Interpolation. Chapter Interpolation. 7.2 Existence, Uniqueness and conditioning

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2 Chapter 7 Interpolation 7.1 Interpolation Definition Interpolation of a given function f defined on an interval [a,b] by a polynomial p: Given a set of specified points {(t i,y i } n with {t i} [a,b], the polynomial p of degree n 1 satisfying p(t i ) = y i, i = 1,,n is called an polynomial interpolation. The points {t i } are called nodes. 7.2 Existence, Uniqueness and conditioning be any linearly inde- More generally, we can use other bases. Let {φ j (t)} n j=1 pendent set of functions and set f(t) = n x j φ j (t). If we require f to interpolate the data {(t i,y i )} m, then we have f(t i ) = j=1 n x j φ j (t i ) = y i, i = 1,2,,m, j=1 which gives a system of equations Ax = y. 77

3 78 CHAPTER 7. INTERPOLATION 7.3 Polynomial Interpolation Monomial Interpolation We let P n (x) be the set of all polynomial of degree less or equal to n. To interpolate n+1 data by polynomials, we let k = n. Most naive choice is to let φ j (t) = t j, j = 0,1,,n Theorem For a given set of data {(t i,y i } n i=0, there exist a unique polynomial p P n (x) such that p(t i ) = y i, i = 1,2,,n. Proof. We want to find a polynomial of the form p = x 0 + x 1 t + + x n t n such that p(t i ) = y i, for i = 0,...,n, i.e, p(t 0 ) = x 0 +x 1 t 0 + +x n t n 0 = y 0 p(t n ) = x 0 +x 1 t n + +x n t n n = y n. In matrix form, 1 t 0 t n 0 1 t 1 t n x 0 y =. (7.1) 1 t n t n x n y n n The equation (7.1) involves a Van der Monde matrix whose determinant is i>j (t i t j ). Thus a unique solution exists as long as the interpolation points are distinct. Use nested multiplication(horner s method) for polynomial evaluations. A basis for V = P n A naive basis for P n (x) is {1,x,x 2,...,x n }. Is it convenient? No! There are other bases, and usually other bases are better. (1) To compute the coefficients in (7.1), one has to solve a linear system. (2) Moreover, the Van der Monde matrix is ill-conditioned.

4 7.3. POLYNOMIAL INTERPOLATION 79 Example Consider x y = 1+ǫ ( )x y = 1. (7.2) Without ǫ, the exact solution is x = 0,y = 1, while with ǫ perturbation, x = 10 9 ǫ, y = 10 9 ǫ 1 ǫ! Why Polynomial Interpolation? For example consider e x or sinx. How to evaluate or integrate them? (1) On a computer we approximate a given function by only arithmetic operations which is done by polynomial interpolation. Taylor series is rarely used. So Find a polynomial p(x) s.t. p(x i ) = e x i for i = 1,2,. This is better than Taylor series which loses accuracy when x is large. (non uniform error) Taylor series requires higher order derivatives which may be difficult to obtain or unavailable. (2) It is easier to handle Polynomial interpolations systematically (in solving the p.d.e. or integrations) Lagrange basis Suppose we wish to construct a linear interpolation with two points t 1 and t 2. Define Then L 1 (t) = t t 2 t 1 t 2, L 2 (t) = t t 1 t 2 t 1. L 1 (t 1 ) = 1, L 1 (t 2 ) = 0 L 2 (t 1 ) = 0, L 2 (t 2 ) = 1. Now given any data f(t 1 ) = y 1, f(t 2 ) = y 2, we can construct a linear interpolant by l(t) = y 1 L 1 (t)+y 2 L 2 (t). (7.3) We now generalize this. Given distinct nodes t 1,t 2,,t n, we construct polynomials L n,i (t)(i = 1,,n) such that L n,i (t j ) = δ ij, j = 1,,n. We can

5 80 CHAPTER 7. INTERPOLATION easily see that L n,i has the following form: L n,i (t) = C We set n (t t j ), for some C. j i L n,i (t i ) = C j i(t i t j ) = 1. Then we obtain C = 1/ j i (t i t j ) and hence L n,i (t) = j i (t t j ) (t i t j ). These are the Lagrange basis polynomials. Now using these, one can readily construction a polynomial interpolation. Proposition Let f C[a,b] and let p n 1 be the unique element of P n 1 (x) such that f(t i ) = p n (t i ), i = 1,1,...,n. Then the Lagrange interpolating polynomial is given by p n (t) = n f(t i ) (t t j ) (t i t j ). j i Newton s form of interpolating polynomial We now describe an efficient way of calculating the coefficients of an interpolating polynomial. Assume {(t i,y i } n is the given data set. In the Newton form of interpolating polynomial the following basis is taken: {1,(t t 1 ),(t t 1 )(t t 2 ),,(t t 1 )(t t 2 ) (t t n 1 )} We compute p n 1 (t) by induction. Since p n 1 (t) is one degree higher than p n 2 (t), one can set p n 1 (t) = p n 2 (t)+x n π n 1 (t), where j 1 π j (t) = (t t k ), j = 1,,n 1. k=1

6 7.3. POLYNOMIAL INTERPOLATION 81 Hence n 1 p n 1 (t) = p n 2 (t)+x n (t t i ) n 2 n 1 = p n 3 (t)+x n 1 (t t i )+x n (t t i ) n 1 = x 1 +x 2 (t t 1 )+x 3 (t t 1 )(t t 2 )+ +x n (t t i ). If we denote f[t 1,...,t k ] for x k, then p n 1 (t) = n k=1 (t t i ) x k k 1 n k=1 k 1 f[t 1,...,t k ] (t t i ). (7.4) This is called the Newton s form of interpolation and f[t 1,...,t k ] are called the divided difference. Now we study how to compute x k. Computing the divided difference Now we show how to compute f[t 1,...,t k ] efficiently. Assume p j (t) is constructed. For any j, p j+1 (t) = p j (t)+x j+1 π j+1 (t) is a polynomial of degree j that interpolates the same j points as p j. The free parameter x j+1 is chosen to so that p j+1 interpolates (j +1)-st points, i.e., Hence p j+1 (t j+1 ) = p j (t j+1 )+x j+1 π j+1 (t j+1 ) = y j+1. x j+1 = y j+1 p j (t j+1 ). π j+1 (t j+1 ) But in practice this formula is neither efficient nor stable for large k. It is known that x k = f[t 1,...,t k ] = f[t 2,...,t k ] f[t 1,...,t k 1 ] t k t 1. (7.5) Thus Newton s formula is useful when a new point of interpolation is added to an existing interpolation. Example Suppose we are given t 1,...,t n and p n. If we have one more

7 82 CHAPTER 7. INTERPOLATION point t n+1. Then p n is constructed by adding one more term to p n 1 : By (7.5) n 1 p n (t) = p n 1 (t)+f[t 1,...,t n 1,t n ] (t t i ). (7.6) Example (1) f[t i ] = f(t i ). (2) f[t 1,t 2 ] = f[t 2] f[t 1 ] t 2 t 1 = f(t 2) f(t 1 ) t 2 t 1. (3) f[t 1,t 2,t 3 ] = f[t 2,t 3 ] f[t 1,t 2 ] t 3 t 1. (4) p 2 (t) = f(t 1 )+f[t 1,t 2 ](t t 1 )+f[t 1,t 2,t 3 ](t t 1 )(t t 2 ). i=0 t 1 f[t 1 ] f[t 1,t 2 ] f[t 1,t 2,t 3 ] f[t 1,t 2,t 3,t 4 ] t 2 f[t 2 ] f[t 2,t 3 ] f[t 2,t 3,t 4 ] t 3 f[t 3 ] f[t 3,t 4 ] t 4 f[t 4 ] Orthogonal Polynomial Let <, > be the weighted inner product defined by < p,q >= b a p(t)q(t)w(t) dt. We say p,q are orthogonal w.r.t w if < p,q >= 0. If w = 1 on [ 1,1] then the orthogonal polynomials are Legendre polynomials. p 0 (t) = 1 2 (3t2 1), p 3 (t) = 1 2 (5t3 3t) p 4 (t) = 1 8 (35t4 30t 2 +3), p 5 (t) = 1 8 (63t5 70t 3 +15t) p 6 (t) = 1 24 (7 33t6 63 5t t 2 5)... Three-term recurrence relation of the form: p k+1 (t) = (α k t+β k )p k (t) γ k p k 1 (t) For Legendre polynomials, we have (k +1)P k+1 (t) = (2k +1)tP k (t) kp k 1 (t).

8 7.3. POLYNOMIAL INTERPOLATION 83 Table 7.1: Orthogonal Polynomials Name Symbol Interval Weight Legendre P k [ 1,1] 1 Chebysheff 1st T k [ 1,1] (1 t 2 ) 1/2 Chebysheff 2nd U k [ 1,1] (1 t 2 ) 1/2 Jacobi (α,β > 1) J k [ 1,1] (1 t) α (1+t) β Laguerre L k [0, ) e t Hermite H k (, ) e t Interpolating continuous function-chebysheff polynomials Thus we may assume f(x) = x n+1 and consider the following problem: We wish to have equi-oscillation property: To minimize n i=0 (x x i), we want to enforce uniform oscillations. This suggest a polynomial whose behavior is similar to a trigonometric function. Assume [a, b] = [ 1, 1]. Note that coskθ oscillates between 1 and 1 (k +1 times in [0,π]), i.e., coskθ j = ( 1) j, for θ j = jπ k, j = 0,,k. We let θ = cos 1 t and consider coskcos 1 t, i.e., T k (t) = cos[kcos 1 t], x [ 1,1]. Recall the addition formula for cosine function: coskθ +cos(k 2)θ = 2cosθcos(k 1)θ. Thus we obtain a recurrence relation T k+1 (t)+t k 1 (t) = 2tT k (t), k = 1,2,... From this we get T 0 = 1, T 1 (t) = t, T 2 (t) = 2t 2 1, T 3 (t) = 4t 3 3t,8t 4 8t 2 +1,16t 5 20t 3 +5t, So the coefficient of T k is 2 k 1, its maximum oscillation. T k (t) is called

9 84 CHAPTER 7. INTERPOLATION Chebysheff polynomial of the first kind. T k (t) = cos[kcos 1 x] = 2 k 1 (t t 0 )(t t 1 ) (t t n 1 ) (7.7) and hence max (t t 0 )(t t 1 ) (t t n 1 ) 2 1 k. (7.8) This is useful in interpolating(if the error term is given by the Chebysheff polynomial(up to certain constant factor), then the maximum error is distributed uniformly over the interval. Thus, we use these points to construct a polynomial interpolation for general f, and expect the error term f (k) (ξ) (k)! k 1 (t t i ) i=0 of this approximation is reasonably small in the maximum norm. Such points are called Chebysheff points. Since T k (t) = 0 at points where kcos 1 t = (2i 1)π/2, i.e., the Chebysheff points on [ 1,1] are ( ) (2i 1)π t k = cos, i = 1,...,k. 2k (Sometimes they use the extrema, t k = cos ( ) iπ k, i = 0,...,k. These are also called Chebysheff points) then If f C n [ 1,1] and p n 1 (x) is a interpolating polynomial at t 1,,t n f(t) p n 1 (t) = f(n) (ξ) n! n (t t i ). (7.9) Assume f (n) (ξ) M forallt [t 1,t n ]andh = max(t i+1 t i ), then(roughly) In particular, if f(t) = t n, then max f(t) p n 1 (t) Mhn t 4n. (7.10) f(t) p n 1 (t) = n (t t i ) and f(t) p n 1 (t) = 2 1 k. Polynomial of degree n has the tendency of n 1 oscillation(has n 1 local extreme points). Thus, if the interval is fixed i=0

10 7.3. POLYNOMIAL INTERPOLATION Figure 7.1: Chebysheff points on [ 1,1], n = 2 and n = 3 and n gets higher(as a result the interpolating points will get close together), it may oscillate unnecessarily. Indeed the following example by Runge shows it. Example (Runge). Given {(x i,f(x i )) i = 0,1,...,n}, let f(x) = 1 1+x 2 on [ 5,5] h = b a 2n, x k = kh, k = n,...,n. Interpolate f(x) by polynomial of degree 2n. p 2n ( 5k n ) = f(5k n ), k = 0,±1,...,±n. Runge showed lim n f(x) p 2n (x) =. Thus Runge s example shows higher degree polynomial is not always good for interpolation. This suggests us to use lower degree polynomial on each subinterval.

11 86 CHAPTER 7. INTERPOLATION 7.4 Piecewise Polynomial Interpolation Hermite Cubic Interpolation Cubic Spline Interpolation A spline is a piecewise polynomial of degree k that is continuous differentiable k 1-times. Take cubic C 2 here. Let y = f C 2 [a,b], a = t 1 <...,< t n = b. We wish to construct s(x), a C 2 [a,b]- interpolation which is piecewise cubic polynomial on I k = [t k 1,t k ], k = 1,,n 1. Conditions : (1) 2 function values on each intervals, so 2(n 1) conditions. (2) k = 1,,n 2, (interior derivative) (3) k = 1,,n 12. (interior second derivative) Since we have cubic polynomial in each interval I k = [t k 1,t k ], k = 1,,n 1, we have 4(n 1) unknowns. So we have two free parameters. We have the following options : Specify first derivatives at two end points Specify second derivatives(to zero- we get natural spline) at two end points Either t 2 or t n 1 is Not a knot. (Treat either of the two end intervals as one) B-Splines There are basis functions for splines of all degrees.