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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 $S(x)$?

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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 $S=P$.

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  • $\begingroup$ How did you get from "fit up to third order in all inner knots" to "p1 = p2 = p3"? $\endgroup$ – Yilun Zhang May 16 '14 at 14:17

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