I'm new to Octave and Matlab and I have a problem. I need to make a program, which solves a Triangular system of linear equations and making a triangular factorization with pivoting.
For example I need to do next exercise
That is my lufact function
function X = lufact (A, B)
[N, N] = size(A);
X = zeros(N, 1);
Y = zeros(N, 1);
C = zeros(1, N);
R = 1:N;
for p=1:N-1
[max1,j]=max(abs(A(p:N,p)));
C = A(p, :);
A(p, :) = A(j + p - 1, :);
A(j + p - 1, :) = C;
d = R(p);
R(p) = R(j + p - 1);
R(j + p - 1) = d;
if A(p,p) == 0
'A is singular. No unique solution'
break
endif
endfor
for k = p + 1: N
mult = A(k,p)/A(p,p);
A(k,p) = mult;
A(k, p + 1: N) = A(k, p + 1: N) - mult*A(p, p+1: N);
endfor
endfunction
Y(1) = B(R(1));
for k = 2: N
Y(k) = B(R(k)) - A(k, 1:k - 1)*Y(1:k - 1);
endfor
X(N) = Y(N)/A(N,N);
for k = N-1: -1: 1
X(k) = (Y(k) - A(k, k + 1: N)*X(k + 1: N))/A(k, k);
endfor
And that is my main function
A = [2 4 -6; 1 5 3; 1 3 2];
BA = [-4; 10; 5]
BB = [20; 49; 32]
XA = lufact(A, BA);
XB = lufact(A, BB);
disp(XA);
disp(XB);
The output of my program
What am I doing wrong and what should I do to fix that?
You don't seem to be attempting to answer the question you're asking, so your question doesn't make much sense...
But, in any case. The exercise asks you to show that given the following matrices / vectors:
A = [ 2, 4, -6; 1, 5, 3; 1, 3, 2 ];
L = [ 1, 0, 0; 1/2, 1, 0; 1/2, 1/3, 1 ];
U = [ 2, 4, -6; 0, 3, 6; 0, 0, 3 ];
show that LY = B, UX = Y, AX = B for a given B.
Ok. In octave, the best way to solve a matrix equation of the form aX = b w.r.t X for a given a and b, is to use the "matrix left-division" operator, \, i.e. x = a\b. Type help mldivide in your octave console for details.
So,
Y = L\B;
X = U\Y;
A*X % the result should be the same as B. QED.
Then there is the separate question of how can I perform LU decomposition in octave? Well, octave provides a function for this: lu.
[L, U] = lu(A)
% L = 1.0000 0 0
% 0.5000 1.0000 0
% 0.5000 0.3333 1.0000
%
% U = 2 4 -6
% 0 3 6
% 0 0 3
Then there's the implied further question of "I would like to perform LU decomposition by hand."
Great. Good luck. Your code here is a bit messy, no comments, no self-explanatory variable names ... I won't attempt to debug it here in detail. One thing to note though is, LU decomposition only takes a matrix as an input. I'm not sure why you are trying to pass the 'B' matrix to it as an input. Plus it doesn't seem like you're using it anywhere in the function. Or creating L or U matrices for that matter. Also, if the whole top part is inside an lufact.m file, then you should know that your function terminates well before the for loops; these get ignored completely. What were you trying to do exactly?
Related
I know that MATLAB works better when most or everything is vectorized. I have two set of vectors X and T. For every vector x in X I want to compute:
this is because I want to compute:
which can be easily expressed as MATLAB linear algebra operations as I wrote above with a dot product. I am hoping that I can speed this up by having those vectors, instead of computing each f(x) with a for loop. Ideally I could have it all vectorized and compute:
I've been think about this for some time now, but it doesn't seem to be a a nice way were a function takes two vectors and computes the norm between each one of them, with out me having to explicitly write the for loop.
i.e. I've implemented the trivial code:
function [ f ] = f_start( x, c, t )
% Computes f^*(x) = sum_i c_i exp( - || x_i - t_i ||^2)
% Inputs:
% x = data point (D x 1)
% c = weights (K x 1)
% t = centers (D x K)
% Outputs:
% f = f^*(x) = sum_k c_k exp( - || x - t_k ||^2)
[~, K] = size(t);
f = 0;
for k=1:K
c_k = c(k);
t_k = t(:, k);
norm_squared = norm(x - t_k, 2)^2;
f = f + c_k * exp( -1 * norm_squared );
end
end
but I was hoping there was a less naive way to do this!
I think you want pdist2 (Statistics Toolbox):
X = [1 2 3;
4 5 6];
T = [1 2 3;
1 2 4;
7 8 9];
result = pdist2(X,T);
gives
result =
0 1.0000 10.3923
5.1962 4.6904 5.1962
Equivalently, if you don't have that toolbox, use bsxfun as follows:
result = squeeze(sqrt(sum(bsxfun(#minus, X, permute(T, [3 2 1])).^2, 2)));
Another method just for kicks
X = [1 2 3;
4 5 6].';
T = [1 2 3;
1 2 4;
7 8 9].';
tT = repmat(T,[1,size(X,2)]);
tX = reshape(repmat(X,[size(T,2),1]),size(tT));
res=reshape(sqrt(sum((tT-tX).^2)).',[size(T,2),size(X,2)]).'
I want to write this matrix in matlab,
s=[0 ..... 0
B 0 .... 0
AB B .... 0
. . .
. . .
. . . 0 ....
A^(n-1)*B ... AB B ]
I have tried this below code but giving error,
N = 50;
A=[2 3;4 1];
B=[3 ;2];
[nx,ny] = size(A);
s(nx,ny,N) = 0;
for n=1:1:N
s(:,:,n)=A.^n;
end
s_x=cat(3, eye(size(A)) ,s);
for ii=1:1:N-1
su(:,:,ii)=(A.^ii).*B ;
end
z= zeros(1,60,1);
su1 = [z;su] ;
s_u=repmat(su1,N);
seems like the concatenation of matrix is not being done.
I am a beginner so having serious troubles,please help.
Use cell arrays and the answer to your previous question
A = [2 3; 4 1];
B = [3 ;2 ];
N = 60;
[cs{1:(N+1),1:N}] = deal( zeros(size(B)) ); %// allocate space, setting top triangle to zero
%// work on diagonals
x = B;
for ii=2:(N+1)
[cs{ii:(N+2):((N+1)*(N+2-ii))}] = deal(x); %//deal to diagonal
x = A*x;
end
s = cell2mat(cs); %// convert cells to a single matrix
For more information you can read about deal and cell2mat.
Important note about the difference between matrix operations and element-wise operations
In your question (and in your previous one) you confuse between matrix power: A^2 and element-wise operation A.^2:
matrix power A^2 = [16 9;12 13] is the matrix product of A*A
element-wise power A.^2 takes each element separately and computes its square: A.^2 = [4 9; 16 1]
In yor question you ask about matrix product A*b, but the code you write is A.*b which is an element-by-element product. This gives you an error since the size of A and the size of b are not the same.
I often find that Matlab gives itself to a coding approach of "write what it says in the equation". That also leads to code that is easy to read...
A = [2 3; 4 1];
B = [3; 2];
Q = 4;
%// don't need to...
s = [];
%// ...but better to pre-allocate s for performance
s = zeros((Q+1)*2, Q);
X = B;
for m = 2:Q+1
for n = m:Q+1
s(n*2+(-1:0), n-m+1) = X;
end
X = A * X;
end
I am trying to use minimization to calculate coefficients of the polynomial p(x) = 1 + c(1)x + c(2)x^2 to approximate e^x. I need to use points xi = 1 + i/n for natural numbers i on [1,n], first for n=5, then n=10, etc. The approach is to minimize the 1, 2, and inf norm(p(x) - e^x) using fminsearch. So the output should be the 2 coefficients for the 3 p(x)'s. Any suggestions is appreciated.
Well, if anyone was wondering, I did figure the question out finally.
for l = [1 2 inf]
fprintf('For norm %d\n', l)
fprintf('Coefficients c1 c2\n')
for n = [5 10 100]
i = 1:n ;
x = 1 + i/n ;
c = [1 1] ;
%Difference function that we want to minimize
g = #(c) x.*c(1) + x.^2.*c(2) + 1 - exp(x);
f_norm = #(c) norm(g(c), l) ;
C = fminsearch(f_norm, c);
fprintf('n = %d ', n)
fprintf('%f %f\n', C(1), C(2))
% Compare plot of e^x and p(x).
p = #(x) C(1)*x + C(2)*x.^2 + 1;
xx = linspace(1,2,1e5);
figure;
plot(xx, p(xx), '--r', xx, exp(xx));
str = sprintf('Plot with n = %d, for norm %d', n,l);
title(str,'FontSize',24)
xlabel('x','FontSize',20)
ylabel('y','FontSize',20)
legend('p2 approximation','exponential');
end
end
This worked in the end to answer the question.
Consider a difference equation with its initial conditions.
5y(n) + y(n-1) - 3y(n-2) = (1/5^n) u(n), n>=0
y(n-1) = 2, y(n-2) = 0
How can I determine y(n) in Matlab?
Use an approach similar to this (using filter), but specifying initial conditions as done here (using filtic).
I'm assuming your initial conditions are: y(-1)=2, y(-2)=0.
num = 1; %// numerator of transfer function (from difference equation)
den = [5 1 -3]; %// denominator of transfer function (from difference equation)
n = 0:100; %// choose as desired
x = (1/5).^n; %// n is >= 0, so u(n) is 1
y = filter(num, den, x, filtic(num, den, [2 0], [0 0]));
%// [2 0] reflects initial conditions on y, and [0 0] those on x.
Here's a plot of the result, obtained with stem(n,y).
The second line of your code does not give initial conditions, because it refers to the index variable n. Since Matlab only allows positive integer indices, I'll assume that you mean y(1) = 0 and y(2) = 2.
You can get an iteration rule out of your first equation by simple algebra:
y(n) = ( (1/5^n) u(n) - y(n-1) + 3y(n-2) ) / 5
Code to apply this rule in Matlab:
n_max = 100;
y = nan(n_max, 1);
y(1) = 0;
y(2) = 2;
for n = 3 : n_max
y(n) = ( (1/5^n) * u(n) - y(n-1) + 3 * y(n-2) ) / 5;
end
This code assumes that the array u is already defined. n_max specifies how many elements of y to compute.
I need to evaluate following expression (in pseudo-math notation):
∑ipi⋅n
where p is a matrix of three-element vectors and n is a three-element vector. I can do this with for loops as follows but I can't figure out
how to vectorize this:
p = [1 1 1; 2 2 2];
n = [3 3 3];
s = 0;
for i = 1:size(p, 1)
s = s + dot(p(i, :), n)
end
Why complicate things? How about simple matrix multiplication:
s = sum(p * n(:))
where p is assumed to be an M-by-3 matrix.
I think you can do it with bsxfun:
sum(sum(bsxfun(#times,p,n)))
----------
% Is it the same for this case?
----------
n = 200; % depending on the computer it might be
m = 1000*n; % that n needs to be chosen differently
A = randn(n,m);
x = randn(n,1);
p = zeros(m,1);
q = zeros(1,m);
tic;
for i = 1:m
p(i) = sum(x.*A(:,i));
q(i) = sum(x.*A(:,i));
end
time = toc; disp(['time = ',num2str(time)]);