Question

need help doing this in matlab. 1.2-1.4

Problem #1. For five data points listed in Table 1, you are asked to do the following: Write down the form of a Newton's interpolating polynomial function of 4th-order with five constants (b, i = 1:5). 1.2 Calculate the following three divided differences (Newton bracket) (showing the detailed steps and numbers on a white paper): [xy, x,]= [xx, x,]= S[X2, x3, x,]= Plot both the data points and the interpolating function by using my m-function file, “Newton Intı Pol" given in In-class activity #7. 1.4 Determine the value of the function at x=0.4. درا 1 Table 1. Data points. 2 X -0.5 0 1) 0.444 1.0 3 0.5 0.444 4 1.0 0.167 5 1.5 0.082

Answer 1

I have done hand calculation as well as matlab. I hope you find it useful.

Hand calculations:

Matlab:

x = [-0.5 0 0.5 1 1.5];
y = [0.444 1 0.444 0.167 0.082];

%part-1.2
%Finding f(x2,x3)
a1=(y(3)-y(2))/(x(3)-x(2));
%Finding f(x3,x4)
a2=(y(4)-y(3))/(x(4)-x(3));
%Finding f(x2,x3,x4)
a3=(a2-a1)/(x(4)-x(2));

fprintf('Value of f[x2,x3] : %.4f \n',a1)
fprintf('Value of f[x3,x4] : %.4f \n',a2)
fprintf('Value of f[x2,x3,x4] : %.4f \n',a3)

%part-1.3
plot(x,y)

%part-1.4
f=@(x) 0.558*x^2-1.67*x+1;
ans=f(0.4);
fprintf('\n\n f(0.4) = %.4f\n',ans)

Editor - C:\Users\SANTOSH\Documents\MATLAB\div_diff1.m* div_diff1.m* * + 1 - x = (-0.5 0 0.5 1 1.5]; 2 - y = [0.444 1 0.444 0.167 0.082]; 3 4 part-1.2 5 Finding f(x2, x3) 6 - al=(v (3) -Y (2))/(x(3) -x (2)); 7 Finding f(x3, x4) 8 - a2=(y (4)-y (3))/(x(4) -x (3)); 9 Finding f(x2, x3, x4) 10 - a3=(a2-al)/ (x(4) -x (2)); 11 12 fprintf ("Value of f[x2, x3] : $.4f \n', al) 13 - fprintf("Value of f[x3, x4] : $.4f \n', a2) 14 fprintf("Value of f[x2, x3, x4] : $.4f \n', a3) 15 16 Spart-1.3 17 - plot (x, y) 18 19 Spart-1.4 20 - f=@ (x) 0.558*x^2-1.67*x+1; 21 ans=f(0.4); 22 - fprintf("\n\n f(0.4) = $.4f\n', ans) TI Command Window Value of f[x2,x3] : -1.1120 Value of f[x3, x4] : -0.5540 Value of f(x2, x3, x4] : 0.5580 f(0.4) = 0.4213 fx >>

0.9 0.8 0.7 0.6 0.5 01 0.4 0.3 0.2 0.1 -0.5 0.5 1 1.5

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