Some notes on Chapter 8: Polynomial and Piecewise-polynomial Interpolation

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1 Some notes on Chapter 8: Polynomial and Piecewise-polynomial Interpolation See your notes. 1. Lagrange Interpolation (8.2) 1

2 2. Newton Interpolation (8.3) different form of the same polynomial as Lagrange (!!! the polynomial is unique!) The set of polynomials (1) 1, x x 0, (x x 0 )(x x 1 ),..., (x x 0 )... (x x n 1 ) form a basis for the set of all polynomials of degree n: linearly indep., any polyn. of deg. n can be written as linear combination: (2) P n (x) = a 0 + a 1 (x x 0 ) + a 2 (x x 0 )(x x 1 ) + + a n (x x 0 )... (x x n 1 ) where a 0, a 1,..., a n are constants to be determined from p n (x i ) = f(x i ) at x 0 : a 0 = P n (x 0 ) = f(x 0 ) at x 1 : f(x 0 ) + a 1 (x x 0 ) = P n (x 1 ) = f(x 1 ) a 1 = f(x 1) f(x 0 ) = f(x 1) a 0 x 1 x 0 x 1 x 0 at x 2 : a 2 = f(x 2) a 0 a 1 (x 2 x 0 ) (x 2 x 0 )(x 2 x 1 ) Notice: a 2 = a 2 (x 0, x 1, x 2, a 0, a 1, f(x 2 )), adding a new point x 3 does not change the previously calculated a i Divided differences 0-th: f[x i ] = f(x i ) 1-st: f[x i, x j ] = f[x j] f[x i ] x j x i f (ξ), ξ [x i, x j ] 2-nd: f[x i, x j, x k ] = f[x j, x k ] f[x i, x j ] x k x i f (ξ), ξ [x i, x k ] k-th: f[x i,..., x i+k ] = f[x i+1,..., x i+k ] f[x i,..., x i+k 1 ] x i+k x i f (k) (ξ), ξ [x i, x i+k ] Difference analog for Tailor polynomial f(x) = f(ξ) + (x ξ)f (ξ) + 1 2! (x ξ)2 f (ξ) +... In formula for P n (x): a k = f[x 0,..., x k ] Newton s form of interpolation polynomial (N. dd.): (3) P n (x) = f[x 0 ] + n f[x 0,..., x k ](x x 0 )... (x x k 1 ) k=1 2

3 Lagrange: nodes are fixed, interpolate many diff. functions Newton: interpolate the same functions, increas # of nodes 3. Interpolation Error (8.4) r n (x) = f(x) P n (x) - error of interpolation/remainder Theorem. Let f(x) C n+1 [a, b], x k [a, b], k = 0,..., n. Then x [a, b] ξ(x) [a, b]: Alternative form. thus we have r n (x) = f (n+1) (ξ(x)) (n + 1)! ω(x), ω(x) = r n (x) = f[x 0,..., x n, x] ω(x), ω(x) = f[x 0,..., x n, x] = f (n+1) (ξ(x)) (n + 1)! n (x x k ) k=0 n (x x k ) k=0 Examples. f(x) = 1 x, nodes x 0 = 2, x 1 = 2.75, x 2 = 4 2nd Lagrange polynomial P (x) = 1 64 (x 2.75)(x 4) (x 2)(x 4) (x 2)(x 2.75) = 1 22 x x Determine the error form for this polynomial and the maximum error when the polynomial is used to approximate f(x) for x [2, 4]. Solution: Because f(x) = x 1, we have f (x) = x 2, f (x) = 2x 3, andf (x) = 6x 4. As a consequence, the second Lagrange polynomial has the error form f (ξ(x)) (x x 0 )(x x 1 )(x x 2 ) = (ξ(x)) 4 (x 2)(x 2.75)(x 4), 3! The maximum value of ξ(x) 4 (4 2) 2 = 1/16. We now need to deterthe maximum value on this interval of the absolute value of the polynomial Then critical points: ( 7, 25 ) and ( 7, Max. error: g(x) = (x 2)(x 2.75)(x 4) = x x x ) g (x) = 1 (3x 7)(2x 7) 2 f (ξ(x)) (x x 0 )(x x 1 )(x x 2 ) 1 3! =

4 4. Piecewise-polynomial interpolation: Cubic Splines? Is the interpolation project convergent? Namely, if # of nodes n, can we expect the interpolation error r n (x) = f(x) p n (x) 0? Answer: NO Why? High-degree polynomials (i.e. interpolations on large number of nodes) can oscillate erratically (they are too smooth), that is, a minor fluctuation over a small portion of the interval can induce large fluctuations over the entire range. (depends on the function and choice of nodes) What can we do? Reduce the smoothness of interpolant, i.e. consider piecewise-polynomial approximation. Namely, we divide the approximation interval into a collection of subintervals and construct a (generally) different approximating polynomial on each subinterval. I. Easiest: piecewise-linear, joining a set of data points (x i, f i ), i = 0,..., n with f i = f(x i ) by a series of straight lines (that s how Matlab creates continuous smooth graphs). : sharp endpoints, no differentiability, interpolating func. is not smooth (contradiction with physics) II. piecewise-quadratic interpolant - fitting one quadratic polynomial between each successive pair of nodes: interpolant is continious, its derivative is not 4

5 Figure 1. linear-piecewise inerpolation III. Cubic spline interpolation - piecewise cubic polynomial with two cont. derivatives. What the strange name spline is all about? Mathematical splines originated in the CAD software developed by the aircraft and automobile design industry in the late 1950s and early 1960s and were named after a special wooden or metal drafting tool used in the manual design of ship hulls: a spline. Those were used to manually interpolate the curve using the known points. 5

6 Figure 2. linear (red) and cubic (magenta) spline interpolation with 5 nodes Grid (set of nodes) on [a, b] a = x 0 < x 1 < < x n = b. Cubic spline s(x) corresponding to f(x) and x i : (1) s(x) is cubic polynomial, denoted s i (x) = a i + b i (x x i ) + c i (x x i ) 2 + d i (x x i ) 3 on each [x i 1, x i ], i = 1,..., n [4n unknowns] (2) s(x i ) = f(x i ) - condition of interpolation: s i (x i 1 ) = f(x i 1 ); s i (x i ) = f(x i ) i = 1,..., n [2n eq s] (3) s(x), s (x), s (x) are continuous on [a, b]: s i (x i ) = s i+1 (x i ), (follows from 2) s i(x i ) = s i+1(x i ), i = 1,..., n 1 [2n 2 eq s] s i (x i ) = s i+1(x i ), i = 1,..., n 1 [2n 2 eq s] Total: 4n unknowns vs. 4n 2 eq s: leaves 2 degrees of freedom for conditions on the endpoints: s (x 0 ) = s (x n ) = 0 (natural (or free) boundary) (zero curvature at end, thin rod); s (x 0 ) = f (x 0 ) and s (x n ) = f (x n ) (clamped boundary). Constructing spline: s i (x) = a i + b i (x x i ) + c i 2 (x x i ) 2 + d i (x x 6 i) 3 on [x i, x i+1 ], i = 0,..., n 1 why a i, b i, c i, d i : (4) (5) (6) thus s i(x) = b i + c i (x x i ) + d i 2 (x x i) 2 s i (x) = c i + d i (x x i ) s i (x) = d i a i = s i (x i ), b i = s i(x i ), c i = s i (x i ), d i = s i (x i ) 6

7 Interpolation conditions s(x i ) = f(x i ): a i = f(x i ) = f i, i = 0,..., n Continiuity of s(x): s i (x i ) = s i+1 (x i ), i = 1,..., n 1: a i = a i+1 + b i+1 (x i x i+1 ) + c i+1 2 (x i x i+1 ) 2 + d i+1 6 (x i x i+1 ) 3 Denoting h i = x i x i 1 h i b i h2 i 2 c i + h3 i 6 d i = f i f i 1, i = 1,..., n Continiuity of s (x): s i(x i ) = s i+1(x i ), i = 1,..., n 1: c i h i d i 2 h2 i = b i b i 1, i = 2,..., n Continiuity of s (x): s i (x i ) = s i+1(x i ), i = 1,..., n 1: d i h i = c i c i 1, i = 2,..., n Rewriting we get the three-diagonal system of linear equations: b i = h i 2 c i h2 i 6 d i f i f i 1 h i ; d i = c i c i 1 h i and ( fi+1 f i h i c i 1 + 2(h i + h i+1 )c i + h i+1 c i+1 = 6 h i+1 f i f i 1 h i i = 1,..., n 1, ), c 0 = c n = 0 [= s (x 0 ); = s (x n )= 0 for natural splines] 7

8 FINALLY: We get cubic spline: on each [x i, x i+1 ], i = 0,..., n 1 For h i = x i x i 1 s i (x) = a i + b i (x x i ) + c i 2 (x x i) 2 + d i 6 (x x i) 3 a i = f i, b i = h i 2 c i h2 i 6 d i f i f i 1 h i ; d i = c i c i 1 h i and where α 1 β β 1 α 2 β β n β n 2 α n 1 c 1 c 2. c n 1 = z 1 z 2. z n 1 ( fi+1 f i α i = 2(h i + h i+1 ); β i = h i+1, z i = 6 f ) i f i 1 h i+1 h i 8

9 Special case for uniformly distributed nodes and some adjustements for MATLAB The cubic spline s i (x) = a i + b i (x x i ) + c i 2 (x x i) 2 + d i 6 (x x i) 3 MATLAB: for k=1:n-1 x1 = x((x<=nodes(k+1))&(x>=nodes(k)))-nodes(k+1); Spline((x<=Nodes(k+1))&(x>=Nodes(k)))=a(k)+b(k)*x1+c(k)./2*x1. 2+ d(k)./6*x1. 3; end For h i = x i x i 1 (mesh step) we have a i = f i+1, b i = h ( c i h ) 2 3 d i + f i+1 f i ; d i = c i c i 1 h i h and linear system for c 1, c 2,..., c n 1 (remember, c 0 = c n = 0) 4h h c 1 h 4h h c h h 4h c n 1 = z 1 z 2. z n 1 where z i = 6 h (f i 2f i+1 + f i+2 ) 9

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