The operator $\frac{D}{e^D-1}$ in definition of Bernoulli polynomials

Question

I am trying to understand Bernoulli polynomials, and so I came across this abomination in the article: $$ B_n(x) = \frac{D}{e^D-1} x^n $$ where $D$ is the differentiation operator and the fraction is "expanded out as a formal power series"

Since we have $$\frac{t}{e^t-1} = \sum_{n=0}^\infty B_n \frac{t^n}{n!}$$ Does this imply that I should interpret the fraction as $$ \frac{D}{e^D-1} = \sum_{n=0}^\infty B_n \frac{D^n}{n!} $$

Answer 1

I'm so silly $$ \begin{align*} &\frac{D}{e^D-1}x^n \\ &= \sum_{k=0}^\infty B_k \frac{D^k(x^n)}{k!} \\ &= \sum_{k=0}^n B_k \frac{(n)_k}{k!} x^{n-k} \\ &= \sum_{k=0}^n \binom{n}{k} B_k x^{n-k} \\ &= \sum_{k=0}^n \binom{n}{k} B_{n-k} x^k \\ &= B_n(x) \end{align*} $$

Answer 2

We are looking for solutions to

$$B_n(x+1) - B_n(x) = n x^{n-1}$$

that is

$$(e^D -1) B_n(x) = D x^n$$

So take

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

Note: $D$ is the derivative. $B(x+1) = e^D B\,(x)$ from Taylor's formula.

$\bf{Added:}$

The Euler polynomials, satisfying

$$\frac{1}{2}( E_n(x+1) + E_n(x))= x^n$$

are given by

$$E_n(x) = \frac{2}{e^D+1} x^n$$

They are also useful in calculating alternating sums

$$\sum_{k=0}^{m-1} (-1)^k (x+k)^n$$