I need N variables for my equation system:
X = cell(N,1)
for k=1:N
X(k) = {sym('X(k)')};
end
After creating these variables I want to utilize them in an equation system:
for i=1:N
for j=1:N
if i~=j
S(i)=sum(X(j))
end
end
f(i)=x(i)+2*S(i)+3
end
I get the error Undefined function 'sum' for input arguments type 'cell'. How should I define the variables X(1),...X(N) without using 'cell'?
According to sym documentation you can use eg. A = sym('A%d%d', [2 2]); to create a symbolic matrix.
Is this what you mean?
N = 5;
% Initialize symbolic matrices with proper size
X = sym('x%d', [N 1]);
S = sym(zeros(N, 1));
f = sym(zeros(N, 1));
for i=1:N
for j=1:N
if i~=j
S(i) = S(i) + X(j);
end
end
f(i)=X(i)+2*S(i)+3;
end
You can convert them to a matrix using the cell2mat function.
http://www.mathworks.com/help/matlab/ref/cell2mat.html
Related
so what im trying to do is to refer to different symbolics by indexing as you would with a vector
syms k_1 k_2 k_3
for i= 1:3
'expression to get k_i'
end
Desired answer would be k1 k2 k3.
I do not intend to do this with a vector created by the sym function unless there's a way to assign a matrix to a vector position without getting this error
.
You can add them all to a struct, then use dynamic referencing to the ith field name:
% Create symbolic vars
syms k_1 k_2 k_3
% Assign to struct
K.k_1 = k_1;
K.k_2 = k_2;
K.k_3 = k_3;
% Loop through
for ii = 1:3
kstr = sprintf('k_%d', ii); % = 'k_1', 'k_2', ...
ki = K.(kstr); % equivalent to ki = K.k_1 etc
end
Equivalently, once you have the struct you can just use the field names, although it's a bit less explicit which field you're accessing each iteration.
f = fieldnames(K);
for ii = 1:numel(f)
ki = K.(f{ii});
end
Similarly to #Wolfie's answer, you can use a cell array:
syms k_1 k_2 k_3
k = {k_1, k_2, k_3};
for ii = 1:3
ki = k{ii};
end
This is somewhat simpler than the struct solution, but doesn't preserve the variable names. In this case, k{1} is similar enough to k_1, so maybe this doesn't matter.
I need to model negative, positive and simple regulation of a gene for my systems biology class using MATLAB. The problem is that the functions for negative and simple regulation work but the positive regulation function is only outputting zeros.
My script is as follows:
% Simulation of simple regulation, negative autoregulation and positive
% autoregulation
% Define constants
global a b K n
a = 1;
b = 1;
K = 0.5;
n = 2; % Hill coefficient
% Simulation time
tspan = [0,10];
% Initial condition
X0 = 0;
% Run simulations
[t1,X1] = ode45(#autoregulation_f0,tspan,X0); % Simple regulation
[t2,X2] = ode45(#autoregulation_f1,tspan,X0); % Negative autoregulation
[t3,X3] = ode23(#autoregulation_f2,tspan,X0); % Positive autoregulation
% Plot results
figure;
plot(t1,X1,t2,X2,t3,X3);
legend('simple','negative','Location','southeast');
And my functions are:
function dxdt = autoregulation_f0(t,X)
global a b
dxdt = b - a*X;
end
function dxdt = autoregulation_f1(t,X)
global a b K n
dxdt = b/(1+(X^n)/(K^n)) - a*X;
end
function dxdt = autoregulation_f2(t,X)
global a b K n
dxdt = b*X.^n./(K.^n+X.^n) + a*X;
end
The third function "autoregulation_f2(t,X)" is the one that outputs zeros and therefore when plotting the graph I just get a straight line.
Does anyone know what could be causing this?
Thanks in advance!
It looks to be the correct result for the given function. Your provided dxdt has an X in every term. The initial X0=0 will result in dxdt=0, giving you no change in X. As a result you just end up with a flat line.
I tried writing an Ordinary Differential Equation in MATLAB.
I wrote this code:
function [y] = odefun(t,y)
t = [0:0.01:10];
y = [0 0]';
y(1) = y(2);
y(2) = sin(2*t)-2*y(2)-2*y(1); % I get an error here
end
I get an error in the last line in this code. MATLAB doesn't show me what the error is. It just tells me I have an error in that line.
Why do I get this error and how to resolve it?
What you want is to carefully read the documentation of the diverse ode solvers and the examples there and then correct your code to something like
% Solve ODE y''(t)+2*y'(t)+2*y(t) = sin(2*t), y(0)=y'(0)=0
function ydot = odefun(t,y)
ydot = zeros_like(y)
ydot(1) = y(2);
ydot(2) = sin(2*t)-2*y(2)-2*y(1);
end
% or
% odefun = #(y,t) [ y(2); sin(2*t)-2*y(2)-2*y(1) ]
% define sample points
tspan = [0:0.01:10];
% define initial value to t=tspan(1)
y0 = [0 0]';
[ t, y ] = ode45(odefunc, tspan, y0)
% t,y now contain the times and values that
% the solution was actually computed for.
You try to assign to y(2) a vector of 1001 elements:
>> size(sin(2*t)-2*y(2)-2*y(1))
ans =
1 1001
The error message is pretty clear:
In an assignment A(:) = B, the number of elements in A and B must be the same.
Also, y and t are never use since you redefined them in the function.
I'm writing a function of the Gauss Seidel method of solving a linear system of equations of the form Ax=b, x being the unknown we are looking for.
I am having a problem with the while loop in my function, it seems that it runs infinitely. I can't seem to figure out why.
This is my function for creating the coefficient matrix A and the column vectors x and b, all with the same number of rows of course. No problem with this one.
function [A, b, x0] = test_system(n)
u = ones(n, 1);
A = spdiags([u 4*u u], [-1 0 1], n, n);
b = zeros(n, 1);
b(1) = 3;
b(2 : 2 : end-2) = -2;
b(3 : 2 : end-1) = 2;
b(end) = -3;
x0 = ones(n, 1);
This is my function for solving the system. I have included all of it just in case, but I believe the real problem is within the while loop at the very end which runs infinitely when I execute the function. The counter doesn't break away from it either. I can't really see what its problem is. Any clues?
Be gentle, I'm new at Matlab :)
function [x] = GaussSeidel(A,b,x0,tol)
% implementation of the GaussSeidel iterative method
% for solving a linear system of equations Ax = b
%INPUTS:
% A: coefficient matrix
% b: column vector of constants
% x0: setup for the unknown vector (using vector of ones)
% tol: result must be within 'tol' of correct answer.
%OUTPUTS:
% x: unknown
%check that A is a matrix
if ~(ismatrix(A))
error('A is not a matrix');
end
%check that A is square
[m,n] = size(A);
if m ~= n
error('Matrix A is not square');
end
%check that b is a column vector
if ~(iscolumn(b))
error('b is not a column vector');
end
%check that x0 is a column vector
if ~(iscolumn(x0))
error('x0 is not a column vector');
end
%check that A, b and x0 agree in size
[rowA,colA] = size(A);
[rowb,colb] = size(b);
[rowx0,colx0] = size(x0);
if ~isequal(colA,rowb)||~isequal(rowb,rowx0)
error('matrix dimensions of A, b and xo do not agree');
end
%check that A and b have real entries
if ~isreal(A) || ~isreal(b)
error('matrix A or vector b do not have real entries');
end
%check that the provided tolerance is positive
if tol <= 0
error('tolerance must be positive');
end
%check that A is strictly diagonally dominant
absoluteA = abs(A);
row_sum=sum(absoluteA,2);
diagonal=diag(absoluteA);
if ~all(2*diagonal > row_sum)
warning('matrix A is not strictly diagonally dominant');
end
L = tril(A,-1);
U = triu(A,+1);
D = diag(diag(A));
x = x0;
M1 = inv(D).*L;
M2 = inv(D).*U;
M3 = D\b;
k = 0; %iterations counter
disp(size(M1));
disp(size(M2));
disp(size(M3));
disp(size(x));
while (norm(A*x - b) > tol)
for i=1:n
x(i) = - M1(i,:).*x - M2(i,:).*x + M3(i,:);
end
k=k+1;
if(k >= 10e4)
error('too many iterations carried out');
end
end
end %end function
Coding Discussion
The source of the coding error is the use of element-wise operations instead of matrix-matrix and matrix-vector operations.
These operations produce zero-matrices, so no progress is made.
The M matrices should be defined by
M1 = inv(D)*L; % Note that D\L is more efficient
M2 = inv(D)*U; % Note that D\U is more efficient
M3 = D\b;
and the iterator performing the update should be
x(i) = - M1(i,:)*x - M2(i,:)*x + M3(i,:);
Method Discussion
I also think it is worth mentioning that the code is currently implementing the Jacobi Method since the updater has the form
while a Gauss-Seidel update has the form
.
I don’t have 50 reputation so I can not comment this.
The line if(k >= 10e4), I think this does not make what you think it would. 10e4 is 100,000 and 1e4 is 10,000. This will be the reason why you think your counter does not work. Matlab ist still running because it’s running further than you think it will. Also I run in the same Problem knedlsepp already pointed out.
I am trying to write a function that implements Newton's method in two dimensions and whilst I have done this, I have to now adjust my script so that the input parameters of my function must be f(x) in a column vector, the Jacobian matrix of f(x), the initial guess x0 and the tolerance where the function f(x) and its Jacobian matrix are in separate .m files.
As an example of a script I wrote that implements Newton's method, I have:
n=0; %initialize iteration counter
eps=1; %initialize error
x=[1;1]; %set starting value
%Computation loop
while eps>1e-10&n<100
g=[x(1)^2+x(2)^3-1;x(1)^4-x(2)^4+x(1)*x(2)]; %g(x)
eps=abs(g(1))+abs(g(2)); %error
Jg=[2*x(1),3*x(2)^2;4*x(1)^3+x(2),-4*x(2)^3+x(1)]; %Jacobian
y=x-Jg\g; %iterate
x=y; %update x
n=n+1; %counter+1
end
n,x,eps %display end values
So with this script, I had implemented the function and the Jacobian matrix into the actual script and I am struggling to work out how I can actually create a script with the input parameters required.
Thanks!
If you don't mind, I'd like to restructure your code so that it is more dynamic and more user friendly to read.
Let's start with some preliminaries. If you want to make your script truly dynamic, then I would recommend that you use the Symbolic Math Toolbox. This way, you can use MATLAB to tackle derivatives of functions for you. You first need to use the syms command, followed by any variable you want. This tells MATLAB that you are now going to treat this variable as "symbolic" (i.e. not a constant). Let's start with some basics:
syms x;
y = 2*x^2 + 6*x + 3;
dy = diff(y); % Derivative with respect to x. Should give 4*x + 6;
out = subs(y, 3); % The subs command will substitute all x's in y with the value 3
% This should give 2*(3^2) + 6*3 + 3 = 39
Because this is 2D, we're going to need 2D functions... so let's define x and y as variables. The way you call the subs command will be slightly different:
syms x, y; % Two variables now
z = 2*x*y^2 + 6*y + x;
dzx = diff(z, 'x'); % Differentiate with respect to x - Should give 2*y^2 + 1
dzy = diff(z, 'y'); % Differentiate with respect to y - Should give 4*x*y + 6
out = subs(z, {x, y}, [2, 3]); % For z, with variables x,y, substitute x = 2, y = 3
% Should give 56
One more thing... we can place equations into vectors or matrices and use subs to simultaneously substitute all values of x and y into each equation.
syms x, y;
z1 = 3*x + 6*y + 3;
z2 = 3*y + 4*y + 4;
f = [z1; z2];
out = subs(f, {x,y}, [2, 3]); % Produces a 2 x 1 vector with [27; 25]
We can do the same thing for matrices, but for brevity I won't show you how to do that. I will defer to the code and you can see it then.
Now that we have that established, let's tackle your code one piece at a time to truly make this dynamic. Your function requires the initial guess x0, the function f(x) as a column vector, the Jacobian matrix as a 2 x 2 matrix and the tolerance tol.
Before you run your script, you will need to generate your parameters:
syms x y; % Make x,y symbolic
f1 = x^2 + y^3 - 1; % Make your two equations (from your example)
f2 = x^4 - y^4 + x*y;
f = [f1; f2]; % f(x) vector
% Jacobian matrix
J = [diff(f1, 'x') diff(f1, 'y'); diff(f2, 'x') diff(f2, 'y')];
% Initial vector
x0 = [1; 1];
% Tolerance:
tol = 1e-10;
Now, make your script into a function:
% To run in MATLAB, do:
% [n, xout, tol] = Jacobian2D(f, J, x0, tol);
% disp('n = '); disp(n); disp('x = '); disp(xout); disp('tol = '); disp(tol);
function [n, xout, tol] = Jacobian2D(f, J, x0, tol)
% Just to be sure...
syms x, y;
% Initialize error
ep = 1; % Note: eps is a reserved keyword in MATLAB
% Initialize counter
n = 0;
% For the beginning of the loop
% Must transpose into a row vector as this is required by subs
xout = x0';
% Computation loop
while ep > tol && n < 100
g = subs(f, {x,y}, xout); %g(x)
ep = abs(g(1)) + abs(g(2)); %error
Jg = subs(J, {x,y}, xout); %Jacobian
yout = xout - Jg\g; %iterate
xout = yout; %update x
n = n + 1; %counter+1
end
% Transpose and convert back to number representation
xout = double(xout');
I should probably tell you that when you're doing computation using the Symbolic Math Toolbox, the data type of the numbers as you're calculating them are a sym object. You probably want to convert these back into real numbers and so you can use double to cast them back. However, if you leave them in the sym format, it displays your numbers as neat fractions if that's what you're looking for. Cast to double if you want the decimal point representation.
Now when you run this function, it should give you what you're looking for. I have not tested this code, but I'm pretty sure this will work.
Happy to answer any more questions you may have. Hope this helps.
Cheers!