L7 - Approximation by Polynomials

# Alternatively a quick informal approach is to equate

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Unformatted text preview: ne) that interpolates the data (x0 , y0 ) and (x1 , y1 ). Answer. Let p(x) = a0 + a1 x. Then solve the simultaneous equations p ( x0 ) = y 0 p ( x1 ) = y 1 =⇒ a0 + a1 x0 = y0 =⇒ a0 + a1 x1 = y1 to obtain a0 and a1 . Alternatively, a quick informal approach is to equate the ratios y1 − y0 p ( x) − y0 = , x − x0 x 1 − x0 and then rearrange to obtain y1 − y0 ( x − x0 ) . p ( x) = y 0 + x 1 − x0 Example. Find the interpolating polynomial through the points (−2, 4), (−1, 2), (0, −3), (1, 5), (3, 4). Answer. We have 5 data points. The interpolating polynomial has degree 4. See Matlab polinterp1.m. 2 Polynomial interpolation in Lagrange form n + 1 data points: (xj , yj ) for j = 0, . . . , n Lagrange polynomials of degree n as basis functions: n i ( x) x − xj xi − xj = j =0 j =i i ( xj ) = 1 for i = j , and i ( xj ) for i = 0, . . . , n = 0 if i = j Interpolating polynomial of degree n: p(x) = No linear system to solve, but i ( x) n i=0 y i i ( x) diﬃcult to evaluate If f ∈ C n+1 ([a, b]), i.e., n + 1 times continually diﬀerentiable in [a, b], then the error in approximating f (x) by p(x) is f (n+1) (ξ ) f ( x) − p ( x) = (n + 1)! n ( x − xj ) j =1 for some unknown ξ ∈ [a, b] depending on x. Mat...
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## This note was uploaded on 05/26/2013 for the course MATHEMATIC 2089 taught by Professor Briancox during the Fall '13 term at National University of Singapore.

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