interpol - Week 5 Divided Difference Lagrange Interpolation...

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Week 5 - Divided Difference - Lagrange Interpolation - Error of interpolation
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f(1) = 1, f(2) = 2, f(4) = -2
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Newton form Overview: Solve equations to expand the polynomial.
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Newton form x: 1 2 4 f(x): 1 2 -2 p 0 (x) = 1 p 1 (x) = 1 + c(x - 1), what is c? 2 = p 1 (2) = 1 + c*(2 - 1), so c = 1
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Newton form x: 1 2 4 f(x): 1 2 -2 p 1 (x) = 1 + (x-1) p 2 (x) = 1 + (x-1) + c(x-1)(x-2), c = ? -2 = p 2 (4) = 1 + (4-1) + c(4-1)(4-2) -2 = 4 + 6c -> c = -1 p 2 (x) = 1 + (x-1) - (x-1)(x-2)
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Lagrange Interpolation x = [ x , x , x , . .. x ] poly = ∑ y * ( ∏ ) f(x) = [ y , y , y , . .. y ] 1 2 3 n 2 1 3 n i x - x x - x j i j j=1 n j=i i=1 n
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Lagrange Interpolation x: 1 2 3 y: 4 5 7 y1 * (x - x2) * (x - x3) (x1 - x2) * (x1 - x3) + y2 * (x - x1) * (x - x3) (x2 - x1) * (x2 - x3) + y3 * (x - x1) * (x - x2) (x3 - x1) * (x3 - x2)
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Lagrange Interpolation x: 1 2 3 y: 4 5 7 4 * (x - 2) * (x - 3) (1 - 2) * (1 - 3) + 5 * (x - 1) * (x - 3) (2 - 1) * (2 - 3) + 7 * (x - 1) * (x - 2) (3 - 1) * (3 - 2)
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Divided-Difference Overview: Fill out the table, then use an outside diagonal to get a polynomial
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This note was uploaded on 10/21/2011 for the course CSCI 2031 taught by Professor Meyer during the Spring '08 term at Minnesota.

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interpol - Week 5 Divided Difference Lagrange Interpolation...

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