Using kron is a sensible option for this cross product of Td1 and Td2 and creating Td as a function?
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function Td1 = compute_Td1(M)
% Create a matrix of binomial coefficients
[I, J] = ndgrid(1:M, 1:M);
binomials = zeros(M, M);
for i = 1:M
binomials(i, 1:i) = arrayfun(@(ii, jj) nchoosek(ii-1, jj-1), I(i, 1:i), J(i, 1:i));
end
% Create a matrix for powers
powers = ((- (M - 1) / 2) .^ (I - J)) .* (J <= I);
% Element-wise multiplication to get Td1
Td1 = binomials .* powers;
end
function Td2 = compute_Td2(M)
% Initialize Td2 with identity
Td2 = eye(M+1);
for n = 2:M+1
for k = 2:n-1
Td2(n, k) = Td2(n-1, k-1) - (n-2) * Td2(n-1, k);
end
end
Td2 = Td2(2:M+1, 2:M+1);
end
M= 3;
Td1 = compute_Td1(M);
disp('T_d^1:');
disp(Td1);
Td2 = compute_Td2(M);
disp('T_d^2:');
disp(Td2);
M= 5;
Td1 = compute_Td1(M);
disp('T_d^1:');
disp(Td1);
Td2 = compute_Td2(M);
disp('T_d^2:');
disp(Td2);
Td = kron(Td1,Td2);
disp('T_d:');
disp(Td);
T_d:
1 0 0 0 0 0 0 0 0
-1 1 0 0 0 0 0 0 0
2 -3 1 0 0 0 0 0 0
-1 0 0 1 0 0 0 0 0
1 -1 0 -1 1 0 0 0 0
-2 3 -1 2 -3 1 0 0 0
1 0 0 -2 0 0 1 0 0
-1 1 0 2 -2 0 -1 1 0
2 -3 1 -4 6 -2 2 -3 1
Using Kron is a suitable choice or not please explain is it possible to recreate Td as a function using cross product of Td1 and Td2 ?
1 Comment
But Td1 and Td2 are 5x5 matrices, as you've defined them. There is no such thing as a cross-product of 5x5 matrices.
Please give the necessary context and mathematical aims of this problem, so that it may be better understood.
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