Interpolation Theory

Transcription

1 Numerical Analysis Massoud Malek Interpolation Theory The concept of interpolation is to select a function P (x) from a given class of functions in such a way that the graph of y P (x) passes through the given data points (x i, y i ), i 1,,, n The points may arise as measurements in a physical problem, or they may be obtained from a known function The interpolating function is usually chosen from a restricted class of functions, with polynomials being the most commonly used class Theorem 1 (Weierstrass Approximation Theorem) If f(x) is a continuous function on [a, b] and ɛ > is given, then there exists a polynomial P (x), defined on [a, b], with the property that f(x) P (x) < ɛ for all x [a, b] Theorem Given n + 1 distinct points x, x 1,, x n and n + 1 ordinates y, y 1,, y n, there is a polynomial P (x) of degree n that interpolates to y i at x i, i, 1,,, n This polynomial P (x) is unique among the set of all polynomials of degree at most n Proof Consider the following linear system of n + 1 equations and n + 1 unknowns a + a 1 x + a x + a n x n y a + a 1 x n + a x n + a n x n n y n The unknowns are (a, a 1,, a n ) The proof follows from the fact that this system has a unique solution Taylor Polynomials Let f(x) be n + 1 times differentiable near a point x Then the n-degree Taylor polynomial of f(x) at x is The function P n (x) f(x ) + f (x )(x x ) + f (x ) (x x ) + f (n) (x ) (x x ) n! n! R n (x) P n (x) f(x) f (n+1) (ξ(x)) (x x ) (n+1) (n + 1)! is called the Remainder function or the Error function Example Approximate ln 11 using a Taylor polynomial of degree 3 Note that if f(x) 1 1+x, then 1 f(x)dx ln 11 Thus we need to integrate the polynomial P 3 (x) 1 x + x x 3 from zero to 1 The error is 1 (1 x + x x 3 )dx 1 1 [ f(x) P 3 (x) x 4 dx x 5 5 ] 1 [x x + x3 3 x ] 1 Bernstein Polynomials Given a function defined on [, 1], the Bernstein polynomial of degree n for f(x) is given by n ( ) n B n (x) f( k k n )xk (1 x) n k k

2 Massoud Malek Interpolation Theory Page where ( ) n k denotes n! k!(n k)! It can be shown that, if f(x) is continuous on [, 1] and x [, 1], then lim B n(x ) f(x ) n The polynomial can be used in a constructive proof of Weierstrass Theorem Lagrange Polynomial Given two points (x, y ) and (x 1, y 1 ) with x x 1, we can draw a straight line through them The line is the graph of the linear polynomial P 1 (x) (x 1 x) x 1 x y + (x x ) x 1 x y 1 Given n + 1 distinct points x, x 1,, x n and n + 1 ordinates y, y 1, ldots, y n, according to Theorem, there is a unique polynomial P n (x) that interpolates to y k at x k, k, 1,,, n For k 1,,, n, define the polynomial L n,k (x) (x x n ) (x x k 1 )(x x k+1 ) (x x n ) (x k x ) (x k x k 1 )(x k x k+1 ) (x k x n ) Then the polynomial L n (x) n y k L n,k (x) k is called the Lagrange Interpolation Polynomial Note that L n,k (x j ) δ k,j, where δ k,j is called Kronecker delta function, { 1, j δ k,j k;, j k k, j n j k n (x x j ) (x k x j ) Theorem 3 If x, x 1, x n are distinct numbers in the interval [a, b] and if f(x) C n+1 [a, b], then, for each x [a, b], there exist a number ξ(x) (a, b) such that f(x) L n (x) + f (n+1) (ξ(x)) (x x )(x x 1 ) (x x n ) (n + 1)! Example Let f(x) 1 1+x, the find the Lagrange polynomial of degree two passing through the points (, f()), (1, f(1)) and (, f()) Hence We have f() 1, f(1) 1, f() 1 3 and L (x) k (x 1)(x ) L, (x) ( 1)( ) L,1 (x) L, (x) (x 1)(x ) x 3x + (x )(x ) (1 )(1 ) (x )(x ) x + x (x )(x 1) ( )( 1) L,k (x)f(k) x 3x + (x )(x 1) x x (1) + ( x + x)( 1 ) + x x ( 1 3 ) x 6 x 3 + 1

3 Massoud Malek Interpolation Theory Page 3 Inverse Interpolation Suppose f(x) C 1 [a, b], f (x) on [a, b] and f(x) has one zero x in [a, b] Then clearly f(x) is a one to one function on [a, b] By choosing few numbers x k s in [a, b], we create the points (y k, x k ) s and using these points, we define the Lagrange polynomial n L n (y) L n,k (y)x k This way L() becomes an approximation to x k Newton Divided Differences The Lagrange form of the interpolation polynomial can be used for interpolating a function But there are other forms that are much more convenient Suppose P n 1 (x) is a polynomial of degree n that interpolate f(x) We would like to pass from P n 1 (x) to another interpolation polynomial of degree one greater; so we need to find P n (x) P n 1 (x) + C(x) C(x) correction term Then in general C(x) is a polynomial of degree n, since usually degree(p n 1 ) n 1 and degree(p n ) n Also we have C(x k ) P n (x k ) P n 1 (x k ) f(x k ) f(x k ) k, 1,, n 1 Since P n (x) f(x n ), we have f(x n ) P n 1 (x n ) a n (x n x ) (x n x n 1 ) This coefficient a n is called the n-th order Newton Divided Difference of f(x), and it is denoted by a n f[x, x 1,, x n ] Thus our interpolation formula becomes P n (x) P n 1 (x) + (x x ) (x x n 1 )f[x,, x n ] Using the Lagrange formula, we obtain f[x, x 1,, x n ] f[x 1, x,, x n ] f[x, x 1,, x n 1 ] x n x We have P (x) f(x ) P 1 (x) f(x ) + (x x )f[x, x 1 ] P n (x) f(x ) + (x x )f[x, x 1 ] + (x x )(x x 1 )f[x, x 1, x ] + (x x ) (x x n 1 )f[x, x 1,, x n ] This is called Newton s divided difference formula for the interpolation polynomial It is much better for computation that the Lagrange formula To construct the divided differences, use the format shown in the following table x i f(x i ) f[x i, x i+1 ] D f[x i ] D 3 f[x i ] D 4 f[x i ] x y f[x, x 1 ] x 1 y 1 f[x, x 1, x ] f[x 1, x ] f[x, x 1, x, x 3 ] x y f[x 1, x, x 3 ] f[x, x 4 ] f[x, x 3 ] f[x 1, x, x 3, x 4 ] x 3 y 3 f[x, x 3, x 4 ] f[x 1,, x 5 ]

4 Massoud Malek Interpolation Theory Page 4 Example We construct a divided difference table for f(x) x, shown in the next table x i f(x i ) f[x i, x i+1 ] D f[x i ] D 3 f[x i ] D 4 f[x i ] The polynomial P 4 (x) (x )[(3494) + (x 1)[( 411) + (x )[(9167) + (x 3)[( 84)]]]] is known as the Newton Forward Divided-Difference form The following polynomial is known as the Newton Backward Divided-Difference form: Q 4 (x) (x 4)[(3618)+(x 3)[( 3585)+(x )[(8333)+(x 1)[( 84)]]]] Algorithm DIVDIF(N,X,D) Remark: On entrance, X (X(), X(N)) and D (F (X 1 ) F (X n )) On exit, D(J) will contain F [X, X 1, X J ] STEP1 INPUT N a positive integer; STEP INPUT X a vector of points x(), X(N); STEP3 INPUT D a vector of values of some function evaluated at the nodes in X; STEP4 FOR I : 1, N; FOR J : N, I, 1; D(J) : D(J) D(J 1) X(J) X(J I) RETURN; STEP5 STOP Algorithm FORWDINTERP(N,X,F) Remark: On entrance, X (X(), X(N)) and D (D(), D(1), D(n) On exit, P will contain the value P N (T ) of the N-th degree polynomial interpolation of F (X) STEP1 P : D N ; STEP FOR K : N 1,, 1; P : D(K) + (T X(K) )P RETURN; STEP3 STOP Note By using Divided-difference and Inverse Interpolation, we may approximate the zero of certain functions Hermite Interpolation For a variety of applications, it is convenient to consider polynomials P (x) that interpolate to a function f(x) at certain points x, x 1, x n and in addition to have the derivative polynomial P (x) interpolate to the derivative function f (x) Since the first derivative of f(x) and P (x) agree at the interpolation points, they have basically the same shape To construct the Hermite interpolation polynomial we use the functions L n,k (x) defined in the Lagrange formula Consider the function f(x) and n + 1 points x, x 1, x n We use the Newton s Divided Differences by choosing the following points: (x, y ), (x, y ), (x 1, y 1 ), (x 1, y 1 ), (x, y ), (x, y ),, (x n, y n ), (x n, y n )

5 Massoud Malek Interpolation Theory Page 5 and replacing f[x i, x i ] with f (x i ) The following table illustrates the interpolation method: x i f(x i ) f[x i, x i+1 ] D f[x i ] D 3 f[x i ] D 4 f[x i ] x y f (x ) x y f[x, x, x 1 ] f[x, x 1 ] f[x, x, x 1, x 1 ] x 1 y 1 f[x, x 1, x 1 ] f[x, x, x 1, x 1, x ] f (x 1 ) f[x, x 1, x 1, x ] x 1 y 1 f[x 1, x 1, x ] f[x, x 1, x 1, x, x ] f[x 1, x ] f[x, x 1, x 1, x ] x y f[x 1, x, x ] f[x 1, x 1, x, x, x 3 ] f (x ) f[x 1, x 1, x, x ] x y f[x, x, x 3 ] f[x 1, x, x, x 3, x 3 ] Interpolation Using Spline Functions The simplest method of interpolation is to connect the points points by straight line segments; this is called Piecewise Linear Interpolation, and the associated interpolation function is denoted by S(x) Although S(x) agrees with the data, but it has the disadvantage of not having a smooth graph Another choice is to connect the data points by using succession of quadratic interpolation polynomials Although the resulting curve is smoother than the first method, but the graph can still have some corners For many applications, this interpolating function may be completely adequate there, but we would like to obtain a smooth curve We shall see that a cubic spline produces the desired curve Cubic Spline Suppose n + 1 data points (x, y ), (x 1, y 1 ),, (x n, y n ) are given For simplicity, assume x < x 1 < x n and let a x, b x n We seek a function S(x) defined on [a, b] that interpolates the data S(x i ) y i, i, 1,, n For smoothness of S(x), we need to require that S (x) and S (x) be continuous In addition, we would like to have the curve follow the general shape given by the piecewise linear function connecting the data points (x i, y i ) The standard way in which this has been done is to ask that the derivative S (x) not change too rapidly between note points This has been carried out by requiring the second derivative S (x) to be as small as possible and, more precisely, by requiring that b a [S (x)] dx be also as small as possiblethis may not be a perfect mathematical realization of the idea of a smooth shape-preserving interpolation function for the data {(x i, y i )}, but it usually gives a very good interpolating function There is a unique solution S(x) to this problem, and it satisfies the following: S1 S(x) is a cubic polynomial, denoted S j (x) on each [x j, x j+1 ], for j, 1, n 1; S S(x j ) f(x j ) for each j, 1,, n; S3 S j+1 (x j+1 ) S j (x j+1 ) for each j, 1,, n ;

6 Massoud Malek Interpolation Theory Page 6 S4 S j+1 (x j+1) S j (x j+1) for each j, 1,, n ; S5 S j+1 (x j+1) S j (x j+1) for each j, 1,, n ; S6 one of the following set of boundary conditions is satisfied: (i) S (x ) S (x n ) (Free boundary) (ii) S (x ) f (x ) and S (x n ) f (x n ) (Clamped boundary) To construct the cubic spline interpolant for a given function f(x), the conditions (S 1 through S 6 ) can be applied to the cubic polynomials S j (x) A j + B j (x x j ) + C j (x x j ) + D j (x x j ) 3 for each j, 1,, n 1 For j, 1,, n 1, define h j x j+1 x j, then we have A j S j (x j ) f(x j ) and A j+1 A j + B j h j + C j h j + D j h 3 j B f (x ), B j S j(x j ) (j > ) and B j+1 B j + C j h j + 3D j h j B j S (x j ), C n S (x n ) C j+1 C j + 3D j h j Since h j s and A j s are given, once the values of C j s are known, then it is a simple matter to find B j s and D j s Free Boundary When S (a) S (b), then the vector C (C, C 1,, C n ) t is the unique solution to the equation 1 h (h + h 1 ) h 1 h 1 (h 1 + h ) h h n (h n + h n 1 ) h n 1 1 C C 1 C n 1 C n 3(A A 1) h 1 3(A n A n 1) h n 1 3(A1 A) h 3(An 1 An ) h n Clamped Boundary For S (a) f (a) and S (b) f (b), the vector C (C, C 1,, C n ) t is the unique solution to the equation h h h (h + h 1 ) h 1 h 1 (h 1 + h ) h h n (h n + h n 1 ) h n 1 h n 1 h n 1 C C 1 C n 1 C n 3(A 1 A ) h 3(A A 1) h 1 3(A n A n 1) h n 1 3f (a) 3(A1 A) h 3(An 1 An ) h n 3f (b) 3(An An 1) h n 1