If we apply operator $D\Delta^{-1}$ to a function, we will get the (Bernoulli) umbral analog of the function. Particularly, applying it to $x^n$ we will get the Bernoulli polynomials $B_n(x)$. Evaluating them at zero we will get the moments of Bernoulli umbra, the Bernoulli numbers $1, -1/2, 1/6,...$ .

But what if we consider the operator $\Delta D^{-1}$? It is the inverse of the aforementioned operator.

What is interesting about it, it corresponds to an umbra with moments $1,1/2,1/3,1/4,...$.

This looks quite trivial, but what are the properties of this "anti-Bernoulli" umbra (let's denote it as $\overline B$)? Denoting the index-lowering operator as $\operatorname{eval}$, the following identities hold:

$\operatorname{eval} \psi(\overline{B}+x)=\ln (x)$

$\operatorname{eval} \ln(\overline{B}+x)=-x \ln (x)+x \ln (x+1)+\ln (x+1)-1$

$\operatorname{eval} \tan(\overline{B}+x)=\frac1\pi\ln \left(\frac{\frac{1}{2}+\frac{x}{\pi }}{\frac{1}{2}-\frac{x}{\pi }}\right)$

Thus, this operator also links exponential functions to logarithms.

Of interest also is the umbra $\overline{B}-1/2$, its moments are reciprocals of https://oeis.org/A001787, with even elements replaced by zeros.

Thus I wonder, what are other properties of umbra $\overline{B}$ and polynomials of it? Are there any known power series expansions with coefficients with denominators from https://oeis.org/A001787 ?

P.S., the following interesting identity holds:

$\operatorname{eval}\frac{x}{\overline{B} x+1}=\ln (x+1)$

  • $\begingroup$ In your notation, $\frac{x}{1+\bar{B}x} = x - \bar{B}^1 x^2 + \bar{B}^2 x^3 - \bar{B}^3 x^4 + \cdots$ umbrally evaluated gives $x - \bar{B}_1 x_2 + \bar{B}_2 x^3 - \bar{B}_3 x^4 + \cdots = x - x_2/2 + x^3/3 - x^4 /4+ \cdots = \ln(1+x)$, so you can drop the qualifier 'it seems'. $\endgroup$ Feb 3, 2023 at 17:00
  • $\begingroup$ From the sliding average formula in my answer, in my notation, $\psi(\bar{b}.+x) = \psi(\bar{B}.(x)) = \int_x^{x+1} \psi(t) dt = \ln(x)$ (from Wolfram Alpha). $\endgroup$ Feb 3, 2023 at 17:07

2 Answers 2


I'll sketch here some of the general Sheffer Appell-umbral calculus relationships between the Bernoulli numbers / polynomials and the reciprocal (natural) numbers / polynomials. Perhaps when I have more time, I'll point to numerous sets of notes I've written over the years on this topic (some are included below).

The Bernoulli numbers

$(b.)^n = b_n$

are the moments defined by the e.g.f.

$e^{b.t} = \frac{t}{e^t-1}$

for the Appell Sheffer sequence of Bernoulli polynomials

$(B.(x))^n=B_n(x) = (b. + x)^n$

with the e.g.f.

$B(x,t) = B(t) e^{xt} = e^{b.t}e^{xt}= e^{(b.+x)t} = e^{B.(x)t}= \frac{t}{e^t-1}e^{xt}$.

Note that all umbral quantities are flagged with a period in the subscript to simplify and collate the notation and that umbral evaluation, i.e., the lowering of powers to the period subscript position, is understood to occur only after the analytic expressions are expressed as a power series in the umbral quantity of interest.

The reciprocal natural numbers

$(\bar{b}.)^n = \bar{b}_n = \frac{1}{n+1}$

are the moments with the e.g.f.

$e^{\bar{b}.} = \frac{e^t-1}{t}$

for the Appell Sheffer sequence of reciprocal polynomials

$\bar{B}_n(x) =(\bar{b}.+x)^n = \frac{(1+x)^{n+1}-x^{n+1}}{n+1}$

with the e.g.f.

$\bar{B}(x,t) = \bar{B}(t) e^{xt} = e^{\bar{b}.t}e^{xt}= e^{(\bar{b}.+x)t} = e^{\bar{B}.(x)t}= \frac{e^t-1}{t}e^{xt}$.

The operational generators, with the derivative op $D = \partial_x$, for the dual pair of polynomials are then the Todd operator

$B(D)x^n= \frac{D}{e^D-1}x^n = e^{b.D} x^n = (b.+x)^n = B_n(x)$

and the reciprocal op

$\bar{B}(D)x^n= \frac{e^D-1}{D}x^n = e^{\bar{b}.D} x^n = (b.+x)^n = \bar{B}_n(x)$.

Since the product of the two ops is the identity, i.e.,

$ 1 = \bar{B}(D) B(D) = B(D)\bar{B}(D),$

the pair of Appel sequences are mutually inverse under umbral composition, i.e.,

$x^n = B(D)\bar{B}(D)x^n =B(D)\bar{B}_n(x) = \bar{B}_n(b.+x)= \bar{B}_n(B.(x))= (\bar{b}.+b.+x)^n,$

implying the Kronecker delta relation for the pair of moments

$(\bar{b}.+b.)^n = 0^n = \delta_n$,

and, conversely (naturally for inverses),

$x^n =\bar{B}(D)B(D) x^n =\bar{B}(D)B_n(x) = B_n(\bar{b}.+x)= B_n(\bar{B}.(x))= (\bar{b}.+b.+x)^n.$

This also follows, consistently, from the reciprocal relation of the moment e.g.f.s

$1 = B(t) \frac{1}{B(t)} = B(t)\bar{B}(t) = e^{\bar{b}.t}e^{b.t}= e^{(b. + \bar{b}.)t}.$

This implies recursion relations among the moments and relations to important sequences of compositional partition polynomials (see, e.g., A133314 ).

Umbral composition of a pair of Sheffer sequences corresponds to matrix multiplication of their (infinite) lower triangular matrices of coefficients--in this case, for the pair of Appell Sheffer sequences $[B]$ with $B_{n,k} = \binom{n}{k}B_{n-k}$ and $[\bar{B}]$ with $\bar{B}_{n,k} = \binom{n}{k}\bar{B}_{n-k}$, so

$[B][\bar{B}]=[\bar{B}][B] = [I]$, the identity matrix.

See A178252 on $[\bar{B}]$.

The ops above can be used to find the raising $R$ and lowering $L$ ops for the pair of Appell sequences such that for the polynomials

$L P_n(x) = n \; P_{n-1}(x)$ and $R \; P_n(x) = P_{n+1}(x).$

Note the Graves-Pincherle-Weyl commutator relation $[L,R]P_n(x) = (LR-RL)P_n(x) = P_n(x)$, or $[L,R] = I$, is satisfied.

All Appell polynomial sequences $A_n(x)$ share the same lowering op $L =D$ since

$D \; A_n(x) = D \; (a.+x)^n = n (a.+x)^{n-1} = n\; A_{n-1}(x) = L \; A_n(x) .$

The raising ops are easily found from the conjugation of the raising op $R_P=x$ of the prototypical Appell sequence $P_n(x) = x^n$. For a generic Appell sequence $A_n(x)$ and its umbral inverse sequence $\bar{A}_n(x)$ (sometimes called reciprocal sequence),

$R_A = A(D) \; x\; (A(D))^{-1} = A(D) \; x \; \bar{A}(D) = x + \frac{d}{dt} \ln(A(t)) |_{t = D} = x + \frac{d}{dD} \ln(A(D))$


$R_A A_n(x) = A(D) \; x\; \bar{A}(D)A_n(x) = A(D) \; x\; x^{n} = A(D) \; x^{n+1} = A_{n+1}(x).$


$R_\bar{A} = \bar{A}(D) \; x \;(\bar{A}(D))^{-1} = \bar{A}(D)\; x \; A(D) = x + \frac{d}{dD} \ln(\bar{A}(D)) = x - \frac{d}{dD} \ln(A(D)) $

(that one sign makes a big difference).

One defining property of the Bernoulli polynomials that can be couched in terms the umbral inversion duality of the pair $B_n(x)$ and $\bar{B}_n(x)$ is

$x^n =\bar{B}_n(B.(x)) = \frac{(B.(x)+1)^{n+1} - (B.(x))^{n+1}}{n+1}$

$ = \frac{(b.+x+1)^{n+1} - (B.(x))^{n+1}}{n+1}$

$ = \frac{(B.(x+1))^{n+1} - (B.(x))^{n+1}}{n+1} ,$


$\frac{(B.(x+1))^{n+1} - (B.(x))^{n+1}}{n+1} = x^n = D \; \frac{x^{n+1}}{n+1}$ ,

implying the term by term derivative of a formal Taylor series

$T(x) = c_0 + c_1 x + c_2 \frac{x^2}{2!} + \cdots$


$D\; T(x) = T(B.(x+1)) - T(B.(x))$. (See this MO-Q also.)

The term-by-term action on a formal power series of umbral composition with the reciprocal polynomials is evident in their form and, naturally, reflected in their differential generator--$D^{-1}$ suggests integration and $e^D-1$ generates a finite difference.


$\bar{B}(D) x^n=\frac{e^D-1}{D}x^n = e^{\bar{b}.D}\; x^n = (\bar{b}.+x)^n = (\bar{B}.(x))^n = \frac{(x+1)^{n+1}-x^{n+1}}{n+1}$

$ = \int_{x}^{x+1} t^n dt= \int_{0}^{1}(x+ t)^n dt,$

so umbral composition w.r.t. $\bar{B}_n(x)$ of a power series gives a sliding average of the function represented by the power series, i.e.,

$T(\bar{B}.(x)) = \int_{x}^{x+1} T(t) dt= \int_{0}^{1}T(x+ t) dt.$

Other identities, such as Faulhaber's formula, related to the Bernoulli numbers / polynomials can be fairly easily derived from the umbral inverse relationship between $B_n(x)$ and $\bar{B}_n(x)$. More advanced general algebraic relations presented by other authors can be couched in terms of this duality as well. See, e.g., "Bernoulli Appells".

Search (on the phrase within the quotes, i.e., don't include the quotes) "Copeland Bernoulli polynomials umbral" in the OEIS for possible info on the reciprocal polynomials.

See my response to the MO-Q Intuitive explanation why "shadow operator" $\frac{D}{e^D-1}$ connects logarithms with trigonometric functions? and the links therein.

Peruse my WordPress site "Shadows of Simplicity" for numerous sets of notes on the Bernoulli and reciprocal polynomials. Pretty much any entry with the keyword Bernoulli has some notes on the reciprocal polynomials as well. (Caveat: the term 'reciprocal polynomials' is used by me and others to refer to other types of sequences of polynomials as well as the above.)

Some OEIS sequences related to general Appell sequences are A133314 and A263634. All the generic compositional partition polynomials--the Bell / Faa di Bruno / refined Stirling polynomials of the second kind of A036040, the cycle index polynomials of the symmetric groups / the refined Stirling polynomials of the first kind of A036039, and the refined Lah polynomials of A036039 as well as the classic noncrossing partition / refined Narayana / Voiculescu (of free probability theory) polynomials of A134264 are Appell polynomial sequences in the distinguished initial indeterminate, e.g., $c_1$ or $c_0$. (To spot check these assertions, take the derivative of the first few partition polynomials with respect to the distinguished indeterminate. The proof with e.g.f.s is simple. Be aware that for the full panoply of the Appell operator calculus to apply, it must be that $A_0(x) = a_0 = 1$.)

  • $\begingroup$ To avoid problems in umbral eval in other contexts, note $(c+a.x)^m = \sum_{n=0}^m \binom{m}{n} c^{m-n} (a.x)^n=\sum_{n=0}^m \binom{m}{n} c^{n} (a.x)^{m-n} $ umbrally evaluated gives $\sum_{n=0}^m \binom{m}{n} c^{m-n} a_nx^n =\sum_{n=0}^m \binom{m}{n} c^{n} a_{m-n}x^{m-n} $. Then generalizing, $\frac{x}{1+ax} = x(1+ax)^{-1} = x\sum_{n \geq 0} \binom{-1}{n} a^n x^n =x\sum_{n \geq 0} (-1)^n a^n x^n = a^0x - a^1x^2 + a^2x^3 -\cdots $ umbrally evaluated gives $a_0 x- a_1x^2 + a_2x^3 -\cdots$. With $a_0= 1$, this detail can be ignored as is certainly the case for $a_n=1$ for all $n \geq 0$. $\endgroup$ Feb 3, 2023 at 17:45

It turns out that this umbra has quite trivial representation.

Consider a set of integrable functions on the interval $(0,1)$ with evaluation operator defined as simple functional $\operatorname{eval} f= \int_0^1 f(x)dx$, which is the mean value of the function. On this set, the function $f(x)=x$ serves as $\overline{B}$ and has the property $\operatorname{eval}{\overline{B}^n}=\int_0^1 x^n dx=\frac1{n+1}$, which gives the moments $1,1/2,1/3,1/4,...$.


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