I'm trying to evaluate the following type of problem with Matlab.
where the summation is taken over wn which is a vector, and the variable of integration is wPrime. w, wn and Ri are given vectors.
What I've tried is the following: using two for-loops.
integral = zeros(size(wn));
sumF = zeros(size(w));
for j = 1:length(w);
for k = 1:length(wn);
integral(k) = trapz(wPrime,1./(w(j)-1i*wn(k)-wPrime));
end
sumF(j) = sum(integral.*wn./sqrt(Ri));
end
but I've understood that for loops are very slow in Matlab, so I´m wondering if there is a better approach to this problem?
Related
I am working on an assignment that requires me to use the trapz function in MATLAB in order to evaluate an integral. I believe I have written the code correctly, but the program returns answers that are wildly incorrect. I am attempting to find the integral of e^(-x^2) from 0 to 1.
x = linspace(0,1,2000);
y = zeros(1,2000);
for iCnt = 1:2000
y(iCnt) = e.^(-(x(iCnt)^2));
end
a = trapz(y);
disp(a);
This code currently returns
1.4929e+03
What am I doing incorrectly?
You need to just specify also the x values:
x = linspace(0,1,2000);
y = exp(-x.^2);
a = trapz(x,y)
a =
0.7468
More details:
First of all, in MATLAB you can use vectors to avoid for-loops for performing operation on arrays (vectors). So the whole four lines of code
y = zeros(1,2000);
for iCnt = 1:2000
y(iCnt) = exp(-(x(iCnt)^2));
end
will be translated to one line:
y = exp(-x.^2)
You defined x = linspace(0,1,2000) it means that you need to calculate the integral of the given function in range [0 1]. So there is a mistake in the way you calculate y which returns it to be in range [1 2000] and that is why you got the big number as the result.
In addition, in MATLAB you should use exp there is not function as e in MATLAB.
Also, if you plot the function in the range, you will see that the result makes sense because the whole page has an area of 1x1.
I am completely new to Matlab. I am trying to simulate a Wiener and Poisson combined process.
Why do I get Subscripted assignment dimension mismatch?
I am trying to simulate
Z(t)=lambda*W^2(t)-N(t)
Where W is a wiener process and N is a poisson process.
The code I am using is below:
T=500
dt=1
K=T/dt
W(1)=0
lambda=3
t=0:dt:T
for k=1:K
r=randn
W(k+1)=W(k)+sqrt(dt)*r
N=poissrnd(lambda*dt,1,k)
Z(k)=lambda*W.^2-N
end
plot(t,Z)
It is true that some indexing is missing, but I think you would benefit from rewriting your code in a more 'Matlab way'. The following code is using the fact that Matlab basic variables are matrices, and compute the results in a vectorized way. Try to understand this kind of writing, as this is the way to exploit Matlab more efficiently, along with writing shorter and readable code:
T = 500;
dt = 1;
K = T/dt;
lambda = 3;
t = 1:dt:T;
sqdtr = sqrt(dt)*randn(K-1,1); % define sqrt(dt)*r as a vector
N = poissrnd(lambda*dt,K,1); % define N as a vector
W = cumsum([0; sqdtr],1); % cumulative sum instead of the loop
Z = lambda*W.^2-N; % summing the processes element-wiesly
plot(t,Z)
Example for a result:
you forget index
Z(k)=lambda*W.^2-N
it must be
Z(k)=lambda*W(k).^2-N(k)
I'm having trouble with implementing double integral in Matlab.
Unlike other double integrals, I need the result of the first (inside) integral to be an expression of the second variable, before going through the second (outside) integral, as it must be powered by k.
For example:
In the example above, I need the result of the inside integral to be expressed as 2y, so that I can calculate (2y)^k, before doing the second (outside) integral.
Does anyone know how to do this in Matlab?
I don't like doing things symbolically, because 99.9% of all problems don't have a closed form solution at all. For 99.9% of the problems that do have a closed-form solution, that solution is unwieldy and hardly useful at all. That may be because of my specific discipline, but I'm going to assume that your problem falls in one of those 99.9% sets, so I'll present the most obvious numerical way to do this.
And that is, integrate a function which calls integral itself:
function dbl_int()
f = #(x,y) 2.*x.*y + 1;
k = 1;
x_limits = [0 1];
y_limits = [1 2];
val = integral(#(y) integrand(f, y, k, x_limits), ...
y_limits(1), y_limits(2));
end
function val = integrand(f, y, k, x_limits)
val = zeros(size(y));
for ii = 1:numel(y)
val(ii) = integral(#(x) f(x,y(ii)), ...
x_limits(1), x_limits(2));
end
val = val.^k;
end
I'm trying to use the meshgrid "function" in Matlab to generate the value of the characteristic function phi of a bivariate normal distribution. Currently I am using two-for loops to do this.
for i = 1:M
for j = 1:M
U = [u1(i); u2(j)];
phi(i,j) = exp(-1i*mu*U-0.5*U'*SIGMA*U);
end
end
Here u1 and u2 are the two vectors spanning the space for which values phi can take. However, this double for-loop is very time consuming, and I would like to do this using a meshgrid approach.
U = meshgrid(u1,u2);
PHI = exp(-1i*mu*[U1(:) U2(:)]-0.5*[U1(:) U2(:)]'*SIGMA.*[U1(:) U2(:)]);
I am trying to do this using the code above. But this is definitely not working due to the different dimensions of mu and [U1(:) U2(:)]. Do anyone have any hints on how this can be done?
This should do the trick:
[U1, U2] = meshgrid(u1,u2);
U = [U1(:) U2(:)]';
PHI = reshape(exp(-1i*mu*U-0.5*sum(U.*(SIGMA*U),1)),M,M).';
PS: Using ndgrid seems to be preferable to meshgrid also in this situation, because we don't need to transpose afterwards.
[U1, U2] = ndgrid(u1,u2);
U = [U1(:) U2(:)]';
PHI = reshape(exp(-1i*mu*U-0.5*sum(U.*(SIGMA*U),1)),M,M);
Solved it by just writing out the matrix multiplications:
[U1,U2]= meshgrid(u1,u2);
PHI = exp(-1i.*(mu_adj(1)*U2+mu_adj(2)*U1)...
-0.5*(SIGMA(1,1)*U2.^2+(SIGMA(2,1)+SIGMA(1,2))*U1.*U2+SIGMA(2,2)*U1.^2));
We have an equation similar to the Fredholm integral equation of second kind.
To solve this equation we have been given an iterative solution that is guaranteed to converge for our specific equation. Now our only problem consists in implementing this iterative prodedure in MATLAB.
For now, the problematic part of our code looks like this:
function delta = delta(x,a,P,H,E,c,c0,w)
delt = #(x)delta_a(x,a,P,H,E,c0,w);
for i=1:500
delt = #(x)delt(x) - 1/E.*integral(#(xi)((c(1)-c(2)*delt(xi))*ms(xi,x,a,P,H,w)),0,a-0.001);
end
delta=delt;
end
delta_a is a function of x, and represent the initial value of the iteration. ms is a function of x and xi.
As you might see we want delt to depend on both x (before the integral) and xi (inside of the integral) in the iteration. Unfortunately this way of writing the code (with the function handle) does not give us a numerical value, as we wish. We can't either write delt as two different functions, one of x and one of xi, since xi is not defined (until integral defines it). So, how can we make sure that delt depends on xi inside of the integral, and still get a numerical value out of the iteration?
Do any of you have any suggestions to how we might solve this?
Using numerical integration
Explanation of the input parameters: x is a vector of numerical values, all the rest are constants. A problem with my code is that the input parameter x is not being used (I guess this means that x is being treated as a symbol).
It looks like you can do a nesting of anonymous functions in MATLAB:
f =
#(x)2*x
>> ff = #(x) f(f(x))
ff =
#(x)f(f(x))
>> ff(2)
ans =
8
>> f = ff;
>> f(2)
ans =
8
Also it is possible to rebind the pointers to the functions.
Thus, you can set up your iteration like
delta_old = #(x) delta_a(x)
for i=1:500
delta_new = #(x) delta_old(x) - integral(#(xi),delta_old(xi))
delta_old = delta_new
end
plus the inclusion of your parameters...
You may want to consider to solve a discretized version of your problem.
Let K be the matrix which discretizes your Fredholm kernel k(t,s), e.g.
K(i,j) = int_a^b K(x_i, s) l_j(s) ds
where l_j(s) is, for instance, the j-th lagrange interpolant associated to the interpolation nodes (x_i) = x_1,x_2,...,x_n.
Then, solving your Picard iterations is as simple as doing
phi_n+1 = f + K*phi_n
i.e.
for i = 1:N
phi = f + K*phi
end
where phi_n and f are the nodal values of phi and f on the (x_i).