Example: Cubic Splines with Known Derivatives


The example for Cubic Splines with Known Data Points produced the second derivatives at each entered point. The following example uses those derivatives to arrive at the same result


Example

From the second derivatives listed below, interpolate the function at 0.8.

x

f''(x)


0.6 -0.1999347
0.7 -0.2200526828
0.9 -0.2409746016
1.0 -0.2422470246
1.1 -0.2396373



Procedure

Press

Display


Select the program [ RUN ] { MTH }
{ INT } { CUB }
{ NEW }
D_eLsbAL
Enter the number of points 5 { m }
{ EOD }
D_w2jA0F
Enter second derivatives .1999347 [ +/- ]
{ ENT }
D_qgAR5J
.2200526828 [ +/- ]
{ ENT }
D_SXqccq
.2409746016 [ +/- ]
{ ENT }
D_VOkAa5
.2422470246 [ +/- ]
{ ENT }
D_Ct3e1u
.2396373 [ +/- ]
{ ENT }
D_1QsPpH
Interpolate at .8 { NO }
.8 { x }
D_yG8qsQ
Repeat the problem { ESC } { ESC }
{ OLD }
D_eLsbAL
Recalculate the interpolation, using zero for the endpoint second derivatives.

Procedure

Press

Display


Enter the number of points 5 { m }
{ EOD }
D_w2jA0F
Enter second derivatives 0
{ ENT }
D_qgAR5J
[ CE ]
{ ENT }
D_SXqccq
[ CE ]
{ ENT }
D_VOkAa5
[ CE ]
{ ENT }
D_Ct3e1u
0
{ ENT }
D_1QsPpH
Interpolate at .8 { NO }
.8 { x }
D_yG8qsQ

The result is the same as for nonzero endpoint second derivatives. The endpoints are too distant from the point of interpolation for a small change in the endpoint second derivative to have any effect on the result.


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