You can derive the formula for this using the Chinese Remainder Theorem.
Consider the problem of finding a polynomial p(x) such that p(x_i) = y_i for some finite set of (x_i) and (y_i). The classic approach to this is Lagrange interpolation, which can be seen as a special case of the Chinese Remainder Theorem. Essentially, you are trying to find a p which satisfies the following modular congruences:
p(x) == y_i (mod (x - x_i))
The constructive proof of CRT, specialised to this problem, yields an algorithm equivalent to Lagrange interpolation. To extend Lagrange interpolation so as to include derivative information y'_i at each x_i, change that congruence to:
p(x) == y_i + y'_i (x - x_i) (mod (x - x_i)^2)
And to give second derivative information, change the modular congruence to:
Bezier curves are actually my favorite way to teach recursion. They give a nice visual representation of what's happening, the base case is pretty straight forward, and so is the math. The time-step element adds just enough extra complexity to take it beyond "let's sum a list" or "check if a string is a palindrome".
If you happen to desire to implement your own NURBS library, (and/or the bspline and bezier simplifications) take the nurbs book c reference implementation, and modernize, convert to you language of choice.
Grant Sanderson (3blue1brown) had a video submission contest and the winner that blew me away did a video on bezier curves. https://youtu.be/aVwxzDHniEw
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[ 231 ms ] story [ 756 ms ] threadConsider the problem of finding a polynomial p(x) such that p(x_i) = y_i for some finite set of (x_i) and (y_i). The classic approach to this is Lagrange interpolation, which can be seen as a special case of the Chinese Remainder Theorem. Essentially, you are trying to find a p which satisfies the following modular congruences:
The constructive proof of CRT, specialised to this problem, yields an algorithm equivalent to Lagrange interpolation. To extend Lagrange interpolation so as to include derivative information y'_i at each x_i, change that congruence to: And to give second derivative information, change the modular congruence to: And so on...https://link.springer.com/book/10.1007/978-3-642-97385-7