numarical question interpolating polynomial code

what are the results of the folowing code based on the table below and how do i plot interpolating polynomial
n = input('Enter n for (n+1) nodes, n: ');
x = zeros(1,n+1);
y = zeros(n+1,n+1);
for i = 0:n
fprintf('Enter x(%d) and f(x(%d)) on separate lines: \n', i, i);
x(i+1) = input(' ');
y(i+1,1) = input(' ');
end
x0 = input('Now enter a point at which to evaluate the polynomial, x = ');
n = size(x,1);
if n == 1
n = size(x,2);
end
for i = 1:n
D(i,1) = y(i);
end
for i = 2:n
for j = 2:i
D(i,j)=(D(i,j-1)-D(i-1,j-1))/(x(i)-x(i-j+1));
end
end
fx0 = D(n,n);
for i = n-1:-1:1
fx0 = fx0*(x0-x(i)) + D(i,i);
end
fprintf('Newtons iterated value: %11.8f \n', fx0)

2 Kommentare

Rik
Rik am 11 Mai 2021
Bearbeitet: Rik am 11 Mai 2021
I recovered the removed content from the Google cache (something which anyone can do). Editing away your question is very rude. Someone spent time reading your question, understanding your issue, figuring out the solution, and writing an answer. Now you repay that kindness by ensuring that the next person with a similar question can't benefit from this answer.
image-pro
image-pro am 12 Apr. 2022
how can apply above same method on mri image

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Antworten (1)

DGM
DGM am 10 Mai 2021
% don't waste the user's time and invite error by making them
% retype every single number every time the script runs
x = 1.4:0.2:2.6;
y = [2.151 2.577 3.107 4.015 5.105 6.314 7.015];
x0 = 2.2
n = numel(x);
D = zeros(n);
D(:,1) = y'; % no loop needed
for i = 2:n
for j = 2:i
D(i,j)=(D(i,j-1)-D(i-1,j-1))/(x(i)-x(i-j+1));
end
end
% find val at query point
fx0 = D(n,n);
for k=(n-1):-1:1
fx0 = fx0*(x0-x(k)) + D(k,k);
end
% find poly coefficients to meet points
C = D(n);
for k = n:-1:1
fx0 = fx0*(x0-x(k)) + D(k,k);
C = conv(C,poly(x(k)));
nc = numel(C);
C(nc) = C(nc) + D(k,k);
end
% outputs
fprintf('Newtons iterated value: %11.8f \n', fx0)
clf
plot(x,y); hold on; grid on
xx = linspace(1.4,2.6,50);
plot(xx,polyval(C,xx));
C

1 Kommentar

The given x from the table is a uniformly increasing vector with a step size of 0.2.
x = 1.4:0.2:2.6
x =
1.4000 1.6000 1.8000 2.0000 2.2000 2.4000 2.6000

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