I'm following a Numerical Methods course and I made a small MATLAB script to compute integrals using the trapezoidal method. However my script uses a FOR loop and my friend told me I'm doing something wrong if I use a FOR loop in Matlab. Is there a way to convert this script to a Matlab-friendly one?
%Number of points to use
N = 4;
%Integration interval
a = 0;
b = 0.5;
%Width of the integration segments
h = (b-a) / N;
F = exp(a);
for i = 1:N-1
F = F + 2*exp(a+i*h);
end
F = F + exp(b);
F = h/2*F
Vectorization is important speed and clarity, but so is using built-in functions whenever possible. Matlab has a built in function for trapezoidal numerical integration called trapz. Here is an example.
x = 0:.125:.5
y = exp(x)
F = trapz(x,y)
It is recommended to vectorize your code.
%Number of points to use
N = 4;
%Integration interval
a = 0;
b = 0.5;
%Width of the integration segments
h = (b-a) / N;
x = 1:1:N-1;
F = h/2*(exp(a) + sum(2*exp(a+x*h)) + exp(b));
However, I've read that Matlab is no longer slow at for loops.
Related
I am currently involved in a group project where we have to conduct portfolio selection and optimisation. The paper being referenced is given here: (specifically page 5 and 6, equations 7-10)
http://faculty.london.edu/avmiguel/DeMiguel-Nogales-OR.pdf
We are having trouble creating the optimisation problem using M-Portfolios, given below
min (wrt w,m) (1/T) * sum_(rho)*(w'*r_t - m) (Sorry I couldn't get the formatting to work)
s.t. w'e = 1 (just a condition saying that all weights add to 1)
So far, this is what we have attempted:
function optPortfolio = portfoliofminconM(returns,theta)
% Compute the inputs of the mean-variance model
mu = mean(returns)';
sigma = cov(returns);
% Inputs for the fmincon function
T = 120;
n = length(mu);
w = theta(1:n);
m = theta((n+1):(2*n));
c = 0.01*ones(1,n);
Aeq = ones(1,(2*n));
beq = 1;
lb = zeros(2,n);
ub = ones(2,n);
x0 = ones(n,2) / n; % Start with the equally-weighted portfolio
options = optimset('Algorithm', 'interior-point', ...
'MaxIter', 1E10, 'MaxFunEvals', 1E10);
% Nested function which is used as the objective function
function objValue = objfunction(w,m)
cRp = (w'*(returns - (ones(T,1)*m'))';
objValue = 0;
for i = 1:T
if abs(cRp(i)) <= c;
objValue = objValue + (((cRp(i))^2)/2);
else
objValue = objValue + (c*(abs(cRp(i))-(c/2)));
end
end
The problem starts at our definitions for theta being used as a vector of w and m. We don't know how to use fmincon with 2 variables in the objective function properly. In addition, the value of the objective function is conditional on another value (as shown in the paper) and this needs to be done over a rolling time window of 120 months for a total period of 264 months.(hence the for-loop and if-else)
If any more information is required, I will gladly provide it!
If you can additionally provide an example that deals with a similar problem, can you please link us to it.
Thank you in advance.
The way you minimize a function of two scalars with fmincon is to write your objective function as a function of a single, two-dimensional vector. For example, you would write f(x,y) = x.^2 + 2*x*y + y.^2 as f(x) = x(1)^2 + 2*x(1)*x(2) + x(2)^2.
More generally, you would write a function of two vectors as a function of a single, large vector. In your case, you could rewrite your objfunction or do a quick hack like:
objfunction_for_fmincon = #(x) objfunction(x(1:n), x(n+1:2*n));
I'm trying to solve a system of ordinary differential equations in MATLAB.
I have a simple equation:
dy = -k/M *x - c/M *y+ F/M.
This is defined in my ode function test2.m, dependant on the values X and t. I want to trig 'F' with a signal, generated by my custom function squaresignal.m. The output hereof, is the variable u, spanding from 0 to 1, as it is a smooth heaviside function. - Think square wave. The inputs in squaresignal.m, is t and f.
u=squaresignal(t,f)
These values are to be used inside my function test2, in order to enable or disable variable 'F' with the value u==1 (enable). Disable for all other values.
My ode function test2.m reads:
function dX = test2(t ,X, u)
x = X (1) ;
y = X (2) ;
M = 10;
k = 50;
c = 10;
F = 300;
if u == 1
F = F;
else
F = 0,
end
dx = y ;
dy = -k/M *x - c/M *y+ F/M ;
dX = [ dx dy ]';
end
And my runscript reads:
clc
clear all
tstart = 0;
tend = 10;
tsteps = 0.01;
tspan = [0 10];
t = [tstart:tsteps:tend];
f = 2;
u = squaresignal(t,f)
for ii = 1:length(u)
options=odeset('maxstep',tsteps,'outputfcn',#odeplot);
[t,X]=ode15s(#(t,X)test2(t,X,u(ii)),[tstart tend],[0 0],u);
end
figure (1);
plot(t,X(:,1))
figure (2);
plot(t,X(:,2));
However, the for-loop does not seem to do it's magic. I still only get F=0, instead of F=F, at times when u==1. And i know, that u is equal to one at some times, because the output of squaresignal.m is visible to me.
So the real question is this. How do i properly pass my variable u, to my function test2.m, and use it there to trig F? Is it possible that the squaresignal.m should be inside the odefunction test2.m instead?
Here's an example where I pass a variable coeff to the differential equation:
function [T,Q] = main()
t_span = [0 10];
q0 = [0.1; 0.2]; % initial state
ode_options = odeset(); % currently no options... You could add some here
coeff = 0.3; % The parameter we wish to pass to the differential eq.
[T,Q] = ode15s(#(t,q)diffeq(t,q,coeff),t_span,q0, ode_options);
end
function dq = diffeq(t,q,coeff)
% Preallocate vector dq
dq = zeros(length(q),1);
% Update dq:
dq(1) = q(2);
dq(2) = -coeff*sin(q(1));
end
EDIT:
Could this be the problem?
tstart = 0;
tend = 10;
tsteps = 0.01;
tspan = [0 10];
t = [tstart:tsteps:tend];
f = 2;
u = squaresignal(t,f)
Here you create a time vector t which has nothing to do with the time vector returned by the ODE solver! This means that at first we have t[2]=0.01 but once you ran your ODE solver, t[2] can be anything. So yes, if you want to load an external signal source depending on time, then you need to call your squaresignal.m from within the differential equation and pass the solver's current time t! Your code should look like this (note that I'm passing f now as an additional argument to the diffeq):
function dX = test2(t ,X, f)
x = X (1) ;
y = X (2) ;
M = 10;
k = 50;
c = 10;
F = 300;
u = squaresignal(t,f)
if u == 1
F = F;
else
F = 0,
end
dx = y ;
dy = -k/M *x - c/M *y+ F/M ;
dX = [ dx dy ]';
end
Note however that matlab's ODE solvers do not like at all what you're doing here. You are drastically (i.e. non-smoothly) changing the dynamics of your system. What you should do is to use one of the following:
a) events if you want to trigger some behaviour (like termination) depending on the integrated variable x or
b) If you want to trigger the behaviour based on the time t, you should segment your integration into different parts where the differential equation does not vary during one segment. You can then resume your integration by using the current state and time as x0 and t0 for the next run of ode15s. Of course this only works of you're external signal source u is something simple like a step funcion or square wave. In case of the square wave you would only integrate for a timespan during which the wave does not jump. And then exactly at the time of the jump you start another integration with altered differential equations.
So, currently I am trying to find the numerical solutions for the berger equation, $u_t+u∗u_x=0$. The numerical solution to this equation is:
$u^{n+1}_j=u_n^j−\frac{Δx}{Δt}u^n_j(u^n_j−u^n_{j−1})$
I wrote code on Matlab that computes this. As you can see, there is only one for-loop in the code. However, my u matrix extremely large, the algorithm becomes slow. Is there a way for me to use no loops at all. I was thinking about using the command cumsum, but I am not sure how to integrate that to my program. Please look at my code. Thanks.
dx = 0.9;
dt = 0.9;
tf = 10;
l = 10;
xstep = [-1:dx:l];
tstep = [0:dt:tf];
uinit = zeros(length(tstep),length(xstep));
%%Setting Initial and Boundary Conditions
bc.left = #(t) 2;
bc.right = #(t) -1;
ic = #(x) ...
2*(x<=0) ...
-1*(x>0);
uinit(1, :) = ic(xstep);
uinit(:, 1) = bc.left(tstep);
uinit(:, end) = bc.right(tstep);
%% Numerical method part
for c=1:(length(tstep)-1)
uinit(2:end -1, c+1) = uinit(2:end -1, c) - dx/dt*(uinit(2:end -1, c).*(uinit(2:end-1, c) - uinit(1:end-2, c)));
end
surf(xstep,tstep,uinit)
I've written a matlab m-file to draw a double integral as below. Does everybody can show me its equivalent in mathematica???
tetha = pi/4;
lamb = -1;
h = 4;
tetha0 = 0;
syms x y l
n = [h.*((cos(tetha)).^2)./sin(tetha); h.*abs(cos(tetha)); 0];
ft = ((tetha - pi/2)./sin(tetha)).^4;
Rt = [cos(tetha) -sin(tetha); sin(tetha) cos(tetha)];
zt = [cos(tetha0) -sin(tetha0); sin(tetha0) cos(tetha0)];
lt = [x;y];
integrand = #(x,y)(ft.*h.*((abs(cos(tetha)).* (x.*cos(tetha)-y.*sin(tetha)))-((cos(tetha)).^2/sin(tetha)).*(x.*sin(tetha)+y.*cos(tetha))));
PhiHat = #(a,b)(dblquad(integrand,0,a,0,b));
ezsurfc(PhiHat,[0,5,0,5])
Here you go (only minimal changes made), but you'll have to do your homework to understand function definitions, integration, plotting etc. in Mathematica. Also, this is not idiomatic Mathematica, but let's not go there...
tetha=Pi/4;
lamb=-1;
h=4;
tetha0=0;
n={h*((Cos[tetha])^2)/Sin[tetha],h*Abs[Cos[tetha]],0};
ft=((tetha-Pi/2)/Sin[tetha])^4;
Rt={{Cos[tetha], -Sin[tetha]}, {Sin[tetha], Cos[tetha]}};
zt={{Cos[tetha0], -Sin[tetha0]}, {Sin[tetha0], Cos[tetha0]}};
integrand[x_,y_]:= (ft*h*((Abs[Cos[tetha]]*(x*Cos[tetha]-y*Sin[tetha]))-((Cos[tetha])^2/Sin[tetha])*(x*Sin[tetha]+y*Cos[tetha])));
PhiHat[a_,b_]:=NIntegrate[integrand[x,y],{x,0,a},{y,0,b}];
Plot3D[PhiHat[x,y],{x,0,5},{y,0,5}]
If I'm to integrate a function
y = -((F+h)M^3(cosh(h*M)+M*beta*sinh(h*M)))/(h*M*cosh(h*M)+(-1+h*M^2*beta)*sinh(h*M))- (alpha*(M^2*(F+h)*(-1+2*h^2*M^2+ cosh(2*h*M)-2*h*M*sinh(2*h*M)))/(8*(h*M*cosh(h*M)+(-1+h*M^2*beta)*sinh(h*M))^2));
with respect to x, where
phi = 0.6;
x = 0.5;
M = 2;
theta = -1:0.5:1.5;
F = theta - 1;
h = 1 + phi*cos(2*pi*x);
alpha = 0.2;beta = 0.0;
I have written an Mfile
function r = parameterIntegrate(F,h,M,beta,alpha,theta,phi)
% defining a nested function that uses one variable
phi = 0.6;
x = 0.5;
r = quad(#testf,0,1 + phi*cos(2*pi*x));
% simpson's rule from 0 to h
function y = testf(x)
h = 1 + phi*cos(2*pi*x);
theta = -1:0.5:1.5;
F = theta - 1;
M = 2;
beta = 0;
alpha = 0;
y = -((F+h)*M^3*(cosh(h*M)+M*beta*sinh(h*M)))/(h*M*cosh(h*M)+(-1+h*M^2*beta)*sinh(h*M))- (alpha*(M^2*(F+h)*(-1+2*h^2*M^2+ cosh(2*h*M)-2*h*M*sinh(2*h*M)))/(8*(h*M*cosh(h*M)+(-1+h*M^2*beta)*sinh(h*M))^2));
end
end
and called the function by
tol = [1e-5 1e-3];
q = quad(#parameterIntegrate, 0, h,tol)
or
q = quad(#parameterIntegrate, 0,1 + phi*cos(2*pi*0.5),tol)
its not working its giving me
Error using ==> plus
Matrix dimensions must agree.
What your error message means is that for some line of code, there are 2 matrices, but the dimensions don't match, so it can't add them. What I suggest you do to solve this is as follows:
Figure out exactly which line of code is causing the problem.
If the line has large numbers of variables, simplify them some.
Remember that if there are any matrixs at all, and you don't want to do matrix multiplication/division, use the .*, ./, and .^.
I suspect that if you change your multiplies/divides with step 3, your problem will go away.