Polynomial Interpolation
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1 Capter 4 Polynomial Interpolation In tis capter, we consider te important problem of approximatinga function fx, wose values at a set of distinct points x, x, x,, x n are known, by a polynomial P x suc tat P x i = fx i, i =,,,, n Suc a polynomial is known as an interpolating polynomial An approximation of tis function is desirable if fx is difficult to evaluate or manipulate like te error function erfx = x e t 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 fx i may be replaced by a sort table and a compact interpolation subroutine 4 Linear Interpolation Let fx be given at two distinct points x i and x i+ Now, given fx i = f i and fx i+ = f i+, we wis to determine te interpolating polynomial P x of degree ie, a straigt line tat approximates fx on te interval [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 Terefore, fi+ f i P x = f i + x x i x i+ x i Example 4 Using linear interpolation, approximate sin 65 from a table wic gives a sin 6 = 453; sin 7 = 87 b sin = ; sin = 7365
2 Cap 4 Polynomial Interpolation CS44 Class Notes 57 Solution a P 65 = b P 65 = Te first answer is correct to 5 decimals wereas te second answer is correct only to decimals! Conclusion: Linear interpolation is suitable only over small intervals 4 Polynomial Interpolation Since linear interpolation is not adequate unless te given points are closely spaced, we consider iger order interpolating polynomials Let fx be given at te selected sample of n + points: x < x < < x n, ie, we ave n + pairs x i, f i, i =,,,, n Te objective now is to find te lowest degree polynomial tat passes troug tis selected sample of points fx P n x fx x x x x 3 x n Figure 4: Interpolating te function fx by a polynomial of degree n, P n x Consider te nt degree polynomial P n x = a + a x + a x + + a n x n We wis to determine te coefficients a j, j =,,, n, suc tat Tese n + conditions yield te linear system P n x j = fx j, j =,,,, n a + a x + a x + + a nx n = f a + a x + a x + + a nx n = f a + a x + a x + + a n x n = f a + a x n + a x 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 + matrix given by x x x n x x x n V = x x x n x n x n x n n
3 Cap 4 Polynomial Interpolation CS44 Class Notes 58 Te matrix V is known as te Vandermonde matrix It is nonsingular wit nonzero determinant For n = 3, for example, n detv = x i x j j= i>j detv = x x x x x 3 x x x x 3 x x 3 x 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 and Qx be two interpolating polynomials of degree at most n, for te same set of points x < x < < x n Also, let Rx = P x Qx Ten Rx is also a polynomial of degree at most n Since P x i = Qx i = f i, we ave, Rx i = for i =,,, n In oter words Rx as n + roots From te Fundamental Teorem of Algebra owever, Rx cannot ave more tan n roots A contradiction! Tus, Rx, and P x Qx Example 4 polynomial Given te following table for te function fx, obtain te lowest degree interpolating x k 4 f k 4 4 Solution Since te number of nodes = 3 = n +, terefore n = Let P x = a + a x + a x, tat satisfies te following conditions at te points x = ; x =, x = 4: P = ; P = 4; P 4 = 4 4 a a = a 4 Using Gaussian elimination wit partial pivoting, we compute te solution a 4 a = a a =, Terefore, te interpolating polynomial is given by P x = 4 x + x We can use tis approximation to estimate te root of fx tat lies in te interval [, 4]: γ = = + 5
4 Cap 4 Polynomial Interpolation CS44 Class Notes 59 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 = fx L x + fx L x + + fx 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 we see tat P n x i = fx i, i n, L j x i = Te above property is certainly satisfied if we coose ie, { i j i = j 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 fx at te n + nodes x < x < < x n Tus, for tis given set of nodes, bot forms yield te same polynomial! Vandermonde Approac P n x = n 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 On aritmetic operations rater tan On 3 operations because of te special form of te Vandermonde matrix Lagrange Approac n P n x = f j L j x = n j= j= g j n x x i i= x x j = n g j ν j j= were g j = fx 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 n ν j = x x i x x j i=
5 Cap 4 Polynomial Interpolation CS44 Class Notes 6 for a given x Te cost of tese operations for eac j =,,,, n are: ux = n x x i n + ops i= ν j = ux x x j g j = f j n x j x i i= i j Now, te cost for evaluating te polynomial n + ops nn + ops evaluated once only n P n x = g j ν j j= n + ops Terefore, te total number of operations are 5n + 3 once g j s are evaluated 43 Error in Polynomial Interpolation Let e n x be te error in polynomial interpolation given by e n x = fx P n x Since P n x i = fx i, we ave 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 write fx P n x fx x x x x 3 x n Figure 4: Computing te error in polynomial interpolation e n x = x x x x x x n x = πx x were x is a function of x Let us define gz = e n z πzx were x x j, for j n Tus, gx =, gx j =, j n,
6 Cap 4 Polynomial Interpolation CS44 Class Notes 6 ie, gz as n+ distinct roots Assuming tat fx as at least n+ continuous derivatives, te n+t derivative of gz, ie, d n+ dz n+ gz gn+ z as at least one root x Hence, gz x x x x3 x 4 g! z g!! z g!!! z g iv z x* Figure 43: Te n + t derivative of a function gx wit n + roots n = 3 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 = fz P n z, terefore, e n+ n z = f n+ z P n 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 xn +! 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 +! Consequently, we ave te following representation of te error in polynomial interpolation e n x = fx 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
7 Cap 4 Polynomial Interpolation CS44 Class Notes 6 Example 43 Obtain te Lagrange interpolating polynomial from te data and use it to evaluate sin π and sin π 4 sin =, sin π 6 =, sin π 3 = 3, sin π =, Solution Te interpolating polynomial is given by At x = π, Terefore, P 3 x = f L x + f L x + f L x + f 3 L 3 x = 3 L x + L x + L 3 x L = π 3 π π 6 6 π 3 6 π = 5 6 L = π 6 π π 3 3 π 6 3 π = 5 6 L 3 = π 6 π 3 π π 6 π = 6 3 P 3 = 66 Since sin π = 588, we ave error = 8 At x = π 4, L = 565, L = 565, L 3 = Terefore P 3 = Since sin π 4 = 77, we ave error = Let us estimate te error e n x at x = π 4 : Since fx = sin x; f 4 x = sin x, and tus, e 3 = π 6 4 π 3 4 π f 4 x 4! f 4 x Terefore, Similarly, e n x at x = π π π e π π π 4 4! = 76 3 is bounded by π π e 3 π π 4 5π 4! = 94 3
8 Cap 4 Polynomial Interpolation CS44 Class Notes 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 [, ] fx = + 5x using te interpolation polynomial P n x on te equidistant nodes x j = + j, j =,,, n n Figure?? sows ow te polynomials P 4 x, P 8 x, P x, and P 6 x approximate fx on [, ] We see n = 4 n = n = n = Figure 44: Equal spaced interpolation of Runge s function 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 fx 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?? Te roots of te Cebysev polynomial of degree n + are given by k + x k = cos n + π, k =,,, n
9 Cap 4 Polynomial Interpolation CS44 Class Notes 64 n = 4 n = n = n = Figure 45: Cebysev interpolation of Runge s function 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 < < x n were fx j = f j, j =,,, n Let te interpolating polynomial be Using te n conditions P n x = c + c x x + c x x x x + c 3 x x x x x x + + 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 = 3, for example, tis yields a system of linear equations of te form Lc = f, were L is given by, L = x x x x x x x x x 3 x x 3 x x 3 x x 3 x x 3 x x 3 x
10 Cap 4 Polynomial Interpolation CS44 Class Notes 65 Note tat te evaluation of L for n nodes requires On aritmetic operations Assuming tat te nodes are equidistant, ie, x i+ x i =, te system Lc = f is given by c f c c = f f c 3 f 3 Step Ten we ave te system order 3: Step Ten we get te system of order : Step 3 c = f f = f f / f = f f / f 3 = f 3 f /3 c c c 3 = c = f f = f f / f 3 = f 3 f / c = c 3 f f 3 c = f f 3 3 = f 3 f / f f f 3 Step 4 c 3 = f 3 3 MATLAB function c = InterpolateNewtonx,f % x: interpolating nodes % f: function values at x % c: coef of Newton s form of interpolating polynomial n = lengtx; for k = :n- fk+:n = fk+:n-fk/xk+:n-xk; end c = y;
11 Cap 4 Polynomial Interpolation CS44 Class Notes 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 x x x 3 x 4 Figure 46: Spline interpolation tat tis mecanical spline is given by a cubic polynomial degree 3 polynomial in eac subinterval [x i, x i+ ], i =,,, n, wit adjacent cubics joint continuously wit continuous first and second derivatives Let sx = a + a x + a x + a 3 x 3, be te form of eac cubic spline Since we ave n subintervals [x i, x i+ ], i n, ten we ave 4n unknowns; 4 parameters for eac subinterval: x,a,a, and a 3 Suppose s i x denotes te cubic spline te it interval Te conditions to be satisfied by tese cubic splines are: Continuity at eac interior node: s i x i = s i x i, i =, 3,, n Continuity of first derivative at eac interior node: s i x i = s ix i, i =, 3,, n 3 Continuity of second derivative at eac interior node: s i x i = s i x i, i =, 3,, n 4 Interpolation of function at eac node: sx j = fx j, j =,,, n Tese provide 3n + n = 4n 6 conditions; owever, in order to completely specify te cubic splines, we still need 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 =
12 Cap 4 Polynomial Interpolation CS44 Class Notes 67 Let te nodes x j, j =,,, n, be equidistant, ie, x j+ x j = and let, ie, for all i satisfying condition?? Since σ j = 6 s x j s i x i = s i x i = 6σ i sx = a + a x + a x + a 3 x 3, we ave s x = a + a x + 3a 3 x s x = a + 6a 3 x, ie, s x is a straigt line Tus, s i x takes te following form see Fig??: s i x = x i+ x 6σ i + x x i 6σ i+, i =,,, n s'' i x 6σ i+ 6σ i x i x xi+ Integrating once, we ave and integrating once more, we ave Figure 47: Computing spline witin an interval s ix = 6σ i x i+ x + 6σ i+ x x i + β s i x = σ i x i+ x 3 + σ i+ x x i 3 + β x + β were β and β are constants Clearly we can write s i x as follows, Using conditions?? and??, s i x = σ i x i+ x 3 + σ i+ x x i 3 + γ x x i + γ x i+ x s i x i = f i f i = σ i + γ s i x i+ = f i+ f i+ = σ i+ + γ
13 Cap 4 Polynomial Interpolation CS44 Class Notes 68 Terefore, and γ = f i+ σ i+, γ = f i σ i, s i x = σ i x i+ x 3 + σ i+ fi + σ i x x i 3 fi+ + x i+ x s i x = 3σ i x i+ x + 3σ i+ x x i + fi σ i σ i+ x x i fi+ σ i+ Using condition??, ie, s i x i = s i x i, we get fi+ f i fi f i 3σ i + σ i+ σ i = 3σ i + σ i σ i or were σ i+ + 4σ i + σ i = i i i = f i+ f i Using te additional conditions σ = σ n =, we ave Now, assigning we ave te linear system i =, 3,, n, 4σ + σ 3 = / σ i + 4σ i + σ i+ = i i / i = 3, 4,, n σ n + 4σ n = n n / g j = j j = f j+ f j + f j, σ σ 3 σ n σ n = g g 3 g n g n Te tridiagonal system above, denoted by [, 4, ], may be factored as follows: µ λ µ λ µ 3 λ n 3 µ n Te following conditions must be satisfied: µ = 4, λ i µ i =, λ i + µ i+ = 4 Te algoritm for computing L and U is given below
14 Cap 4 Polynomial Interpolation CS44 Class Notes 69 Triangular decomposition for system [,4,] µ = 4 for i =,,, n 3 λ i = µ i µ i+ = 4 λ i end; Once L and U are obtained, te above system is solved as sown in Capter??
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