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3. [Polynomial interpolation versus least squares fitting, 10pts] Recall how Q7 in HW3 required you to find the cubic best fit to six given data points. This led to a least squares optimization problem. We are given the same points as in HW3: i 01 | 2 | 3 | 4 | 5 X 0.0 0.5 1.0 1.5 2.0 2.5 Y 0.0 0.20 0.27 0.30 0.32 0.33 (a) Write down the least squares problem associated to finding the cubic best fit using (i) all six points, (ii) only the data for i = 0, 1, 2, 3, 4, and (iii) i = 0,1,2,3. In each case solve the system and plot both the data and the polynomial. Why is case (iii) not a least squares problem? (b) What is the degree of the polynomial you would have to use so that the solution interpolates (i.e., goes through) all six data points?

Answer 1

3) دیا Let the cubic polynomial be y=ax' + bx? +cx+d The following the Least square we are to minimise E’=Ě fax. + bx,* + cx + d-y.) for given (x,y:) The normal equations are yra{ x'+b% x'+ex+nd Exy=aX x* +b x'+c x'+d2 x Ex’y=a2 x'+b% x*+c x'+d2 x? Xx'y=a xº+ bx+c x*+d2 x (1) a) i) For the given 6 values of (x;,y:) Î 0.0 0.20 0.27 0.30 X 0.0 0.5 1.0 1.5 2.0 2.5 0.32 0.33 Solving the set of normal equations (1) we get a=.05111 ,b=-2738 ,c=.497 and d=.0033 Hence the fitted polynomial is y= .0511 x3 - .2738 x + 497 x+ .0033 ii) For the given 5 values of (x,y:) X 0.0 0.5 1.0 1.5 Y 0.0 0.20 0.27 0.30 2.0 0.32 Solving the set of normal equations (1) we get a= .08 ,b=-.3543,c=.549 and d= .00086 Hence the fitted polynomial is y= .08 x .3543 x + 549 X+.00086 iii) For the given 4 values of (x,y:) Х Y 0.0 0.5 1.0 1.5 0.0 0.20 0.27 0.30 Solving the set of normal equations (1) we get a= .12 ,b=-.44 ,C=.59 and d=0 Hence the interpolating polynomial is y= .12 x - ,44 x² + .59 X This case (iii) is not a least square solution because for four given non coplanar points There exists unique polynomialm of degree three. So in this case the polynomial will pass through the four given points giving the error to be zero. So in this case we can call the polynomial to be an interpolating polynomial rather than fitted polynomial. b) Just like Case-(iii) of the previous Example we conclude that the degree of the polynomial will be 5 so that the solution interpolates all the six points.