L7 - Approximation by Polynomials

We solve a0 a0 114 1 5 25 1 a1 8 1 7 49 a1 23 3 1

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Unformatted text preview: lab lagpoly.m Example. Find the interpolating polynomial through the points (5, 1), (−7,−23) (−6, −54) using standard basis functions and Lagrange polynomials. Answer. 3 data points = polynomial of degree 2 Standard basis functions: Let p(x) = a0 + a1 x + a2 x2 . We solve ⎤ ⎤⎡ ⎤ ⎡ ⎤ ⎡⎤ ⎡ ⎡ a0 a0 −114 1 5 25 1 ⎣a1 ⎦ = ⎣ −8 ⎦ ⎣1 −7 49⎦ ⎣a1 ⎦ = ⎣−23⎦ =⇒ 3 1 −6 36 −54 a2 a2 Thus p(x) = −114 − 8x + 3x2 . Lagrange polynomials: (x − (−7))(x − (−6)) 1 = (x + 7)(x + 6) (5 − (−7))(5 − (−6)) 132 (x − 5)(x − (−6)) 1 = (x − 5)(x + 6) 1 ( x) = (−7 − 5)(−7 − (−6)) 12 (x − 5)(x − (−7)) 1 = − (x − 5)(x + 7) 2 ( x) = (−6 − 5)(−6 − (−7)) 11 0 ( x) Thus p(x) = = 0 (x) − 23 1 (x) − 54 2 (x). 3 (This equals p(x) = −114 − 8x +3x2 .) Equally spaced points (or nodes) On [a, b], xj = a + jh for j = 0, . . . , n, with h = Runge’s example: f (x) = 1 1+25x2 b −a n on [−1, 1] Large error near end points Error grows as degree increases! Chebyshev points Choose xj to minimize maxx∈[−1,1] n j =1 (x − xj ) On [−1, 1], tj = cos( 2n2+1−2j π ) for j = 0, . . . , n n+2 On [a, b], xj = a+b 2 + b −a t 2j for j = 0, . . . , n Chebyshev nodes are the zeros of the Chebyshev polynomial Tn+1 (x) of the ﬁrst kind of degree n + 1 (details omitted) Interpolation error is minimized by choosing Chebyshev nodes We can also represent the interpolating polynomial using Chebyshev polynomials Ti (x) for i = 0, . . . , n as basis functions (details omitted) Example. See the plots on the bottom of page 1 and Matlab finterp.m. 4...
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