Data Interpolation and Its Effects on Digital Sound Quality - McMurry ...

152.3.1 Example Suppose we want to do a Cubic Spline <strong>Interpolation</strong> to thefollowing 5 points: (-1, -1), (0, 0), (1, 1), (2, 0), (3, -1). First we find a cubicfunction that passes through the first three points. The easy choice is f 1 (x) = x 3 .Let the second piece of the function be defined as f 2 (x) = c 1 x 3 + c 2 x 2 + c 3 x + c 4 .Now, the derivative of the first function evaluated at 1 will be equal to thederivative of our second cubic at 1. This gives us the following:3f 1( x) x2f 1'( x) 3xf21'(1) 3(1) 3From this we know that the derivative of our second cubic function must equal 3when evaluated at 1. Taking the derivative of f 2 , we find that f 2 ’(x) = 3c 1 x 2 + 2c 2 x+ c 3 . We also know that the function itself must equal 1 when evaluated at 1,must equal 0 when evaluated at 2, <strong>and</strong> must equal –1 when evaluated at 3.Given this information, we obtain the following system of equations:3c1 2c2 c3 3c1 c2 c3 c418c1 4c2 2c3 c4 027c1 9c2 3c3 c4 1To obtain the cubic satisfying the above, we solve the system of equations forour constants. Using a TI-83 calculator <strong>and</strong> Gauss-Jordan elimination, the resultyields that c 1 =2, c 2 =-12, c 3 =21 <strong>and</strong> c 4 =-10. Thus the second part of ourinterpolant can be expressed as 2x 3 -12x 2 +21x-10. So we say our interpolant