3.1 Global polynomial interpolation Properties of global polynomial interpolation Splines... 22

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1 Chapter 3 Interpolation Contents 31 Global polynomial interpolation Properties of global polynomial interpolation Splines 22 Suppose we have some table of discrete data, the population of a country at ten yearly intervals, say, and we need to evaluate the population at a date where no data are available The estimation of the unknown data, using some or all of the available data, is a process known as interpolation Similarly, interpolation allows us to approximate a complicated or unknown function, f, with a simple function that agrees with f at a nite number of points, x k i If f is very computationally expensive to calculate, approximating f in some calculations by a simpler interpolating function reduces this cost This was a common practice before calculators and computers became widely available ii Exact calculations (eg derivation, quadrature) may become possible if the interpolating function is used instead of the complicated or unknown function f This is a building block for advanced numerical methods (eg nite element and spectral-collocation methods) In this section we shall consider the approximation of a function by polynomials or piecewise polynomial functions (splines) only 31 Global polynomial interpolation Fit a polynomial of degree n 1, P(x) = a n 1 x n 1 + a n 2 x n a 1 x + a 0 through n known points (x k, f k ), k = 1,,n Hence, the interpolating polynomial satises P(x k ) = f k (The coordinates x k are not necessarily equally spaced) 17

2 18 31 Global polynomial interpolation 311 Lagrange polynomials A practical way to construct the polynomial P(x) is to use the basis formed by Lagrange polynomials Let us dene the Lagrange polynomials of degree n 1, L k (x) (k = 1,,n), such that L k (x) = { 1 at x = xk, 0 at x = x i for i k, ie L k (x i ) = δ ik (Kronecker delta) In other words, L k (x) is equal to 1 at the point x k and 0 at all the other given points For n = 2, (ie {x 1, x 2 } given), the two Lagrange polynomials L 1 (x) and L 2 (x) satisfy L 1 (x 1 ) = 1, L 2 (x 1 ) = 0, L 1 (x 2 ) = 0, L 2 (x 2 ) = 1, from which we obtain the linear functions (ie polynomials of degree 1), L 1 (x) = x x 2 x 1 x 2, L 2 (x) = x x 1 x 2 x 1 For n > 2, L 1 (x 1 ) = 1 and L 1 (x k ) = 0 for k = 2,,n From above we see that { x x 2 1 at x = x1, = x 1 x 2 0 at x = x 2, while { x x 3 1 at x = x1, = x 1 x 3 0 at x = x 3, etc Thus, by multiplying n 1 such factors together, we obtain the Lagrange polynomial of degree n 1, ( ) ( ) ( ) x x2 x x3 x xn L 1 (x) =, x 1 x 2 x 1 x 3 x 1 x n written in compact form as Similarly, L k (x) = L 1 (x) = ( x xi i=2 ( x xi i=1 i k x k x i x 1 x i ) ), k = 1,,n (31) The n Lagrange polynomials L k (x) of degree n 1 form a complete set of independent polynomials, so {L 1 (x), L 2 (x),,l n (x)} forms a basis of polynomials of degree n Lagrange form of the interpolating polynomial Having obtained L k (x), we can write the Lagrange form of the polynomial interpolating the set of points {(x k, f k ) k = 1,,n}, P(x) = n f k L k (x) = n n ( x xi f k x k x i i=1 i k ) (32) (It is the Lagrange interpolating polynomial of degree n 1 satisfying P(x k ) = f k )

3 Chapter 3 Interpolation 19 For n = 2, which rearranges to P(x) = f 1 L 1 (x) + f 2 L 2 (x), ( ) ( ) x x2 x x1 = f 1 + f 2 x 1 x 2 x 2 x 1 P(x) = f 1(x 2 x 1 ) + (f 2 f 1 )(x x 1 ) x 2 x ( ) 1 f2 f 1 = f 1 + (x x 1 ) (33) x 2 x 1 This is the equation of a straight line between the two points (x 1, f 1 ) and (x 2, f 2 ) (linear interpolation) Example 31 Let us consider the function f(x) = 1/x and look for approximations to f(3) using polynomial interpolation First, let us consider only the two points (25, f(25)) and (4, f(4)) So, using the linear interpolation formula (33) we have an approximation for f(3) P(3) where f(4) f(25) P(3) = f(25) + (3 25) = = 7/20 = The absolute error is E = 1/3 7/20 = 1/ Next, consider the quadratic polynomial interpolating the three points (x i, f(x i )), where x 1 = 2, x 2 = 25 and x 3 = 4, P(x) = f(2)l 1 (x) + f(25)l 2 (x) + f(4)l 3 (x) = 05 L 1 (x) + 04 L 2 (x) L 3 (x) So, f(3) P(3) = 05 L 1 (3) + 04 L 2 (3) L 3 (3) where (3 25)(3 4) L 1 (3) = (2 25)(2 4) = 05 (3 2)(3 4) L 2 (3) = (25 2)(25 4) = 4/3 L 3 (3) = (3 2)(3 25) (4 2)(4 25) = 1/6 Thus, P(3) = 05 ( 05) / /6 = 13/40 = 0325 The absolute error is now E = 1/3 13/40 = 1/

4 20 32 Properties of global polynomial interpolation /x x Note that, in both interpolations, linear and quadratic, n L k(3) = 1 Why? 32 Properties of global polynomial interpolation 321 Uniqueness The interpolating polynomial is unique in that there is only one polynomial of degree at most n 1, P, that satises P(x k ) = f k for n distinct points (x k, f k ) (This seems intuitively true, since we have n equations, n 1 P(x k ) = a i x i k = f k, i=0 k = 1,,n, with n unknowns, a i, the coecients of the polynomial P ) Proof Suppose there exists two polynomials of degree n 1, P and Q, such that P(x k ) = Q(x k ) = f k, k = 1n Then P(x k ) Q(x k ) = 0, so the polynomial R(x) = P(x) Q(x) has n distinct roots, x k However, R(x) is a polynomial of degree n 1 and so can have at most n 1 distinct roots, unless R(x) 0 Hence P(x) = Q(x), so the interpolating polynomial is unique 322 The error term in global polynomial interpolation In the section, interpolation is used to approximate a known function (not to interpolate tabulated data in which case error cannot be evaluated) If f C n [a,b] (ie f is a function n times dierentiable in [a,b] with continuous derivatives) then, x [a,b], there exists ξ (a,b), that depends on x, such that f(x) P(x) = f(n) (ξ) n! (x x k ) (34)

5 Chapter 3 Interpolation 21 Proof This result is trivially true for x = x i since both sides are zero For x x i, let us dene so that Let us now dene the function c(x) = f(x) = P(x) + c(x) f(x) P(x) n (x x k), (x x k ) w(t) = f(t) P(t) c(x) (t x k ), which satises w(t) = 0 for t = x and t = x 1,,x n So w(t) has n + 1 distinct roots in [a,b] By Rolle's theorem, there must be at least one root of w(t) between two successive roots of w(t) So, w (t) has at least n distinct roots in (a,b) Apply Rolle's theorem to derivatives of increasing order successively, to show that w (n) (t) must have at least one root in (a,b) Let us call this point ξ: w (n) (ξ) = 0 The n th derivative of w(t) is w (n) (t) = f (n) (t) P (n) (t) c(x)n! but P is a polynomial of degree n 1, so P (n) 0 Let t = ξ, So, f (n) (ξ) c(x)n! = 0 c(x) = f(n) (ξ) n! f(x) P(x) = f(n) (ξ) n! (x x k ) 323 Accuracy of global polynomial interpolation Using Equation (34) we can obtain bounds on the accuracy of Lagrange interpolation, provided we can bound the n th derivative, f (n) Since the error term involves f (n), Lagrange interpolation is exact for polynomial of degree n-1 at most In the example of sin(3x) (see appendix E1), f (n) 3 n while, eg, for the ve points {x 1 = 1, x 2 = 13, x 3 = 16, x 4 = 19, x 5 = 22} and x = 15, 7 (x x i ) = = 2500 = i=1

6 22 33 Splines So, the error in the approximation to f(15) is f(15) P(15) 35 5! = = The actual error is f(15) P(15) However, the function f(x) = 1/(1+x 2 ) cannot be approximated accurately by an interpolation polynomial, using equidistant points on the interval [ 5,5] (see appendix E2) Although, f is dierentiable, its derivatives grow with n! ( f (n) (0) = n!, for n even) So, f (n) (ξ)/n! does not converge to 0 while n i=1 (x x i) increases (since interval width > 1) 33 Splines Alternatively, to interpolate through n points (x k, f k ) we can use piecewise polynomial functions called splines, ie S(x) = S k (x) where the low degree polynomials S k (x) are dened on the intervals [x k, x k+1 ], k = 1,,n Piecewise linear interpolation The simplest spline is composed of linear functions of the form S k (x) = a k x + b k, for x [x k, x k+1 ] (ie a straight line between the two successive points x k and x k+1 ) S n 1 S 1 S k x n 1 x 1 x 2 x k x n The coeicients a k and b k are determined by the conditions (i) S k (x k ) = f k and (ii) S k (x k+1 ) = f k+1, k = 1,,n 1 Thus, x k+1 S k (x) = f k + (x x k ) f k+1 f k x k+1 x k, x [x k, x k+1 ], k = 1,,n 1 (35) The interpolating function is continuous but it is not dierentiable, ie not smooth, at the interior points (S k (x k+1) S k+1 (x k+1)) To retain the smoothness of Lagrange interpolation without producing large oscillations, higher order splines are needed 332 Quadratic splines Consider the spline composed of n 1 quadratic polynomials of the form S k (x) = a k x 2 +b k x+c k, for x [x k, x k+1 ], k = 1,,n 1 We need 3(n 1) equations to determine the 3(n 1) coeicients {a k, b k, c k } The conditions S k (x k ) = f k and S k (x x+1 ) = f k+1 provide 2(n 1) equations while the continuity of the derivative at the interior points, S k (x k+1) = S k+1 (x k+1), k = 1,,n 2, provides n 2 extra equations

7 Chapter 3 Interpolation 23 In total we have 3(n 1) unknowns and 3(n 1) 1 equations So, only one degree of freedom is left, which is unsatisfactory if one needs to impose conditions on the derivatives of the interpolating function at both ends 333 Cubic splines In practical applications, cubic splines are preferred Consider the spline composed of n 1 cubic polynomials of the form S k (x) = a k (x x k ) 3 + b k (x x k ) 2 + c k (x x k ) + d k, (36) so that, S k (x) = 3a k(x x k ) 2 + 2b k (x x k ) + c k, (37) S k k(x x k ) + 2b k (38) for x [x k, x k+1 ], k = 1,,n 1 Properties The interpolating function must pass through all the data points, be continuous, and have continuous derivative at the interior points Furthermore, there are enough degrees of freedom to impose continuous curvature as well (continuity of second derivative) S k (x k ) = f k, k = 1,,n 1, (39) S k (x k+1 ) = f k+1, k = 1,,n 1, (310) S k (x k+1) = S k+1 (x k+1), k = 1,,n 2, (311) S k k+1) = S k+1 k+1), k = 1,,n 2 (312) We have 4(n 1) unknowns and 4(n 1) 2 equations (ie conditions) Hence, we have the freedom to impose two extra conditions Let us dene, h k = x k+1 x k, the width of the k th interval, and C k = S k (x k) the curvature in x k (C n = S n 1 (x n)) Next, we shall write all the equations in terms of h k, f k (given) and C k (unknown) only From equation (38): From equations (38) and (312): From equations (36) and (39): C k = 2b k b k = C k, k = 1,,n 1 (313) 2 6a k h k + C k = C k+1 a k = 1 C k+1 C k, k = 1,,n 1 (314) 6 h k From equations (36) and (310): d k = f k, k = 1,,n 1 (315) a k h 3 k + b kh 2 k + c kh k + d k = f k+1 1 C k+1 C k h 3 k 6 h + C k k 2 h2 k + c kh k + f k = f k+1 c k = f k+1 f k h k 1 6 h k(c k+1 + 2C k ), k = 1,,n 1 (316)

8 24 33 Splines So far we have obtained expressions for the coecients a k, b k, c k and d k in terms of h k, C k and f k All that remains is to match the slopes at the interior points From equations (37) and (311): 3a k h 2 k + 2b kh k + c k = c k+1, k = 1,,n 2, 1 2 C k+1 C k h 2 k h + C kh k + f k+1 f k 1 k h k 6 h k(c k+1 + 2C k ) = f k+2 f k+1 1 h k+1 6 h k+1(c k+2 + 2C k+1 ) After some simple algebra, we obtain a system of n 2 equations to solve for C k, k = 1,,n 2 ( fk+2 f k+1 h k C k + 2(h k + h k+1 )C k+1 + h k+1 C k+2 = 6 f ) k+1 f k (317) h k+1 h k So, the problem as been reduced from nding 4(n 1) coecients a k, b k, c k and d k to nding n values of the curvature C k We only have n 2 equations for C k but we can obtain two additional equations by specifying end conditions on the curvature, ie conditions involving C 1 and C n i Natural or free boundaries: take C 1 = C n = 0, ie the end cubic splines have no curvature at the end points ii Clamped boundaries: x the slopes at each end to specied values (Note that, the slope at x 1, say, involves c 1 and thus relates to C 1 and C 2 through (316)) iii Periodic boundaries: take C 1 = C n Equation (317) can be written in matrix form, MC = F (318) In the case of natural splines (ie for C 1 = C n = 0), the matrix is non-singular and has a tri-diagonal structure So, the solution for C k is unique and can easily be obtained using rapid tri-diagonal matrix solvers 2(h 1 + h 2 ) h h 2 2(h 2 + h 3 ) h M = 0 h k 2(h k + h k+1 ) h k+1 0, h n 3 2(h n 3 + h n 2 ) h n h n 2 2(h n 2 + h n 1 ) C = C 2 C 3 C n 2 C n 1 and F = 6 f 3 f 2 h 2 f 2 f 1 h 1 f k+2 f k+1 h k+1 f n f n 1 h n 1 f k+1 f k h k f n 1 f n 2 h n 2 So, equation (318) can be solved for C k and the coecients of the splines can be recovered from equations (313), (314), (315), (316)

9 Chapter 3 Interpolation 25 Example 32 Find the natural spline for the following data: k x k f k So, {h 1 = 1, h 2 = 2, h 3 = 1} and equation (317) becomes in matrix form, ( ) ( ) 2(1 + 2) 2 C2 = 6 2 2(2 + 1) C 3 ( ) (3 4)/2 (4 2)/1 (4 3)/1 (3 4)/2 ( ) ( ) C2 = C 3 ( ) 15 9 Inverting gives C 2 = 27/8 and C 3 = 21/8 while C 1 = C 4 = 0 Thus, using equations (313), (314), (315), (316) we obtain a 1 = 9/16, a 2 = 1/2, a 3 = 7/16; b 1 = 0, b 2 = 27/16, b 3 = 21/16; c 1 = 41/16, c 2 = 7/8, c 3 = 1/8; d 1 = 2, d 2 = 4, d 3 = S 1 (x) S 2 (x) S 3 (x) 3 S(x) x

10 26 33 Splines