# Error: Matrix dimensions must agree.

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Kutlu Yigitturk on 26 Dec 2020
Commented: Cris LaPierre on 26 Dec 2020
% dening data points, vectors X and Y
X_VALUES=[0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5];
Y_VALUES=[0 0.247466462 0.881373587 1.550157957 2.094712547 2.532068064 2.893443986 3.200334942 3.466711038 3.70191108 3.912422766];
% dening x in range 0 to 5 with total 100 values
x = linspace(0,5,100);
ls = lagrange_self(X_VALUES,Y_VALUES,x);
sp = spline(X_VALUES,Y_VALUES,x);
plot(X_VALUES,Y_VALUES,'O',x,ls,'.',x,sp,'-')
title("Interpolating curves");
legend('Original','Cubic Polynomial Lagrange interpolating','Cubic Spline Interpolation')
% Calculate total error
% error for polynomal interpolation
ls = lagrange_self(X_VALUES,Y_VALUES,x);
% error
I get the 'Matrix dimensions must agree.' error on line 16
sp = sum(sqrt((Y_VALUES-ls).^2));
% displaying difference
fprintf("Total error for Cubic Spline Interpolation is %f\n",sp);
% error
s_cubic = sum(sqrt((Y_VALUES-y_cubic).^2));
% displaying difference
fprintf("Total error for Cubic Polynomial Lagrange interpolating is %f\n",s_cubic);
function v = lagrange_self(x,y,u)
n = 4;
v = zeros(size(u));
for k = 1:n
w = ones(size(u));
for j = [1:k-1 k+1:n]
w = (u-x(j))./(x(k)-x(j)).*w;
end
v = v+w*y(k)
end
end

Cris LaPierre on 26 Dec 2020
Edited: Cris LaPierre on 26 Dec 2020
Y_VALUES is a 1x11 matrix while ls is 1x100. In order to subtract them, they must either both have the same size, or one must be a scalar (single number). Because they are not, MATLAB can't subtract them, and you get this error message.
The simplest fix is create to have the same number of points as Y_VALUES.
x = linspace(0,5,length(Y_VALUES));
##### 2 CommentsShowHide 1 older comment
Cris LaPierre on 26 Dec 2020
It looks like you don't compute the error using every point. Just the points that correspond to .

R2020b

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