I am given the Hermite interpolation formula directly in my text book without ANY explanations about how it was first made (obviously it was somehow constructed for the first time with some sort of intuition ) .
the formula for n+1 data from $x_0$ till $x_n$ with $f(x_0)$ till $f(x_n)$
and with $ f^{\prime}(x_0)$ till $f^{\prime}(x_n)$
$$H_{2n+1}(x) = \sum_{j=0}^n f(x_j)H_{n,j}(x) + \sum_{j=0}^n f^{\prime}(x_j)\hat H_{n,j}(x)$$
where
$$H_{n,j} = [1 − 2(x − x_j)L^{\prime}_{n,j}(x_j)]L_{n,j}^2(x) $$
$$ \hat H_{n,j}(x) = (x-x_j) L_{n,j}^2(x) $$
I DO understand the proof and why the polynomial agrees with data and their derivatives.
i DO understand the intuition behind Lagrange polynomials.
so I am looking for the intuition behind the formula (how it was made) specially the construction of $H$ and $\hat H$. so instead of memorizing it i can learn it!
Best Answer
Both kinds of interpolation formulas rely on the superposition principle (the sum of the effect of individuals causes is the effect of the sum of the causes), and achieve a decomposition such that every point brings its own contribution. Actually, you form a basis of polynomials and linear combinations thereof.
In the case of Lagrange, consider the special case $f(x_i)=\delta_{ij}$: all ordinates but the $j^{th}$ are zero, and the latter is one. This is easily achieved by forming the product of $(x-x_i)$ and normalizing to one at $x_j$. From these $n$ basis polynomial, you can construct the interpolant for any ordinates.
The generalization to Hermite follows the same idea. You will form two families of polynomials: the first family carries the ordinates ($f(x_i)=\delta_{ij}, f'(x_i)=0$), and the second one carries the derivatives ($f(x_i)=0,f'(x_i)=\delta_{ij}$).
The rest is technical trickery, based on the idea that by squaring a Lagrange polynomial, the simple roots become double roots and the derivative vanishes at the roots, preparing candidates for the first and second family.
More precisely, $L^2_j$ achieves $f(x_i)=\delta_{ij}$, and $f'(x_i)=0$, except at $x_j$.
Let us introduce the polynomial $Z_j=(x-x_j)L_j$, such that $Z_j(x_i)=0$ and $Z_j'(x_j)=1$.
To obtain the first family, we cancel the derivative of $L_j^2$ at $x_j$ by introducing a corrective term $-Z_j(L_j^2)'$, that derives as $-Z_j'(L_j^2)'-Z_j(L_j^2)''$, i.e. $-(L_j^2)'$ at $x_j$ and $0$ elsewhere: $$H_j=L_j^2-Z_j(L_j^2)'=(1-2(x-x_j)L_j')L_j^2.$$
To obtain the second family, we cancel $f(x_j)$ using the product $Z_jL_j$, and we have $(Z_jL_j)'=Z_j'L_j+Z_jL_j'=\delta_{ij}$ as desired. Hence: $$\hat H_j=Z_jL_j=(x-x_j)L_j^2.$$