## Cubic Splines

Cubic splines are preferred for many applications. Consider the spline composed of cubic polynomials of the form (1)

so that, (2) and (3) for The interpolating function must pass through all the data points, be continuous,

and have continuous derivative at the interior points. Furthermore, there are enough degrees

of freedom to impose continuous curvature as well (continuity of second derivative) at the interior points. (4) (5) (6) (7)

We have unknowns and equations ( from each of (1) and (2) and from each of (3) and (4). Hence, we have the freedom to impose two extra conditions.

We can say the width of the interval, and Next, we shall write all the equations in terms of (given) and C-k (unknown) only. From equations (3) and (7) From equations (1) and (4), From equations (1) and (5)  So far we have obtained expressions for the coefficients and in terms of and All that remains is to match the slopes at the interior points.

From equations (2) and (6): Substitute into this (8) and (9) to give After some simple algebra, we obtain a system of equations to solve for  The problem as been reduced from finding coefficients and to finding values of the curvature We only have equations for but we can obtain two additional equations by specifying

end conditions on the curvature, i.e. conditions involving and 1. Natural or free boundaries: take i.e. the end splines have no curvature at the end points.

2. Clamped boundaries: fix the slopes at each end to specified values.

3. Periodic boundaries: take The last equations can be written in matrix form, In the case of natural splines (i.e. for ), the matrix is non-singular and has a tri-diagonal structure. The solution for is unique and can easily be obtained using rapid tri-diagonal matrix solvers.  We can solve for and from this construct the splines Example: Find the natural spline for the following data: 1 2 3 4 0 1 3 4 2 4 3 4

From this MC=F becomes  Hence Thus, using equations (8), (9), and we obtain      