Question

NOTE: Algebraic expressions follow FORTRAN conventions. Use full calculator precision for intermediate values. To find the least squares polynomial of degree 2 to approximate points (X,Y) given in the table 10 . 5.1 the normal equations to be solved have the form AC = v, where A is the matrix (NOT NECESSARILY DISPLAYED TO THE PRECISION USED FOR THE CALCULATIONS) m'o'o and c is the coefficient vector for the polynomial; and v is the transpose of the vector: _ ____11_ __12_ The solution vector c (FOUND USING DOUBLE PRECISION CALCULATIONS with the ORIGINAL DATA above) is the transpose of vector: _ ____14_ - 15-

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Answer 1

0.8 24 1.8 16 3 43 23.9 35 15.1-41 Quadratic (degre. 2) Y = ax + bx+c. Y=2x whene d = and 0 Xਜ x x 4 X: 0.64 0.8 1 3.24 1.8 9 3 1 1 = (xx) x x 45.22 39 1 is the solution. 126.0126.01 1 5.1 (xx) = 999 225 54.09 225 54.1 14.6 54.09 T 14.5 - 80.21 104.4 3 77 y 10.84 + 5311 L-22.56