I have the following equation that I want to solve using MATLAB:
X is the unknown variable. I am trying to solve it using MATLAB solve, but I find it hard to code the left part of the equation.
Is it possible to use solve? Are there any other options?
EDIT
Since A and B depends respectively on j and i I have tried to put them in vectors as follows:
A = [A(1) ... A(j) ... A(N)]
B = [B(1) ... B(i) ... B(N)]
I was trying to have something that looks like this:
eqn = sum(A ./ sum(B .* D)) == C;
solve(eqn);
but the whole difficulty is in this part:
because it depends on both i and j.
To write equation you can use this code:
syms x real
C = 1;
beta = 10;
alph = 0.5;
N = 10;
lenA = N;
lenB = N;
A = rand(1,N);
B = rand(1,N);
eq = 0;
for j=2:N
eqaux = 0;
for i=1:N
eqaux = eqaux+B(i)/((alph+beta*x)^(i+j+1));
end
eq = eq+A(j)/eqaux;
end
eq = simplify(eq==C);
If x must be a complex number, delete real of syms x real.
To solve the equation use this code:
sol = solve(eq,x);
sol = vpa(sol);
Of course yu must use your own values of C, alph, beta, A, B and N.
Related
I'm having some issues getting my RK2 algorithm to work for a certain second-order linear differential equation. I have posted my current code (with the provided parameters) below. For some reason, the value of y1 deviates from the true value by a wider margin each iteration. Any input would be greatly appreciated. Thanks!
Code:
f = #(x,y1,y2) [y2; (1+y2)/x];
a = 1;
b = 2;
alpha = 0;
beta = 1;
n = 21;
h = (b-a)/(n-1);
yexact = #(x) 2*log(x)/log(2) - x +1;
ye = yexact((a:h:b)');
s = (beta - alpha)/(b - a);
y0 = [alpha;s];
[y1, y2] = RungeKuttaTwo2D(f, a, b, h, y0);
error = abs(ye - y1);
function [y1, y2] = RungeKuttaTwo2D(f, a, b, h, y0)
n = floor((b-a)/h);
y1 = zeros(n+1,1); y2 = y1;
y1(1) = y0(1); y2(1) = y0(2);
for i=1:n-1
ti = a+(i-1)*h;
fvalue1 = f(ti,y1(i),y2(i));
k1 = h*fvalue1;
fvalue2 = f(ti+h/2,y1(i)+k1(1)/2,y2(i)+k1(2)/2);
k2 = h*fvalue2;
y1(i+1) = y1(i) + k2(1);
y2(i+1) = y2(i) + k2(2);
end
end
Your exact solution is wrong. It is possible that your differential equation is missing a minus sign.
y2'=(1+y2)/x has as its solution y2(x)=C*x-1 and as y1'=y2 then y1(x)=0.5*C*x^2-x+D.
If the sign in the y2 equation were flipped, y2'=-(1+y2)/x, one would get y2(x)=C/x-1 with integral y1(x)=C*log(x)-x+D, which contains the given exact solution.
0=y1(1) = -1+D ==> D=1
1=y1(2) = C*log(2)-1 == C=1/log(2)
Additionally, the arrays in the integration loop have length n+1, so that the loop has to be from i=1 to n. Else the last element remains zero, which gives wrong residuals for the second boundary condition.
Correcting that and enlarging the computation to one secant step finds the correct solution for the discretization, as the ODE is linear. The error to the exact solution is bounded by 0.000285, which is reasonable for a second order method with step size 0.05.
As you probably guessed from the title, I'm attempting to do tridiagonal GaussJordan elimination. I'm trying to do it without the default solver. My answers aren't coming out correct and I need some assistance as to where the error is in my code.
I'm getting different values for A/b and x, using the code I have.
n = 4;
#Range for diagonals
ranged = [15 20];
rangesd = [1 5];
#Vectors for tridiagonal matrix
supd = randi(rangesd,[1,n-1]);
d = randi(ranged,[1,n]);
subd = randi(rangesd,[1,n-1]);
#Creates system Ax+b
A = diag(supd,1) + diag(d,0) + diag(subd,-1)
b = randi(10,[1,n])
#Uses default solver
y = A/b
function x = naive_gauss(A,b);
#Forward elimination
for k=1:n-1
for i=k+1:n
xmult = A(i,k)/A(k,k);
for j=k+1:n
A(i,j) = A(i,j)-xmult*A(k,j);
end
b(i) = b(i)-xmult*b(k);
end
end
#Backwards elimination
x(n) = b(n)/A(n,n);
for i=n-1:-1:1
sum = b(i);
for j=i+1:n
sum = sum-A(i,j)*x(j);
end
x(i) = sum/A(i,i)
end
end
x
Your algorithm is correct. The value of y that you compare against is wrong.
you have y=A/b, but the correct syntax to get the solution of the system should be y=A\b.
I need to create a symbolic matrix in MATLAB. It can be done statically as
syms a11 a12 a21 a22;
A = [a11 a12; a21 a22];
or using compact syntax as
A = sym('A%d', [2 2]);
However i don't see how any of these syntax's should allow dynamic initialization. I would like to initialize each matrix element individually using two loops. One shot at a possible syntax (it's NOT valid MATLAB syntax) could be
syms a x1;
P = sym(zeros(n,n));
for l = 1:n
for m = 1:n
kl = l; km = m;
P(l,m)(x1) = a*sin(kl*x1)*sin(km*x1);
end
end
where I used the (invalid) syntax P(l,m)(x1) to specify that each element in P is a function of x1. Hence P(2) would be an (n,n) matrix with elements P(l,m) = a*sin(kl*2)*sin(km*2). I know it's possible to construct the P by building the matrix string (A=[...]) on run time and evaluating using eval. Something like
syms a x1;
command = [];
for i = 1:n
for j = 1:n
kl = i; km = j;
command = [command ' + a*sin(' num2str(kl) '*x1)*sin(' num2str(km) '*x1)'];
if(j < n) command = [command ',']; end
end
if(i < n) command = [command ';']; end
end
eval(['P(x1) = [' command '];'])
However, using eval is bad practice, so I see this solution only as a last resort. Is there any correct way to achieve my goal?
NB: I have written the elements P(l,m) = a*sin(kl*x1)*sin(km*x1) to simplify the code. The actual expression is P(l,m) = sin(kl*x1)*sin(km*x1)*epsilon + kl*km*cos(kl*x1)*cos(km*x1).*b + km*cos(km*x1)*sin(kl*x1)-kl*cos(kl*x1)*sin(km*x1)/(kl^2-km^2).
If you're just trying to avoid the for loops, you can use meshgrid, which hides them from you:
syms a x1
n = 3;
x = meshgrid(1:n)*x1; % multiplying by the symbolic x1 makes x symbolic
P = a*sin(x).*sin(x.');
which returns
P =
[ a*sin(x1)^2, a*sin(2*x1)*sin(x1), a*sin(3*x1)*sin(x1)]
[ a*sin(2*x1)*sin(x1), a*sin(2*x1)^2, a*sin(2*x1)*sin(3*x1)]
[ a*sin(3*x1)*sin(x1), a*sin(2*x1)*sin(3*x1), a*sin(3*x1)^2]
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.
I write a code for linear feedback shift register.My code is in below:
X=5712;
D1(1)=0;
D2(1)=0;
D3(1)=0;
D4(1)=0;
D5(1)=0;
D6(1)=1;
for i=1:X-1
D6(i+1)=D1(i);
D5(i+1)=xor(D1(i),D6(i));
D4(i+1)=D5(i);
D3(i+1)=D4(i);
D2(i+1)=D3(i);
D1(i+1)=D2(i);
end
In my code i can only use 6 shift register.I know for degree,n=2,3,4,6,7,15,22, the polynomial is x^n+x+1.As the polynomial is same for those degrees so i want to write a common code for all.
Matlab experts Please need your help.
Your problem is that you are making separate vectors for each register. Rather make a single matrix (i.e. D replaces all of your D1, D2, ..., Dn) so that you can loop:
X = 20;
n = 6;
D = zeros(X, n);
D(1,n) = 1;
for ii = 1:X-1
D(ii+1, 1:n-2) = D(ii, 2:n-1);
D(ii+1, n-1) = xor(D(ii,1), D(ii,n));
D(ii+1, n) = D(ii, 1);
end
E = D(:, end:-1:1)