I am having some trouble understanding what the question below is asking. What does the given polynomial $P(x)$ have to do with deriving the not-a-knot spline interpolant for $S(x)$? Also, since not-a-knot is a boundary condition, what does it mean to derived it for $S(x)$?
For general data points $(x_1, y_1), (x_2, y_2),…,(x_N , y_N )$, where $x_1 < x_2 < . .. < x_N$ and $N \geq 4$,
Assume that S(x) is a cubic spline interpolant for four data points $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$, and $(x_4, y_4)$
$$
S(x) =
\begin{cases}
p_1(x), & [x_1,x_2] \\
p_2(x), & [x_2,x_3] \\
p_3(x), & [x_3,x_4] \\
\end{cases}
$$
Suppose $P (x) = 2x^3 + 5x +7$ is the cubic interpolant for the same four points $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$, $(x_4, y_4)$ where $x_1 < x_2 < x_3 < x_4$ are knots. What is the not-a-knot spline
interpolant $S(x)$?
Best Answer
If $S$ is a N-a-K spline with knots $x_1, \dotsc, x_4$ then it satisfies the spline conditions: twelve equations in twelve unknowns. (Twelve coefficients, six equations to prescribe values at the knots and six more to force continuity of derivatives up to third order at $x_2$ and $x_3$.) Since $p_1, p_2, p_3$ fit up to third order in all (two) inner knots, it follows that $p_1 = p_2 = p_3$. So $S$ and $P$ are both Lagrange interpolation polynomials and therefore $S=P$.