Question

Consider the following table of data points:

f(x) i Xi 0 0.000 1.00000 1 0.125 0.98450 2 0.250 0.93941 0.375 0.86882 4 0.500 0.77880 5 0.625 0.67663 6 0.750 0.56978 0.875 0.46504 8 1.000 0.36788

Using least squares fitting, find the polynomial Q(x) of degree 2 that fits the data points given in the table above. Approximate f(0.3) using Q(0.3).

Answer 1

Matlab code for 1least square fit of degree 2 polynomial clear all close all All data given 0.250 0. 375 0.500 0.625 0.750 .00017 xx0. 0.98450 o 93941 0.86882 0.77880 0.67663 0.56978 0.46504 0.87 .367881 %displaying the data fprintf\t\t\t\tx £(x) \n' disp(xx' fx'] ); finding degree 2 polynomi al %creating A matrix for i-1:length (xx A(i,1) 1 A(i, 2)-xx(i); A(i,3) (xx (i) ).^2 end %displaying A matrix fprintf\ndisplaying A matrix\n\n') disp (vpa (A,6) ) fprintf\ndisplaying AT*A matrix\n\n' ) disp (vpa (A'*A, 6)) fprintf\ndisplaying AT*fx matrix \n\n' disp (vpa (A'*fx',6)) fprintfThe coefficient for degree 2 polynomials are\n' c (inv (AA)) *A' *fx; disp (c) fprintfOrder 2 polynomial function is Q(x)= syms x Q(x)c(1)+c (2 ). *x+c (3).*x.^2; disp(vpa (Q,6)) fprintf \tHence f(0.3) using (0.3) is %f\n',Q(0.3)) %plotting the data and polynomial for j-1:length xx) double (Q (xx())); fp(j end

hold on plot (xx, fx, 'r*') plot (xx, fp) xlabelx') ylabelf(x) ') title f(x) vs x plot') legendActual data','fitted data') End of Code %8888ssusss g f(x) 0.125 0.9845 0.25 0.93941 0.375 0.86882 0.5 0.7788 0.625 0.67663 0.75 0.56978 0.875 0.46504 0.36788 displaying A matrix 0. 1.0, 0 i [1.0, 0.125, 0.015625] 1.0, [1.0, 0.375, 0.140625] 1.0, 0.25, 0.0625] 0.5 0.25] [1.0, 0.625, 0.390625] 1.0 1.0, 0.8 75, 0.765625 ] 1.0 0.75 0.5625 1.0] 1.0 displaying AT*A matrix 9.0 3.1875 4.5 4.5 3.1875, 2.53125] 3.1875, 2.53125, 2.1416 displaying ATfx matrix 6.65086 2.69814 1.69971 The coefficient for degree 2 polynomials are 1,01823793939394 -0.289904536 796539 -0.37920346320 34 5 7 2

Order 2 polynomial function is Q(x)= - 0.379203 *x*2 - 0.289905 *x 1.01824 symbolic function inpu ts: x Hence f(0.3) using Q(0.3) is 0.89 7138 fx) vs. x plot 1.1 Actual data fitted data n 9 0 B 0.7 06 0.5 0.4 03 0.1 02 03 04 0.5 0,6 0.7 0 8 0.9 1 Published with MATLAB® R2018a

%%Matlab code for least square fit of degree 2 polynomial
clear all
close all

%All data given
xx=[0.000 0.125 0.250 0.375 0.500 0.625 0.750 0.875 1.000];
fx=[1.00000 0.98450 0.93941 0.86882 0.77880 0.67663 0.56978 0.46504 0.36788];

%displaying the data
fprintf(' \t\t\t\tx f(x) \n')
disp([xx' fx']);

%finding degree 2 polynomial
%creating A matrix
for i=1:length(xx)
  
    A(i,1)=1;
    A(i,2)=xx(i);
    A(i,3)=(xx(i)).^2;
  
end

%displaying A matrix
fprintf('\ndisplaying A matrix\n\n')
disp(vpa(A,6))

fprintf('\ndisplaying AT*A matrix\n\n')
disp(vpa(A'*A,6))

fprintf('\ndisplaying AT*fx matrix \n\n')
disp(vpa(A'*fx',6))

fprintf('The coefficient for degree 2 polynomials are\n')
c=(inv(A'*A))*A'*fx';
disp(c)

fprintf('Order 2 polynomial function is Q(x)= ')
syms x
Q(x)= c(1)+c(2).*x+c(3).*x.^2;
disp(vpa(Q,6))

fprintf('\tHence f(0.3) using Q(0.3) is %f\n',Q(0.3))

%plotting the data and polynomial
for j=1:length(xx)
  
    fp(j)=double(Q(xx(j)));
  
end

hold on
plot(xx,fx,'r*')
plot(xx,fp)

xlabel('x')
ylabel('f(x)')
title('f(x) vs. x plot')
legend('Actual data','fitted data')

%%%%%%%%%%%%%%%%%%%%% End of Code %%%%%%%%%%%%%%%%%%