I am having some trouble with implementing the following equation in Matlab:
The trouble is with using numerical/symbolic variables/implementation.
Can someone please write down the code to help me. Implementation would be great.
The constants for the equation are:
m=1; rho=0.5; H=1; I=1877; sigma=20;
For example if N=2, then:
for n=1, An=0.257, Zn-1=inf, Zn=0.4146;
for n=2, An=1, Zn-1=0.4146, Zn=0.1066;
Thanks for the help.
In numerical methods all infinite quantities are approximated by finite ones. Therefore, you have to study this expression for convergence (analytically or by means of numerical experiment). Once you know the radius of convergence you know which number to pick up to approximate the sign $\infinity$ in sums and integrals. The numerical evaluation of integrals is a large subject itself (you may read any book on numerical methods or specifically on numerical quadratures https://en.wikipedia.org/wiki/Numerical_integration). The simplest numerical approximation of an integral is based on the rectangular rule for a regular uniform grid:
$$\int_a^b f(x) dx=\sum_j f(x_j) \Delta x$$, where $\Delta x = x_{j+1}-x_j$
Related
I am trying to solve two equations with complex coefficients using ode45.
But iam getting an error message as "Inputs must be floats, namely single or
double."
X = sym(['[',sprintf('X(%d) ',1:2),']']);
Eqns=[-(X(1)*23788605396486326904946699391889*1i)/38685626227668133590597632 + (X(2)*23788605396486326904946699391889*1i)/38685626227668133590597632; (X(2)*23788605396486326904946699391889*1i)/38685626227668133590597632 + X(1)*(- 2500000 + (5223289665997855453060886952725538686654593059791*1i)/324518553658426726783156020576256)] ;
f=#(t,X)[Eqns];
[t,Xabc]=ode45(f,[0 300*10^-6],[0 1])
How can i fix this ? Can somebody can help me ?
Per the MathWorks Support Team, the "ODE solvers in MATLAB 5 (R12) and later releases properly handle complex valued systems." So the complex numbers are the not the issue.
The error "Inputs must be floats, namely single or double." stems from your definition of f using Symbolic Variables that are, unlike complex numbers, not floats. The easiest way to get around this is to not use the Symbolic Toolbox at all; just makes Eqns an anonymous function:
Eqns= #(t,X) [-(X(1)*23788605396486326904946699391889*1i)/38685626227668133590597632 + (X(2)*23788605396486326904946699391889*1i)/38685626227668133590597632; (X(2)*23788605396486326904946699391889*1i)/38685626227668133590597632 + X(1)*(- 2500000 + (5223289665997855453060886952725538686654593059791*1i)/324518553658426726783156020576256)] ;
[t,Xabc]=ode45(Eqns,[0 300*10^-6],[0 1]);
That being said, I'd like to point out that numerically time integrating this system over 300 microseconds (I assume without units given) will take a long time since your coefficient matrix has imaginary eigenvalues on the order of 10E+10. The extremely short wavelength of those oscillations will more than likely be resolved by Matlab's adaptive methods, and that will take a while to solve for a time span just a few orders greater than the wavelength.
I'd, therefore, suggest an analytical approach to this problem; unless it is a stepping stone another problem that is non-analytically solvable.
Systems of ordinary differential equations of the form
,
which is a linear, homogenous system with a constant coefficient matrix, has the general solution
,
where the m-subscripted exponential function is the matrix exponential.
Therefore, the analytical solution to the system can be calculated exactly assuming the matrix exponential can be calculated.
In Matlab, the matrix exponential is calculate via the expm function.
The following code computes the analytical solution and compares it to the numerical one for a short time span:
% Set-up
A = [-23788605396486326904946699391889i/38685626227668133590597632,23788605396486326904946699391889i/38685626227668133590597632;...
-2500000+5223289665997855453060886952725538686654593059791i/324518553658426726783156020576256,23788605396486326904946699391889i/38685626227668133590597632];
Eqns = #(t,X) A*X;
X0 = [0;1];
% Numerical
options = odeset('RelTol',1E-8,'AbsTol',1E-8);
[t,Xabc]=ode45(Eqns,[0 1E-9],X0,options);
% Analytical
Xana = cell2mat(arrayfun(#(tk) expm(A*tk)*X0,t,'UniformOutput',false)')';
k = 1;
% Plots
figure(1);
subplot(3,1,1)
plot(t,abs(Xana(:,k)),t,abs(Xabc(:,k)),'--');
title('Magnitude');
subplot(3,1,2)
plot(t,real(Xana(:,k)),t,real(Xabc(:,k)),'--');
title('Real');
ylabel('Values');
subplot(3,1,3)
plot(t,imag(Xana(:,k)),t,imag(Xabc(:,k)),'--');
title('Imaginary');
xlabel('Time');
The comparison plot is:
The output of ode45 matches the magnitude and real parts of the solution very well, but the imaginary portion is out-of-phase by exactly π.
However, since ode45's error estimator only looks at norms, the phase difference is not noticed which may lead to problems depending on the application.
It will be noted that while the matrix exponential solution is far more costly than ode45 for the same number of time vector elements, the analytical solution will produce the exact solution for any time vector of any density given to it. So for long time solutions, the matrix exponential can be viewed as an improvement in some sense.
When solving the log likelihood expression for autoregressive models, I cam across the variance covariance matrix Tau given under slide 9 Parameter estimation of time series tutorial. Now, in order to use
fminsearch
to maximize the likelihood function expression, I need to express the likelihood function where the variance covariance matrix arises. Can somebody please show with an example how I can implement (determinant of Gamma)^-1/2 ? Any other example apart from autoregressive model will also do.
How about sqrt(det(Gamma)) for the sqrt-determinant and inv(Gamma) for inverse?
But if you do not want to implement it yourself you can look at yulewalkerarestimator
UPD: For estimation of autocovariance matrix use xcov
also, this topic is a bit more explained here
I don't know how to set the intevals of a integral to get the best precise result.
For example, this is the orginal definition of the formula.
y=integral(#(x) log2((f1(x))./(f2(x))), -inf, inf).
Note: f1(x)->0 and f2(x)->0 when x->-inf or inf, and the decreasing speeds are different.
If I use [-inf, inf] Matlab gives me NaN.
If I narrow down the inteval, Matlab gives a number. But if I increas the inteval a little bit, I get another number. So I am wondering how to deal this kind of integral calculation? How to make it as precise as possible without NaN?
Thanks a lot.
I don't think your integral converges for the definitions you have given. For example, for N=1 the integrand simplifies to (1/2 - 2*x)/log(2), which is clearly nonconverging at infinity. For larger N the integrand goes to -inf for x->inf and to inf for x->-inf, and I don't think the integral converges either, though I do not have a full proof at the moment.
It is good practice to examine mathematical functions analytically before running numerical analysis. If this is not possible, then try first plotting the function itself over the relevant range to get an idea of its behavior. A good way to plot functions over many orders of magnitude is by using the logspace function for x values.
I'm not too familiar with MATLAB or computational mathematics so I was wondering how I might solve an equation involving the sum of squares, where each term involves two vectors- one known and one unknown. This formula is supposed to represent the error and I need to minimize the error. I think I'm supposed to use least squares but I don't know too much about it and I'm wondering what function is best for doing that and what arguments would represent my equation. My teacher also mentioned something about taking derivatives and he formed a matrix using derivatives which confused me even more- am I required to take derivatives?
The problem that you must be trying to solve is
Min u'u = min \sum_i u_i^2, u=y-Xbeta, where u is the error, y is the vector of dependent variables you are trying to explain, X is a matrix of independent variables and beta is the vector you want to estimate.
Since sum u_i^2 is diferentiable (and convex), you can evaluate the minimal of this expression calculating its derivative and making it equal to zero.
If you do that, you find that beta=inv(X'X)X'y. This maybe calculated using the matlab function regress http://www.mathworks.com/help/stats/regress.html or writing this formula in Matlab. However, you should be careful how to evaluate the inverse (X'X) see Most efficient matrix inversion in MATLAB
This question has already confused me several days. While I referred to senior students, they also cannot give a reply.
We have ten ODEs, into which each a noise term should be added. The noise is defined as follows. since I always find that I cannot upload a picture, the formula below maybe not very clear. In order to understand, you can either read my explanation or go the this address: Plos one. You could find the description of the equations directly above the Support Information in this address
The white noise term epislon_i(t) is assumed with Gaussian distribution. epislon_i(t) means that for equation i, and at t timepoint, the value of the noise.
the auto-correlation of noise are given:
(EQ.1)
where delta(t) is the Dirac delta function and the diffusion matrix D is defined by
(EQ.2)
Our problem focuses on how to explain the Dirac delta function in the diffusion matrix. Since the property of Dirac delta function is delta(0) = Inf and delta(t) = 0 if t neq 0, we don't know how to calculate the epislonif we try to sqrt of 2D(x, t)delta(t-t'). So we simply assume that delta(0) = 1 and delta(t) = 0 if t neq 0; But we don't know whether or not this is right. Could you please tell me how to use Delta function of diffusion equation in MATLAB?
This question associates with the stochastic process in MATLAB. So we review different stochastic process to inspire our ideas. In MATLAB, the Wienner process is often defined as a = sqrt(dt) * rand(1, N). N is the number of steps, dt is the length of the steps. Correspondingly, the Brownian motion can be defined as: b = cumsum(a); All of these associate with stochastic process. However, they doesn't related to the white noise process which has a constraints on the matrix of auto-correlation, noted by D.
Then we consider that, we may simply use randn(1, 10) to generate a vector representing the noise. However, since the definition of the noise must satisfy the equation (2), this cannot enable noise term in different equation have the predefined partial correlation (D_ij). Then we try to use mvnrnd to generate a multiple variable normal distribution at each time step. Unfortunately, the function mvnrnd in MATLAB return a matrix. But we need to return a vector of length 10.
We are rather confused, so could you please give me just a light? Thanks so much!
NOTE: I see two hazy questions in here: 1) how to deal with a stochastic term in a DE and 2) how to deal with a delta function in a DE. Both of these are math related questions and http://www.math.stackexchange.com will be a better place for this. If you had a question pertaining to MATLAB, I haven't been able to pin it down, and you should perhaps add code examples to better illustrate your point. That said, I'll answer the two questions briefly, just to put you on the right track.
What you have here are not ODEs, but Stochastic differential equations (SDE). I'm not sure how you're using MATLAB to work with this, but routines like ode45 or ode23 will not be of any help. For SDEs, your usual mathematical tools of separation of variables/method of characteristics etc don't work and you'll need to use Itô calculus and Itô integrals to work with them. The solutions, as you might have guessed, will be stochastic. To learn more about SDEs and working with them, you can consider Stochastic Differential Equations: An Introduction with Applications by Bernt Øksendal and for numerical solutions, Numerical Solution of Stochastic Differential Equations by Peter E. Kloeden and Eckhard Platen.
Coming to the delta function part, you can easily deal with it by taking the Fourier transform of the ODE. Recall that the Fourier transform of a delta function is 1. This greatly simplifies the DE and you can take an inverse transform in the very end to return to the original domain.