Spline Representations III
================================ Ok-I can’t quite get the mistake, so I’m going to take a more straightforward approach. We want to specify coefficients for two cubic functions meeting the zero and first order continuity conditions:
Now, and . Then, plugging x = 3 into the first cubic and its derivative, we have
which is easy to solve: and
Similarly, plugging x = 3 into the second equation and its derivative produces and while plugging x = 6 into these produces the system
whose solution is and . This produces a spline (shown in green and shifted up one unit from its initial position) as shown here This is well and good, except that it doesn’t meet the second order condition since y”(3) would be –4/3 for the first cubic and –22/9 for the second.
Let’s suppose we know the slope at all control points except the last and that the 2nd order continuity condition is satisfied. So
Clearly d1 = 0, d2 = 4, c1 = 1 and c2 = ½. So we need 4 independent equations for a1, a2, b1 and b2: The value of the first cubic at its right endpoint; the slope of the first cubic at its right endpoint yield this 2x2 system
which yields b1 = ½ and a1 = –7/54.
The 2nd order continuity condition yields the value of b2:
…and the value of the second cubic at its right endpoint yields the final parameter:
So the spline is
whose green graph overlays the red and blue cubics below:
So now I get! To do a cubic-splines thing, one usually starts with more than 4 points (here I have only 3) and one finds the slopes that satisfy the 1st and 2nd order continuity conditions. Think about degrees of freedom in a cubic: there are 4.
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