Need help with the cope (Lagrange interpolation)

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Sanjamdeep Logani
Sanjamdeep Logani on 13 Mar 2021
Edited: Jan on 13 Mar 2021
clc;
%a
x =[1 2 3 5 7 8];
fx =[3 6 19 99 291 444];
f1 = interp1(x,fx,2.8,'nearest');
f2 = interp1(x,fx,4.4,'nearest');
f3 = interp1(x,fx,7.1,'nearest');
fprintf("\nf(x) at x = 2.8: %d",f1)
fprintf("\nf(x) at x = 4.4: %d",f2)
fprintf("\nf(x) at x = 7.1: %d",f3)
%b
x =[1 2 3 5 7 8];
fx =[ 3 6 19 99 291 444];
sum = 0;
a = 2;
for i = 1:length(x)
u = 1;
l = 1;
for j = 1:length(x)
if j ~= i
u = u * (a - x(j));
l = l * (x(i) - x(j));
end
end
sum= sum + u / l * fx(i);
end
disp(sum);
sum = 0;
b = 3;
for i = 1:length(x)
u = 1;
l = 1;
for j = 1:length(x)
if j ~= i
u = u * (b - x(j));
l = l * (x(i) - x(j));
end
end
sum= sum + u / l * fx(i);
end
disp(sum);
c = 4;
for i = 1:length(x)
u = 1;
l = 1;
for j = 1:length(x)
if j ~= i
u = u * (c - x(j));
l = l * (x(i) - x(j));
end
end
sum= sum + u / l * fx(i);
end
disp(sum);
%c
x = [ 2.8, 4.4 ,7.1];
fx = [19, 99, 291];
sum = 0;
a = 2;
for i = 1:length(x)
u = 1;
l = 1;
for j = 1:length(x)
if j ~= i
u = u * (a - x(j));
l = l * (x(i) - x(j));
end
end
sum= sum + u / l * fx(i);
end
disp(sum);
sum = 0;
b = 3;
for i = 1:length(x)
u = 1;
l = 1;
for j = 1:length(x)
if j ~= i
u = u * (b - x(j));
l = l * (x(i) - x(j));
end
end
sum= sum + u / l * fx(i);
end
disp(sum);
c = 4;
for i = 1:length(x)
u = 1;
l = 1;
for j = 1:length(x)
if j ~= i
u = u * (c - x(j));
l = l * (x(i) - x(j));
end
end
sum= sum + u / l * fx(i);
end
disp(sum);
I need to double check that AM I doing it right?
You are given the following errorless dataset:x= [1 2 3 5 7 8]
f(x) = [3 6 19 99 291 444 ]
a.Use nearest neighbour interpolation to find f(x) for the following x values: 2.8, 4.4, and 7.1
b.Implement a MATLABfunctionor scriptto perform interpolation based on the closest 2, 3, and 4 points using Lagrangian interpolating polynomials
c.Run your function/scripton the x values given in a)for the closest 2-, 3-, and 4-point Lagrangian interpolating polynomia

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