I am studying Hermite Interpolation and the most common practice I came across was the use of Lagrange Polynomials. I tried to construct a Hermite Polynomial for 2 points $x_0$ and $x_1$ but instead I did it by constructing one polynomial term at a time like this :
Constraints :
- $f(x_0)=p(x_0)$
- $f^{'}(x_0)=p^{'}(x_0)$
- $f(x_1)=p(x_1)$
- $f^{'}(x_1)=p^{'}(x_1)$
To satisfy constraints (1) and (2) $\space p(x)$ should be equal to : $p(x)=f(x_0)+(x-x_0)*f^{'}(x_0)$
We now need to add one term so that constraint (3) stands and constraints (1)-(2) are not broken so $ p(x)$ should look like this : $$p(x)=f(x_0)+(x-x_0)*f^{'}(x_0)+a*(x-x_0)^2$$
The $(x-x_0)^2$ term is there so that $p(x_0)=f(x_0)$ and it's squared so that $f^{'}(x_0)=p^{'}(x_0)$ still stands.
Solving for $a$ as we demand that constraint (3) holds : $$f(x_1)=p(x_1)$$ $$f(x_1)=f(x_0)+(x_1-x_0)*f^{'}(x_0)+a*(x_1-x_0)^2 $$ $$a=\frac{f(x_1)-f(x_0)}{(x_1-x_0)^2}-\frac{f^{'}(x_0)}{(x_1-x_0)}$$
Now we need to add one final term so that constraint (4) stands and constraints (1)-(2)-(3) are not broken so $ p(x)$ should be : $$p(x)=f(x_0)+(x-x_0)*f^{'}(x_0)+a*(x-x_0)^2+b*(x-x_0)^2(x-x_1)$$ The addition of terms $(x-x_0)^2$ and $(x-x_1)$ are there again so that the previous constraints are not broken and now we demand that $f^{'}(x_1)=p^{'}(x_1)$ to solve for $b$ : $$f^{'}(x_1)=p^{'}(x_1)$$ $$f^{'}(x_1)=f^{'}(x_0)+2a(x_1-x_0)+2b*(x_1-x_0)(x_1-x_1)+b(x_1-x_0)^2 $$ $$b=\frac{f^{'}(x_1)-f^{'}(x_0)}{(x_1-x_0)^2}-\frac{2a}{(x_1-x_0)}$$
I am pretty sure that this is a valid way to do the exact same thing and it's way more efficient than then Lagrange approach as you can add as many points as you want. So my questions are :
- Why does every textbook refer to the Lagrange approach and what's the intuition behind it?
The approach is : $$ p(x)=\sum_{i=0}^{i=n} H_i(x)*f(x_i)+ H^*_i(x)f^{'}(x_i) $$
To calculate the $ H_i(x)$ is it is first defined as $H_i(x) = W_i(x)*L^2_i(x)$ while $H^*_i(x)$ is defined as $H^*_i(x)=Z_i(x)*L^2_i(x)$ where $L^2_i(x)$ is the $i_{th}$ Lagrange term.
By applying some constraints these 2 terms end up being : $$H_i(x) = (1-2L^{'}_i(x)(x-x_i))*L^2_i(x)$$ $$H^*_i(x)=(x-x_i)*L^2_i(x)$$
I managed to prove it but I still can't get the intuition behind it. The first approach seems much more elegant to me and one more thing I was wondering was if the first approach can be transformed to the Lagrange one by some algebraic manipulation.