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November 18, 2005

Spline Representations X

 

An Example to Motivate and Illustrate Hildebrand Interpolation.

 

Say you’re conducting a scientific experiment with data acquisition tools , a finite sequence of data points is obtained through data acquisition and it is desired to connect these points with a smooth curve.  Isobars on weather maps, for instance.  Cubic splines are one way to do this, but in Connecting the Dots Parametrically: An Alternative to Cubic Splines by Wilbur Hildebrand in the College Mathematics Journal, Vol. 21, No. 3 (May, 1990), 208-215, an alternative approach is suggested.

 

Given a sequence of n ordered points , the method cubic splines determines n  1 cubic polynomials (possibly degenerate) chosen to meet the 2nd order continuity condition at the knots where they are “tied” together.  This makes the curve pleasingly smooth throughout.  Additional conditions are applied at the first and last points P1 and Pn to determine a unique spline meeting all conditions. 

 

As we’ve seen, this can lead to significant computations, but a bigger problem may be that, if you take a spline apart, the pieces don’t meet the same end conditions as the whole spline itself. That’s what motivated Hildebrand to seek an alternate approach to interpolation.  His method uses a parametrically defined curve to make a smooth connection between two points by considering the positions of the two nearest neighbors before and after these points. At a given point the tangent to the curve exists and is parallel to the chord connecting the preceding point to the succeeding point.  For this method, there is no great benefit in equally spaced x-values and no requirement that y is a function of x. 

 

Hildebrande motivates his method with an example.  Consider the four consecutive points

A(6, 7), B(0,0),C(12, 9) and D(14,2).  The method for constructing a smooth curve connecting B and C is described in the following steps

  1. Determine the parameterization of a curve with both x(t) and y(t) in the form of a quadratic function  and such that the parameterization passes through A when t=1, through B when t=0 and through C when t=1.  Since 3 points determine a parabola, this is possible.  In the example here,  and  
  2. Similarly, determine the parametrization of a curve with x(t) and y(t) again in quadratic form, this time passing through B when t=0 and through C when t=1 and through D when t=2.  In the example,  and

At this stage we have two different arcs connecting B and C whose parameterizations both correspond to the interval .  We can take a weighted average of these over the interval, giving linearly more weight to one and less to the other, like so:
                 
and 
               
 If we now want to add a fifth point, E, say, at (18, 1), then you could re-center at C and just repeat fitting parabolic parametric functions  on [1,1] and   on [0,2].  Taking the sum of weighted averages we get, on the interval ,

                                                   

The first derivatives of the averaged parameterizations from the left and right agree at C:

                                        

 

                                        

However, the second derivatives are close, but do not agree: