Question

Question 9 1 pts Problem 9: Interpolation, least squares, and finite difference Consider the following data table: 0 2 0.2 2.018 = f(0) = 0.4 2.104 0.6 L2.306 The resulting linear least-squares fit is (Chop after 2 decimal places) None of the above. yx) - 1.95+4,50 x yix)- 4+2.50 x yox)- 1.95+0.50 x yox) 4.34-2.50 x

Answer 1

Given data, 2 O f(xi) - Yi a 2018 i. 2 ri: 3 0.2 0.4 0.6 2.104 2.306 mathematical expression expression for a straight line using Least squares Regression be fit can given ay Y= oota,ate written as Now the sum of sallares of resideus can be (9; -Q0-Q12:12 391 - E il Now, differentiate sa wr.to unknown Coefficients dser -2 € (9; -2 0-012;) dao dsr до, - 2 Ex{ ;-ag-0,2;) These equations set علام to to Wto Obtain the final equations Eyj = nao +Q, Exi Ediyi = xi +(€) a solving these two equations We can obtain the values of ao & a,

ai Yi at Xi yi o o 2 0.2 2.018 0.04 04036 3 0.4 2.104 016 0.8416 T 06 2.306 0:36 1.3836 n=4 Exi= 1.2 Eyi -8.428 EX - 056 Exiyi - 2.6288 substitute these values the cauations above, we have 8.428 - 400+(1:2)a, 2.6288 1.2) 20+ (0-56), solving these two equations we can have the values for aoda, By solving; Qo1.9564 a = 0.502 Rounding to to a decimal places 20= 1.95 a, = 0.50 Therefore we can write 9031) 1.95 + 0-50%