I'm having some troubles wrapping my head around multi-index notation of a derivative operator on a basis. The following texts are summarised from a paper I'm reading.
"Let $\mathbf{P} = \{ \phi_1, \cdots, \phi_Q \}$ be the basis of interest, where each function of the basis is defined as Taylor monomials. Select at node $i$, the local polynomial basis $\mathbf{P}_i (x)$ is shown through multi-index notation:
$$
\phi_{\alpha}(x) = \frac{1}{\alpha!} \left( \frac{x – x_i}{\epsilon} \right)^{\alpha}
$$
we note that the derivative operator is only non-zero when the multi-index of the derivative is the same as that of the monomial, i.e.:
$$
D^{\beta} \mathbf{P}_i = \hat{e}_{\beta} \epsilon^{-\beta}
$$
where $\hat{e}_{\beta}$ is the canonical basis vector consisting of zeros with a one in the entry corresponding to multi-index $\beta$."
I am not that familiar with multi-index, and through Wikipedia, I have learned that they are n-tuples. I'm trying to learn through a 1D simple example. Say that I have my basis function up to $Q = 2$, i.e.
\begin{align}
\mathbf{P}_i(x) &= [\phi_0(x), \phi_1(x), \phi_2(x)]^T \\
{} &= [1, \frac{1}{\epsilon}(x – x_i), \frac{1}{2 \epsilon^2} (x – x_i)^2]
\end{align}
I can't quite see how the italic statement above really applied. Say if I choose $\beta = 1$, then:
\begin{align}
\frac{d}{dx}\mathbf{P}_i (x) = [0, \frac{1}{\epsilon}, \frac{1}{\epsilon^2}(x – x_i)]
\end{align}
which contradicts with the statement. I feel like I'm missing some key understandings, but can't seem to wrap my head around. Any comments?
Best Answer
The "derivative operator" they talked about isn't $f\mapsto D^\beta f$, but $f\mapsto (D^\beta f)\rvert_{x_i}$ with the evaluation at node. So any $\beta\neq\alpha$ operating on $\phi_\alpha$ either