7. Piecewise Polynomial (Spline) Interpolation

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1 Piecewise Polynomial (Spline Interpolation Single polynomial interpolation has two major disadvantages First, it is not computationally efficient when the number of data points is large When the number of data points is large, degree of the interpolating polynomial becomes large too and, therefore, it takes much more time to compute the value of the interpolating polynomial at an intermediate point Second, it generates relatively large numerical error when the number of data points is large because the number of arithmetic operations involved in the computation of a high degree polynomial is large too We will present an interpolation technique that overcomes the problems of simple polynomial interpolation in this chapter Given a set of m points (x 1, f 1, (x, f,, (x m, f m, and (x m, f m with a x 1 < x < < x m < x m b, the idea is to construct a function g (x on [a,b ] such that g ( f i, i 1,,, m (71 and g (x (x, x [ ]; i 1,,, m (7 where (x are polynomials (of low degree In general, (x are different and they are usually chosen to be of degree three g (x constructed this way is called a piecewise polynomial and is a continuous function on [a, b ] (since ( g ( (, i 1,,, m 1 Advantages of piecewise polynomial interpolation are disadvantages of single polynomial interpolation Two types of piecewise polynomial interpolation techniques will be discussed 71 Hermite Interpolation For each data point (, f i, if the derivative of the unknown function f at, f (, is also given, and the piecewise interpolant is required to interpolate the derivatives as well, then the resulting piecewise interpolant is called a piecewise Hermite polynomial This means that for each interval [ ], (x needs to satisfy four conditions: If (x is a cubic polynomial, ie, ( f i ; ( f i ; ( f i ; ( f i (73 (a i,, c i, d i to be determined then we have (x a i + x + c i x + d i x 3

2 a i + + c i + d i 3 f i a i + + c i + c i + c i 3 + d i + 3d i + 3d i f i f i f i (74 or a i c i d i f i f i f i f i (75 Therefore by solving this system, say using Gaussian elimination with partial pivoting, for a i,, c i, and d i, we get the polynomial (x An algorithm can be given as follows Algorithm: Piecewise Cubic Hermite Interpolation (PCHI Input: (, f i, f i, i 1,,,m, a x 1 < x < < x m < x m b, m : number of segments x : a number between a and b Output: the value of the piecewise cubic Hermite polynomial at x var: C : array[1m,14] of real; {to store coefficients of all } k, i : integer; 1 [Construct the piecewise cubic Hermite polynomial] for i :1 to m do begin Gauss(, f i, f i, f i, f i, a i,, c i, d i ; D [i, 1] : a i ; D [i, ] : ; D [i, 3] : c i ; D [i, 4] : d i ;

3 end; [Compute the value of the piecewise polynomial at x] k : 1; while (x > x k do k : k ; {Find the interval I i [ ] that contains x } b : D [k,4]; {b d k } for i : 3 downto 1 do b : D [k,x ] + x*b ; 3 Return b p k (x In step 1 of the above algorithm, Gauss(, f i, f i, f i, f i, a i,, c i, d i is a procedure that solves the equation in (75 for a i,, c i, and d i using Gaussian elimination with partial pivoting (x, however, can be computed in a more efficient way Note that if h 1 (t, h (t, h 3 (t and h 4 (t are defined as follows: h 1 (t 1 3t + t 3 ; h (t 3t t 3 ; h 3 (t t t + t 3 ; t [0, 1] (76 h 4 (t t + t 3 ; h 1 (t h (t 0 1 Figure 71 Graphs of h 1 (t and h (t then it is easy to see that the value of h 1 (t at t 0 is 1 and its value at t 1 and its derivatives at t 0 and t 1 are all equal to 0, while the value of h (t at t 1 is 1 and its value at t 0 and its derivatives at t 0 and t 1 are all equal to 0 (see Figure 71 for the graphs of h 1 (t and h (t On the other hand, the derivative of h 3 (t at t 0 is 1 and its derivative at t 1 and its values at

4 h 3 (t h 4 (t 0 1 Figure 7 Graphs of h 3 (t and h 4 (t t 0 and t 1 are all equal to 0, while the derivative of h 4 (t at t 1 is 1 and its derivative at t 0 and its values at t 0 and t 1 are all equal to 0 (see Figure 7 for the graphs of h 3 (t and h 4 (t Namely, h 1 (0 1; h 1 (1 0; h 1 (0 0; h 1 (1 0; h (0 0; h (1 1; h (0 0; h (1 0; h 3 (0 0; h 3 (1 0; h 3 (0 1; h 3 (1 0; h 4 (0 0; h 4 (1 0; h 4 (0 0; h 4 (1 1; Hence, if the values and derivatives of a function g (t are given at t 0 and t 1, and denoted by A, B, C, and D, respectively, ie, A g (0, B g (1, C g (0 and D g (1, then a polynomial of degree three which interpolates g (t and g (t at t 0 and t 1 can be defined as follows: p (t A h 1 (t + B h (t + C h 3 (t + D h 4 (t (77 It is easy to see that p (0 A, p (1 B, p (0 C and p (1 D h 1 (t and h (t are called Hermite basis functions of the first kind and h 3 (t and h 4 (t are called Hermite basis functions of the second kind To construct the polynomial (x which satisfies the conditions in (73, one first defines H 1 (x, H (x, H 3 (x and H 4 (x as follows: x H 1 (x h 1 ( x 1 3( x + ( 3 ; x x x 3 ; H (x h ( x H 3 (x h 3 ( 3( x 3( x ( x + ( 3 ; (78 x H 4 (x h 4 ( x ( x + ( 3 ;

5 where x [ ] H 1 (x, H (x, H 3 (x and H 4 (x satisfy the following conditions: H 1 ( 1; H 1 ( 0; H 1 ( 0; H 1 ( 0; H ( 0; H ( 1; H ( 0; H ( 0; H 3 ( 0; H 3 ( 0; H 3 ( 1; H 3 ( 0; H 4 ( 0; H 4 ( 0; H 4 ( 0; H 4 ( 1; Namely, H 1 (x, H (x, H 3 (x and H 4 (x are Hermite basis functions defined on the interval [ ] Therefore, following the p (t defined in (77, one can define the (x which satisfies the conditions in (73 as follows: (x f i H 1 (x + f i H (x + f i H 3 (x + f i H 4 (x (79 It is easy to see that ( f i, ( f i, ( f i and ( f i 7 Spline Interpolation A piecewise cubic interpolating polynomial g (x is called a cubic spline if it has continuous first and second derivatives The process of determining such a piecewise polynomial is called cubic spline interpolation Given a set of n points (x 1, f 1, (x, f,, (x m, f m, and (x m, f m with a x 1 < x < < x m < x m b, a cubic spline g (x that interpolates these points must satisfy the following conditions: (1 g ( f i, i 1,,, m ; ( for each subinterval [ ], i 1,,, m, g (x is equal to some cubic polynomial (x for x [ ]; (3 for each, i, 3,, m, the left derivative of g at equals the right derivative of g at, ie, g ( g ( +, and the left second derivative of g at equals the right second derivative of g at, ie, g ( g ( + Condition (3 is equivalent to the following conditions: 1 ( (, 1 ( ( ; i, 3,, m (710 For each i, 3,, m, by using ( 1, f i 1, ( 1, f i 1, (, f i and (, f i as interpolation points for the interval [ 1 ], we can put 1 (x in Newton s form as follows: 1 (x f [ 1 ] + f [ 1 1 ](x 1 + f [ 1 1 ](x 1 (x 1 + f [ 1 1 ](x 1 (x 1 (x (711 By (418 we know that if g (x is a (continuous function on the interval [a, b ] then the second

6 divided difference of g (x, g [a, b ], can be expressed as follows If g (x is differentiable at x a then we have This follows from the fact that g [b ] g [a ] g (b g (a g [a, b ] b a b a g [a, a ] g (a (71 g [a, a ] lim g [a, b ] lim b a b a g (b g (a b a g (a Therefore by using (418, (71 and the fact that f [ 1 ] f ( 1 f i 1, (711 can be expressed as 1 (x f i 1 + f ( 1 (x f [ 1 ] f [ 1 1 ] 1 f [ 1 ] f ( 1 1 (x 1 (x 1 [(x 1 ( 1 ] f i 1 + f ( 1 (x 1 + f [ 1 ] f ( 1 1 (x 1 + (f [ 1 ] f [ 1 1 ] (x 1 + f [ 1 ] f [ 1 1 ] 1 (x 1 3 By combining like terms and using the notation, 1 (x can be expressed as 1 (x a i (x 1 + c i 1 (x 1 + d i 1 (x 1 3 ; i, 3,,m (71 for where a i 1 f i 1 (713 1 f ( 1 (714 c i 1 f [ 1 ] 1 d i 1 (715

7 d i 1 f [ 1 ] f [ 1 1 ] f [ 1 ] + 1 ( (716 Note that in (71, 1 are the only unknowns; c i 1 and d i 1 are computable once we have for all i 1,,, m, m To find a solution for (71, recall that g (x must have continuous second derivative at, i, 3,, m, which is equivalent to the the second part of (710, ie, Since 1 ( ( ; i, 3,, m (717 and 1 ( c i 1 + 6d i 1 ( c i (717 is equivalent to c i 1 + 3d i 1 c i ; i, 3,, m (718 Using the definition of c k in (715, (718 can be written as f [ 1 ] 1 d i 1 + 3d i 1 f [ ] d i By combining like terms and using the definition of d k in (716, we have f [ 1 ] 1 f [ 1 ] ( f [ ] f [ ] + ( or Therefore, we have f [ 1 ] f [ 1 ] f [ ] + f [ ] 1 + ( + + 3(f [ 1 ] + f [ ] ; i, 3,, m (719

8 where, i 1,,, m, are to be determined (Note that f ( But we don t know the values of f ( (719 can also be put in in matrix form, as follows x ( x 1 + x x 3 x 1 ( x + x 3 x m 1 x ( x m 1 + x m x m x m ( x m + x m 1 x m 1 b 1 b b ṁ b m e e 3 e m 1 e m (70 where e i 3(f [ 1 ] + f [ ] ; i, 3,, m (71 With m unknows and m 1 equations, this is an underdetermined system We need two extra conditions (equations to solve this system We will consider two cases here Case 1: f 1 and f m are given In this case, the system of equations in (70 becomes of the following form: ( x 1 + x x 3 x 1 ( x + x 3 x m 1 x ( x m 1 + x m x m x m ( x m + x m 1 b b 3 b m 1 b m e x b 1 e 3 e m 1 e m x m 1 b m (7 where e i, i, 3,, m, are defined in (71 This is a system of m 1 equations in m 1 unknowns and it is strictly diagonally dominant Therefore, (7 is solvable The process of constructing the spline function g (x can be done in four steps (1 Construct e i, i, 3,, m, using the following foumula: e i 3(f [ 1 ] + f [ ] f [ ] f [ 1 ] f [x 3( x i ] f [ ] i + x i 1 1 f i f 3( x i 1 i f i f + x i i 1

9 ( Solve (7 using a special form of Gaussian elimination without partial pivoting for b, b 3,, b m (3 Use (716 to compute d i for i 1,,, m d i f [ 1 ] + 1 ( f i f i ( 1 (4 Use (715 to compute c i for i 1,,, m, and them form (x for i 1,,, m An algorithm can ge given as follows For m given points (x 1, f 1, (x, f,, (x m, f m, and the tangents b 1 and b m at x 1 and x m, this algorithm constructs a cubic spline that interpolates (, f i, i 1,,, m, b 1, and b m, and computes the value of the cubic spline at x z Algorithm: CUBIC_SPLINE_1 Input: m : number of interpolation points (, f i, i 1,,, m, a x 1 < x < < x m < x m b b 1, b m : tangents at x 1 and x m z : a point different from all Output: the value of the cubic spline at z var: X, {array to store all } F, {array to store all f i } B : array[1m] of real; {array to store all } C, {array to store all c i } D, {array to store all d i } : array[1m] of real; {array to store all } 1 [Construct the cubic spline] for i :1 to m do begin X [i ]: ; F [i ]: f i ; end; B [1]: b 1 ; B [m ]: b m ; SPECIAL_GAUSS(X, F, B ;

10 {This procedure solves (7 for, i, 3,, m, and returns these values in B [i ], i, 3,, m } for i :1 to m do [i ]: X [i ] X [i ]; for i :1 to m do D [i ]: (B [i ] * (F [i ] F [i ]/ [i ] + B [i ]/( [i ] ; for i :1 to m do C [i ]: ((F [i ] F [i ]/ [i ] B [i ]/ [i ] [i ]*D [i ] ; [Compute the value of the cubic spline at x z ] k : 1; while (z > X [k ] do k : k ; {Find the interval I k [x k, x k ] that contains z } g : F [k ] + B [k ]* (z X [k ] + C [k ]* (z X [k ] + D [k ]* (z X [k ] 3 ; 3 Reture g To design a special form of Gaussian elimination without partial pivoting, first observe that, since most of the entries in the coefficient matrix of (7 are zero, we only need a (m 1* 3 array to store all the non-zero entries of the coefficient matrix (see Figure and an one dimensional array to store the constants Besides, each equation, except the last one, will be used to eliminate the equation below it only For instance, equation 1 will be used to eliminate equation only Therefore, procedure SPECIAL_GAUSS can be designed as follows ( x + x 1 x 1 x 3 ( x 3 + x x x 4 ( x 4 + x 3 x 3 * * * * * * * * * x m 1 ( x m 1 + x m x m x m ( x m + x m 1 Figure 73 An (m 1 3 array for the coefficient matrix

11 Procedure SPECIAL_GAUSS (X, F : array[1m] of real; var B :array[1m] of real; var: M : array[1m-1,13] of real; H : array[1m-1] of real; a : real; 1 [Forward Elimination] for i :1 to m 1 do begin M [i,1]: X [i +] X [i ]; M [i,3]: X [i ] X [i ]; M [i,]: * (M [i,1] + M [i,3]; H [i ]: 3* (M [i,1]* (F [i ] F [i ]/M [i,3] + M [i,3]* (F [i +] F [i ]/M [i,1]; end; H [1]:H [1] M [1,1]*B [1]; H [m 1]:H [m 1] M [m 1,3]*B [m ]; for i :1 to m do begin a : M [i,1]/m [i,]; {Multiplier} M [i,]: M [i,] a*m [i,3]; H [i ]: H [i ] a*h [i ]; end; [Back Substitution] B [m ]: H [m 1]/M [m 1,]; for i :m downto 1 do B [i ]: (H [i ] M [i,3]*b [i +]/M [i,];

12 References 1 Ward Cheney and David Kincaid, Numerical Mathematics and Computing, 3rd Edition, Brooks/Cole, Pacific Grove, California, 1994 Samuel D Conte and Carl de Boor, Elementary Numerical Analysis: An Algorithmic Approach, 3rd Edition, McGraw-Hill, New York, Carl-Erik Froberg, Introduction to Numerical Analysis, nd edition, Svenska Bokforlaget Bonniers, Sweden, Shoichiro Nakamura, Applied Numerical Methods in C, Prentice Hall, Englewood Cliffs, New Jersey, Pat H Sterbenz, Floating-Point Computation, Prentice Hall, Englewood Cliffs, New Jersey