Consider the Hermite problem $$p^{(r)}(x_i)=y_i^{(r)},\ i=1,2 ;\ r=0,1,2$$ with $p(x)$ a polynomial of degree $\leq 5$.
a - Give a Lagrange type of formula for $p(x)$.
b - Give a Newton divided difference formula and also derive an error formula.
Attempt at a solution:
What I know up to this point is for the basis functions $l(x_2)=l'(x_2)=l''(x_2)=0$, I will then have to use $l(x)=(x-x_2)^3g(x)$, where $g(x)$ is of degree $\leq 2$, then I will have to find $g(x)$.
Your notation is confusing to me, so I used something that makes more sense to me, please forgive that.
The Lagrange Polynomial is given by:
$\tag 1 \displaystyle P(x) = f(x_0)L_{n,0}(x) + \cdots + f(x_n)L_{n,n}(x) = \sum_{k=0}^{n} f(x_k)L_{n,k}(x), \text{where}$
$\tag 2 \displaystyle L_{n,k}(x) = \frac{(x-x_0)(x-x_1)\cdots (x-x_{k-1})(x-x_{k+1}) \cdots (x-x_n)}{(x_k - x_0)(x_k - x_1) \cdots (x_k-x_{k-1})(x_k-x_{k+1}) \cdots (x_k-x_n)} = \prod_{i=0, i \ne k}^n \frac{(x-x_1)}{(x_k-x_i)}$
for each $k = 0, 1, \cdots, n.$
So, for the three values of $r$, we have:
$\tag 3 \displaystyle L_{2,0}(x) = \frac{(x-x_1)(x-x_2)}{(x_0-x_1)(x_0-x_2)}$ and we can calculate $L_{2,0}'(x)$
$\tag 4 \displaystyle L_{2,1}(x) = \frac{(x-x_0)(x-x_2)}{(x_1-x_0)(x_1-x_2)}$ and we can calculate $L_{2,1}'(x)$
$\tag 5 \displaystyle L_{2,2}(x) = \frac{(x-x_0)(x-x_1)}{(x_2-x_0)(x_2-x_1)}$ and we can calculate $L_{2,2}'(x)$
The Hermite polynomials are given by:
$\tag 6 H_{2n+1}(x) = \sum_{j=0}^{n} f(x_j)H_{n,j}(x) + \sum_{j=0}^{n} f'(x_j) \hat H_{n,j}(x)$, where
$\tag 7 H_{n,j}(x) = [1 - 2(x-x_j)L'_{n,j}(x_j)]L^2_{n,j}(x)$, and
$\tag 8 \hat H_{n,j}(x) = (x - x_j)L^2_{n,j}(x).$
Using all of this information, we can write a Lagrange type formula as (the function-values, $f(x_k)$) below represent the six data points you have, that is, three values each for $f(x_k)$ and $f'(x_k)$:
$\tag 9 \displaystyle H_5(x) = f(x_0)H_{2,0}(x) + f(x_1)H_{2,1}(x) + f(x_2)H_{2,2}(x) + f'(x_0) \hat H_{2,0}(x) + f'(x_1) \hat H_{2,1}(x) + f'(x_2) \hat H_{2,2}(x).$
From the above, you can see that this is a complete description of the Hermite polynomials, however, we needed to determine and evaluate the Lagrange polynomials and their derivatives, which is tedious, even for these small values of $n$.
An alternative method for generating the Hermite approximations is to use the Newton interpolatory divided-difference formula for the Lagrange polynomials at $x_0, x_1, \ldots, x_n:$
$\tag {10} \displaystyle P_n(x) = f[x_0] + \sum_{k=1}^{n} f[x_o, x_1, \ldots, x_k](x-x_0) \cdots (x-x_{k-1})$
We can write out a divided-difference table to find all of the coefficients as (you can fill out the three missing columns):
$$\begin{array}{c|c|c} \text{z} & \text{f(z)} & \text{1st divided differences} & \text{2nd divided differences}\\ \hline \\z_0 = x_0 & f[z_0] = f(x_0) & & \\ & & f[z_0, z_1] = f'(x_0) & \\z_1 = x_0 & f[z_1] = f(x_0) & & f[z_0, z_1, z_2] = \frac{f[z_1, z_2]-f[z_0,z_1]}{z_2-z_0} \\ & & f[z_1, z_2] = \frac{f[z_2]-f[z_1]}{z_2-z_1} & \\z_2 = x_1 & f[z_2] = f(x_1) & & f[z_1, z_2, z_3] = \frac{f[z_2, z_3]-f[z_1,z_2]}{z_3-z_1} \\ & & f[z_2, z_3] = f'(x_1) & \\z_3 = x_1 & f[z_3] = f(x_1) & & f[z_2, z_3, z_4] = \frac{f[z_3, z_4]-f[z_2,z_3]}{z_4-z_2} \\ & & f[z_3, z_4] = \frac{f[z_4]-f[z_3]}{z_4-z_3} & \\z_4 = x_2 & f[z_4] = f(x_2) & & f[z_3, z_4, z_5] = \frac{f[z_4, z_5]-f[z_3,z_4]}{z_5-z_3} \\ & & f[z_4, z_5] = f'(x_2) & \\z_5 = x_2 & f[z_5] = f(x_2) & & \end{array}$$
We can now form $H_5(x) = P(x)$ using the values in this table with the Newton divided-differences formula.
I'll leave it to you to derive the error term, all you need to do is look at the Lagrange and Newton error term with a slight change.
Hope that all makes sense and too bad we didn't have real data to play with and actually solve.
I'm wondering, for this Hermite problem it seems that you are given 2 nodes and the polynomial must interpolate through the second derivates.
I could be wrong but wouldn't we need to have a formula that is different than the usual hermite problem with just going through the first derivative?
