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HOME › AREAS OF EXPERTISE › Interpolation Applications › ~ Cubic Spline Method

 "Interpolation Solutions" Cubic Spline Method
 X = (3, 5, 7)
 Y = (8.92857142857143, 24.9285714285714, 46.8571428571429)

 Array X = { , , , , } Array Y = { , , , , } Specify New X values = { , , } [ Initial ArrayValues X: {0,2,4,6,8} ] [ Initial ArrayValues Y: {0,4,16,36,64} ] [ Initial SplineX Values specified: {3,5,7} ]

IMPLEMENTATION
Cubic Spline Interpolation

This method provides a great deal of smoothness for interpolations with significantly varying data. It uses weight coefficients on the cubic polynomials to interpolate the data. These coefficients bend the line so that it passes through each of the data points without any erratic behavior or breaks in continuity.

Algorithm Creation

The basic idea of the cubic spline is actually based on a third degree polynomial defined by,

 si(x) = ai(x - xi)3 + (x - xi)2 + ... for i=1,2,...n-1

Testing the Cubic Spline Method

In order to test the Spline method as defined above, a new TestSpline() static method has been added and executed. Supporting code and methods are not shown.

static void TestSpline();
{
ListBox1.Items.Clear();
ListBox2.Items.Clear();
double[] xarray = new double[] { t1, t2, t3, t4, t5 };
double[] yarray = new double[] { t6, t7, t8, t9, t10 };
double[] x = new double[] { t11, t12, t13 };
double[] y = Interpolation.Spline(xarray, yarray, x);
VectorR vx = new VectorR(x);
VectorR vy = new VectorR(y);
ListBox1.Items.Add(" " + vx.ToString());
ListBox2.Items.Add(" " + vy.ToString());
}

We first defined a set of data points as xarray and yarray. We then compute the y values at the xSpline values specified. The user can manipulate all values and try variations on the arrays themselves as well as specifying new xSpline values.

 Other Implementations...

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