Is a cubic Lagrange interpolation tensor product the same as bicubic interpolation?
i don't need the points, but the Community keeps resurrecting unanswered questions (or answered questions that have no accepted answer) until i go batty seeing them on the list, time and time-again.

:-\

i think the bicubic interpolation is more likely 3rd-order Hermite polynomial than 3rd-order Lagrange polynomial interpolation. the former guarantees continuity of both the zeroeth and first derivative. the latter only guarantees continuity of the zeroeth derivative (the interpolated function itself).  

this previous answer deals pretty completely with the necessary math of it.