Question

matlab code

Task 2 (30 points +4 for comments): Write computer code that uses Lagrange's nterpolating Polynomials to fit an interpolating polynomial to the data for Task 2 in the excel file and use the polynomials to interpolate the data set and evaluate the function value at x values of 8.4, 7.6, and 6.7 What are the first through 3rd order estimations for the interpolated function value at> 8.4 and x 7.6? a) b) Extrapolate the data set to determine a value of the function at x 6.7 using your first c) through 3rd order polynomials. Plot the original data points as circles of a given color, and plot the first through 3rd order interpolated estimations atx- 8.4 andx 7.6 and the extrapolated values at x- 6.7 as squares of a different color from the original data points. Add the lines for the polynomials too if you want. Based on your plot, how accurate do you think each of your interpolations were? What about the extrapolation? Explain? d) 4.275 8 5.280799 3.242674 10 3.087088 11 4.314521

Answer 1

x- [7 89 10 11]: Y-[4.275 5.280799 3.242674 3.087088 4.314521]: % See Function at the bottom for interpolation y1 84Langrange (X, Y,1,8.4); y2-84 Langrange (X, Y, 2, 8-0 ; y3 84Langrange (X, Y, 3,8.4); printf ("nAt x8.4") printf ("\n Using 1st order Lagrange polynomial,y.4",y1 84) fprintf("\n Using 2nd order Lagrange polynomial, y 8.4f",y2-84); fprintf("\n Using 3rd order Lagrange polynomial, y-%.4f\n",y3-84); % Using Function defined to calculate y y1_76 -Langrange (X, Y,1,7.6); y2_76 -Langrange (X, Y, 2,7.6) y3 76 -Langrange (X, Y, 3,7.6) printf ("nAt x7.6") printf ("\n Using 1st order Lagrange polynomial,y.4",y1_76): fprintf("\n Using 2nd order Lagrange polynomial, y 8.4f", y2-76); fprintf("\n Using 3rd order Lagrange polynomial, y-%.4f\n",y3-76); y1-67 Langrange (X, Y, 1, 6。7) ; y2_67-Langrange (X, Y, 2,6.7); ИЗ-67 Langrange (X, Y, 3, 6 , 7) ; printf ("nAt x6.7") printf ("\n Using 1st order Lagrange polynomial,y.4",y1 67): fprintf("\n Using 2nd order Lagrange polynomial, y 8.4f", y2-67); fprintf("\n Using 3rd order Lagrange polynomial, y-%.4f\n",y3-67); Plotting Arrangements x query-[8.4,7.6,6.71:

Plotting Arrangements x_query - 8.4,7.6,6.7]: p-plot (X,Y,-ok',x_query,yl_q,-.bs,x_query, y2_q, -.r3, x query,y3_g,'-ms' p (1).MarkerSize8 p(2) .MarkerSize = 8; p (3) .MarkerSize8 p (4) .MarkerSize8 legend (["Original Data", "1st order interpolation", "2nd order interpolation" "3rd order interpolation"], 'Location', 'Best) grid on xlabel ("x":ylabel("y"):title ("Lagrange Interpolatiom") function [y] = Langrange (X, Y, n, x) % Function for lagrange interpolation of order n % Finds y at given x L-one 3 (n+1, 1); for j = 1 : n+1 % Initialization or i-1:n+1 if i ~= j し(j) = L(j)"(x x(1)) / (X (j)-x(1) ); % lagrange Polynomials ( L1 to Ln+1) - end end end y = dot (L, Y (1:n+1)); end % interpolation formula (L1 *Y1 + L2-Y2 + Ln+1 *Yn+1))

Command Window At x = 8.4 Using 1st order Lagrange polynomial, y-5.6831 Using 2nd order Lagrange polynomial, y = 4.8308 Using 3rd order Lagrange polynomial, y = 4.5549 At x7.6 Using 1st order Lagrange polynomial, y -4.8785 Using 2nd order Lagrange polynomial, y = 5.2438 Using 3rd order Lagrange polynomial, y = 5.5196 Using 1st order Lagrange polynomial, y = 3.9733 Using 2nd order Lagrange polynomial, y = 3.3797 Using 3rd order Lagrange polynomial, y-2.6432

Lagrange Interpolatiom 6 Original Data -一日一1st order interpolation 日-2nd order interpolation B-3rd order interpolation 5.5 5 4.5 4 3.5 2.5 6.5 7 7.5 8 8.5 99.510 10.5 11

Text :-

X = [7 8 9 10 11];
Y = [4.275 5.280799 3.242674 3.087088 4.314521];
% See Function at the bottom for interpolation
y1_84 = Langrange(X,Y,1,8.4);       % Using Function defined to calculate y
y2_84 = Langrange(X,Y,2,8.4);
y3_84 = Langrange(X,Y,3,8.4);
fprintf("\nAt x = 8.4");
fprintf("\n Using 1st order Lagrange polynomial, y = %.4f",y1_84);
fprintf("\n Using 2nd order Lagrange polynomial, y = %.4f",y2_84);
fprintf("\n Using 3rd order Lagrange polynomial, y = %.4f\n",y3_84);

y1_76 = Langrange(X,Y,1,7.6);
y2_76 = Langrange(X,Y,2,7.6);
y3_76 = Langrange(X,Y,3,7.6);
fprintf("\nAt x = 7.6");
fprintf("\n Using 1st order Lagrange polynomial, y = %.4f",y1_76);
fprintf("\n Using 2nd order Lagrange polynomial, y = %.4f",y2_76);
fprintf("\n Using 3rd order Lagrange polynomial, y = %.4f\n",y3_76);

y1_67 = Langrange(X,Y,1,6.7);
y2_67 = Langrange(X,Y,2,6.7);
y3_67 = Langrange(X,Y,3,6.7);
fprintf("\nAt x = 6.7");
fprintf("\n Using 1st order Lagrange polynomial, y = %.4f",y1_67);
fprintf("\n Using 2nd order Lagrange polynomial, y = %.4f",y2_67);
fprintf("\n Using 3rd order Lagrange polynomial, y = %.4f\n",y3_67);

% Plotting Arrangements
x_query = [8.4,7.6,6.7];
y1_q = [y1_84,y1_76,y1_67];
y2_q = [y2_84,y2_76,y2_67];
y3_q = [y3_84,y3_76,y3_67];

p = plot(X,Y,'-ok',x_query,y1_q,'-.bs',x_query,y2_q,'-.rs',x_query,y3_q,'-.ms');
p(1).MarkerSize = 8;
p(2).MarkerSize = 8;
p(3).MarkerSize = 8;
p(4).MarkerSize = 8;
legend(["Original Data","1st order interpolation","2nd order interpolation",...
        "3rd order interpolation"],'Location','Best')
grid on
xlabel("x");    ylabel("y");    title("Lagrange Interpolatiom");

function [y] = Langrange(X,Y,n,x)
% Function for lagrange interpolation of order n
% Finds y at given x

L = ones(n+1,1);    % Initialization
for j = 1 : n+1
    for i = 1:n+1
        if i ~= j
            L(j) = L(j)*(x - X(i))/(X(j)-X(i)); % lagrange Polynomials ( L1 to Ln+1)
        end
    end
end
y = dot(L,Y(1:n+1));    % interpolation formula (L1*Y1 + L2*Y2 + .... Ln+1*Yn+1))
end