Explanation Given: The data points are: ( 1 , 10 ) ; ( 2 , 20 ) ; ( 3 , 10 ) . Approach: Let the data points be ( x 1 , y 1 ) ; ( x 2 , y 2 ) ; ... ; ( x m , y m ) . Use the following steps to obtain the equation of the least squares parabola: The required equation is of the form y = a x 2 + b x + c . Write A = [ x 1 2 x 1 1 x 2 2 x 2 1 ⋮ ⋮ ⋮ x m 2 x m 1 ] , x = [ a b c ] , and b = [ y 1 y 2 ⋮ y m ] . The least square solution is given by x 0 = ( A T A ) − 1 A T b . Calculation: Let the equation of the least squares parabola for the given data points be y = a x 2 + b x + c . The data points are: ( 1 , 10 ) ; ( 2 , 20 ) ; ( 3 , 10 ) . Then, A = [ 1 1 1 4 2 1 9 3 1 ] , x = [ a b c ] , b = [ 10 20 10 ] . A T A = [ 1 4 9 1 2 3 1 1 1 ] [ 1 1 1 4 2 1 9 3 1 ] = [ 1 ⋅ 1 + 4 ⋅ 4 + 9 ⋅ 9 1 ⋅ 1 + 4 ⋅ 2 + 9 ⋅ 3 1 ⋅ 1 + 4 ⋅ 1 + 9 ⋅ 1 1 ⋅ 1 + 2 ⋅ 4 + 3 ⋅ 9 1 ⋅ 1 + 2 ⋅ 2 + 3 ⋅ 3 1 ⋅ 1 + 2 ⋅ 1 + 3 ⋅ 1 1 ⋅ 1 + 1 ⋅ 4 + 1 ⋅ 9 1 ⋅ 1 + 1 ⋅ 2 + 1 ⋅ 3 1 ⋅ 1 + 1 ⋅ 1 + 1 ⋅ 1 ] = [ 98 36 14 36 14 6 14 6 3 ] ( A T A ) − 1 = [ 98 36 14 36 14 6 14 6 3 ] − 1 Augment the matrix [ 98 36 14 36 14 6 14 6 3 ] with a 3 × 3 identity matrix. = [ 98 36 14 ∣ 1 0 0 36 14 6 ∣ 0 1 0 14 6 3 ∣ 0 0 1 ] Reduce the matrix to row-echelon form. = [ 98 36 14 ∣ 1 0 0 0 38 49 6 7 ∣ − 18 49 1 0 14 6 3 ∣ 0 0 1 ] R 2 ← R 2 − 18 49 ⋅ R 1 = [ 98 36 14 ∣ 1 0 0 0 38 49 6 7 ∣ − 18 49 1 0 0 6 7 1 ∣ − 1 7 0 1 ] R 3 ← R 3 − 1 7 ⋅ R 1 = [ 98 36 14 ∣ 1 0 0 0 6 7 1 ∣ − 1 7 0 1 0 38 49 6 7 ∣ − 18 49 1 0 ] R 2 ↔ R 3 = [ 98 36 14 ∣ 1 0 0 0 6 7 1 ∣ − 1 7 0 1 0 0 − 1 21 ∣ − 5 21 1 − 19 21 ] R 3 ← R 3 − 19 21 ⋅ R 2 Further reduce to reduced row-echelon form

Differential Equations and Linear Algebra (4th Edition)

4th Edition

ISBN: 9780321964670

Author: Stephen W. Goode, Scott A. Annin

Publisher: PEARSON

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Chapter 5.4, Problem 8P

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The equation of the least squares parabola for the data points (1,10); (2,20); (3,10).

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Differential Equations and Linear Algebra (4th Edition)

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The equation of the least squares parabola for the data points ( 1 , 10 ) ; ( 2 , 20 ) ; ( 3 , 10 ) .

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