function [V,H] = Arnoldi(A,v,m)
[n,~] = size(A);
V = zeros(n,m+1);
H = zeros(n,n);
V(:,1) = v/norm(v);
for k = 2:m
V(:,k) = A*V(:,k-1);
for j = 1:(k-1);
H(j,k-1) = V(:,j).'*V(:,k);
V(:,k) = V(:,k)- H(j,k-1)*V(:,j);
end
H(k,k-1) = norm(V(:,k));
V(:,k) = V(:,k)/H(k,k-1);
end
end
This is my implementation of the Arnoldi algorithm. We already used wikipedia but did not find an answer there.
A is a square matrix, v is our starting vector and m is the number of iterations/ dimension of Krylov subspace. It does not give the wanted Hessian H and we can not figure out where we go wrong. Can anybody help us?
Related
I am trying to convert my code over to run with parfor, since as it is it takes a long time to run on its own. However I keep getting this error. I have search around on the website and have read people with similar problems, but none of those answers seem to fix my problem. This is my code:
r = 5;
Mu = 12.57e-9;
Nu = 12e6;
I = 1.8;
const = pi*Nu*Mu*r*I;
a = 55;
b = 69;
c = 206;
[m,n,p] = size(Lesion_Visible);
A = zeros(m,n,p);
parpool(2)
syms k
parfor J = 1:m
for I = 1:n
for K = 1:p
if Lesion_Visible(J,I,K) ~= 0
Theta = atand((J-b)/(I-a));
Rho = abs((I-a)/cosd(Theta))*0.05;
Z = abs(c-K)*0.05;
E = vpa(const*int(abs(besselj(0,Rho*k)*exp(-Z*k)*besselj(0,r*k)),0,20),5);
A (J,I,K) = E;
end
end
end
end
I'm trying to calculate the electric field in specific position on an array and matlab give me the error "The variable A in a parfor cannot be classified". I need help. Thanks.
As classification of variables in parfor loop is not permitted, you should try to save the output of each loop in a variable & then save the final output into the desired variable, A in your case!
This should do the job-
parfor J = 1:m
B=zeros(n,p); %create a padding matrix of two dimension
for I = 1:n
C=zeros(p); %create a padding matrix of one dimension
for K = 1:p
if Lesion_Visible(J,I,K) ~= 0
Theta = atand((J-b)./(I-a));
Rho = abs((I-a)./cosd(Theta))*0.05;
Z = abs(c-K).*0.05;
E = vpa(const.*int(abs(besselj(0,Rho.*k).*exp(-Z.*k).*besselj(0,r.*k)),0,20),5);
C(K) = E; %save output of innnermost loop to the padded matrix C
end
end
B(I,:)=C; % save the output to dim1 I of matrix B
end
A(J,:,:)=B; save the output to dim1 J of final matrix A
end
Go through the following for better understanding-
http://www.mathworks.com/help/distcomp/classification-of-variables-in-parfor-loops.html
http://in.mathworks.com/help/distcomp/sliced-variable.html
Similar problem to this question matrix that forms an orthogonal basis with a given vector but I'm having problems with my implementation. I have a given vector, u, and I want to construct an arbitrary orthonormal matrix V such that V(:,1) = u/norm(u). The problem I'm getting is that V'*V isn't I (but is diagonal). Here's my code:
function V = rand_orth(u)
n = length(u);
V = [u/norm(u) rand(n,n-1)];
for i = 1:n
vi = V(:,i);
rii = norm(vi);
qi = vi/rii;
for j = i+1:n
rij = dot(qi,V(:,j));
V(:,j) = V(:,j) - rij*qi;
end
end
end
Simply normalize columns of V once you've finished the orthogonalization. Note there's also no need to normalize u (first column) initially, since we will normalize the whole matrix at the end:
function V = rand_orth(u)
n = length(u);
V = [u, rand(n,n-1)];
for i = 1:n
vi = V(:,i);
rii = norm(vi);
qi = vi/rii;
for j = i+1:n
rij = dot(qi,V(:,j));
V(:,j) = V(:,j) - rij*qi;
end
end
V = bsxfun(#rdivide, V, sqrt(diag(V'*V))');
end
My script is supposed to run Runge-Kutta and then interpolate around the tops using polyfit to calculate the max values of the tops. I seem to get the x-values of the max points correct but the y-values are off for some reason. Have sat with it for 3 days now. The problem should be In the last for-loop when I calculate py?
Function:
function funk = FU(t,u)
L0 = 1;
C = 1*10^-6;
funk = [u(2); 2.*u(1).*u(2).^2./(1+u(1).^2) - u(1).*(1+u(1).^2)./(L0.*C)];
Program:
%Runge kutta
clear all
close all
clc
clf
%Given values
U0 = [240 1200 2400];
L0 = 1;
C = 1*10^-6;
T = 0.003;
h = 0.000001;
W = [];
% Runge-Kutta 4
for i = 1:3
u0 = [0;U0(i)];
u = u0;
U = u;
tt = 0:h:T;
for t=tt(1:end-1)
k1 = FU(t,u);
k2 = FU(t+0.5*h,u+0.5*h*k1);
k3 = FU((t+0.5*h),(u+0.5*h*k2));
k4 = FU((t+h),(u+k3*h));
u = u + (1/6)*(k1+2*k2+2*k3+k4)*h;
U = [U u];
end
W = [W;U];
end
I1 = W(1,:); I2 = W(3,:); I3 = W(5,:);
dI1 = W(2,:); dI2 = W(4,:); dI3 = W(6,:);
I = [I1; I2; I3];
dI = [dI1; dI2; dI3];
%Plot of the currents
figure (1)
plot(tt,I1,'r',tt,I2,'b',tt,I3,'g')
hold on
legend('U0 = 240','U0 = 1200','U0 = 2400')
BB = [];
d = 2;
px = [];
py = [];
format short
for l = 1:3
[H,Index(l)]=max(I(l,:));
Area=[(Index(l)-2:Index(l)+2)*h];
p = polyfit(Area,I(Index(l)-2:Index(l)+2),4);
rotp(1,:) = roots([4*p(1),3*p(2),2*p(3),p(4)]);
B = rotp(1,2);
BB = [BB B];
Imax(l,:)=p(1).*B.^4+p(2).*B.^3+p(3).*B.^2+p(4).*B+p(5);
Tsv(i)=4*rotp(1,l);
%px1 = linspace(h*(Index(l)-d-1),h*(Index(l)+d-2));
px1 = BB;
py1 = polyval(p,px1(1,l));
px = [px px1];
py = [py py1];
end
% Plots the max points
figure(1)
plot(px1(1),py(1),'b*-',px1(2),py(2),'b*-',px1(3),py(3),'b*-')
hold on
disp(Imax)
Your polyfit line should read:
p = polyfit(Area,I(l, Index(l)-2:Index(l)+2),4);
More interestingly, take note of the warnings you get about poor conditioning of that polynomial (I presume you're seeing these). Why? Partly because of numerical precision (your numbers are very small, scaled around 10^-6) and partly because you're asking for a 4th-order fit to five points (which is singular). To do this "better", use more input points (more than 5), or a lower-order polynomial fit (quadratic is usually plenty), and (probably) rescale before you use the polyfit tool.
Having said that, in practice this problem is often solved using three points and a quadratic fit, because it's computationally cheap and gives very nearly the same answers as more complex approaches, but you didn't get that from me (with noiseless data like this, it doesn't much matter anyway).
I am writing a graphical representation of numerical stability of differential operators and I am having trouble removing a nested for loop. The code loops through all entries in the X,Y, plane and calculates the stability value for each point. This is done by finding the roots of a polynomial of a size dependent on an input variable (length of input vector results in a polynomial 3d matrix of size(m,n,(lenght of input vector)). The main nested for loop is as follows.
for m = 1:length(z2)
for n = 1:length(z1)
pointpoly(1,:) = p(m,n,:);
r = roots(pointpoly);
if isempty(r),r=1e10;end
z(m,n) = max(abs(r));
end
end
The full code of an example numerical method (Trapezoidal Rule) is as follows. Any and all help is appreciated.
alpha = [-1 1];
beta = [.5 .5];
Wind = 2;
Wsize = 500;
if numel(Wind) == 1
Wind(4) = Wind(1);
Wind(3) = -Wind(1);
Wind(2) = Wind(4);
Wind(1) = Wind(3);
end
if numel(Wsize) == 1
Wsize(2) = Wsize;
end
z1 = linspace(Wind(1),Wind(2),Wsize(1));
z2 = linspace(Wind(3),Wind(4),Wsize(2));
[Z1,Z2] = meshgrid(z1,z2);
z = Z1+1i*Z2;
p = zeros(Wsize(2),Wsize(1),length(alpha));
for n = length(alpha):-1:1
p(:,:,(length(alpha)-n+1)) = alpha(n)-z*beta(n);
end
for m = 1:length(z2)
for n = 1:length(z1)
pointpoly(1,:) = p(m,n,:);
r = roots(pointpoly);
if isempty(r),r=1e10;end
z(m,n) = max(abs(r));
end
end
figure()
surf(Z1,Z2,z,'EdgeColor','None');
caxis([0 2])
cmap = jet(255);
cmap((127:129),:) = 0;
colormap(cmap)
view(2);
title(['Alpha Values (',num2str(alpha),') Beta Values (',num2str(beta),')'])
EDIT::
I was able to remove one of the for loops using the reshape command. So;
for m = 1:length(z2)
for n = 1:length(z1)
pointpoly(1,:) = p(m,n,:);
r = roots(pointpoly);
if isempty(r),r=1e10;end
z(m,n) = max(abs(r));
end
end
has now become
gg = reshape(p,[numel(p)/length(alpha) length(alpha)]);
r = zeros(numel(p)/length(alpha),1);
for n = 1:numel(p)/length(alpha)
temp = roots(gg(n,:));
if isempty(temp),temp = 0;end
r(n,1) = max(abs(temp));
end
z = reshape(r,[Wsize(2),Wsize(1)]);
This might be one for loop, but I am still going through the same number of elements. Is there a way to use the roots command on all of my rows at the same time?
I need to code the Gauss Seidel and Successive over relaxation iterative methods in Matlab. I have created the below code for each of them, however my final solution vector does not return the correct answers and i'm really struggling to figure out why. Could anyone please help me?
In both cases, x is the final solution vector and i returns the number of iterations.
Thanks in advance
Gauss Seidel Method:
function [x,i] = gaussSeidel(A,b,x0,tol)
x2 = x0;
count = 0;
D = diag(diag(A));
U = triu(A-D);
disp(U);
L = tril(A-D);
disp(L);
C = diag(diag(A));
disp(C);
Inv = inv(C+D);
error = inf;
while error>tol
x1 = x2;
x2 = Inv*(b-(U*x1));
error = max(abs(x2-x1)/abs(x1));
count = count + 1;
end
x = x2;
i = count;
end
SOR Method:
function [x,i] = sor(A,b,x0,tol,omega)
[m,n] = size(A);
D = diag(diag(A));
U = triu(A-D);
L = tril(A-D);
count = 1;
xtable = x0;
w = omega;
if size(b) ~= size(x0)
error('The given approximation vector does not match the x vector size');
elseif m~=n
error('The given coefficient matrix is not a square');
else
xnew = (inv(D+w*L))*(((1-w)*D-w*U)*x0 +w*b);
RelError = (abs(xnew-x0))/(abs(xnew));
RelErrorCol = max(max(RelError));
while RelErrorCol>tol
xnew = (inv(D+w*L))*(((1-w)*D-w*U)*x0 +w*b);
RelError = (abs(xnew-x0))/(abs(xnew));
RelErrorCol = max(max(RelError));
x0 = xnew;
count = count+1;
xtable = [xtable, xnew];
end
disp(xtable);
x = xnew;
i = count;
end
Gauss-Seidel: Your line that describes C is wrong. Actually it shouldn't be there. Also for the Inv line, it should be inv(D+L), not inv(C+D).
As for the SOR method, in hindsight it seems right. To double check, compare with this method:
http://www.netlib.org/templates/matlab/sor.m. This method relies on http://www.netlib.org/templates/matlab/split.m
Edit: April 4, 2014 - Also check: https://www.dropbox.com/s/p9wlzi9x9evqj5k/MTH719W2013_Assn4_Part1.pdf?dl=1 . I taught a course on Applied Linear Algebra and have MATLAB code that implements Gauss-Seidel and SOR. Check slides 12-20 for the theory and how to implement Gauss-Seidel and slides 35-37 for the SOR method.
Let me know how it goes.