Question

3) (30 pts) In this problem, the last two digits of your student number will be significant. Let N, be a number equal to the last digit of your student number, N, be a number equal to the last two digits of your student number. That is, if the last three digits of your student number is 83 then, N, = 3, N2 = 83. Consider a function f(x)=cos (11 x°) defined and continuous on the interval [N1, N2]. Using the Least Squares method find a polynomial of degree two approximating this function. For the solutions, you must only use a calculator and you have to show all the details.

The last three digits of my student id is 183

Answer 1

Here [No, N2] = (3,83] f(x) = col (623) Let y = not a, x + azt is required second degree polynomial where ao, a, e az are to be determined By Method of least squares, equations satisfied ho ley a gaz are G 383 3 83 2 as leda ta by²dx + ay afted 3 3 3 (3 x² cos (28²) da 83 83 83 naman fake en fest - jameda föda + at falde = Faces (oc) di oferim un film in fide for enter Sudra { cassins fontos, how could fermenya's Now dx = 80 80 ; ada=3440; [x²dx 5 1.9059 x 10 3 83 83 cos fre x3) dx = -9.26x10°; [ 23 d6 = 116 11864560 83 NATARAJ Classic * FINE - 1.389x104 3 o! Equations satisfied by aw, ali a, & az are

: Equations satisfied by aw, a, & az are +3440 av +1.90 59x16% az = -9.26 X165 3440 do + 1.9059x16 a, + 1:186 X107 ay = -1.389x109 80 ao 5 5 7 8 1.9059x10 +1:186 Xloat 7.878 xloº az 30 ao Asso ciated 9 Matrix of as , ai & az is - 1.9059x105 105 1 3440 80 ao 7 3440 1.18 GX 10 al 1.9059x16 1-188x167 1,959 x 5 7 8 ar 7.878 X10 -9.26 X109 = -1.389 X 184 0 Solving these eqr gives -5 -1.257 X10 - ao = -0.000 1257 -7 = a 5.8436 X10 0 05 az - -5.7555 Xio Y = (-1.257 X10%) + (5.8436 xjół) x + (-5.756 X169)