CubicSplinesProof-1 - 280 C HAP 5 C URVE F ITTING Piecewise...

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280 C HAP . 5 C URVE F ITTING Piecewise Cubic Splines The fi tting of a polynomial curve to a set of data points has applications in CAD (computer-assisted design), CAM (computer-assisted manufacturing), and computer graphics systems. An operator wants to draw a smooth curve through data points that are not subject to error. Traditionally, it was common to use a french curve or an ar- chitect s spline and subjectively draw a curve that looks smooth when viewed by the eye. Mathematically, it is possible to construct cubic functions S k ( x ) on each inter- val [ x k , x k + 1 ] so that the resulting piecewise curve y = S ( x ) and its fi rst and second derivatives are all continuous on the larger interval [ x 0 , x N ] . The continuity of S ( x ) means that the graph y = S ( x ) will not have sharp corners. The continuity of S ( x ) means that the radius of curvature is de fi ned at each point. Definition 5.1. Suppose that { ( x k , y k ) } N k = 0 are N + 1 points, where a = x 0 < x 1 < · · · < x N = b . The function S ( x ) is called a cubic spline if there exist N cubic polynomials S k ( x ) with coef fi cients s k , 0 , s k , 1 , s k , 2 , and s k , 3 that satisfy the following properties: I. S ( x ) = S k ( x ) = s k , 0 + s k , 1 ( x x k ) + s k , 2 ( x x k ) 2 + s k , 3 ( x x k ) 3 for x ∈ [ x k , x k + 1 ] and k = 0, 1, . . . , N 1. II. S ( x k ) = y k for k = 0, 1, . . . , N . III. S k ( x k + 1 ) = S k + 1 ( x k + 1 ) for k = 0, 1, . . . , N 2. IV. S k ( x k + 1 ) = S k + 1 ( x k + 1 ) for k = 0, 1, . . . , N 2. V. S k ( x k + 1 ) = S k + 1 ( x k + 1 ) for k = 0, 1, . . . , N 2. Property I states that S ( x ) consists of piecewise cubics. Property II states that the piecewise cubics interpolate the given set of data points. Properties III and IV require
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S EC . 5.3 I NTERPOLATION BY S PLINE F UNCTIONS 281 that the piecewise cubics represent a smooth continuous function. Property V states that the second derivative of the resulting function is also continuous. Existence of Cubic Splines Let us try to determine if it is possible to construct a cubic spline that satis fi es proper- ties I through V. Each cubic polynomial S k ( x ) has four unknown constants ( s k , 0 , s k , 1 , s k , 2 , and s k , 3 ); hence there are 4 N coef fi cients to be determined. Loosely speaking, we have 4 N degrees of freedom or conditions that must be speci fi ed. The data points supply N + 1 conditions, and properties III, IV, and V each supply N 1 conditions. Hence, N + 1 + 3 ( N 1 ) = 4 N 2 conditions are speci fi ed. This leaves us two ad- ditional degrees of freedom. We will call them endpoint constraints : they will involve either S ( x ) or S ( x ) at x 0 and x N and will be discussed later. We now proceed with the construction. Since S ( x ) is piecewise cubic, its second derivative S ( x ) is piecewise linear on [ x 0 , x N ] . The linear Lagrange interpolation formula gives the following representation for S ( x ) = S k ( x ) : (4) S k ( x ) = S ( x k ) x x k + 1 x k x k + 1 + S ( x k + 1 ) x x k x k + 1 x k . Use m k = S ( x k ) , m k + 1 = S ( x k + 1 ), and h k = x k + 1 x k in (4) to get (5) S k ( x ) = m k h k ( x k + 1 x ) + m k + 1 h k ( x x k ) for x k x x k + 1 and k = 0, 1, . . . , N 1. Integrating (5) twice will introduce two constants of integration, and the result can be manipulated so that it has the form (6) S k ( x ) = m k 6 h k ( x k + 1 x ) 3
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