Show that S is a cubic spline (natural or clamped)

Question

Please see question.

I believe the answer should be:

$S_0(2)=\frac12(x^3-3x+2)=2$

$S'_0(2)=\frac12(3x^2-3)=\frac{9}{2}$

$S''_0(2)=\frac12(6x)=6$

$S_1(2)=\frac12(x^3-12x^2+45x-46)=2$

$S'_1(2)=\frac12(3x^2-24x+45)=\frac{9}{2}$

$S''_1(2)=\frac12(6x-24)=-6$

And since $S''_0(2) \neq S''_1(2)$, S cannot be a cubic spline.

Question b is also irrelevant for the same reason.

Please let me know if you agree or disagree (and why), thank you! :-)

Answer 1

I disagree as cubic spline does not require second derivative to be continuous. A cubic spline is basically a piecewise continuous curve with each segment being defined as a cubic polynomial. In general, there is no requirement on the continuity at the joint of each two consecutive cubic segments. The famous Catmull-Rom spline is a cubic spline with C1 continuity only.

The so-called natural cubic spline is a special case of cubic spline where the joints of each two consecutive cubic segments have 2nd derivative continuity (in addition to continuity in position and first derivative). It is also forced to have zero 2nd derivative at the start and end.