# Lagrange Interpolation singleton error

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Mitul Dattani on 6 Mar 2019
Hi, I've been give code and have to add a line in to complete a Lagrange Interpolation, I googled it and found a couple threads tried them out but got the error saying:
Assignment has more non-singleton rhs dimensions than non-singleton subscripts
Error in cw1>my_interpolation (line 404)
L_k(k,:)=L_k(k,:).*(x-x(j))/(x(k)-x(j));
The code causing the error is:
function P = my_interpolation(x,y,N)
N_points = length(x); %the number of interpolation points from the input data
X = linspace(min(x),max(x),N); % creates a uniform vector of points from min(x) to max(x) with resolution N of the interpolating polynomial (not to be confused by the number of data points)
X=X'; % row vector to column vector
P = zeros(N,1); % creates an empty vector of zeros for interpolating polynomial
for k=1:N_points % loop over data points
L_k = ones(N,1); % defines initial Lagrange basis polynomial
for j=1:N_points % second loop over data points
if(k~=j) % conditional statement
% You need to complete the next line of code
% This line generates the Lagrange Basis Polynomials L_k used in the algorithm for the Lagrange Interpolating Polynomial
L_k(k,:)=L_k(k,:).*(x-x(j))/(x(k)-x(j));
%***************************** end of task 3 ***********************************
end
end
P = P + y(k)*L_k;
end
end
The line in between the comments is the one ive added. Any help is much appreciated even a push in the right direction.
Thanks
Mitul

Vijaya Lakshmi Chagi on 13 Mar 2019
Hi,
The error message occurs when you try to assign to a variable, but the indices of the left- and right-hand side of the assignment are incompatible. For example,
A(1) = [1 2 3];
throws this error because the size of the left-hand side is 1-by-1, but the size of the right-hand side is 1-by-3.
In your case, the RHS 'L_k(k,:).*(x-x(j))/(x(k)-x(j))' might be returning an array or vector which is not compatible with the LHS. Could you try storing the result of the RHS in a new varibale and check for its dimensions.