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# Lecture23 - Engineering Analysis ENG 3420 Fall 2009 Dan C...

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1 Engineering Analysis ENG 3420 Fall 2009 Dan C. Marinescu Office: HEC 439 B Office hours: Tu-Th 11:00-12:00

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2 2 Lecture 23 Lecture 23 Attention: The last homework HW5 and the last project are due on Tuesday November 24!! Last time: Linear regression versus sample mean. Coefficient of determination Polynomial least squares fit Multiple linear regression General linear squares More on non-linear models Interpolation (Chapter 15) Today Lagrange interpolating polynomials Splines Cubic splines Searching and sorting. Next Time More on Splines Numerical integration (chapter 17)
3 Newton interpolating polynomial of degree n-1 In general, an ( n -1) th Newton interpolating polynomial has all the terms of the ( n -2) th polynomial plus one extra. The general formula is: where and the f […] represent divided differences . f n - 1 x ( 29 = b 1 + b 2 x - x 1 ( 29 + L + b n x - x 1 ( 29 x - x 2 ( 29 L x - x n - 1 ( 29 b 1 = f x 1 ( 29 b 2 = f x 2 , x 1 [ ] b 3 = f x 3 , x 2 , x 1 [ ] M b n = f x n , x n - 1 , L , x 2 , x 1 [ ]

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4 Divided differences Divided difference are calculated as follows: Divided differences are calculated using divided difference of a smaller number of terms: f x i , x j [ ] = f x i ( 29 - f x j ( 29 x i - x j f x i , x j , x k [ ] = f x i , x j [ ] - f x j , x k [ ] x i - x k f x n , x n - 1 , L , x 2 , x 1 [ ] = f x n , x n - 1 , L , x 2 [ ] - f x n - 1 , x n - 2 , L , x 1 [ ] x n - x 1
5

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6 Lagrange interpolating polynomials Another method that uses shifted value to express an interpolating polynomial is the Lagrange interpolating polynomial . The differences between a simply polynomial and Lagrange interpolating polynomials for first and second order polynomials is: where the L i are weighting coefficients that are functions of x. Order Simple Lagrange 1 st f 1 ( x ) = a 1 + a 2 x f 1 ( x ) = L 1 f x 1 ( 29 + L 2 f x 2 ( 29 2 nd f 2 ( x ) = a 1 + a 2 x + a 3 x 2 f 2 ( x ) = L 1 f x 1 ( 29 + L 2 f x 2 ( 29 + L 3 f x 3 ( 29
7 First-order Lagrange interpolating polynomial The first-order Lagrange interpolating polynomial may be obtained from a weighted combination of two linear interpolations, as shown. The resulting formula based on known points x 1 and x 2 and the values of the dependent function at those points is: f 1 ( x ) = L 1 f x 1 ( 29 + L 2 f x 2 ( 29 L 1 = x - x 2 x 1 - x 2 , L 2 = x - x 1 x 2 - x 1 f 1 ( x ) = x - x 2 x 1 - x 2 f x 1 ( 29 + x - x 1 x 2 - x 1 f x 2 ( 29

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8 Lagrange interpolating polynomial for n points In general, the Lagrange polynomial interpolation for n points is: where L i is given by: f n - 1 x i ( 29 = L i x (29 f x i ( 29 i = 1 n L i x ( 29 = x - x j x i - x j j = 1 j i n
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10 Inverse interpolation Interpolation find the value f ( x ) for some x between given independent data points. Inverse interpolation find the argument x for which f ( x ) has a certain value.
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Lecture23 - Engineering Analysis ENG 3420 Fall 2009 Dan C...

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