2

$$ \tilde{B}_n(x) = \frac{(-1)^{n + 1}}{n!} \left( \delta^{(n - 1)}(x - 1) - \delta^{(n - 1)}(x) \right) $$

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'

Jose Garcia
  • 8,074
  • 2
  • 27
  • 51

2 Answers2

2

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*}

Herman Jaramillo
  • 2,568
  • 21
  • 25
2

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 $$ \int_0^1 \tilde{B}_i(x) B_j(x)\, dx = \delta_{ij}. $$

Andrew
  • 11,375
  • 1
  • 28
  • 51