Say I have the following problem of finding an interpolating polynomial using Hermite interpolation.
Given $(n+1)$ distinct points $x_i$, $i=0,\ldots,n$ in the interval $[a,b]$ with corresponding $f_i$, $f_i'$. So in total $(2n+2)$ data points. Find a polynomial $p_{2n+1}(x)$ of degree $\le (2n+1)$ such that $p_{2n+1}(x_i)=f_i$ and $p_{2n+1}'(x_i)=f_i'$, $i=0,\ldots,n$.
So the polynomial is of the form
$$p_{2n+1}(x)=\sum_{i=0}^nH_i(x)f_i+\sum_{i=0}^nK_i(x)f_i'$$
It says in the text that $H_i$'s and $K_i$'s are the of the form
\begin{align*} H_i(x)&=(L_i(x))^2(1-2L_i'(x_i)(x-x_i))\\ K_i(x)&=(L_i(x))^2(x-x_i) \end{align*}
where $L_i$'s are the Lagrange basis functions. Now I understand that these are set this way so that they coincide with the Kronecker Delta. But what are the calculations involved in getting these?
So suppose you $n+1$ points, that is, $\{x_0,...,x_n\}$. You want to find a polynomial such that $p(x_i) = f(x_i)$ and $p'(x_i) = f'(x_i)$, for $i=0,...,n$. This means you have $(2n + 2)$ conditions and the resulting polynomial will have degree $\leq 2n+1$. Assume that
$p(x) = \sum^n_{i=0} \alpha_i(x)f(x_i) + \sum^n_{i=0}\beta_i f'(x_i)$
In order to $p(x_i) = f(x_i)$ and $p'(x_i) = f'(x_i)$ for $i=0,...,n$, your basis must satisfy:
$\alpha_i(x_j)=\beta'_i(x_j) = \delta_{ij}$ and $\alpha'_i(x_j)=\beta_i(x_j) = 0$
Notice however that $l_i(x_j) = \delta_{ij}$. The basis will be of the form:
$\alpha_i(x) = (ax+b) \ l_i^2(x)$ and $\beta_i(x) = (cx+d) \ l_i^2(x)$
You want the Vandermonde matrix to be the identity, so you need to have:
$\alpha_i(x_i) = 1$, $\alpha'_i(x_i) = 0$, $\beta_i(x_i)=0$, $\beta'_i(x_i)=1$
If you solve this system you'll get your $H_i(x)$ and $K_i(x)$