I have seen the following cubic interpolation formula in multiple books on multigrid:
$$f(x) = -\frac{1}{16}f(x-3h) + \frac{9}{16}f(x-h) + \frac{9}{16}f(x+h) -\frac{1}{16}f(x+3h).$$
Applying the above to some vector $v_i = f(x + (1+2i)h)$, such that $v_i = \delta^k_i$ yields: $(-\frac{1}{16}, 0, \frac{9}{16}, 1, \frac{9}{16}, 0, -\frac{1}{16})$. I do not understand how the initial expression for $f(x)$ was derived however. The best I could find was a paper on pseudo-splines Multigrid methods: grid transfer operators and subdivision schemes where they call the above "the classical cubic interpolation $p_{2,1}$" though I must say the pseudo-splines explanation goes way over my head.
At the very least I could not derive it in the same way I derived the weights for linear interpolation:
$$\frac{f(x_0+t) – f(x_0)}{f(x_1)-f(x_0)} = \frac{x_0+t-x_0}{x_1-x_0} \implies f(x_0+t) = f(x_0) + \frac{t}{x_1-x_0}(f(x_1)-f(x_0))$$
$$\text{let}\, x_1-x_0 = 1 \implies f(x_0+t) = (1-t)f(x_0) + tf(x_1)\implies \\ f\left(x_0+\frac{1}{2}\right) = \frac{1}{2}f(x_0) + \frac{1}{2}f(x_1)$$
For the cubic case I understand that the function $f$ is constructed to interpolate the points $(x\pm h, \frac{9}{16}f(x\pm h)),\, (x\pm 3h, -\frac{1}{16}f(x\pm 3h))$, but I do not understand what the idea was behind making it interpolate those. If someone told me that $f$ should be a cubic interpolating function, then I would have made it interpolate $(x\pm 3h, f(x\pm 3h))$ and $(x\pm h, f(x\pm h))$ similar to the linear case, but this is clearly not the function that is given above.
Best Answer
Those coefficients are the values at 0 of cubic Lagrange polynomials for points -3,-1,1 and 3.Those values will be the same for arbitrary $x$ and $x_k = x + (2k-3)h$