Need help with the cope (Lagrange interpolation)

15 vues (au cours des 30 derniers jours)
Sanjamdeep Logani
Sanjamdeep Logani le 13 Mar 2021
Modifié(e) : Jan le 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

Réponses (0)

Catégories

En savoir plus sur Interpolation dans Help Center et File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by