}{e^{-t}-1}\right) dt = \ln(z) + \int_0^{\infty} e^{-zt} \frac{(e^{-t}-1+t)}{t(e^{-t}-1)} dt$ .

Be aware for numerical checks that the online Demos app has some difficulty calculating and plotting the first Laplace transform but not the second; however, the first is Binet’s expression (22) on page 18 (pdf frame 44) of section 1.7.2 of Higher Transcendental Functions Vol. 1 of the Bateman Manuscript Project of Erdelyi (editor) and Magnus, Oberhettinger, and Tricomi (research associates), valid for $Re(z) > 0$ .

I’ve already presented the Hasse-type formula

$\displaystyle\ ln(B.(z)) := T_{z \to (z+b.)=B.(z)} \ln(z) = \psi(z)$ .

This leads to the operator interpretation/derivation of Binet’s formula as

$\displaystyle \psi(z) = T_{z \to (z+b.)=B.(z)} \ln(z)$

$\displaystyle = \ln(z) - \ln(z) + T_{z \to (z+b.)=B.(z)} \ln(z)$

$\displaystyle = D_z^{-1} \frac{1}{z} - D_z^{-1} \frac{1}{z} + T_{z \to (z+b.)=B.(z)} D_z^{-1} \frac{1}{z}$

$\displaystyle = [D_z^{-1} - D_z^{-1} + T_{z \to (z+b.)=B.(z)} D_z^{-1}] \int_0^{\infty}e^{-zt}dt$

$\displaystyle = \ln(z) + \int_0^{\infty}[ - D_z^{-1} + T_{z \to (z+b.)=B.(z)} D_z^{-1}] e^{-zt} dt$

$\displaystyle = \ln(z) + \int_0^{\infty}\left( - \frac{1}{D_z} + e^{b.D_z} \frac{1}{D_z}\right) e^{-zt} dt$

$\displaystyle = \ln(z) + \int_0^{\infty}\left(- \frac{1}{D_z} + \frac{D_z}{e^{D+z}-1} \frac{1}{D_z}\right) e^{-zt} dt$

$\displaystyle = \ln(z) + \int_0^{\infty}\left(- \frac{1}{D_z} + \frac{1}{e^{D+z}-1} \right) e^{-zt} dt$

$\displaystyle = \ln(z) + \int_0^{\infty} e^{-zt} \left(\frac{1}{t}+ \frac{1}{e^{-t}-1}\right) dt$ .

(See below for arguments that $D^{-1} \frac{1}{z} = \ln(z)$ .)

____________

Letting $u= e^{-t}$ , or $t=-\ln(u)$ , Mellin transform and fractional integral reps are obtained from the Laplace transform rep as

$\displaystyle \psi(z) = \ln(z) - \int_0^1 u^{z-1} \left(\frac{1}{\ln(u)} + \frac{1}{1-u} \right)du = \ln(z) + \int_0^{1} u^{z-1} \frac{-u+1+\ln(u)}{\ln(u)(u-1)} du$

$\displaystyle = \ln(z) - \int_0^{1} (1-u)^{z-1} \left(\frac{1}{\ln(1-u)}+ \frac{1}{u}\right) du= \ln(z) - \int_0^{1} (1-u)^{z-1} \frac{u+\ln(1-u)}{u\ln(1-u)} du$

$\displaystyle = \ln(z) - (z-1)!D_{t=1}^{-z} \left(\frac{1}{\ln(1-t)}+ \frac{1}{t}\right)= \ln(z) - (z-1)!D_{t=1}^{-z} \frac{t+\ln(1-t)}{t\ln(1-t)}$ .

(For numerical checks with Desmos, the limits of the integrals might need to be expressed as $1 -\epsilon$ and $0+\epsilon$ for $\epsilon$ positive and almost $0$ .)

____________

For some more reps / identities of the digamma function, see my posts here

“A Diorama of the Digamma” and

“Appells and Roses: Newton, Leibniz, Euler, Riemann and Symmetric Polynomials“.

From this last post (see linked MS doc):

The Graves-Pincherle commutator derivatives for general lowering and raising operators (think $D$ and $x$ for the prototypical Sheffer polynomial sequence the power monomials $x^n$ ) that satisfy

$\displaystyle ad_L R = [L,R] = LR - RL = 1$

are, for a wide class of functions,

$[f(L), R] = D_L f(L)= f'(L)$

and

$[L,f(R)] = D_R f(R)= f'(R)$ .

The infinitesimal generator $IG$ for the Heaviside-Pincherle fractional calculus operator $D^t$ is the negative of the raising op $R_z$ , i.e.,

$IG = \ln(D_z) = -\ln(x) + \psi(1+zD_z) =-R_z$ ,

for the Taylor series coefficients $f_n(z)$ of the Taylor series of the entire function $\frac{z^t}{t!}$ in $t$ for $Re(z) > 0$ ; i.e.,

$\displaystyle \frac{z^t}{t!} = e^{tR_z}1 = \sum_{n \geq 0} R_z^n 1 \frac{t^n}{n!} = \sum_{n \geq 0} f_n(z) \frac{t^n}{n!},$

where $f_0(z) =1$ and, for $n >0$ ,

$\displaystyle R_z \; f_{n-1}(z) = f_n(z) = \partial_{t=0}^n \frac{z^t}{t!}.$

With $x = \ln(z)$ , or $z = e^x$ , we have $zD_z = D_x$ and

$\displaystyle R_z = \ln(z) - \psi(1+zD_z) = x - \psi(1+D_x) = R_x$

with $R_x$ the raising operator of the Appell Sheffer polynomial sequence

$\displaystyle f_n(e^x) = p_n(x) = R_x^n 1 = \partial_{t=0}^n \frac{e^{xt}}{t!}$

with the e.g.f. (Taylor series entire in $t$ for all $x$ )

$\displaystyle e^{p.(x)t} = \frac{1}{t!}e^{xt}= e^{a.t}e^{xt} = e^{(a.+x)t}$

with

$\displaystyle (a.)^n = a_n = p_n(0) = f_n(1) = \partial_{t=0}^n \frac{1}{t!}$ ,

so the lowering op for $p_n(x)$ is $\partial_x = D_x$ (as for all Appell Sheffer polynomials sequences) since

$\displaystyle L_x \; p_n(x) = D_x (a.+x)^n = n \; (a.+x)^{n-1} = n\; p_{n-1}(x)$ .

The lowering op for the function sequence $f_n(z)$ must then be $L_z = zD_z = D_x = L_x$ which is corroborated by

$\displaystyle ad_{L_x} R_x = [L_x, Rxz] = [D_x, x-\psi(1+D_x)] = [D_x,x] = 1$

and

$\displaystyle ad_{L_z} R_x = [zD_z, R_z] = [zD_z, \ln(z)-\psi(1+zD_z)] = [zD_z,\ln(z)]$

$\displaystyle = zD_z\ln(z) - \ln(z)zD_z = z(D_z\ln(z))+ z\ln(z)D_z - \ln(z)zD_z = z (1/z) = 1$ .

The Graves-Pincherle commutator derivative then gives

$\displaystyle [\ln(D_z),z] = \partial_{D_z}\ln(D_z) = D_z^{-1}$

$\displaystyle = [-R_z,z] = [-\ln(z)+\psi(1+zD_z),z)] = [\psi(1+zD_z),z]$

and, using either

$\displaystyle \psi(z) = -\gamma + \sum_{n \geq 1} \left( \frac{1}{n} - \frac{1}{n+z} \right)$

or the recurrence relation

$\displaystyle \delta_z \psi(z) = \psi(z+1) - \psi(z) = \frac{1}{z}$ ,

we have the action

$\displaystyle D_z^{-1}z^s = [\psi(1+zD_z),z]z^s = \psi(1+zD_z)z^{s+1} - z \psi(1+zD_z)z^s$

$\displaystyle = (\psi(s+1) - \psi(s)) z^{s+1} = \frac{z^{s+1}}{s+1}.$

For $s =-1$ , use the limiting case

$\displaystyle \frac{z^{s+1}}{s+1} = \lim_{\epsilon \to 0} \left(\frac{z^{s+1+\epsilon}}{s+1+\epsilon} + \frac{z^{s+1-\epsilon}}{s+1-\epsilon} \right)$ ,

$\displaystyle D_z^{-1} \frac{1}{z} = \ln(z).$

This is consistent, of course, with other reps of $D^t$ and $\ln(D)$ for the Heaviside-Pincherle fractional calculus.

Note then

$\displaystyle \ln([z,R_z])= \ln(1/D_z) = -\ln(D_z) = R_z,$

so again appears the logarithm in juxtaposition with the digamma as

$\displaystyle \ln([z, \psi(1+zD_z)]) =-\ln(z) + \psi(1+zD_z)$ .

Exponentiating gives

$\displaystyle e^{t\ln([z,R_z])}= e^{t\ln(1/D_z) }= e^{-t\ln(D_z)} = D_z^{-t} = e^{tR_z}.$

My Math StackExchange question “Lie group heuristics for a raising operator for $(-1)^n \frac{d^n}{d\beta^n}\frac{x^\beta}{\beta!}|_{\beta=0}$ ” and self-answer check the consistency of this methodology in a few ways. One example is by noting with

$\displaystyle \triangledown_n^s c_n = \sum_{n=0}^\infty (-1)^n \binom{s}{n}c_n,$

the Newton series formulation gives

$\displaystyle D_z^{-\beta}\frac{x^{\alpha}}{\alpha!}= (1- (1-D^{-1}))^{\beta}\frac{z^{\alpha}}{\alpha!} =\triangledown^{\beta}_n \triangledown_j^n D^{-j} \frac{z^{\alpha}}{\alpha!}$

$\displaystyle =\triangledown^{\beta}_n \triangledown_j^n \frac{z^{j+\alpha}}{(j+\alpha)!}=\frac{z^{\alpha+\beta}}{(\alpha+\beta)!}.$

Then

$\displaystyle \frac{d}{d\beta} D_z^{-\beta}\frac{z^{\alpha}}{\alpha!} |_{\beta=0} = \frac{d}{d\beta} e^{\beta R_z}\frac{x^{\alpha}}{\alpha!} |_{\beta=0} = R_z\frac{x^{\alpha}}{\alpha!}$ $

$\displaystyle =\frac{d}{d\beta}\frac{x^{\alpha+\beta}}{(\alpha+\beta)!}|_{\beta=0}=\ln(1- (1-D_z^{-1} ))\frac{x^{\alpha}}{\alpha!} = \ln(D_z^{-1})\frac{x^{\alpha}}{\alpha!} = -\ln(D_z)\frac{x^{\alpha}}{\alpha!}$

$\displaystyle =\frac{d}{d\beta} \triangledown^{\beta}_{n} \triangledown^{n}_{j} \frac{z^{j+\alpha}}{(j+\alpha)!} |_{\beta=0} = \sum_{n \geq 0}(-1)^n \frac{d}{d\beta}\binom{\beta}{n} \triangledown^{n}_{j} \frac{z^{j+\alpha}}{(j+\alpha)!} |_{\beta=0}$

$\displaystyle =-\sum_{n=1}^{\infty }\frac{1}{n} \triangledown^n_j \frac{x^{j+\alpha}}{(j+\alpha)!}=-\sum_{n=1}^{\infty}\frac{1}{n} \sum^n_{j=0}(-1)^j \binom{n}{j} e^{j\partial_\alpha}\frac{z^{\alpha}}{\alpha!}$

$\displaystyle =-\sum_{n=1}^{\infty }\frac{1}{n} (1-e^{\partial_\alpha})^n \frac{x^{\alpha}}{\alpha!} = \sum_{n=1}^{\infty }\frac{(-1)^{n+1}}{n} \delta_\alpha^n \frac{x^{\alpha}}{\alpha!}= \frac{\ln(1+\delta_\alpha)}{\delta_\alpha}\delta_\alpha \frac{z^{\alpha}}{\alpha!}$

$\displaystyle = \frac{\partial_\alpha}{e^{\partial_\alpha}-1}\delta_\alpha \frac{z^{\alpha}}{\alpha!}= e^{b.\partial_\alpha} \left(\frac{z^{\alpha+1}}{(\alpha+1)!} -\frac{z^{\alpha}}{\alpha! }\right) = T_{\alpha \to (\alpha+b.)=B.(\alpha) }\left(\frac{z^{\alpha+1}}{(\alpha+1)!} -\frac{z^{\alpha}}{\alpha! }\right)$

$\displaystyle = \frac{z^{\alpha+1+b.}}{(\alpha+1+b.)!} -\frac{z^{\alpha+b.}}{(\alpha+b.)! }$

$\displaystyle = \frac{z^{B.(\alpha+1)}}{(B.(\alpha+1))!} -\frac{z^{B.(\alpha)}}{(B.(\alpha))! }$

$\displaystyle = \frac{\partial_\alpha}{e^{\partial_\alpha}-1}(e^{\partial_\alpha}-1)\frac{z^{\alpha}}{\alpha!} =\partial_\alpha \frac{z^{\alpha}}{\alpha!}$

$\displaystyle = (\ln(z) - \psi(1+\alpha)) \frac{z^{\alpha}}{\alpha!} = (\ln(z) - \psi(1+zD_z)) \frac{z^{\alpha}}{\alpha!} .$

Then extracting operators, we have our rep for the IG

$\displaystyle \ln(D_z) = -R_z = -\ln(z) + \psi(1+zD_z) .$

The raising and lowering ops for the function sequence

$\displaystyle f_n(z) = p_n(\ln(z))$

are

$\displaystyle R_z = -\ln(D_z) = \ln(z) - \psi(1+zD_z)$

and

$\displaystyle L_z = zD_z$ ,

so the Graves-Pincherle commutator derivative gives

$\displaystyle [\ln(D_z), L_z^n] = -[R_z, L_z^n] = [L_z^n, R_z] = \partial_{L_z} L_z^n = n \; L_z^{n-1}$

$\displaystyle = [-\ln(z)+\psi(1+zD_z), (zD_z)^n] = -[\ln(z),(zD_Z)^n]= -ad_{\ln(z)}(zD_z)^n$

$\displaystyle = [-\ln(z)+\psi(1+ST2.(:zD_z:), ST2_n(:zD_z:)] = -[\ln(z),ST2_n(:zD_z:)]$

$\displaystyle = -ad_{\ln(z)}ST2_n(:zD_z:)$

$\displaystyle = n \; (zD_z)^{n-1} = n \; ST2_{n-1}(:zD_z:) = \ln(1+\partial_{:zD_z:}) ST2_n(:zD:)$

since the lowering operator of the Stirling polynomials of the second kind $ST2_n(z)$ is $\displaystyle L_{ST2} =\ln(1+\partial_z)$ ,

$\displaystyle \partial_{L_z} = \partial_{(zD_z)} = \ln(1+\partial_{:zD_z:}) = -ad_{R_z} = ad_{\ln(D_z)} = -ad_{\ln(z)}.$

Then the conjugation-commutatior operator identity of Lie theory (see “Adjoint representation“-Wikipedia) for operator $A$ acting on operator $C$

$\displaystyle e^{t\; ad_A} C= Ad_{e^{tA}}C = e^{tA}Ce^{-tA}$

along with

$\displaystyle \partial_{zD_z} = \ln(1+\partial_{:zD_z:})$

and the inverse

$\displaystyle \partial_{:zD_z:} = e^{\partial_{zD}} -1$

implies

$\displaystyle e^{t\; \partial_{L_z}} = e^{t\partial_{(zD_z)}} =(1+\partial_{:zD_z:})^t= e^{-t \; ad_{R_z}}$

$\displaystyle = Ad_{e^{-tR_z}} = e^{t \; ad_{\ln(D_z)}} = Ad_{e^{t\ln(D_z)}} = Ad_{D_z^t} = e^{-t ad_{\ln(z)}} = Ad_{e^{-t\ln(z)}} = Ad_{z^{-t}}$

$\displaystyle e^{t\partial_{(zD_z)}} (zD_z)^\alpha = (zD_z+t)^{\alpha}$

$\displaystyle = e^{t \; ad_{\ln(D_z)}}(zD_z)^\alpha = Ad_{D_z^t} (zD_z)^\alpha = D_z^t (zD_z)^\alpha D_z^{-t}$

$\displaystyle = e^{-t ad_{\ln(z)}} (zD_z)^\alpha = Ad_{z^{-t}} (zD_z)^\alpha = z^{-t} (zD_z)^\alpha z^t,$

$\displaystyle =(1+\partial_{:zD_z:})^t (zD_z)^\alpha = \sum_{n \geq 0} \binom{t}{n} \partial_{:zD_z:}^n(zD_z)^\alpha =\sum_{n \geq 0} \binom{t}{n} (e^{\partial_{zD}}-1)^n(zD_z)^\alpha$

$\displaystyle = \sum_{n \geq 0}(-1)^n \binom{t}{n} \sum_{k=0}^n (-1)^k \binom{n}{k} e^{k\partial_{zD_z}}(zD_z)^\alpha$

$\displaystyle = \sum_{n \geq 0}(-1)^n \binom{t}{n} \sum_{k=0}^n (-1)^k \binom{n}{k} (zD_z+k)^\alpha$

consistent with the actions

$\displaystyle (s+t)^{\alpha} z^s = (zD_z+t)^\alpha z^s$

$\displaystyle = D_z^t (zD_z)^\alpha D_z^{-t}z^s= D_z^t (zD_z)^\alpha \frac{s!}{(s+t)!} z^{s+t}$

$\displaystyle = D_z^t (s+t)^\alpha \frac{s!}{(s+t)!} z^{s+t} =(s+t)^\alpha \frac{s!}{(s+t)!} D^t z^{s+t} = (s+t)^\alpha \frac{s!}{(s+t)!} \frac{(s+t)!}{s!} z^{s}$

$\displaystyle = z^{-t} (zD_z)^\alpha z^t z^s = z^{-t} (zD_z)^\alpha z^{t+s} = z^{-t}z^{t+s}(s+t)^{\alpha}$

$\displaystyle =(1+\partial_{:zD_z:})^t (zD_z)^\alpha z^s = \sum_{n \geq 0}(-1)^n \binom{t}{n} \sum_{k=0}^n (-1)^k \binom{n}{k} (zD_z+k)^\alpha z^s$

$\displaystyle = \sum_{n \geq 0}(-1)^n \binom{t}{n} \sum_{k=0}^n (-1)^k \binom{n}{k} (s+k)^\alpha z^{s}.$

Also when acting on suitable operator functions $h(zD_z)= h(St2.(:zD_z:))$ ,

$\displaystyle (\delta_{(zD_z)} + 1)^t = e^{t\partial_{(zD_z)}} = e^{ -t\; ad_{R_z}} = e^{t \; ad_{\ln(D_z)}} = Ad_{D^t} = e^{-t\;ad_{\ln(z)}} = Ad_{x^{-t}}$

$\displaystyle = Ad_{D_z^t} = Ad_{x^{-t}} .$

Then

$\displaystyle \ln(1+\delta_{zD_z}) = \partial_{(zD_z)} = -ad_{R_z} = ad_{\ln(D_z)} = -ad_{\ln(z)} =\frac{\ln(Ad_{D^t})}{t} = \frac{\ln(Ad_{x^{-t}})}{t}$

$\displaystyle = \sum_{n \geq 1} (-1)^{n+1} \frac{\delta_{(zD_z)}^n}{n}.$

In addition, as for $\frac{\partial_z}{\delta_z}$ ,

$\displaystyle \frac{\partial_{(zD_z)}}{\delta_{zD_z}} = \frac{\ln(1+\delta_{zD_z})}{\delta_{zD_z}} = e^{q.\delta_{zD_z}}$

$\displaystyle = \frac{zD_z}{e^{zD_z}-1} = e^{b.\partial_{zD_z}} = T_{zD_z \to (zD_z+b.)= B.(zD_z)} = \sum_{n \geq 0} (-1)^n \frac{1}{n+1} \delta_{zD_z}^n$

$\displaystyle = \sum_{n \geq 0} (-1)^n \frac{1}{n+1} \sum_{k=0}^n (-1)^k \binom{n}{k} T_{zD_z \to (zD_z+k)}$ .

Then

$\displaystyle \psi(1+zD_z) = \ln(B.(1+zD_z) = \frac{\partial_{(zD_z)}}{\delta_{zD_z}} \ln(1+zD_z)= T_{zD_z \to (zD_z+b.)} \ln(1+zD_z)$

$\displaystyle = \frac{ad_{\ln(D_z)}}{\delta_{zD_z}} \ln(1+zD_z) = \frac{-ad_{\ln(z)}}{\delta_{zD_z}} \ln(1+zD_z) .$

In my next post, I’ll look at some other ramifications of this methodology, in particular, I’ll review and extend some analysis on the commutator

$\displaystyle [\ln(D_z), :zD_z:^n] = [\ln(D_z), (xD)_{\underline{n}}] = [\ln(D_z), ST1_n(zD)].$

____________

I’ve just belatedly googled “fractional calculus and the Riemann zeta function” for the first time (what an oversight!) and came across the following related, interesting papers (trails to explore);

See “Double integrals and infinite products for some classical constants via analytic continuations of Lerch’s transcendent” by Guillera and Sondow: Theorem 2.2 on p. 5 sketches a proof of the Helmut Hasse formula for the Hurwitz zeta. Theorem 5.1 on p. 18 has the series for $\psi(x)$ convergent for $x >0$ . On p. 42 is the identity $\psi(z) = \lim_{s \to 1} \left[ \frac{-1}{1-z} \Phi(1,s,z)\right] = \lim_{s \to 1} \left[ \frac{-1}{1-z} \zeta(s,z)\right]$ where $\Phi(z,s,u) = \frac{1}{u^s} + \frac{z}{(u+1)^s} + \frac{z^2}{(u+2)^s} + \cdots$ is the Lerch transcendent.

The bottom of p, 20 of “Fractional calculus and its reltionship to the Riemann zeta function” by Jerry Keiper (thesis, 1975) has

$\displaystyle \lim_{s \to 1} \zeta(s) - \zeta(s,z) = \frac{-1}{z} \sum_{ n \geq 1} \frac{1}{n} - \frac{1}{n+x} = \psi(z) + \gamma$ ,

on p. 24,

$\displaystyle \lim_{s \to 1} \zeta(s) - \frac{1}{s-1} = \gamma$ ,

on p. 22, for $Re(z) \geq 1$ ,

$\displaystyle -\ln(z) + \psi(z) = \sum_{n \geq 2}(-1)^{n+1} \frac{1}{n} \zeta(n,z)$ ,

and, on p. 29, Theorem XIV has the Cauchy rep for the fractional derivative.

The set of slides “Fractional calculus and zeta functions” by Fernandez reviews Keiper’s work (wit a brief bio) on the relationship of the Riemann Hurwitz zeta function to fractional derivatives of the digamma function (pdf frame 23) and more current work by others on fractional derivatives of the Riemann zeta.

See also other different approaches to extending the Bernoulli polynomials in “Analytic Continuation of Bernoulli Numbers, a New Formula for the Riemann Zeta Function, and the Phenomenon of Scattering of Zeros” and a different Bernoulli function in “A New Representation of the Riemann Zeta Function ζ(s)” both by J. C. Woon and relations among the Riemann zeta and digamma functions and other families of fractional derivatives in “Analytic Continuation of Operators — operators acting complex s-times — Applications: from Number Theory and Group Theory to Quantum Field and String Theories” also by Woon.

]]> https://tcjpn.wordpress.com/2024/12/31/boil-and-bubble-conjuring-the-digamma-and-logd-by-umbrally-mixing-the-logarithm-with-the-bernoulli-function/feed/ 0 tcjpn Bernoulli number umbral translation / polynomial umbral substitution operator and binomial Sheffer polynomial sequences https://tcjpn.wordpress.com/2024/12/19/bernoulli-number-umbral-translation-polynomial-umbral-substitution-operator-and-binomial-sheffer-polynomial-sequences/ https://tcjpn.wordpress.com/2024/12/19/bernoulli-number-umbral-translation-polynomial-umbral-substitution-operator-and-binomial-sheffer-polynomial-sequences/#respond Thu, 19 Dec 2024 08:38:51 +0000 https://tcjpn.wordpress.com/?p=10197 Continue reading →]]> The Helmut Hasse finite diff rep of Bernoulli number umbral translation, or Bernoulli polynomial umbral substitution, discussed in my previous posts,

$\displaystyle T_{x \to (x+b.) = B.(x)}f(x) = f(b.+x) = f(B.(x))$

$\displaystyle = \sum_{n \geq 0} (-1)^n \binom{b.}{n} \sum_{k =0}^n (-1)^n \binom{n}{k} f(k+x)$

$\displaystyle = \sum_{n \geq 0} \frac{1}{n+1} \sum_{k =0}^n (-1)^n \binom{n}{k} f(k+x),$

is most easily seen as an umbral generalization of Newton series translation, a translation expressed in the basis of falling factorials / generalized binomial coefficients rather than the divided power monomials of Taylor series.

Details, reprising earlier notes, of relations to the operator calculus of binomial Sheffer polynomial sequences are presented in my pdf

Bernoulli number umbral translation / polynomial umbral substitution and binomial Sheffer polynomial sequences.

Caveat: I let the equals symbol do a lot of work here. It would perhaps be best to use the symbols $\leftrightharpoons$ and $\rightleftharpoons$ with, e.g., $e^{b.\partial_x} \rightleftharpoons e^{(b.)_{\underline{.}}\delta_x}$ with the upper lifted/raised tick on the right side of the symbol indicating the op on the RHS gives convergent results when acting on a wider domain (lifted increased width) of functions than the op on the LHS does and vice versa for the lower depressed tick (lowered decreased width) on the left side of the symbol.

]]> https://tcjpn.wordpress.com/2024/12/19/bernoulli-number-umbral-translation-polynomial-umbral-substitution-operator-and-binomial-sheffer-polynomial-sequences/feed/ 0 tcjpn The Joy of Polygammary: Umbral witchcraft and the polygamma functions vis a vis the Bernoulli function https://tcjpn.wordpress.com/2024/12/11/the-joy-of-polygammary-umbral-witchcraft-and-the-polygamma-functions-vis-a-vis-the-bernoulli-function/ https://tcjpn.wordpress.com/2024/12/11/the-joy-of-polygammary-umbral-witchcraft-and-the-polygamma-functions-vis-a-vis-the-bernoulli-function/#respond Thu, 12 Dec 2024 00:52:24 +0000 https://tcjpn.wordpress.com/?p=10176 Continue reading →]]> The objective of this short pdf is to reveal, via the extended umbral operational calculus of Appell Sheffer polynomial sequences, relationships among the Bernoulli function; its umbral inverse function, the reciprocal function; and the digamma function and its derivatives, the polygamma functions:

The Joy of Polygammary: Umbral witchcraft and the polygamma functions vis a vis the Bernoulli function (pdf)

Related stuff:

“On polylogarithms, Hurwitz zeta functions, and the Kubert identities” by Milnor. Note that the umbral Bernoulli number translation op / Bernoulli polynomial subsitution op in my pdf above and earlier ones provides the correct extension of the Bernoulli diff op (Todd op)

$\displaystyle T_{x \to (x+b.)=B.(x)} = B(\partial_x) = \frac{\partial_x}{e^{\partial_x}-1}$

$\displaystyle T_{x \to (x+b.)=B.(x)}=\sum_{n \geq 0} \frac{1}{n+1} \sum_{k=0}^n (-1)^k \binom{n}{k} T_{x \to x+k}$

so that the correct dual identity for

$\displaystyle \frac{e^{\partial_x}-1}{\partial_x} \gamma_1(x) = x \; \ln(x) - x$

on page 314 of Milnor is

$\displaystyle \gamma_1(x) = \ln\left( \frac{(x-1)!}{\sqrt{2 \pi}}\right) = \sum_{n \geq 0} \frac{1}{n+1} \sum_{k=0}^n (-1)^k \binom{n}{k} [(x+k) \; \ln(x+k) - (x+k)]$

for $\displaystyle x > 0$ (see this Demos plot for a numerical check).

Taking the derivative w.r.t. $x$ gives an expansion for the digamma function

$\displaystyle \psi(x) = \frac{d}{dx}\ln\left(\left(x-1\right)!\right) = \sum_{n=0}^{\infty}\frac{1}{n+1}\sum_{k=0}^{n}\left(-1\right)^{k}\ \frac{n!}{k!\left(n-k\right)!}\ln\left(x+k\right)$

for $x > 0$ (see this Demos plot for a numerical check).

Iterating the differentiation generates the polygamma functions, and it is in the sense of the operation above that the umbral subsitution

$\displaystyle \ln(B.(x)) := T_{x \to (x+b.) = B.(x)} \ln(x) = \psi(x)$

can be interpreted, which for action on $x^n$ gives $B.(x)^n = B_n(x) = (b.+x)^n$ , involving the Bernoulli polynomials and numbers.

This is consistent with the operational definition of the Bernoulli number umbral translation / Bernoulli polynomial umbral substitution for a wide class of functions $f(x)$

$\displaystyle f(B.(x+1)) - f(B.(x)) = f(b.+1+x) - f(b.+x) = \partial_x f(x)$

and the recursion formula for the digamma function, that is,

$\displaystyle \ln(B.(x+1)) - \ln(B.(x)) = \psi(x+1) - \psi(x) = \partial_x \ln(x) = \frac{1}{x}.$

This is also consistent with an extension of the umbral inversion relationship between the Bernoulli and the reciprocal polynomials to that between the Bernuolli function and reciprocal function.

For the polynomials for $n=0,1,2, \cdots$ ,

$\displaystyle B_n(x) = e^{b.\partial_x} x^n = (b.+x)^n = \sum_{k=0}^n \binom{n}{k} b_n x^{n-k} = T_{x \to (x+b.)=B.(x)} x^n$

and

$\displaystyle R_n(x) = e^{r.\partial_x} x^n = (r.+x)^n = \sum_{k=0}^n \binom{n}{k} \frac{1}{k+1} x^{n-k}$

$\displaystyle = T_{x \to (r.+x) = R.(x)} x^n = \int_{x}^{x+1}t^n dt = \frac{(x+1)^{n+1}-x^{n+1}}{n+1},$

then

$\displaystyle x^n = \frac{\partial_x}{e^{\partial_x}-1} \frac{e^{\partial_x}-1} {\partial_x}x^n = e^{b.\partial_x}e^{r.\partial_x} x^n = e^{b.\partial_x}R_n(x) = R_n(B.(x))$

$\displaystyle = e^{r.\partial_x}e^{b.\partial_x} x^n = e^{r.\partial_x}B_n(x) = B_n(R.(x)) ;$

and for the functions for $s$ real or complex, with the Hurwitz zeta function $\zeta(s,x)$ ,

$\displaystyle B_s(x) = (b.+x)^s := T_{x \to (x+b.)=B.(x)} x^s = -s \; \zeta(-s+1,x)$

and

$\displaystyle R_s(x) = (r.+x)^s := \int_{x}^{x+1}t^s dt = \frac{(x+1)^{s+1}-x^{s+1}}{s+1}$ ,

then

$\displaystyle R_s(B.(x)) = x^s = B_s(R.(x)).$

For the limiting case $s \to -1$ ,

$\displaystyle R_{-1}(B.(x)) = \lim_{s \to -1} \frac{(B.(x)+1)^{s+1}-B.(x)^{s+1}}{s+1} =\ln(B.(x)+1) - \ln(B.(x) )$

$\displaystyle = \ln(B.(x+1)) - \ln(B.(x))= \partial_x \ln(x) = \frac{1}{x} = \psi(x+1)-\psi(x)$

$\displaystyle = \lim_{s \to -1} \frac{B_{s+1}(x+1)-B_{s+1}(x)}{s+1}= \lim_{s \to -1}[-\zeta(-s,x+1) + \zeta(-s,x)],$

and

$\displaystyle B_{-1}(R.(x)) = \zeta(2, R.(x)) = \sum_{n \geq 0} \frac{1}{(n + x)^2}|_{x \to R.(x)} = \sum_{n \geq 0} \frac{1}{(n + R.(x))^2}$

$\displaystyle = \sum_{n \geq 0} \frac{1}{(R.(x+n))^2} = \sum_{n \geq 0} R_{-2}(x+n)$

$\displaystyle = \sum_{n \geq 0} -[\frac{1}{x+n+1} - \frac{1}{x+n}] = \sum_{n \geq 0} \frac{1}{(x+n+1)(x+n)} = \frac{1}{x}.$

This Desmos plot demonstrates the good fit for $x > 0$ of

$\displaystyle y = \sum_{n=0}^{40}\frac{1}{n+1}\sum_{k=0}^{n}\left(-1\right)^{k}\ \frac{n!}{k!\left(n-k\right)!}\ln\left(\frac{x+1+k}{x+k}\right),$

an approximation of $\displaystyle R_{-1}(B.(x)) = T_{x \to (x+b.)=B.(x)}R_{-1}(x)$ ,

with $\displaystyle y = \frac{1}{x}$ .

Added Dec. 30, 2024:

The set of slides “Fractional calculus and zeta functions” by Arran Fernandez reviews (with a brief bio) Jerry Keiper’s work (pdf frame 23) on the relationship of the Riemann Hurwitz zeta function to fractional derivatives of the digamma function in Keiper’s thesis (1975) “Fractional calculus and its relationship to the Riemann zeta function” and more current work by others on fractional derivatives of the Riemann zeta. Keiper provided a generalization of the the relationship I show above between derivatives of the digamma function and the Hurwitz zeta. Using a different type of fractional derivative than I use, Keiper showed, for $Re(s) >1$ ,

$\displaystyle \zeta(s,x) = (-1)^s \frac{{}_{-\infty}D_x^{s-1}}{(s-1)!} \psi(x).$

Then, with this type of fractional derivative, for $Re(s) > 0$ ,

$\displaystyle B_{-s}(x) = s\zeta(1+s,x) = (-1)^{s+1} \frac{{}_{-\infty}D_x^{s}}{(s-1)!} \psi(x)$

with the limiting case

$\displaystyle B_{0}(x) = 1 = - \lim_{x \to 0^{+}} \frac{\psi(x)}{(x-1)!}$ .

]]> https://tcjpn.wordpress.com/2024/12/11/the-joy-of-polygammary-umbral-witchcraft-and-the-polygamma-functions-vis-a-vis-the-bernoulli-function/feed/ 0 tcjpn On Euler’s derivation of the e.g.f. of the Bernoulli numbers https://tcjpn.wordpress.com/2024/10/08/on-eulers-derivation-of-the-e-g-f-of-the-bernoulli-numbers/ https://tcjpn.wordpress.com/2024/10/08/on-eulers-derivation-of-the-e-g-f-of-the-bernoulli-numbers/#respond Tue, 08 Oct 2024 19:22:50 +0000 https://tcjpn.wordpress.com/?p=10166 Continue reading →]]> I added a supplement to one of my old MSE questions “Original author of an exponential generating function for the Bernoulli numbers?, that elaborates on Euler’s method of deriving the ecponential generating function of the Bernoulli numbers using a modern matrix format rather than Euler’s format. Hopefully, this clarifies Euler’s derivation for the reader.

]]> https://tcjpn.wordpress.com/2024/10/08/on-eulers-derivation-of-the-e-g-f-of-the-bernoulli-numbers/feed/ 0 tcjpn The Thermodynamic 3-D Permutahedron https://tcjpn.wordpress.com/2024/08/26/the-thermodynamic-3-d-permutahedron/ https://tcjpn.wordpress.com/2024/08/26/the-thermodynamic-3-d-permutahedron/#comments Mon, 26 Aug 2024 17:31:49 +0000 https://tcjpn.wordpress.com/?p=10141 Continue reading →]]> I’ve written over several years on the permutahedra / permutohedra–a family of convex polytopes–and the connections among their geometry, in particular, their associated refined Euler characteristics / refined, signed face polynomials; the algebra of multiplicative inversion of formal Taylor series, or, exponential generating functions; a matrix calculus, a Lie infinitesimal generator calculus, and differential operator calculus associated with Appell Sheffer polynomial sequences; symmetric polynomial / function theory; and an infinite dihedral group interweaving mutiplicative and compositional inversions of formal Taylor and Laurent series. But only today have I come across the relation of the 3-dimensional permutahedron, a.k.a. the truncated octahedron, to the Legendre transformations in thermodynamics as explained in the following paper;

Thermodynamic Venn diagrams: Sorting out forces, fluxes, and Legendre transforms by W. C. Kerr and J. C. Macosko

Vignette: I knew of Ronald Fox’s work cited in the paper since he was one of my brilliant and entertaining professors eons ago with, as he was proud to boast, a justifiably “big ego” which he was fond of saying “was other people’s problem, not mine”. Yet, he told his class–I had skipped class that day– “It blew my mind” when I correctly answered for the first time, in his many years of teaching, a nonlinear homework problem in his preferred standard textbook on thermodynamics–the apparent answer he had accepted up to that time was actually wrong. A friend/classmate told me this and that was good for my ego and at the same time a humbling reminder that everyone makes mistakes. (It was through googling ‘Fox thermodynamics Legendre transformation’ that I came across the article above, so yet again he has taught me some interesting mathematical physics.)

]]> https://tcjpn.wordpress.com/2024/08/26/the-thermodynamic-3-d-permutahedron/feed/ 2 tcjpn Umbral witchcraft with Bernoulli and Blissard https://tcjpn.wordpress.com/2024/08/25/umbral-witchcraft-with-bernoulli-and-blissard/ https://tcjpn.wordpress.com/2024/08/25/umbral-witchcraft-with-bernoulli-and-blissard/#comments Sun, 25 Aug 2024 21:06:49 +0000 https://tcjpn.wordpress.com/?p=10116 Continue reading →]]> Galvanized by comments in the recent post Bernoulli Numbers and the Harmonic Oscillator by John Baez on the blog site The n-Category Café, I’ve decided to post a fraction of some sets of notes I’ve developed on the appearance of the Bernoulli numbers, polynomials, and function in diverse areas of mathematics. I find the umbral Sheffer calculus to often be the most comprehensive and elegant way to derive and view the various relationships among the formulas involved in the related concepts, so I pay homage to Blissard in the title who was one of the first to use umbral characters in deriving Bernoulli number identities.

First, I’ll reiterate the relationship between a formal group law and the Bernoulli numbers:

The refs attached to my formulas dated Sep 18 2014 in OEIS A008292 are relevant to the connections between formal group laws and the Bernoulli numbers. As I commented in the Café, the shifted reciprocal of the bivariate e.g.f. for the Eulerian polynomials, the $Exp$ for the hyperbolic formal group law, generates the Bernoulli numbers. In more detail, given the compositional-inverse pair of functions

$Exp_{(Eulerian)}(x;a,b) = A(x,a,b) = \frac{e^{ax}-e^{bx}}{a e^{bx}-b e^{ax}}$

$= x + (a+b) \frac{x^2}{2!} + (a^2+4ab+b^2)\frac{x^3}{3!} + (a^3+11a^2b+11ab^2+b^3) \frac{x^4}{4!} + \cdots$

and

$Log_{(Eulerian)}(x;a,b) = B(x,a,b) = \frac{1}{a-b} \ln\left(\frac{1+ax}{1+bx}\right)$

$= x - (a+b)\frac{x^2}{2} + (a^2+ab+b^2)\frac{x^3}{3} - (a^3+a^2b+ab^2+b^3)\frac{x^4}{4} + ... ,$

the Bernoulli numbers appear in the shifted reciprocal

$\frac{x}{Exp_{(Eulerian)}(x;a,b)} = x \frac{a e^{bx}-b e^{ax}}{e^{ax}-e^{bx}}$

$= 1 + \frac{-1}{2}(a + b) x + \frac{1}{6}(a - b)^2 \frac{x^2}{2!} + \frac{-1}{30} (a - b)^4 \frac{x^4}{4!} + \cdots .$

Relevant refs are “Elliptic formal group laws, integral Hirzebruch genera and Krichever genera” by Victor Buchstaber and Elena Bunkova (the Bernoulli numbers occur on pgs. 36 & 37), “Towards generalized cohomology Schubert calculus via formal root polynomials” by Cristian Lenart and Kirill Zainoulline, and the more recent papers “Theta divisors and permutohedra” and “Chern-Dold character in complex cobordisms and theta divisors” by Buchstaber and A.P. Veselov (the Eulerian polynomials are the h-polynomials of the permutohedra).

$A(x,a,b)$ satisfies

$\frac{dA}{dx} = (1+aA)(1+bA),$

a Ricatti equation, and can be written in terms of a Weierstrass elliptic function (see Buchstaber & Bunkova), so

$A(B(z)+x) = e^{x(1+az)(1+bz)\partial_z} z$

and the formal group law is

$FGL(x,y) = A(B(x,a,b) + B(y,a,b),a,b) = \frac{x+y+(a+b)xy}{1-(ab)xy},$

called the hyperbolic formal group law and related to a generalized cohomology theory by Lenart and Zainoulline.

Note $(1+az)(1+bz)\partial_z = \partial_x + (a+b)z\partial_z + (ab)z^2\partial_x$ is a linear combination of the infinitesimal generators for $SL_2$ .

In my pdf in my post “The Elliptic Lie Triad: Riccati and KdV Equations, Infinigens, and Elliptic Genera” at my Web blog Shadows of Simplicity, I note in detail the relations of the compositional inverse pair $A(x,a,b)$ and $B(x,a,b)$ to the KdV equation and, therefore, to the Burgers’ equation, the heat equation, the Schwarzian derivative, and more.

The coefficients of $\frac{x}{Exp_{(Eulerian)}(x;a,b)}$ containing the Bernoulli numbers can be used to generate $Log_{(Eulerian)}(x;a,b)$ via the partition polynomials of OEIS A134264, the refined Narayana partition polynomials, enumerating / labeling Dyck paths, polygon dissections, forests, and noncrossing partitions; therefore, the Bernoulli numbers can be generated from $Log_{(Eulerian)}(x;a,b)$ by the inverse set of partition polynomials of A350499. This pair of sets of partition polynomials allows conversions between the free cumulants and the free moments of free probability theory and between the coefficients of a compositional-inverse pair of Laurent series. (In fact, at the bottom of p. 11 of “Todd polynomials and Hirzebruch numbers” by Buchstaber and Veselov is an example of the use of A350499 to calculate the Bernoulli numbers, of which they were unaware until I notified them in June.)

For details of the relationship between the Hirzebruch-Todd criterion and the K-P equation mentioned in comments in the Café, see my pdf

The Khovanskii-Pukhliko formula, the Hirzebruch-Todd criterion, the Bernoulli op, and umbral witchcraft.

(See also the answers to the MO-Q “Hirzebruch’s motivation of the Todd class” and “Formal Groups, Witt vectors and Free Probability” by R. Friedrich and J. McKay.)

For the links between the Bernoulli polynomials, Newton series, and the derivative / tangent space of functions, see my pdf

Conjuring the continuous from the discrete with umbral witchcraft: the derivative and the Bernoulli function.

For a review of a natural extension of the Bernoulli polynomials to the Bernoulli function of Milnor, see my pdf

Résumé of the Bernoulli function.

For some relationships among the quantum zeta function, the quantum partition function, heat kernels, statistical mechanics, the harmonic oscillator, Hermite and Bernoulli polynomials, the Bernoulli function and Hurwitz zeta function, and Bose-Einstein condensates, see my pdf

(forthcoming)

]]> https://tcjpn.wordpress.com/2024/08/25/umbral-witchcraft-with-bernoulli-and-blissard/feed/ 3 tcjpn A Schur Thing – Appendix to As Above, So Below: The Up-Down Operators for the (m)-Associahedra Partition Polynomials https://tcjpn.wordpress.com/2023/08/24/a-schur-thing-appendix-to-as-above-so-below-the-up-down-operators-for-the-m-associahedra-partition-polynomials/ https://tcjpn.wordpress.com/2023/08/24/a-schur-thing-appendix-to-as-above-so-below-the-up-down-operators-for-the-m-associahedra-partition-polynomials/#respond Fri, 25 Aug 2023 03:56:01 +0000 https://tcjpn.wordpress.com/?p=10008 Continue reading →]]> This is a pointer to the appendix

A Schur Thing – Appendix to As Above, So Below: The Up-Down Operators for the (m)-Associahedra Partition Polynomials (pdf)

to the earlier set of notes “As Above, So Below . . .”. (A link to this pdf is provided in that post as well.)

In these notes, a Lagrange-Schur-Jabotinsky identity relating different coefficients of a power series raised to different integer powers is used to prove the raising and lowering operations of the sets $[A^{(m)}]$ of (m)-associahedra partition polynomials are related to the set $[N]$ of noncrossing partition polynomials (both sets introduced in various previous posts) by

$[N] [A^{(m)}] = [A^{(m)}][N]^{-1} = [A^{(m+1)}]$

and

$[N]^{-1} [A^{(m)}] = [A^{(m)}][N] = [A^{(m-1)}].$

Consequently, the sets $[N^{(m)}]$ of (m)-noncrossing partition polynomials, which satisfy the identities

$[A^{(m)}] = [N^{(m)}][R] = [N^{(m)}][A^{(0)}]$ ,

also satisfy

$[N^{(m)}] = [N]^m$

for $m$ any integer.

]]> https://tcjpn.wordpress.com/2023/08/24/a-schur-thing-appendix-to-as-above-so-below-the-up-down-operators-for-the-m-associahedra-partition-polynomials/feed/ 0 tcjpn (m)-Associahedra and (m)-Noncrossing Partition Polynomials and the Infinite Dihedral Group https://tcjpn.wordpress.com/2023/07/26/m-associahedra-and-m-noncrossing-partition-polynomials-and-the-infinite-dihedral-group/ https://tcjpn.wordpress.com/2023/07/26/m-associahedra-and-m-noncrossing-partition-polynomials-and-the-infinite-dihedral-group/#comments Wed, 26 Jul 2023 18:31:53 +0000 https://tcjpn.wordpress.com/?p=9978 Continue reading →]]> I was playing around once again yesterday with the basic algebraic relations among the sets of (m)-associahedra partition polynomials $[A^{(m)}]$ and the sets of (m)-noncrossing partitions polynomials $[N^{(m)}] = [N]^m$ , which I’ve presented in several posts over the last year or so, and decided to post a few defining relations for the dual sets of polynomials in a question on Math Stack Exchange in the hope that the group would be recognized by someone. The MSE user Karl pointed out that it sounds like I was describing the infinite dihedral group, linking to the associated Wikipedia article. This led me to the Wiki on the dihedral group and a set of group relations that are shared by my group of partition polynomials. Each set $[N]^m$ plays the role of a rotation $r_m$ in the dihedral group and each set $[A^{(m)}]$ , a reflection $s_m$ . In the following I’ll show the applicability of these relations under the substitution operation I’ve illustrated in previous posts.

An infinite group $\mathcal{N}$ is formed by iterating the substitution operation on $[N]^1 = [N]$ and its inverse $[N]^{-1}$ . The elements of this infinite group are $[N]^{m}$ where $m$ is any integer; $[N]^0=[I]$ , the identity; and, e.g., $[N]^{3} = [N][N][N]$ and $[N]^{-3} = [N]^{-1}[N]^{-1}[N]^{-1}$ under the repeated operation.

The infinite sets of (m)-associahedra partition polynomials satisfy for any integer $m$

(I)

$[A^{(m)}]^2 = [I]$

and

(II)

$[N]^{\pm 1}[A^{(m)}] = [A^{(m\pm1)}].$

That is, $[A^{(m)}]$ is involutive and $[N]$ and $[N]^{-1}$ are the ladder ops–the raising and lowering ops–for the infinite set $\mathcal{A}$ comprised of the infinite sets $[A^{(m)}]$ , where $m$ runs over the infinite set of integers, as well as the ladder ops for the group $\mathcal{N}$ .

For any integers $i$ and $j$ , clearly, by the definition of $\mathcal{N}$ above,

$[N]^i[N]^j = [N]^{i+j},$

and the relation

$[N]^{\pm 1}[A^{(m)}] = [A^{(m\pm1)}]$

implies

$[N]^{i}[A^{(j)}] = [A^{(j+i)}],$

or equivalently

$[N]^{-i}[A^{(j)}] = [A^{(j-i)}].$

Recalling $[A^{(m)}]^2 = [I]$ , so $[A^{(m)}]^{-1} =[A^{(m)}]$ , and taking the inverse of this last equality gives

$([N]^{-i}[A^{(j)}])^{-1} = ([A^{(j-i)}])^{-1},$

implying

$[A^{(j)}] [N]^{i} = [A^{(j-i)}].$

Similarly, since

$[A^{(j)}] = [N]^j[A^{(0)}] = [A^{(j)}]^{-1}= ([N]^j[A^{(0)}])^{-1} = [A^{(0)}] [N]^{-j},$

finally

$[A^{(i)}][A^{(j)}] = [N]^i[A^{(0)}][N]^j[A^{(0)}]$

$= [N]^i[A^{(0)}][A^{(0)}][N]^{-j}$

$= [N]^i[N]^{-j} = [N]^{i-j}.$

Consequently, with $[N]^i \to r_i$ and $[A^{(i)}] \to s_i$ , my group satisfies the four relations

$r_i r_j = r_{i+j}, \; r_i s_j = s_{i+j}, \; s_i r_j = s_{i-j}, \; s_i s_j = r_{i-j}$

presented in the Wikipedia article on the dihedral group.

The group also satisfies the conjugation relations

1) $[A^{(m)}][N]^n[A^{(m)}] = [N]^{-n}$

and

2) $[A^{(0)}][A^{(m)}][A^{(0)}] = [A^{(-m)}].$

An analogous algebraic realization of the infinite dihedral group can be obtained by scaling the indeterminates by starting with multiplicative and compositional inversion of exponential generating functions (Taylor series) rather than ordinary generating function (power series). The extrapolation in both cases includes Laurent series. It would be interesting if there were other multivariate realizations under substitution.

Related stuff:

“Noncrossing partitions under rotation and reflection” by Callan and Smiley (pg.6)

“The power of group generators and relations: an examination of the concept and its applications” by Zhou

The Geometry and Topology of the Coxeter Groups by Davis

Group Theory by Milne

The CRM Winter School on Coxeter groups

Generalized Dihedral Group at GroupProps

“Dihedral group” by K. Conrad

]]> https://tcjpn.wordpress.com/2023/07/26/m-associahedra-and-m-noncrossing-partition-polynomials-and-the-infinite-dihedral-group/feed/ 1 tcjpn Iterative formula for the Kreweras-Voiculescu polynomials–the noncrossing partitions polynomials https://tcjpn.wordpress.com/2023/07/19/iterative-formula-for-the-kreweras-voiculescu-polynomials-the-noncrossing-partitions-polynomials/ https://tcjpn.wordpress.com/2023/07/19/iterative-formula-for-the-kreweras-voiculescu-polynomials-the-noncrossing-partitions-polynomials/#comments Thu, 20 Jul 2023 00:56:56 +0000 https://tcjpn.wordpress.com/?p=9965 Continue reading →]]> The Kreweras-Voiculescu partition polynomials, which I usually call the noncrossing partitions or the refined Narayana partition polynomials of OEIS A134264–flag and enumerate distinct noncrossing partitions and give the free moments in terms of the free cumulants in free probability theory. They are often useful in mathematical / physical analyses involving compositional inversion and are also of broad use in combinatorics characterizing parking functions, trees, Dyck lattice paths, cluster complexes, and other combinatorial geometric constructs. I’ve derived yet another method for iteratively determining these polynomials in the pdf

Iterative formula for the Kreweras-Voiculescu polynomials–the noncrossing partitions polynomials

]]> https://tcjpn.wordpress.com/2023/07/19/iterative-formula-for-the-kreweras-voiculescu-polynomials-the-noncrossing-partitions-polynomials/feed/ 2 tcjpn

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