dual formula to Bernoulli polynomials $$ \tilde{B}_n(x) = \frac{(-1)^{n + 1}}{n!} \left( \delta^{(n - 1)}(x - 1) - \delta^{(n - 1)}(x) \right) $$
Wikipedia says this formulae is DUAL to the Bernoulli POlynomials but dual in what sense ??
thanks
EDIT: link   http://en.wikipedia.org/wiki/Euler_Maclaurin    in the section  'derivation by functinal analysis'  
 A: As is written in the provided link, the dual base $\{\mathbf{e}^i\}$ to a base $\{\mathbf{e}_i\}$ is defined by the relation $\mathbf{e}^i(\mathbf{e}_j) = \delta_{ij}.$ In the case of Bernoulli polynomials as a base in $L_2([0,1])$ it is 
$$
\int_0^1 \tilde{B}_i(x) B_j(x)\, dx = \delta_{ij}.
$$
A: For some reason Wikipedia removed that part.
In a Hilbert space $\mathcal{H}$, let us assume we have  a basis $\{ e_k \}$.
Let us denote the dual of this Hilbert space (The space of functionals
from $\mathcal{H}$ to $\mathbb{R}$) by $\mathcal{H}^*$, and its dual base
vectors as $e^*_k$.
By the Riez Representation Theorem 
the functionals in  $\mathcal{H}^*$ are inner products; and by the definition
of dual basis we have that
\begin{eqnarray}
  e_k^*(e_j) = \langle e_k^* \, , \, e_j \rangle = \delta_{ij}.
\end{eqnarray}
Initially let us assume that we are in the space of polynomials of degree $n$ and
that the basis used are the monomial basis $\{1,x, \cdots x^k, \cdots , x^n \}$.
From $\langle \phi(x), \delta^{(1)}(x) \rangle =- \langle \phi^{(1)}(x),
\delta(x) \rangle= -\phi'(0)$ applied $k$ times we find 
\begin{eqnarray*}
 \langle \delta^{(j)}(x), x^k  \rangle  =
 \left . (-1)^j j! x^{k-j}   \right |_{x=0} = j! \delta_{jk}.
\end{eqnarray*}
Hence the weight $(-1)^k/k!$ serves as a normalization factor, and we say that
\begin{eqnarray*}
  e^*_j = (-1)^j \frac{\delta^{(j)}}{j!}
\end{eqnarray*}
is the dual base vector corresponding to $e_k=x^k$, on the space of polynomials
of order $n$.
Let us now assume that the $e_j=B_j$ vectors are Bernoulli polynomials (it can be shown
that they form a basis for the space of polynomials).  
We use similar ideas here. We consider the periodic Bernoulli polynomials in the
interval $[0,1]$. Since
\begin{eqnarray*}
  \frac{d B_n(x)}{dx} = n B_{n-1}(x)
\end{eqnarray*}
Then
\begin{eqnarray*}
  \langle \delta^{(1)}(x) \; ,  \; B_m(x) \rangle &=& -m B_{m-1}(0) \\
  \langle \delta^{(2)}(x) \; ,  \; B_m(x) \rangle &=& (-1)^2 \, m (m-1) B_{m-2}(0) \\
  \vdots \\
  \langle \delta^{(k)}(x) \; ,  \; B_m(x) \rangle &=& (-1)^k m (m-1) \cdots (m-k+1) 
    B_{m-k}(0) \\
  \vdots \\
  \langle \delta^{(n-1)}(x) \; ,  \; B_m(x) \rangle &=& 
  (-1)^{n-1} \frac{n! B_{m-n+1}(0)}{ (m-n+1)! } 
\end{eqnarray*}
Now, similarly
\begin{eqnarray*}
  \langle \delta^{(n-1)}(x-1) \; ,  \; B_m(x) \rangle &=& 
  (-1)^{n-1} \frac{n! B_{m-n+1}(1)}{(m-n+1)!} 
\end{eqnarray*}
and subtracting the two last equations (last minus previous), and
scaling by $(-1)/n!$, 
\begin{eqnarray*}
  \frac{(-1)^n}{n!} 
  \langle \; \delta^{(n-1)}(x-1) - \delta^{(n-1)}(x)\; ,  \; B_n(x) \; \rangle = 
 \frac{1}{(m-n+1)!} [B_{m-n+1}(1) - B_{m-n+1}(0)] = \delta_{nm}.
\end{eqnarray*}
