(0, 0), (1, ), (2, ), (3, ), (4, ), (5, ), (6, ).

Transcription

1 1 Interpolation: The method of constructing new data points within the range of a finite set of known data points That is if (x i, y i ), i = 1, N are known, with y i the dependent variable and x i [x 1, x N ], then in order to find the value of y at x = x 0, and x 0 [x 1, x N ] then we can use interpolation If x 0 / [x 1, x N ] then we have extrapolation 2 Consider the points (x i, y i ) given by (0, 0), (1, 08415), (2, 09093), (3, 01411), (4, 07568), (5, 09589), (6, 02794) Interpolation means finding a value for y given any x (0, 6) 3 A function y = f(x) is said to interpolate the data points (x 1, y 1 ),,(x n, y n ) if f(x i ) = y i for each 1 i n Since f is a function, this means that the x i s must all be distinct - there is no such restriction on the y i s The main theorem of polynomial interpolation shows that: Ifx 0, x 1, x n are distinct real numbers, then for arbitrary values, y 1,, y n, there is a unique polynomial f n (x), of degree n such that f n (x i ) = y i, (0 i n) a) Example: Find an interpolation ploynomial for the set of points (5, 1), ( 7, 23) b) Find an interpolation polynomial for (x 0, y 0 ), (x 0, y 1 ) We repeat the above arguments to find: Letting f 1 (x) = y 1 x x 0 x x 0 + y 0 x x 1 x 0 x 1 l 0 (x) = x x 1 x 0 x 1, l 1 (x) = x x 0 x 1 x 0, we have f 1 (x) = y 0 l 0 (x) + y 1 l 1 (x) 4 Piecewise constant interpolation - assign the value to be the same as the nearest known datapoint eg at x = 025 the piecewise constant interpolated value would be Obviously not that accurate but can be used in multidimensional problems 5 Linear Interpolation - straight line between two data points If you have (x a, y a ), (x b, y b ), then the interpolant at x [x a, x b ] is y = y a + (y b y a ) x x a x b x a Not that precise though its fast and easy The interpolant is not differentiable at the point x b, x a The error is f(x) g(x) C(x b x a ) 2, 1

2 where C = 1 8 max y [x a,x b ] g (y) Problem - show this The error in linear interpolation is proportional to the square of the distance between the data points 6 Fit N +1 points with a N th degree polynomial f(x) = exact function of which only N + 1 discrete values are known and used to establish an interpolating or approximating function p(x) p(x) is the approximating or interpolating function and will pass through all the N + 1 specified interpolation points or nodes 7 Polynomial Interpolation - replace linear function with a polynomial function If we have N data points, then there is one polynomial of degree N 1 that goes through all the data points So for the 7 data points here, there is just one polynomial of degree 6 that goes through all the data points a) (x i, y i ), i (0, 1,,,, n) b) p(x) = a n x n + a n 1 x n 1 + a n 2 x n a 1 x + a 0, with p(x i ) = y i, i (0, 1,, n) c) Substituting this into the above we get the matrix equation Ax = d, where the matrix A is the Vandermonde matrix Problem Find the form of this matrix equation We can solve this matrix equation for the coefficients a n (contained in the vector x, using Gaussian elimination, but if the matrix A is large, this can lead to large errors in computing the coefficients a i d) For Lagrange Interpolation, note that p(x) = N f i V i (x), i=0 where V i (x) is a polynomil of degree N associated with each node i such that V i (x j ) = 0, i j, V i (x j ) = 1, i = j, and f i = p(x i ) With 5 interpolation nodes, we have p(x 3 ) = f 0 V 0 (x 3 ) + f 1 V 1 (x 3 ) + f 2 V 2 (x 3 ) + f 3 V 3 (x 3 ) + f 4 V 4 (x 3 ), 2

3 which means V 0 (x 3 ) = 0, V 1 (x 3 ) = 0, V 2 (x 3 ) = 0, V 3 (x 3 ) = 1, V 4 (x 3 ) = 0 and so p(x 3 ) = f 3 In order to construct the V i (x) we need a degree N polynomial with roots at x 0, x 1, x i 1, x i+1,, x N and V i (x i ) = 1 The function W i (x) = (x x 0 )(x x 1 )(x x 2 )(x x i 1 )(x x i+1 )(x x N ) has the required roots - it equals zero at the nodes x 0, x 1, x 2,, x N except at node x i The degree of W i (x) is N After the following normalization, we get the Lagrange basis functions V i (x), V i (x) = (x x 0)(x x 1 )(x x 2 )(x x i 1 )(x x i+1 )(x x N ) (x i x 0 )(x i x 1 )(x i x 2 )(x i x i 1 )(x i x i+1 )(x i x N ) and V i (x i ) = 1 Also V i (x j ) = 0, i j e) For Linear Lagrange (N=1), we have where p(x) = f 0 V 0 (x) + f 1 V 1 (x), V 0 (x) = (x x 0) (x 0 x 1 ), V 1(x) = (x x 0) (x 1 x 0 ) For example, given the following data: x 0 = 2, x 1 = 5, f 0 = 15, f 1 = 40, find the linear interpolating function p(x) Lagrange basis functions are V 0 (x) = (5 x), V 3 1 (x) = (x 2) The interpolating polynomial is p(x) = 3 15V V 1 f) For the following data, x 0 = 3, x 1 = 4, x 2 = 5, f 0 = 1, f 1 = 2, f 2 = 4 we find the quadratic interpolating function: Lagrange basis functions are V 0 (x) = (x 4)(x 5) (3 4)(3 5), V 1 (x) = V 2 (x) = (x 3)(x 5) (4 3)(4 5), (x 3)(x 4) (5 3)(5 4), and the interpolating function is p(x) = 10V 0 (x) + 20V 1 (x) + 40V 2 (x) e) For cubic Lagrange interpolation (N = 3), use the following data to construct the interpolating polynomial (040, ), (05, ), (07, ), (080, ) 3

4 Whats the interpolated value at x = 06? g) Using Taylor series type arguments, the error can be shown to be e(x) = f(x) p(x), where p(x) is the interpolating polynomial and f(x) is the exact function you want to interpolate e(x) = L(x)f (N+1) (ξ), x 0 ξ x N, where f N+1 (ξ) is the N + 1 th derivative of f wrt x evaluated at ξ, and L(x) = (x x 0)(x x 1 )(x x N ) (N + 1)! L(x) is minimum at the center of [x 0, x N ] and maximum near the edges, L(x) = 0 at all data points As the size of the interpolating domain [x 0, x N ] increases, so does the maximum error in the interval As N increases, L max decreases But for N > N crit, L max increases Properties of f (N+1) (ξ) will also influence the situation h) Estimate the error made in the previous example knowing that f(x) = ln(x) e(x) L(x)f N+1 (x m ), where N = 3, x m = 06, then e(06) = and E(x) = ln(060) g(060) = d) In terms of Lagrange polynomials, we can write the interpolating polynomial as p(x) = n i=0 Π 0 j n,j i x x j x i x j This interpolating polynomial is unique e) Let f be a continuous function in [a, b] and let P be a ploynomial of degree smaller than or equal to n that interpolates the function f at n+1 distinct points x 0, x 1,, x n [a, n] Then to each x [a, b], there exists a ξ x [a, b] such that f(x) p(x) = 1 (n + 1)! f(n+1) (ξ x )Π n i=0(x x i ) Put another way, if the function f satisfies the bound f (n+1) (x) M for x [a, b], then f(x) p n (x) 4 1 4(n + 1) M(b a n )n+1

5 f) Let f(x) = sinx and approximate it with an interpolation polynomial with 10 equally spaced nodes in [0, 16875]n = 9, a = 0, b = 16875, f (10) (x) = sinx, f (10) (x) 1, forallx [a, b] Then sinx p n (x) 1 40 (1)(16875 ) If all the derivatives of f can be uniformly bounded for all n, then we can force the error to zero by increasing the number of interpolation nodes But if we cannot uniformly bound all the derivatives, then increasing the number of nodes does not necessarily drive the error to zero even if f and all its derivatives are continuous g) Computationally expensive, can exhibit oscillatory artifacts (Runge s phenomena) Can restrict attention to Chebyshev polynomials h) Newton Polynomials: a) Let q 0 (x) = 1, q 1 (x) = x x 0, q 2 = (x x 0 )(x x 1 ),, q n (x) = (x x 0 )(x x 1 )(x x n 1 ), then the q 0 (x), q 1 (x),, q n (x) is a basis span of 1, x, x 2,, x n This leads to the Newton form of the interpolating polynomial: f n (x) = c 0 q 0 (x) + c 1 q 1 (x) + + c n q n (x) Given the data (x 0, f(x 0 )), (x 1, f(x 1 )),,(x n, f(x n )), we have to specify the following coefficients: c 1, i = 0, 1,, n b) Divided difference: The zeroth order divided difference is f[x i ] = f(x i ) Then given x i, x j and f(x i ), f(x j ), the first order divided difference is f[x i, x j ] = f(x i) f(x j ) x i x j Note that f[x i, x j ] is an approximation of the derivatives f (x i ) and f (x j ) if they exist Higher order divided differences are defined recursively: f[x 0, x 1, x 2, x n ] = f[x 0, x 1,,,,, x n 1 ] f[x 1, x 2,, x n ] x 0 x n c) Whats f[x 0, x 1 ], f[x 0, x 1, x 2 ], f[x 0, x 1, x 2, x 3 ]? If we have (3, 1), (1, 3), (5, 2), (6, 4) what are the numerical values of these divided differences? d) For any x, f[x, x 0 ] = f(x) f(x 0) x x 0, and f(x) = f(x 0 ) + f[x, x 0 ](x x 0 ), 5

6 and and So then f[x, x 0, x 1 ] = f[x, x 0] f[x 0, x 1 ] x x 1, f[x, x 0 ] = f[x 0, x 1 ] + f[x, x 0, x 1 ](x x 1 ) f(x) = f(x 0 ) + f[x 0, x 1 ](x x 0 ) + f[x, x 0, x 1 ](x x 0 )(x x 1 ) Carrying on the with the third order divided difference, we have f(x) = f(x 0 )+f[x 0, x 1 ](x x 0 )+f[x 0, x 1, x 2 ](x x 0 )(x x 1 )+f[x, x 0, x 1, x 2, x 3 ](x x 0 )(x Hence f(x) = f(x 0 ) + f[x 0, x 1 ](x x 0 ) + + f[x 0,, x n ](x x 0 )(x x n 1 ) Take the Newton form of the interpolation polynomial as N n (x) = f(x 0 ) + f[x 0, x 1 ](x x 0 ) + + f[x 0,, x n ](x x 0 )(x x n 1 ) e) Check that N n (x i ) = f(x i ), 1 i n f) So c 0 = f(x 0 ), c 1 = f[x 0, x 1 ],, c n = f[x 0, x 1,, x n ] g) Whats the difference between Lagrange and Newton form? h) Example: f(x) = sinx(0, f(0)) = (0, 0), (π/2, f(π/2)) given Then L 1 (x) = 0 x π/2) 0 π/2 + 1 x 0 π/2 0, N 1 (x) = f(x 0 ) + f[x 0, x 1 ](x x 0 ) = (x 0), 0 π/2 This can be used to compute the approximation at x = π/8 Now if we add the point (π/4, f(π/4), we can see that for the Newton form, we only have to add one term but for the Lagrange form we have to start again a) ) Forward Newton-Gregory: Forward differences are defined as follows: 0 f i = f i = f(x i ), f i = f i+1 f i, 2 f i = f i+1 f i, 6

7 2 f i = (f i+2 f i+1 ) (f i+1 f i ) = f i+2 2f i+1 f i, 3 f i = 2 f i+1 2 f i, 3 f i = f i+3 3f i+2 + 3f i+1 f i k f i = k 1 f i+1 k 1 f i N + 1 data points develop upto N th order forward differences i) Develop a forward difference table for (2, 7), (4, 3), (6, 6), (8, 25), (10, 62), (12, 129) b) The Taylor series expansion for f(x) about x 0 is f(x) = f(x 0 )+(x x 0 ) df dx x=x ! (x x 0) 2 d2 f dx 2 x = x ! (x x 0) d3 f dx 3 x=x 0 +O(x x 4 ) 4 f(x) = f 0 + (x x 0 )f (1) ! (x x 0) 2 f (2) ! (x x 0) 4 f (3) 0 + O(x x 0 ) 4, Since f 0 = f 1 f 0 and because f 1 = f(x 1 ), we can write, f 1 = f 0 +(x 1 x 0 )f (1) ! (x 1 x 0 ) 2 f (2) ! (x 1 x 0 ) 3 f (3) 0 +O(x 1 x 0 ) 4, If h = x 1 x 0, then so that f 1 = f 0 + hf (1) ! h2 f (2) ! h3 f (3) 0 + O(h 4 ), f (1) 0 = f 0 h 1 2! f(2) 0 1 3! h2 f (3) 0 O(h 3 ) Now express the second order forward difference in terms of f 0, f (1) 0,, 2 f 0 = f 2 2f 1 + f 0, Noting that f 1 = f(x 1 ) and we already have an expression for it and f 2 = f(x 2 ), f 2 = f 0 +(x 2 x 0 )f (1) ! (x 2 x 0 ) 2 f (2) ! (x 2 x 0 ) 3 f (3) 0 +O(x 2 x 0 4 ) Note that x 2 x 0 = 2h, f 2 = f 0 + 2hf (1) ! h2 f (2) ! f(3) 0 + O(h 4 ), 7

8 and hence f (2) 0 = 2 f 0 h 2 hf (3) 0 + O(h) 2, Now we express the third order difference in terms of f 0, f (1) 0, so that f (3) 0 = 3 f 0 h 3 Noting that f(x) = g(x) + e(x), we have + O(h) g(x) = f 0 + (x x 0 ) f 0 h + 1 2! (x x 0)(x x 1 ) 2 f 0 h ! 3 f 0 h N! (x x 0)(x x 1 )(x x 2 )(x x N 1 ) N f 0 h N, the N th degree polynomial approximation to N + 1 data points and is identical to that derived for Lagrange interpolation l) The error function is e(x) = f(x) g(x) = (x x 0)(x x 1 )(x x N ) f (N+1) (ξ), x 0 ξ x N (N + 1)! This is identical to the error for Lagrange Interpolation But note that f (N+1) (x 0 ) (N+1 )f 0 h N+1 8 Newton Backward Interpolation: This essentially the same as Newton forward interpolation except that backward differences are used a) These are defined as: 0 f i = f i, f i = f i f i 1, 2 f i = f i f i 1, k f i = k 1 f i k 1 f i 1, b) For N + 1 data points fitted with an Nth degree polynomial, we have g(x) = f N +(x x N ) f N h + 1 2! (x x N)(x x N 1 ) 2 f N h ! (x x N)(x x N 1 (x x N 2 ) 8

9 9 These Newtonian formulae are easily developed for non-unifrmly spaced nodes 10 Fourier Interpolation: Interpolation with trigonometric polynomials a) A trigonometric polynomial is of the form p(x) = a 0 + K a k cos(kx) + k=1 K b k sin(kx) There are 2K + 1 coefficients, a 0, a 1, a 2, a K, b 1, b K, and we wish to compute these coefficients so that the function passes through N points, (x i, y i ), 0 i N 1 Since the trigonometric polynomial is periodic with period 2π, the N points can be distributed and ordered in one period as 0 < x 0 < x 1 < x 2 < < x N 1 < 2π, the problem is to find the coefficients such that the trigonometric polynomial p satisfies these conditions We assume N 2K + 1 a) Similar to the formula for Lagrange interpolation, the solution is: p(x) = 2K Π 2K m=0,m k k=0 k=1 sin(05(x x m ) sin(05(x k x m )) b) When the points x n are equally spaced, we have x n = 2π n N, 0 < n < N And Y k = N 1 n=0 y n = p(x n ) = 1 N nk i2π y n e N, N 1 k=0 nk i2π Y k e N This is the discrete Fourier transform 6 Can sometimes reduce errors by Chebyshev Interpolation - where Chebyshev polynomials are used instead of ordinary polynomials 7 Rational Function interpolation - this is interpolation using a rational function - a function which is just a ratio of two polynomials: f(x) = p 0 + p 1 x + + p L x L q 0 + q 1 x + + q N x N Polynomial interpolation methods can sometimes be unstable on equidistant grids If we can choose the grids, we can sometimes get around this with either Chebysehv polynomials or rational function interpolation However though we can always find an interpolating polynomial, not all sets of points have an interpolating rational function and poles can sometimes also be a big problem 9

10 5 Extrapolation: find p(x M ) when x M is outside of [x 0, x N ] 6 Linear extrapolation: y(x M ) y k 1 + (x k x k 1 ) x k x k 1 (y k y k 1 ), where (x k, y k ), (x k 1, y k 1 ) are the two data points nearest x M 7 A polynomial curve using Lagrange or Newton methods can be fitted through the entire data or just near the end and the resulting polynomial used to extrapolate the data 10