Use zero based matrix in MATLAB - matlab

Question

Could the following equation be possible in MATLAB?
Suppose we have given data of length N, and we want to consider a linear equation among the some L nubmers and find coefficients ai. Is this possible? Because if yes, then coefficients can be solved by
a = pinv(D)*d
where D is a given matrix and d is a left vector.
The above equation comes from the following linear models
k = L, L+1, L+2, N-1
I have tested this code with some fixed f.
unction [a] = find_coeficient(y,N,L)
Lp = L + 1;
Np = N + 1;
d = y(L:N-1);
D=[];
for ii=Lp:(Np-1)
% Index into the y vector for each row of D
D = vertcat(D, y(ii:-1:(ii-Lp+1))');
end
a = D\d;
end
Is it correct?

Answer 1

This is absolutely possible in MATLAB. However, 0 indexes are not natively supported. You will need to do a "change of variables" by letting the index in each element be index+1. Here's a bit of an example:
% Generate some data
N = 40;
y = 10 * randn(N,1);
% Select an L value
L = N - 4 + 1;
d = y(L:N);
D = reshape(y,4,10);
% Solve the equation using the '\' rather than the pseudo inverse
b = D\d
For more information on the divide operator, see Systems of Linear Equations.
OK, I've thought through this a bit more. Part of the confusion here is the change of variable limits. The substitution applies to the indexing variable, not the size of the data, so L and N are unchanged, but the index is adjusted to keep it from falling off the edge of the array. So in the formula, just add 1 to every element index.
y[L] = [ y[L-1] y[L-2] ... y[0] ] * a1
.
.
y[N-1] = [ y[N-2] y[N-3] ... y[N-L-1] ] * aL
becomes:
y[L+1] = [ y[L-1+1] y[L-2+1] ... y[0+1] ] * a1
.
.
y[N-1+1] = [ y[N-2+1] y[N-3+1] ... y[N-L-1+1] ] * aL
=
y[L+1] = [ y[L] y[L-1] ... y[1] ] * a1
.
.
y[N] = [ y[N-1] y[N-2] ... y[N-L] ] * aL
Which we can then use to complete our script:
function a = find_coeficient(y,N,L)
d = y((L+1):N);
D=[];
for ii=L:(N-1)
% index into the y vector for each row of D
D = vertcat(D, y(ii:-1:(ii-L+1))');
end
a = D\d;
end