An operator performing the mapping is $$B(\partial_x) = e^{b.\partial_x} =: \frac{\partial_x}{e^{\partial_x}-1},$$ with $\frac{\partial}{\partial x} = \partial_x$ and $(B.(0))^n=B_n(x)|_{x=0}= (b.)^n = b_n$, since the Bernoulli polynomials are an Appell sequence. Edit (6/20/2017): This operator is essentially the Todd operator. See the discussions on pg. 30 and Appendix B of "[Permutohedra, associahedra, and beyond][1]" by Postnikov of the Todd operator as a transform of the homogeneous volume polynomials for classes of polytopes into a generalized Ehrhart polynomial coding the number of lattice points in the polytopes. (Edit 8/2018) For some idea of the importance of this Todd operator in modern mathematics and physics, see [New Models for Veneziano Amplitudes: Combinatorial, Symplectic and Supersymmetric Aspects][2] by Kholodenko. (Edit 9/2024) A rep of the Bernoulli op that is more robust than $$T_{x \to (x+b.)}= e^{b.\partial_x}=: \frac{\partial_x}{e^{\partial_x}-1}$$ is $$T_{x \to (x+b.)} = \sum_{k \geq 0} \frac{1}{k+1} \sum_{j=0}^k (-1)^j \binom{k}{j} T_{x \to x+j}.$$ Define the Bernoulli translation op $T_{x \to (x+b.)}$ as the op that umbrally translates the variable $x$ to $(x+b.) = B.(x)$ such that $$T_{x \to (x+b.)} x^n = (x+b.)^n = \sum_{k=0}^n \binom{n}{k}(b.)^k x^{n-k}= \sum_{k=0}^n \binom{n}{k}b_k x^{n-k} = B_n(x)$$ where $b_n$ is the $n$-th Bernoulli number and $B_n(x)$ the $n$-th Bernoulli polynomial. Then a diff op rep when acting on $x^n$ is $$T_{x \to (x+b.)} = e^{b.\partial_x} =: \frac{\partial_x}{e^{\partial_x}-1}.$$ A more robust, extended rep of the Bernoulli op that allows action on $x^s$ where $s$ is complex can be realized via an operator compositional-inverse pair. Define the finite diff op $\delta_x$ by $$\delta_x x^s = (x+1)^s-x^s = (T_{x \to x+1}-1) x^s.$$ Then its diff op rep is (by analytic continuation) $$\delta_x x^s = (e^{\partial_x}-1) x^s = (x+1)^s -x^s$$ and inverting $$\partial_x = \ln(1+\delta_x),$$ so when acting on $x^n$ $$T_{x \to (x+b.)=B.(x)} = e^{b.\partial_x} = \frac{\partial_x}{e^{\partial_x}-1} = \frac{\partial_x}{\delta_x} = \frac{\ln(1+\delta_x)}{\delta_x}$$ $$ = \sum_{k \geq 0} (-1)^k \frac{1}{k+1} \delta_x^k = \sum_{k \geq 0} \frac{1}{k+1} \sum_{j=0}^k (-1)^j \binom{k}{j} e^{j\partial_x}$$ $$ = \sum_{k \geq 0} \frac{1}{k+1} \sum_{j=0}^k (-1)^j \binom{k}{j} T_{x \to x+j}.$$ Removing the intermediate steps gives the Bernoulli translation rep $$T_{x \to (x+b.)} = \sum_{k \geq 0} \frac{1}{k+1} \sum_{j=0}^k (-1)^j \binom{k}{j} T_{x \to x+j}$$ with action $$T_{x \to (x+b.)}x^s = \sum_{k \geq 0} \frac{1}{k+1} \sum_{j=0}^k (-1)^j \binom{k}{j} (x+j)^s = -s\zeta(-s+1,x),$$ essentially the Helmut Hasse formula for the Hurwitz zeta function $\zeta(s,x)$. Taking this as the interpretation of $$T_{x \to (x+b.)}x^s = (x+b.)^s = (B.(x))^s = B_s(x)$$ gives the Bernoulli function $$B_s(x) = -s\zeta(-s+1,x) = \sum_{k \geq 0} \frac{1}{k+1} \sum_{j=0}^k (-1)^j \binom{k}{j} (x+j)^s = T_{x \to (x+b.)}x^s,$$ and as $$T_{x \to (x+b.)} \frac{(x+1)^{n+1}-x^{n+1}}{n+1} = e^{b.\partial_x} \frac{(x+1)^{n+1}-x^{n+1}}{n+1}$$ $$ = \frac{(b.+x+1)^{n+1}-(b.+x)^{n+1}}{n+1}$$ $$ = \frac{(B.(x+1))^{n+1}-(B.(x))^{n+1}}{n+1} = \frac{B_{n+1}(x+1)-B_{n+1}(x)}{n+1} $$ $$= \frac{\partial_x}{\delta_x}\delta_x \frac{ x^{n+1}}{n+1} = \partial_x \frac{ x^{n+1}}{n+1} = x^n$$ so does $$T_{x \to (x+b.)} \frac{(x+1)^{s+1}-x^{s+1}}{s+1} = \frac{B_{s+1}(x+1)-B_{s+1}(x)}{s+1}$$ $$ = - \zeta(-s,x+1) -(-\zeta(-s,x)) = x^s = \partial_x \frac{x^{s+1}}{s+1}.$$ Note the action of the inverse op $$ \frac{\delta_x}{\partial_x}x^n = \frac{e^{\partial_x}-1}{\partial_x}x^n = \sum_{k \geq 0} \frac{1}{k+1} \frac{\partial_x^k}{k!}x^n = \sum_{k \geq 0} \binom{n}{k} \frac{1}{k+1} x^{n-k}= \int_0^1 (x+t)^ndt $$ $$= \int_{x}^{x+1}t^n dt = \frac{(x+1)^{n+1}-x^{n+1}}{n+1} $$ has a rep as a sliding average, which, with $\hat{b}_k = \frac{1}{k+1}$, gives the action $$T_{ x \to (x+\hat{b}.) = \hat{B}.(x)}x^s = \int_{x}^{x+1}t^s dt = \int_0^1 (x+t)^sdt = \frac{(x+1)^{s+1}-x^{s+1}}{s+1} = \hat{B}_s(x),$$ with a log function in the limiting case $s \to -1$. Thus the umbral inversion relation for the Bernoulli polynomials $$\hat{B}_n(B.(x)) =x^n$$ is generalized to $$\hat{B}_s(B.(x)) =x^s.$$ This is consistent with Borel summation and analytic continuation of the action of $e^{b.\partial_x}$ and $e^{\hat{b}.\partial_x}$ on the Euler hybrid Mellin-Laplace rep of $x^s$ (and the more broadly valid Hankel contour rep of this integral), noting that $$e^{b.\partial_x}e^{\hat{b}.\partial_x} = e^{(b.+\hat{b}.)\partial_x} = 1.$$ For example, the divergent series $$e^{b.\partial_x}x^{-s} = e^{b.\partial_x}\int_0^{\infty} e^{-xt} \frac{t^{s-1}}{(s-1)!}dt$$ is Borel summed via $$ \int_0^{\infty}e^{b.\partial_x} e^{-xt} \frac{t^{s-1}}{(s-1)!}dt = \int_0^{\infty}\frac{-t}{e^{-t}-1}e^{-xt} \frac{t^{s-1}}{(s-1)!}dt = B_{-s}(x)$$ $$ = \int_0^{\infty} e^{-(x+b.)t} \frac{t^{s-1}}{(s-1)!}dt = \int_0^{\infty} e^{-B.(x)t} \frac{t^{s-1}}{(s-1)!}dt$$ $$=: \frac{1}{(B.(x))^s} =: (B.(x))^{-s} := B_{-s}(x),$$ an example of Riemann-Ramanujan modified Mellin transform interpolation / analytic continuation. Then the various actions are reduced to actions on the Laplace kernel $e^{-xt}$. [1]: https://arxiv.org/abs/math/0507163 [2]: https://arxiv.org/abs/hep-th/0503232