I was studying cubic spline interpolation and then I stumbled upon "Cubic runout spline". The idea behind it is that you set the boundary conditions for second derivatives to be:
$f_1''(x_0) = 2f_1''(x_1) - f_2''(x_2)$
$f_n''(x_n) = 2f_n''(x_{n-1}) - f_{n-1}''(x_{n-2})$
As a result, the first two and last two points have their own single curve.
Now, I've figured out the natural, clamped, periodic, not-a-knot type splines and their ideas and proofs, but this one for some reason confused me.
I want to try prove that $M_0 = 2M_1 - M_2$ <=> first 2 curves are the same. I wrote a system of equations for $M_0, M_1$ and $M_2$, where $M_i$ - is a second derivative of our function at point i. I ended up with this:
$S_i(x) = \alpha_i + \beta_i (x - x_i) + \gamma_i (x - x_i)^2 + \delta_i (x - x_i)^3$
$M_0 = 2 \gamma_0$
$M_1 = 2 \gamma_1 = 2 \gamma_0 + 6\delta_0 h_1$
$M_2 = 2 \gamma_1 + 6 \delta_1 h_2$
$h$ is the difference between adjacent x-values
Now, I decided to split the solution into 2 parts: first one is where $h_1 = h_2 = h$ and the second is where they are different.
- for the first part: solving this from right to left (i.e. same curves => equation holds) yielded $\delta_0 = \delta_1 = 0$, which then gave me $M_0 = M_1 = M_2$, which gives me the final equation. Now, the opposite (left to right) proof kinda stopped me. I managed to derive the equality of delta but for gamma I failed to proceed further.
Could anyone give me a hint regarding the proper approach towards this?