Polynomial Interpolation

Capter 4 Polynomial Interpolation In tis capter, we consider te important problem of approximating a function f(x, wose values at a set of distinct points x, x, x 2,,x n are known, by a polynomial P (x suc tat P (x i =f(x i, i =,, 2,,n Suc a polynomial is known as an interpolating polynomial An approximation of tis function is desirable if f(x is difficult to evaluate or manipulate like te error function erf(x = 2 x e t2 dt, π for example Note tat polynomials are easily evaluated, differentiated or integrated Anoter purpose of suc an approximation is tat a very long table of function values f(x i may be replaced by a sort table and a compact interpolation subroutine 4 Linear Interpolation Let f(x be given at two distinct points x i and x i+ Now, given f(x i =f i and f(x i+ =f i+, we wis to determine te interpolating polynomial P (x of degree (ie, a straigt line tat approximates f(x onteinterval[x i,x i+ ] Let P (x =αx + β From te conditions P (x i =f i and P (x i+ =f i+ we obtain te two equations ie, ( xi x i+ αx i + β = f i αx i+ + β = f i+, ( α β = ( fi f i+ Tis system is clearly nonsingular since x i x i+, and it as a unique solution: α = f i+ f i x i+ x i β = f ix i+ f i+ x i x i+ x i 68

Cap 4 Polynomial Interpolation Numerical Metods Class Notes 69 Terefore, Example 4 wic gives ( fi+ f i P (x =f i + (x x i x i+ x i Using linear interpolation, approximate sin 65 from a table (a sin 6 =45; sin 7 =287 (b sin =;sin =765 Solution ( 287 45 (a P (65 = 45 + 7 6 2 (b P (65 = + 765 (65 287 Te first answer is correct to 5 decimals wereas te second answer is correct only to 2 decimals! Conclusion: Linear interpolation is suitable only over small intervals 42 Polynomial Interpolation Since linear interpolation is not adequate unless te given points are closely spaced, we consider iger order interpolating polynomials Let f(x begiven at te selected sample of (n + points: x <x < <x n, ie, we ave (n +pairs(x i,f i, i =,, 2,,n Te objective now is to find te lowest degree polynomial tat passes troug tis selected sample of points f(x P n (x f(x x x x 2 x x n Figure 4: Interpolating te function f(x by a polynomial of degree n, P n (x Consider te nt degree polynomial P n (x =a + a x + a 2 x 2 + + a n x n

Cap 4 Polynomial Interpolation Numerical Metods Class Notes 7 We wis to determine te coefficients a j, j =,,,n, suc tat P n (x j =f(x j, j =,, 2,,n Tese (n + conditions yield te linear system a + a x + a 2 x 2 + + a nx n = f a + a x + a 2 x 2 + + a nx n = f a + a x 2 + a 2 x 2 2 + + a n x n 2 = f 2 a + a x n + a 2 x 2 n + + a nx n n = f n or V a = f, were a T =(a,a,,a n, f T =(f,f,,f n, and V is an (n + (n +matrixgivenby x x 2 x n x x 2 x n V = x 2 x 2 2 x n 2 x n x 2 n x n n Te matrix V is known as te Vandermonde matrix nonzero determinant n det(v = (x i x j j= i>j It is nonsingular wit For n =, for example, det(v =(x x (x 2 x (x x (x 2 x (x x (x x 2 Hence a can be uniquely determined as a = V f Anoter proof of te uniqueness of te interpolating polynomial can be given as follows Teorem 4 Uniqueness of interpolating polynomial Given a set of points x <x < <x n, tere exists only one polynomial tat interpolates a function at tose points Proof Let P (x andq(x be two interpolating polynomials of degree at most n, for te same set of points x <x < <x n Also, let R(x =P (x Q(x Ten R(x is also a polynomial of degree at most n Since P (x i =Q(x i =f i, we ave, R(x i =fori =,,,n In oter words R(x as(n +roots From te Fundamental Teorem of Algebra owever, R(x cannot ave more tan n roots A contradiction! Tus, R(x, and P (x Q(x Example 42 Given te following table for te function f(x, obtain te lowest degree interpolating polynomial

Cap 4 Polynomial Interpolation Numerical Metods Class Notes 7 x k 2 4 f k 4 4 Solution Since te number of nodes = = n +, terefore n =2 Let P 2 (x =a + a x + a 2 x 2, tat satisfies te following conditions at te points x =;x =2,x 2 =4: P ( = ; P (2 = 4; P (4 = 4 2 4 a a = 4 4 6 a 2 4 Using Gaussian elimination wit partial pivoting, we compute te solution a = a a = 4 2, a 2 Terefore, te interpolating polynomial is given by P 2 (x = 4 2x + x 2 We can use tis approximation to estimate te root of f(x tat lies in te interval [2, 4]: γ = 2+ 4+6 =+ 5 2 An important remark is in order One, in general, sould not determine te interpolating polynomial by solving te Vandermonde linear system Tese systems are surprisingly ill-conditioned for n no larger tan For example, for <x <x < <x n = uniformly distributed in [,], large n yields a Vandermonde matrix wit almost linearly dependent columns, and te Vandermonde system becomes almost singular A more satisfactory form of te interpolating polynomial is due to Lagrange, P n (x =f(x L (x+f(x L (x+ + f(x n L n (x Here L j (x, for j n, are polynomials of degree n called fundamental polynomials or Lagrange polynomials From te(n + conditions P n (x i =f(x i, i n, we see tat L j (x i = { i j i = j

Cap 4 Polynomial Interpolation Numerical Metods Class Notes 72 Te above property is certainly satisfied if we coose ie, L j (x = (x x (x x (x x j (x x j+ (x x n (x j x (x j x (x j x j (x j x j+ (x j x n L j (x = n i= i j (x x i (x j x i Te fact tat L j (x is unique follows from te uniqueness of te interpolation polynomial Recall tat tere is one and only one polynomial of degree n or less tat interpolates f(x atte(n +nodesx <x < <x n Tus,fortisgiven set of nodes, bot forms yield te same polynomial! Vandermonde Approac n P n (x = a i x i i= V a = f Operation count: It can be sown tat te Vandermonde system V a = f can be solved in O(n 2 aritmetic operations rater tan O(n operations because of te special form of te Vandermonde matrix Lagrange Approac P n (x = n f j L j (x = j= n j= g j n (x x i i= (x x j = n g j ν j j= were g j = f(x j n (x j x i i= i j Operation count: Te Lagrange form avoids solving te ill-conditioned Vandermonde linear, systems, and evaluates te polynomials ν j = (x x j n (x x i i=

Cap 4 Polynomial Interpolation Numerical Metods Class Notes 7 for a given x Te cost of tese operations for eac j =,, 2,,n are: u(x = n (x x i (2n +ops i= ν j = u(x (x x j f j g j = n (x j x i i= i j (n +ops Now, te cost for evaluating te polynomial P n (x = 2n(n + ops (evaluated once only n g j ν j j= (2n +ops Terefore, te total number of operations are 5n +onceg j s are evaluated 4 Error in Polynomial Interpolation Let e n (x be te error in polynomial interpolation given by Since P n (x i =f(x i, we ave e n (x =f(x P n (x e n (x i = x <x < <x n In oter words e n (x is a function wit at least (n + roots Hence, we can f(x P n (x f(x x x x 2 x x n Figure 42: Computing te error in polynomial interpolation write e n (x =(x x (x x (x x n (x =π(x (x were (x is a function of x Let us define g(z =e n (z π(z(x

Cap 4 Polynomial Interpolation Numerical Metods Class Notes 74 were x x j,for j n Tus, g(x =, g(x j =, j n, ie, g(z as(n + 2 distinct roots Assuming tat f(x as at least n + continuous derivatives, te (n + t derivative of g(z, ie, d (n+ dz g(z (n+ g(n+ (z as at least one root x Hence, g(z x x x 2 x x 4 g ( z g ( z g ( z g iv ( z x* Figure 4: Te (n + t derivative of a function g(x witn +2 roots(n = must ave at least one root g (n+ (z =e (n+ n (z (xπ (n+ (z, were π (n+ (z =(n +! Moreover, since e n (z =f(z P n (z, terefore, e (n+ n (z =f (n+ (z P (n+ (z in wic te last term P n (n+ (z is zero Using te above equations, we ave g (n+ (z =f (n+ (z (x(n +! We ad sown earlier tat g (n+ (z as at least one root x in te interval containing x and x j, j =,,n Tis implies tat g (n+ (x =, and terefore, (x = f (n+ (x (n +! n

Cap 4 Polynomial Interpolation Numerical Metods Class Notes 75 Consequently, we ave te following representation of te error in polynomial interpolation e n (x =f(x P n (x =(x x (x x (x x n f (n+ (x (n +! were x is in te interval containing x and x j, j =,,n Example 4 Obtain te Lagrange interpolating polynomial from te data sin =, sin π 6 = 2, sin π = 2, sin π 2 =, and use it to evaluate sin π 2 and sin π 4 Solution Te interpolating polynomial is given by At x = π 2, Terefore, P (x = f L (x+f L (x+f 2 L 2 (x+f L (x = 2 L (x+ 2 L 2(x+L (x 2 L = 2 π 2 π ( 2 2 π 6 6 π 6 π = 5 6 2 2 L 2 = 2 π 6 2 π ( 2 2 π π 6 π = 5 6 2 2 L = 2 π 6 2 π ( 2 π 2 2 π 6 2 π = 6 P =2662 2 Since sin π 2 =25882, we ave error =8 At x = π 4, L =5625, L 2 =5625, L =625 4 4 4 Terefore P =7589 4 Since sin π 4 =77, we ave error =22 Let us estimate te error e n (x atx = π 4 : e = 4 4 4 π 6 4 π 4 π f (4 (x 2 4! Since f(x =sinx; f (4 (x =sinx, and tus, f (4 (x

Cap 4 Polynomial Interpolation Numerical Metods Class Notes 76 Terefore, ( π π e 4 4 π 2 π 2 π 4 4! =76 Similarly, e n (x atx = π 2 is bounded by ( π π e 2 2 π 2 π 4 5π 2 4! =294 44 Runge s Function and Equidistant Interpolation In tis section, we discuss a problem tat arises wen one constructs a global interpolation polynomial on a fairly large number of equally spaced nodes Consider te problem of approximating te following function on te interval [, ] f(x = +25x 2 using te interpolation polynomial P n (x on te equidistant nodes x j = + 2j n, j =,,,n Figure 44 sows ow te polynomials P 4 (x, P 8 (x, P 2 (x, and P 6 (x approximate f(x on[, ] We see tat as n increases, te interpolation error in te central portion of te interval diminises, wile increasing rapidly near te ends of te interval Suc beavior is typical in equidistant interpolation wit polynomials of ig degree Tis is known as Runge s penomenon, and te function f(x is known as Runge s function If te (n + interpolation nodes are not cosen equally spaced, but rater placed near te ends of te interval at te roots of te so called Cebysev polynomial of degree (n +, te problem wit Runge s function disappears (see Fig 45 Te roots of te Cebysev polynomial of degree (n + are given by ( k + 2 x k = cos n + π, k =,,,n 45 Te Newton Representation Te Lagrange form for te interpolating polynomial is not suitable for practical computations Te Newton form of te interpolating polynomial proves to be muc more convenient Let us obtain a different form of te lowest degree interpolating polynomial at te distinct interpolation nodes given by x <x <x 2 < <x n

Cap 4 Polynomial Interpolation Numerical Metods Class Notes 77 n = 4 n = 8 8 8 6 6 4 4 2 2 2 4 6 2 8 4 8 6 4 2 2 4 6 8 n = 2 8 6 4 2 2 4 6 8 2 n = 6 5 2 5 4 5 6 2 8 25 5 2 4 8 6 4 2 2 4 6 8 4 8 6 4 2 2 4 6 8 Figure 44: Equal spaced interpolation of Runge s function were f(x j =f j, j =,,,n Let te interpolating polynomial be Using te n conditions P n (x = c + c (x x +c 2 (x x (x x + c (x x (x x (x x 2 + + c n (x x (x x (x x n P n (x j =f j, j =,,,n, we can construct a system of equations in te unknown c j For n =,for example, tis yields a system of linear equations of te form Lc = f, werel is given by, L = (x x (x 2 x (x 2 x (x 2 x (x x (x x (x x (x x (x x (x x 2 Note tat te evaluation of L for n nodes requires O(n 2 aritmetic operations Assuming tat te nodes are equidistant, ie, x i+ x i =, te system Lc = f

Cap 4 Polynomial Interpolation Numerical Metods Class Notes 78 2 n = 4 2 n = 8 8 8 6 6 4 4 2 2 2 8 6 4 2 2 4 6 8 n = 2 2 8 6 4 2 2 4 6 8 n = 6 9 9 8 8 7 7 6 6 5 5 4 4 2 2 8 6 4 2 2 4 6 8 8 6 4 2 2 4 6 8 Figure 45: Cebysev interpolation of Runge s function is given by Step 2 2 2 6 2 6 c c c 2 c = f f f 2 f c = f f ( = (f f / f ( 2 = (f 2 f /2 f ( = (f f / Ten we ave te system order : Step 2 c c 2 = 2 2 2 c c = f ( f ( f ( 2 f (

Cap 4 Polynomial Interpolation Numerical Metods Class Notes 79 f (2 2 = (f ( 2 f ( / f (2 = (f ( f ( /2 Ten we get te system of order 2: ( ( c2 c = ( f (2 2 f (2 Step c 2 = f (2 2 f ( = (f (2 f (2 2 / Step 4 c = f ( MATLAB function c = InterpolateNewton(x,f % x: interpolating nodes % f: function values at x % c: coef of Newton s form of interpolating polynomial n = lengt(x; for k = :n- f(k+:n = (f(k+:n-f(k/(x(k+:n-x(k; end c = y; 46 Spline Interpolation Draftsmen used to draw smoot curves troug data points by using splines Tese are tin flexible strips of plastic or wood wic were laid on paper and eld wit weigts so as to pass troug te required data points (or nodes Te weigts are constructed in suc a way tat te spline is free to slip As a result, te flexible spline straigtens out as muc as it can subject to passing over tese points Teory of elasticity suggests tat tis mecanical spline is given by a cubic polynomial (degree polynomial in eac subinterval [x i,x i+ ], i =, 2,,n, wit adjacent cubics joint continuously wit continuous first and second derivatives Let s(x =a + a x + a 2 x 2 + a x, be te form of eac cubic spline Since we ave (n subintervals [x i,x i+ ], i n, ten we ave 4(n unknowns; (4 parameters for eac subinterval: x,a,a 2,anda Suppose s i (x denotestecubicsplineteit interval Te conditions to be satisfied by tese cubic splines are:

Cap 4 Polynomial Interpolation Numerical Metods Class Notes 8 x x 2 x x 4 Figure 46: Spline interpolation Continuity at eac interior node: s i (x i =s i (x i, i =2,,,n 2 Continuity of first derivative at eac interior node: s i (x i=s i (x i, i =2,,,n Continuity of second derivative at eac interior node: s i (x i=s i (x i, i =2,,,n 4 Interpolation of function at eac node: s(x j =f(x j, j =, 2,,n Tese provide (n 2 + n =4n 6 conditions; owever, in order to completely specify te cubic splines, we still need 2 additional conditions Tese are conditions specified at te end points x and x n For te mecanical or natural spline we set s (x =s (x n = Let te nodes x j, j =, 2,,n, be equidistant, ie, x j+ x j = and let, ie, σ j = 6 s (x j s i (x i=s i (x i=6σ i

Cap 4 Polynomial Interpolation Numerical Metods Class Notes 8 for all i satisfying condition Since we ave s(x =a + a x + a 2 x 2 + a x, s (x = a +2a 2 x +a x 2 s (x = 2a 2 +6a x, ie, s (x is a straigt line Tus, s i (x takes te following form (see Fig 47: i (x =x i+ x (6σ i + (x x i (6σ i+, i =, 2,,n s s'' i ( x 6σ i+ 6σ i x i x xi+ Figure 47: Computing spline witin an interval Integrating once, we ave s i (x = 6σ i (x i+ x 2 + 6σ i+ (x x i 2 + β 2 2 and integrating once more, we ave s i (x = σ i (x i+ x + σ i+ (x x i + β x + β 2 were β and β 2 are constants Clearly we can write s i (x as follows, s i (x = σ i (x i+ x + σ i+ (x x i + γ (x x i +γ 2 (x i+ x Using conditions 4 and, Terefore, s i (x i =f i f i = σ i 2 + γ 2 s i (x i+ =f i+ f i+ = σ i+ 2 + γ γ = f i+ σ i+, γ 2 = f i σ i,

Cap 4 Polynomial Interpolation Numerical Metods Class Notes 82 and s i (x = σ i (x i+ x + σ i+ ( fi + σ i ( (x x i fi+ + (x i+ x s i (x = σ i (x i+ x 2 + σ i+ (x x i 2 + ( fi σ i σ i+ (x x i ( fi+ σ i+ Using condition 2, ie, s i (x i=s i (x i, we get ( ( fi+ f i fi f i σ i + (σ i+ σ i =σ i + (σ i σ i or σ i+ +4σ i + σ i = i i i =2,,,n, were i = f i+ f i Using te additional conditions σ = σ n =,weave 4σ 2 + σ = ( 2 / σ i +4σ i + σ i+ = ( i i / i =, 4,,n 2 σ n 2 +4σ n = ( n n 2 / Now, assigning g j = j j we ave te linear system 4 4 4 4 = f j+ 2f j + f j 2, σ 2 σ σ n 2 σ n = g 2 g g n 2 g n Te tridiagonal system above, denoted by [, 4, ], may be factored as follows: µ λ µ 2 λ 2 µ λ n µ n 2

Cap 4 Polynomial Interpolation Numerical Metods Class Notes 8 Te following conditions must be satisfied: µ =4, λ i µ i =, λ i + µ i+ =4 Te algoritm for computing L andu is given below Triangular decomposition for system [,4,] µ =4 for i =, 2,,n λ i = µ i µ i+ =4 λ i end; Once L andu are obtained, te above system is solved as sown in Capter

Cap 4 Polynomial Interpolation Numerical Metods Class Notes 84 47 Appendix: Cubic Spline Interpolation In tis Appendix, we discuss ow to find out te coefficients in te cubic splines in an intuitive way (no integration involved Basically, it follows te 4 conditions to derive te coefficients but will result in a better way for uman beings to calculate Tecubicpolynomialoninterval[x i,x i+ ]is s i (x =a i + b i (x x i +c i (x x i 2 + d i (x x i, i =,,n ( Apply condition (iv, So, we ave all te values of a i s (2 Apply condition (i, Ten, s i (x i =a i = f(x i, i =,,n s i (x i+ =s i+ (x i+ =f(x i+, i =,,n 2 s i (x i+ =a i + b i (x i+ x i +c i (x i+ x i 2 + d i (x i+ x i = f(x i+, were i =,,n 2 Let i = x i+ x i Ten, a i + b i i + c i 2 i + d i i = f(x i+ =a i+, i =,,n 2 Since i is from to n 2, we migt define a n = f(x n Ten, a i+ = a i + b i i + c i 2 i + d i i, i =,,n (A ( Te first derivative of s i is Apply condition (ii, Ten, s i (x =b i +2c i (x x i +d i (x x i 2 s i(x i+ =s i+(x i+ =b i+, i =,,n 2 b i+ = s i(x i+ =b i +2c i i +d i 2 i, i =,,n 2 Since i is from to n 2, we migt define b n = s (x n Ten, b i+ = b i +2c i i +d i 2 i, i =,,n (B

Cap 4 Polynomial Interpolation Numerical Metods Class Notes 85 (4 Te second derivative of s i is s i (x =2c i +6d i (x x i Apply condition (iii, s i (x i+ =s i+(x i+ =2c i+, i =,,n 2 Ten, 2c i+ = s i (x i+ =2c i +6d i i, i =,,n 2 Since i is from to n 2, we migt define c n = s (x n /2 Ten, c i+ = c i +d i i, i =,,n (C (5 We ave all a i s in (, for all i =, n From(C,weknow d i = c i+ c i, i =,,n (D i Introduce (D into (A, ten a i+ = a i + b i i + c i 2 i + d i i = a i + b i i + c i 2 i + c i+ c i i i = a i + b i i + 2 i (c i+ +2c i (E Introduce (D into (B, ten b i+ = b i +2c i i +d i 2 i = b i +2c i i +(c i+ c i i = b i + i (c i+ + c i (F (6 From (E, we get b i i = (a i+ a i 2 i (c i+ +2c i b i = (a i+ a i i i (c i+ +2c i (G Consider (F wit (G using indices i and i (ie (F is written as b i = b i + i (c i + c i, were i =2,,n Ten, (a i+ a i i i (c i+ +2c i = (a i a i i i (c i +2c i + i (c i + c i (a i+ a i (a i a i = i i i (c i+ +2c i i (c i +2c i + i (c i + c i (a i+ a i (a i a i i i = i (c i+ +2c i i (c i +2c i + i (c i + c i = i c i+ +2 i c i i c i 2 i c i + i c i + i c i = i c i+ +2 i c i +2 i c i + i c i = i c i +2c i ( i + i + i c i+,

Cap 4 Polynomial Interpolation Numerical Metods Class Notes 86 were i =2,,n Ten, i c i +2c i ( i + i + i c i+ = (a i+ a i (a i a i, i i were i =2,,n (7 Now, use te natural conditions Since S (x n =, by te definition of c n we gave in (4, c n = On te oter and, because S (x =, s (x =2c +6d (x x = So,c = (8 Consider te results in (6 and (7 We can construct a linear system to solve for all c i s as below c 2( + 2 2 c 2 2 2( 2 + c n 2 2(n n 2 + n n c n c n = 2 (a a 2 (a 2 a (a 4 a 2 (a a 2 n (a n a n n 2 (a n a n 2 Let tis linear system be represented as Ac = b Matriax A is a strictly diagonally dominant matrix; terefore, is nonsingular So te above linear system as an unique solution for c (9 Up to now, we get all a i s and c i s We can obtain b i s and d i s by te equations we ad derived in (D and (G, respectively d i = i (c i+ c i b i = (a i+ a i i i (c i+ +2c i Example 44 Obtain te natural cubic spline interpolation for f(x wose values are given at four points: f( = 2, f( 2 =, f( =, f(4 = Solution Weuseatalbetoelpuscalculate

Cap 4 Polynomial Interpolation Numerical Metods Class Notes 87 = x =( a =2 b =( 25 87 c = d = 4 87 2 = x 2 =( 2 a 2 = b 2 =( 92 87 c 2 = 4 29 d 2 =( 9 = x = a = b =( 76 87 c =( 67 87 d = 67 78 x 4 =4 a 4 = b 4 =x c 4 = d 4 =x 78 We already ave te a i = f i for i =, 2,, 4 Ten, we need to find all c i s One can construct a linear system for c i s as c 8 c 2 2 c = 9 5 c 4 Ten, c = c 4 = Furtermore,c 2 = 4 29 and c =( 67 87 After aving all c i s and a i s, we can decide te d i s and b i s as follow: d = d 2 = d = (c 2 c = 4 29 = 4 87 (c c 2 = 2 9 ( 9 87 =( 9 78 (c 4 c = 67 9 87 = 67 87 Te cubic splines are b = (a 2 a (c 2 +2c =( 25 87 b 2 = (a a 2 2 2 (c +2c 2 =( 92 87 b = (a 4 a (c 4 +2c = 76 87 s = 2+( 25 87 s 2 = +( 92 4 (x ( 2 + 87 4 (x ( + + (x ( 87 29 (x ( 22 +( 9 78 s = + 76 (x + ( 67 87 87 (x 2 + 67 78 (x Te natural cubic spline interpolation S = s s 2 s (x ( 2