since by using a computer we are limited to the use of elementary arithmetic operations


 Brett Norton
 2 years ago
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1 > 4. Interpolation and Approximation Most functions cannot be evaluated exactly: x, e x, ln x, trigonometric functions since by using a computer we are limited to the use of elementary arithmetic operations +,,, With these operations we can only evaluate polynomials and rational functions (polynomial divided by polynomials).
2 > 4. Interpolation and Approximation Interpolation Given points and corresponding values x 0, x 1,..., x n find a function f(x) such that y 0, y 1,..., y n f(x i ) = y i, i = 0,..., n. The interpolation function f is usually taken from a restricted class of functions: polynomials.
3 > 4. Interpolation and Approximation > 4.1 Polynomial Interpolation Theory Interpolation of functions f(x) x 0, x 1,..., x n f(x 0 ), f(x 1 ),..., f(x n ) Find a polynomial (or other special function) such that What is the error f(x) = p(x)? p(x i ) = f(x i ), i = 0,..., n.
4 > 4. Interpolation and Approximation > Linear interpolation Linear interpolation Given two sets of points (x 0, y 0 ) and (x 1, y 1 ) with x 0 x 1, draw a line through them, i.e., the graph of the linear polynomial x 0 x 1 l(x) = x x 1 y 0 + x x 0 y 1 x y 0 y 0 x 1 x 1 x 0 1 l(x) = (x 1 x)y 0 + (x x 0 )y 1 x 1 x 0 (5.1) We say that l(x) interpolates the value y i at the point x i, i = 0, 1, or l(x i ) = y i, i = 0, 1 Figure: Linear interpolation
5 > 4. Interpolation and Approximation > Linear interpolation Example Let the data points be (1, 1) and (4,2). The polynomial P 1 (x) is given by P 1 (x) = (4 x) 1 + (x 1) 2 3 The graph y = P 1 (x) and y = x, from which the data points were taken. (5.2) Figure: y = x and its linear interpolating polynomial (5.2)
6 > 4. Interpolation and Approximation > Linear interpolation Example Obtain an estimate of e using the function values e 0.82 = , e 0.83 = Denote x 0 = 0.82, x 1 = The interpolating polynomial P 1 (x) interpolating e x at x 0 and x 1 is and P 1 (x) = while the true value s (0.83 x) (x 0.82) P 1 (0.826) = (5.3) e = to eight significant digits.
7 > 4. Interpolation and Approximation > Quadratic Interpolation Assume three data points (x 0, y 0 ), (x 1, y 1 ), (x 2, y 2 ), with x 0, x 1, x 2 distinct. We construct the quadratic polynomial passing through these points using Lagrange s folmula P 2 (x) = y 0 L 0 (x) + y 1 L 1 (x) + y 2 L 2 (x) (5.4) with Lagrange interpolation basis functions for quadratic interpolating polynomial L 0 (x) = L 1 (x) = L 2 (x) = (x x1)(x x2) (x 0 x 1)(x 0 x 2) (x x0)(x x2) (x 1 x 0)(x 1 x 2) (x x0)(x x1) (x 2 x 0)(x 2 x 1) Each L i (x) has degree 2 P 2 (x) has degree 2. Moreover L i (x j ) = 0, j i for 0 i, j 2 i.e., L L i (x i ) = 1 i (x j ) = δ i,j = the Kronecker delta function. P 2 (x) interpolates the data P 2 (x) = y i, i=0,1,2 { 1, i = j 0, i j (5.5)
8 > 4. Interpolation and Approximation > Quadratic Interpolation Example Construct P 2 (x) for the data points (0, 1), (1, 1), (2, 7). Then P 2 (x)= (x 1)(x 2) 2 ( 1)+ x(x 2) 1 ( 1)+ x(x 1) 2 7 (5.6) Figure: The quadratic interpolating polynomial (5.6)
9 > 4. Interpolation and Approximation > Quadratic Interpolation With linear interpolation: obvious that there is only one straight line passing through two given data points. With three data points: only one quadratic interpolating polynomial whose graph passes through the points. Indeed: assume Q 2 (x), deg(q 2 ) 2 passing through (x i, y i ), i = 0, 1, 2, then it is equal to P 2 (x). The polynomial has deg(r) 2 and R(x) = P 2 (x) Q 2 (x) R(x i ) = P 2 (x i ) Q 2 (x i ) = y i y i = 0, for i = 0, 1, 2 So R(x) is a polynomial of degree 2 with three roots R(x) 0
10 > 4. Interpolation and Approximation > Quadratic Interpolation Example Calculate a quadratic interpolate to e from the function values e 0.82 = e 0.83 = e 0.84 = With x 0 = e 0.82, x 1 = e 0.83, x 2 = e 0.84, we have P 2 (0.826) = to eight digits, while the true answer e = and P 1 (0.826) =
11 > 4. Interpolation and Approximation > Higherdegree interpolation Lagrange s Formula Given n + 1 data points (x 0, y 0 ), (x 1, y 1 ),..., (x n, y n ) with all x i s distinct, unique P n, deg(p n ) n such that P n (x i ) = y i, i = 0,..., n given by Lagrange s Formula P n (x) = n y i L i (x) = y 0 L 0 (x) + y 1 L 1 (x) + y n L n (x) (5.7) i=0 where L i (x)= n j=0,j i where L i (x j ) = δ ij x x j x i x j = (x x 0) (x x i 1 )(x x i+1 ) (x x n ) (x i x 0 ) (x i x i 1 )(x i x i ) (x i x n ),
12 > 4. Interpolation and Approximation > Divided differences Remark; The Lagrange s formula (5.7) is suited for theoretical uses, but is impractical for computing the value of an interpolating polynomial: knowing P 2 (x) does not lead to a less expensive way to compute P 3 (x). But for this we need some preliminaries, and we start with a discrete version of the derivative of a function f(x). Definition (Firstorder divided difference) Let x 0 x 1, we define the firstorder divided difference of f(x) by f[x 0, x 1 ] = f(x 1) f(x 2 ) x 1 x 2 (5.8) If f(x) is differentiable on an interval containing x 0 and x 1, then the mean value theorem gives f[x 0, x 1 ] = f (c), for c between x 0 and x 1. Also if x 0, x 1 are close together, then usually a very good approximation. f[x 0, x 1 ] f ( x0 + x 1 2 )
13 > 4. Interpolation and Approximation > Divided differences Example Let f(x) = cos(x), x 0 = 0.2, x 1 = 0.3. Then f[x 0, x 1 ] = cos(0.3) cos(0.2) = (5.9) while ( ) f x0 + x 1 = sin(0.25) = so f[x 0, x 1 ] is a very good approximation of this derivative.
14 > 4. Interpolation and Approximation > Divided differences Higherorder divided differences are defined recursively Let x 0, x 1, x 2 R distinct. The secondorder divided difference f[x 0, x 1, x 2 ] = f[x 1, x 2 ] f[x 0, x 1 ] x 2 x 0 (5.10) Let x 0, x 1, x 2, x 3 R distinct. The thirdorder divided difference f[x 0, x 1, x 2, x 3 ] = f[x 1, x 2, x 3 ] f[x 0, x 1, x 2 ] x 3 x 0 (5.11) In general, let x 0, x 1,..., x n R, n + 1 distinct numbers. The divided difference of order n f[x 0,..., x n ] = f[x 1,..., x n ] f[x 0,..., x n 1 ] x n x 0 (5.12) or the Newton divided difference.
15 > 4. Interpolation and Approximation > Divided differences Theorem Let n 1, f C n [α, β] and x 0, x 1,..., x n n + 1 distinct numbers in [α, β]. Then f[x 0, x 1,..., x n ] = 1 n! f (n) (c) (5.13) for some unknown point c between the maximum and the minimum of x 0,..., x n. Example Let f(x) = cos(x), x 0 = 0.2, x 1 = 0.3, x 2 = 0.4. The f[x 0, x 1 ] is given by (5.9), and f[x 1, x 2 ] = cos(0.4) cos(0.3) = hence from (5.11) f[x 0, x 1, x 2 ] = ( ) = (5.14) For n = 2, (5.13) becomes f[x 0, x 1, x 2 ] = 1 2 f (c) = 1 2 cos(c) 1 2 cos(0.3) = which is nearly equal to the result in (5.14).
16 > 4. Interpolation and Approximation > Properties of divided differences The divided differences (5.12) have special properties that help simplify work with them. (1) Let (i 0, i 1,..., i n ) be a permutation (rearrangement) of the integers (0, 1,..., n). It can be shown that f[x i0, x i1,..., x in ] = f[x 0, x 1,..., x n ] (5.15) The original definition (5.12) seems to imply that the order of x 0, x 1,..., x n is important, but (5.15) asserts that it is not true. For n = 1 f[x 0, x 1 ] = f(x 0) f(x 1 ) = f(x 1) f(x 0 ) = f[x 1, x 0 ] x 0 x 1 x 1 x 0 For n = 2 we can expand (5.11) to get f(x 0 ) f[x 0, x 1, x 2 ] = (x 0 x 1 )(x 0 x 2 ) + f(x 1 ) (x 1 x 0 )(x 1 x 2 ) + f(x 2 ) (x 2 x 0 )(x 2 x 1 ) (2) The definitions (5.8), (5.11) and (5.12) extend to the case where some or all of the x i coincide, provided that f(x) is sufficiently differentiable.
17 > 4. Interpolation and Approximation > Properties of divided differences For example, define f(x 0 ) f(x 1 ) f[x 0, x 0 ] = lim f[x 0, x 1 ] = lim x 1 x 0 x 1 x 0 x 1 x 0 f[x 0, x 0 ] = f (x 0 ) For an arbitrary n 1, let all x i x 0 ; this leads to the definition f[x 0,..., x 0 ] = 1 n! f (n) (x 0 ) (5.16) For the cases where only some of nodes coincide: using (5.15), (5.16) we can extend the definition of the divided difference. For example f[x 0, x 1, x 0 ] = f[x 0, x 0, x 1 ] = f[x 0, x 1 ] f[x 0, x 0 ] x 1 x 0 = f[x 0, x 1 ] f (x 0 ) x 1 x 0
18 > 4. Interpolation and Approximation > Properties of divided differences MATLAB program: evaluating divided differences Given a set of values f(x 0 ),..., f(x n ) we need to calculate the set of divided differences f[x 0, x 1 ], f[x 0, x 1, x 2 ],..., f[x 0, x 1,..., x n ] We can use the MATLAB function divdif using the function call divdif y = divdif(x nodes, y values) Note that MATLAB does not allow zero subscripts, hence x nodes and y values have to be redefined as vectors containing n + 1 components: x nodes = [x 0, x 1,..., x n ] x nodes(i) = x i 1, i = 1,..., n + 1 y values = [f(x 0 ), f(x 1 ),..., f(x n )] y values(i) = f(x i 1 ), i = 1,..., n + 1
19 > 4. Interpolation and Approximation > Properties of divided differences MATLAB program: evaluating divided differences function divdif y = divdif(x nodes,y values) % % This is a function % divdif y = divdif(x nodes,y values) % It calculates the divided differences of the function % values given in the vector y values, which are the values of % some function f(x) at the nodes given in x nodes. On exit, % divdif y(i) = f[x 1,...,x i], i=1,...,m % with m the length of x nodes. The input values x nodes and % y values are not changed by this program. % divdif y = y values; m = length(x nodes); for i=2:m for j=m:1:i divdif y(j) = (divdif y(j)divdif y(j1))... /(x nodes(j)x nodes(ji+1)); end end
20 > 4. Interpolation and Approximation > Properties of divided differences This program is illustrated in this table. i x i cos(x i ) D i E E E E E E E2 Table: Values and divided differences for cos(x)
21 > 4. Interpolation and Approximation > Newton s Divided Differences Newton interpolation formula Interpolation of f at x 0, x 1,..., x n Idea: use P n 1 (x) in the definition of P n (x): P n (x) = P n 1 (x) + C(x) } C(x) P n correction term = C(x i ) = 0 i = 0,..., n 1 = C(x) = a(x x 0 )(x x 1 )... (x x n 1 ) P n (x) = ax n +...
22 > 4. Interpolation and Approximation > Newton s Divided Differences Definition: The divided difference of f(x) at points x 0,..., x n is denoted by f[x 0, x 1,..., x n ] and is defined to be the coefficient of x n in P n (x; f). C(x) = f[x 0, x 1,..., x n ](x x 0 )... (x x n 1 ) p n (x; f) = p n 1 (x; f)+f[x 0, x 1,..., x n ](x x 0 )... (x x n 1 ) (5.17) p 1 (x; f) = f(x 0 ) + f[x 0, x 1 ](x x 0 ) (5.18) p 2 (x; f)=f(x 0 )+f[x 0, x 1 ](x x 0 )+f[x 0, x 1, x 2 ](x x 0 )(x x 1 ) (5.19). p n (x; f) = f(x 0 ) + f[x 0, x 1 ](x x 0 ) +... (5.20) +f[x 0,..., x n ](x x 0 )... (x x n 1 ) = Newton interpolation formula
23 > 4. Interpolation and Approximation > Newton s Divided Differences For (5.18), consider p 1 (x 0 ), p 1 (x 1 ). Easily, p 1 (x 0 ) = f(x 0 ), and [ ] f(x1 ) f(x 0 ) p 1 (x 1 )=f(x 0 )+(x 1 x 0 ) =f(x 0 )+[f(x 1 ) f(x 0 )]=f(x 1 ) x 1 x 0 So: deg(p 1 ) 1 and satisfies the interpolation conditions. Then the uniqueness of polynomial interpolation (5.18) is the linear interpolation polynomial to f(x) at x 0, x 1. For (5.19), note that p 2 (x) = p 1 (x) + (x x 0 )(x x 1 )f[x 0, x 1, x 2 ] It satisfies: deg(p 2 ) 2 and p 2 (x i ) = p 1 (x i ) + 0 = f(x i ), i = 0, 1 p 2 (x 2 ) = f(x 0 ) + (x 2 x 0 )f[x 0.x 1 ] + (x 2 x 0 )(x 2 x 1 )f[x 0, x 1, x 2 ] = f(x 0 ) + (x 2 x 0 )f[x 0.x 1 ] + (x 2 x 1 ){f[x 1, x 2 ] f[x 0, x 1 ]} = f(x 0 ) + (x 1 x 0 )f[x 0.x 1 ] + (x 2 x 1 )f[x 1, x 2 ] = f(x 0 ) + {f(x 1 ) f(x 0 )} + {f(x 2 ) f(x 1 )} = f(x 2 ) By the uniqueness of polynomial interpolation, this is the quadratic interpolating polynomial to f(x) at {x 0, x 1, x 2 }.
24 > 4. Interpolation and Approximation > Newton s Divided Differences Example Find p(x) P 2 such that p( 1) = 0, p(0) = 1, p(1) = 4. p(x) = p(x 0 ) + p[x 0, x 1 ](x x 0 ) + p[x 0, x 1, x 2 ](x x 0 )(x x 1 ) x i p(x i ) p[x i, x i+1 ] p[x i, x i+1, x i+1 ] p(x) = 0 + 1(x + 1) + 1(x + 1)(x 0) = x 2 + 2x + 1
25 > 4. Interpolation and Approximation > Newton s Divided Differences Example Let f(x) = cos(x). The previous table contains a set of nodes x i, the values f(x i ) and the divided differences computed with divdif D i = f[x 0,..., x i ] i 0 p n (0.1) p n (0.3) p n (0.5) True Table: Interpolation to cos(x) using (5.20) This table contains the values of p n (x) for various values of n, computed with interp, and the true values of f(x).
26 > 4. Interpolation and Approximation > Newton s Divided Differences In general, the interpolation node points x i need not to be evenly spaced, nor be arranged in any particular order to use the divided difference interpolation formula (5.20) To evaluate (5.20) efficiently we can use a nested multiplication algorithm P n (x) = D 0 + (x x 0 )D 1 + (x x 0 )(x x 1 )D (x x 0 ) (x x n 1 )D n with D 0 = f(x 0 ), D i = f[x 0,..., x i ] for i 1 P n (x) = D 0 + (x x 0 )[D 1 + (x x 1 )[D 2 + (5.21) + (x x n 2 )[D n 1 + (x x n 1 )D n ] ]] For example P 3 (x) = D 0 + (x x 0 )D 1 + (x x 1 )[D 2 + (x x 2 )D 3 ] (5.21) uses only n multiplications to evaluate P n (x) and is more convenient for a fixed degree n. To compute a sequence of interpolation polynomials of increasing degree is more efficient to se the original form (5.20).
27 > 4. Interpolation and Approximation > Newton s Divided Differences MATLAB: evaluating Newton divided difference for polynomial interpolation function p eval = interp(x nodes,divdif y,x eval) % % This is a function % p eval = interp(x nodes,divdif y,x eval) % It calculates the Newton divided difference form of % the interpolation polynomial of degree m1, where the % nodes are given in x nodes, m is the length of x nodes, % and the divided differences are given in divdif y. The % points at which the interpolation is to be carried out % are given in x eval; and on exit, p eval contains the % corresponding values of the interpolation polynomial. % m = length(x nodes); p eval = divdif y(m)*ones(size(x eval)); for i=m1:1:1 p eval = divdif y(i) + (x eval  x nodes(i)).*p eval; end
28 > 4. Interpolation and Approximation > 4.2 Error in polynomial interpolation Formula for error E(t) = f(t) p n (t; f) n P n (x) = f(x j )L j (x) j=0 Theorem Let n 0, f C n+1 [a, b], x 0, x 1,..., x n distinct points in [a, b]. Then f(x) P n (x) = (x x 0 )... (x x n )f[x 0, x 1,..., x n, x] (5.22) = (x x 0)(x x 1 ) (x x n ) f (n+1) (c x ) (5.23) (n + 1)! for x [a, b], c x unknown between the minimum and maximum of x 0, x 1,..., x n.
29 > 4. Interpolation and Approximation > 4.2 Error in polynomial interpolation Proof of Theorem Fix t R. Consider p n+1 (x; f) interpolates f at x 0,..., x n, t p n (x; f) interpolates f at x 0,..., x n From Newton p n+1 (x; f) = p n (x; f) + f[x 0,..., x n, t](x x 0 )... (x x n ) Let x = t. f(t) := p n+1 (t, f) = p n (t; f) + f[x 0,..., x n, t](t x 0 )... (t x) E(t) := f[x 0, x 1,..., x n, t](t x 0 )... (t x n )
30 > 4. Interpolation and Approximation > 4.2 Error in polynomial interpolation Examples f(x) = x n, f[x 0,..., x n ] = n! n! = 1 or p n (x; f) = 1 x n f[x 0,..., x n ] = 1 f(x) P n 1, f[x 0, x 1,..., x n ] = 0 f(x) = x n+1, f[x 0,..., x n ] = x 0 + x x n R(x) = x n+1 p n (x; f) P n+1 R(x i ) = 0, i = 0,..., n R(x i ) = (x x 0 )... (x x n ) = x n+1 (x x n )x n +... }{{} P n = f[x 0,..., x n ] = x 0 + x x n. If f P m, f[x 0,..., x m, x] = { polynomial degree m n 1 if n m 1 0 if n > m 1
31 > 4. Interpolation and Approximation > 4.2 Error in polynomial interpolation Example Take f(x) = e x, x [0, 1] and consider the error in linear interpolation to f(x) using x 0, x 1 satisfying 0 x 0 x 1 1. From (5.23) e x P 1 (x) = (x x 0)(x x 1 ) e cx 2 for some c x between the max and min of x 0, x 1 and x. Assuming x 0 < x < x 1, the error is negative and approximately a quadratic polynomial e x P 1 (x) = (x 1 x)(x x 0 ) e cx 2 Since x 0 c x x 1 (x 1 x)(x x 0 ) 2 e x 0 e x P 1 (x) = (x 1 x)(x x 0 ) e x 1 2
32 > 4. Interpolation and Approximation > 4.2 Error in polynomial interpolation For a bound independent of x and e x 1 e on [0, 1] (x 1 x)(x x 0 ) max = h2 x 0 x x 1 2 8, h = x 1 x 0 e x P 1 (x) h2 8 e, 0 x 0 x x 1 1 independent of x, x 0, x 1. Recall that we estimated e = by e 0.82 and e 0.83, i.e., h = e x P 1 (x) h2 8 e = ex P 1 (x) = , 8 The actual error being , it satisfies this bound.
33 > 4. Interpolation and Approximation > 4.2 Error in polynomial interpolation
34 > 4. Interpolation and Approximation > 4.2 Error in polynomial interpolation Example Again let f(x) = e x on [0, 1], but consider the quadratic interpolation. e x P 2 (x) = (x x 0)(x x 1 )(x x 2 ) e cx 6 for some c x between the min and max of x 0, x 1, x 2 and x. Assuming evenly spaced points, h = x 1 x 0 = x 2 x 1, and 0 x 0 < x < x 2 1, we have as before e x P 2 (x) (x x 0 )(x x 1 )(x x 2 ) 6 e1 while (x x 0 )(x x 1 )(x x 2 ) max = h3 x 0 x x (5.24) hence e x P 2 (x) h3 e h3
35 > 4. Interpolation and Approximation > 4.2 Error in polynomial interpolation For h = 0.01, 0 x 1 e x P 2 (x)
36 > 4. Interpolation and Approximation > 4.2 Error in polynomial interpolation Let w 2 (x) = (x + h)x(x h) 6 = x3 xh 6 a translation along xaxis of polynomial in (5.24). Then x = ± h 3 satisfies 0 = w 2(x) = 3x2 h 2 6 and gives w 2 (± 3 h ) = h3 9 3 Figure: y = w 2 (x)
37 > 4. Interpolation and Approximation > Behaviour of the error When we consider the error formula (5.23) or (5.22), the polynomial ψ n (x) = (x x 0 ) (x x n ) is crucial in determining the behaviour of the error. Let assume that x 0,..., x n are evenly spaced and x 0 x x n. Figure: y = Ψ 6 (x)
38 > 4. Interpolation and Approximation > Behaviour of the error Remark that the interpolation error is relatively larger in [x 0, x 1 ] and [x 5, x 6 ], and is likely to be smaller near the middle of the node points In practical interpolation problems, highdegree polynomial interpolation with evenly spaced nodes is seldom used. But when the set of nodes is suitable chosed, highdegree polynomials can be very useful in obtaining polynomial approximations to functions. Example Let f(x) = cos(x), h = 0.2, n = 8 and then interpolate at x = 0.9. Case (i) x 0 = 0.8, x 8 = 2.4 x = 0.9 [x 0, x 1 ]. By direct calculation of P 8 (0.9), cos(0.9) P 8 (0.9) = Case (ii) x 0 = 0.2, x 8 = 1.8 x = 0.9 [x 3, x 4 ], where x 4 is the midpoint. cos(0.9) P 8 (0.9) = , a factor of 24 smaller than the first case.
39 > 4. Interpolation and Approximation > Behaviour of the error Example Le f(x) = 1 1+x 2, x [ 5, 5], n > 0 an even integer, h = 10 n x j = 5 + jh, j = 0, 1, 2,..., n It can be shown that for many points x in [ 5, 5], the sequence of {P n (x)} does not converge to f(x) as n. Figure: The interpolation to 1 1+x 2
40 > 4. Interpolation and Approximation > 4.3 Interpolation using spline functions x y The simplest method of interpolating data in a table: connecting the node points by straight lines: the curve y = l(x) is not very smooth. Figure: y = l(x): piecewise linear interpolation We want to construct a smooth curve that interpolates the given data point that follows the shape of y = l(x).
41 > 4. Interpolation and Approximation > 4.3 Interpolation using spline functions Next choice: use polynomial interpolation. With seven data point, we consider the interpolating polynomial P 6 (x) of degree 6. Figure: y = P 6 (x): polynomial interpolation Smooth, but quite different from y = l(x)!!!
42 > 4. Interpolation and Approximation > 4.3 Interpolation using spline functions A third choice: connect the data in the table using a succession of quadratic interpolating polynomials: on each [0, 2], [2, 3], [3, 4]. Figure: y = q(x): piecewise quadratic interpolation Is smoother than y = l(x), follows it more closely than y = P 6 (x), but at x = 2 and x = 3 the graph has corners, i.e., q (x) is discontinuous.
43 > 4. Interpolation and Approximation > Spline interpolation Cubic spline interpolation Suppose n data points (x i, y i ), i = 1,..., n are given and a = x 1 <... < x n = b We seek a function S(x) defined on [a, b] that interpolates the data S(x i ) = y i, i = 1,..., n Find S(x) C 2 [a, b], a natural cubic spline such that S(x)is a polynomial of degree 3on each subinterval [x j 1, x j ], j = 2, 3,...,n; S(x i ) = y i, i = 1,..., n, S (a) = S (b). On [x i 1, x i ]: 4 degrees of freedom (cubic spline) 4(n 1) DOF.
44 > 4. Interpolation and Approximation > Spline interpolation S(x i ) = y i : n constraints S (x i 0) = S (x i + 0) continuity 2 = 1,..., n 1 S (x i 0) = S (x i + 0) i = 2,..., n 1 S(x i 0) = S(x i + 0) i = 2,..., n 1 n + (3n 6) = 4n 6 constraints 3n 6 Two more constraints needed to be added, Boundary Conditions: S = 0 at a = x 1, x n = b, for a natural cubic spline. (Linear system, with symmetric, positive definite, diagonally dominant matrix.)
45 > 4. Interpolation and Approximation > Construction of the cubic spline Introduce the variables M 1,..., M n with M i S (x i ), i = 1, 2,..., n Since S(x) is cubic on each [x j 1, x j ] S (x) is linear, hence determined by its values at two points: S (x j 1 ) = M j 1, S (x j ) = M j S (x) = (x j x)m j 1 + (x x j 1 )M j x j x j 1, x j 1 x x j (5.25) Form the second antiderivative of S (x) on [x j 1, x j ] and apply the interpolating conditions we get S(x j 1 ) = y j 1, S(x j ) = y j S(x) = (x j x) 3 M j 1 + (x x j 1 ) 3 M j 6(x j x j 1 ) + (x j x)y j 1 + (x x j 1 )y j 6(x j x j 1 ) 1 6 (x j x j 1 ) [(x j x)m j 1 + (x x j 1 )M j ] (5.26) for x [x j 1, x j ], j = 1,..., n.
46 > 4. Interpolation and Approximation > Construction of the cubic spline Formula (5.26) implies that S(x) is continuous over all [a, b] and satisfies the interpolating conditions S(x i ) = y i. Similarly, formula (5.25) for S (x) implies that is continuous on [a, b]. To ensure the continuity of S (x) over [a, b]: we require S (x) on [x j 1, x j ] and [x j, x j+1 ] have to give the same value at their common pointx j, j = 2, 3,..., n 1, leading to the tridiagonal linear system x j x j 1 6 M j 1 + x j+1 x j 1 3 M j + x j+1 x j M j+1 (5.27) 6 = y j+1 y j x j+1 x j y j y j 1 x j x j 1, j = 2, 3,..., n 1 These n 2 equations together with the assumption S (a) = S (b) = 0: M 1 = M n = 0 (5.28) leads to the values M 1,..., M n, hence the function S(x).
47 > 4. Interpolation and Approximation > Construction of the cubic spline Example Calculate the natural cubic spline interpolating the data {(1, 1), (2, 12 ), (3, 13 ), (4, 14 ) } n = 4, and all x j 1 x j = 1. The system (5.27) becomes and with (5.28) this yields which by (5.26) gives S(x) = Are S (x) and S (x) continuous? 1 6 M M M 3 = M M M 4 = 1 12 M 2 = 1 2, M 3 = x3 1 4 x2 1 3 x + 3 2, 1 x x x2 7 3 x , 2 x x , 3 x 4
48 > 4. Interpolation and Approximation > Construction of the cubic spline Example Calculate the natural cubic spline interpolating the data x y n = 7, system (5.27) has 5 equations Figure: Natural cubic spline interpolation y = S(x) Compared to the graph of linear and quadratic interpolation y = l(x) and y = q(x), the cubic spline S(x) no longer contains corners.
49 > 4. Interpolation and Approximation > Other interpolation spline functions So far we only interpolated data points, wanting a smooth curve. When we seek a spline to interpolate a known function, we are interested also in the accuracy. Let f(x) be given on [a, b], that we want to interpolate on evenly spaced values of x. For n > 1, let h = b a n 1, x j = a + (j 1)h, j = 1, 2,..., n and S n (x) be the natural cubic spline interpolating f(x) at x 1,..., x n. It can be shown that max f(x) S n(x) ch 2 (5.29) a x b where c depends on f (a), f (b) and max a x b f (4) (x). The reason why S n (x) doesn t converge more rapidly (have an error bound with a higher power of h) is that f (a) 0 f (b), while by definition S n(a) = S n(b) = 0. For functions f(x) with f (a) = f (b) = 0, the RHS of (5.29) can be replaced with ch 4.
50 > 4. Interpolation and Approximation > Other interpolation spline functions To improve on S n (x), we look for other interpolating functions S(x) that interpolate f(x) on a = x 1 < x 2 <... < x n = b Recall the definition of natural cubic spline: 1 S(x) cubic on each subinterval [x j 1, x j ]; 2 S(x), S (x) and S (x) are continuous on [a, b]; 3 S (x 1 ) = S (x n ) = 0. We say that S(x) is a cubic spline on [a, b] if 1 S(x) cubic on each subinterval [x j 1, x j ]; 2 S(x), S (x) and S (x) are continuous on [a, b]; With the interpolating conditions S(x i ) = y i, i = 1,..., n the representation formula (5.26) and the tridiagonal system (5.27) are still valid.
51 > 4. Interpolation and Approximation > Other interpolation spline functions This system has n 2 equations and n unknowns: M 1,..., M n. By replacing the end conditions (5.28) S (x 1 ) = S (x n ) = 0, we can obtain other interpolating cubic splines. If the data (x i, y i ) is obtained by evaluating a function f(x) y i = f(x i ), i = 1,..., n then we choose endpoints (boundary conditions) for S(x) that would yield a better approximation to f(x). We require S (x 1 ) = f (x 1 ), S (x n ) = f (x n ) or S (x 1 ) = f (x 1 ), S (x n ) = f (x n ) When combined with( ), either of these conditions leads to a unique interpolating spline S(x), dependent on which of these conditions is used. In both cases, the RHS of (5.29) can be replaced by ch 4, where c depends on max x [a,b] f (4) (x).
52 > 4. Interpolation and Approximation > Other interpolation spline functions If the derivatives of f(x) are not known, then extra interpolating conditions can be used to ensure that the error bond of (5.29) is proportional to h 4. In particular, suppose that x 1 < z 1 < x 2, x n 1 < z 2 < x n and f(z 1 ), f(z 2 ) are known. Then use the formula for S(x) in (5.26) and s(z 1 ) = f(z 1 ), s(z 2 ) = f(z 2 ) (5.30) This adds two new equations to the system (5.27), one for M 1 and M 2, and another equation for M n 1, M n. This form is preferable to the interpolating natural cubic spline, and is almost equally easy to produce. This is the default form of spline interpolation that is implemented in MATLAB. The form of spline formed in this way is said to satisfy the notaknot interpolation boundary conditions.
53 > 4. Interpolation and Approximation > Other interpolation spline functions Interpolating cubic spline functions are a popular way to represent data analytically because 1 are relatively smooth: C 2, 2 do not have the rapid oscillation that sometime occurs with the highdegree polynomial interpolation, 3 are relatively easy to work with on a computer. They do not replace polynomials, but are a very useful extension of them.
54 > 4. Interpolation and Approximation > The MATLAB program spline The standard package MATLAB contains the function rm spline. The standard calling sequence is y = spline(x nodes, y nodes, x) which produces the cubic spline function S(x) whose graph passes through the points {(ξ i, η i ) : i = 1,..., n} with (ξ i, η i ) = (x nodes(i), y nodes (i)) and n the length of x nodes (and y nodes). The notaknot interpolation conditions of (5.30) are used. The point (ξ 2, η 2 ) is the point (z 1, f(z 1 )) of (5.30) and (ξ n 1, η n 1 ) is the point (z 2, f(z 2 )).
55 > 4. Interpolation and Approximation > The MATLAB program spline Example Approximate the function f(x) = e x on the interval [a, b] = [0, 1]. For n > 0, define h = 1/n and interpolating nodes x 1 = 0, x 2 = h, x 3 = 2h,..., x n+1 = nh = 1 Using spline, we produce the cubic interpolating spline interpolant S n,1 to f(x). With the notaknot interpolation conditions, the nodes x 2 and x n are the points z 1 and z 2 in (5.30). For a general smooth function f(x), it turns out that te magnitude of the error f(x) S n,1 (x) is largest around the endpoints of the interval of approximation.
56 > 4. Interpolation and Approximation > The MATLAB program spline Example Two interpolating nodes are inserted, the midpoints of the subintervals [0, h] and [1 h, 1]: x 1 =0, x 2 = 1 2 h, x 3 =h, x 4 =2h,..., x n+1 =(n 1)h, x n+2 =1 1 2 h, x n+3 =1 Using spline results in a cubic spline function S n,2 (x); with the notaknot interpolation conditions conditions, the nodes x 2 and x n+2 are the points z 1 and z 2 of (5.30). Generally, S n,2 is a more accurate approximation than is S n,1 (x). The cubic polynomials produced for S n,2 (x) by spline for the intervals [x 1, x 2 ] and [x 2, x 3 ] are the same, thus we can use the polynomial for [0, 1 2 h] for the entire interval [0, h]. Similarly for [1 h, h]. n E n (1) Ratio E n (2) Ratio E E E E E E E E Table: Cubic spline approximation to f(x) = e x
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