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.
Related
Hi I am trying to solve a linear system of the following type:
A*x=b,
where A is the coefficient matrix,
x is the vectors of unknowns and
b is the vector of solution.
The coefficient matrix (A) is a n-by-n sparse matrix, with even zeros in the diagonal. In order to solve this system in an accurate way I am using an iterative method in Matlab called bicgstab (Biconjugate gradients stabilized method).
This coefficient matrix (A) has a
det(A)=-4.1548e-05 and a rcond(A)= 1.1331e-04.
Therefore the matrix is ill-conditioned. I first try to perform a scaling and the results where:
det(A)= -1.2612e+135 but the rcond(A)=5.0808e-07...
Therefore the matrix is still ill-conditioned... I verify and the sum of all absolute value of the non-diagonal elements where 163.60 and the sum of all absolute value of the diagonal elements where 32.49... Therefore the matrix of coefficient is not diagonally dominant and will not converge using my function bicgstab...
I am looking for someone that can help me with performing a pivoting to the coefficient matrix (A) so it can be diagonally dominant. Or any advice to solve this problem....
Thanks for the help.
First there should be quite a few things noted here:
Don't use the determinant to estimate the "amount of singularity" of your matrix. The determinant is the product of all the eigenvalues of your matrix, and therefore its scaling can be wildly misleading compared to a much better measure like the condition number, leading to the next point..
your conditioning (according to rcond) isn't that bad, are you working with single or double precision? Large problems can routinely get condition numbers in this range and still be quite solvable, but of course this depends on a very complicated interaction of many factors, of which the condition number plays only a small part. This leads to another complicated point:
Diagonal dominance may not help you at all here. BiCGStab as far as I know does not require diagonal dominance for its convergence, and also I don't think diagonal dominance is known even to help it. Diagonal dominance is usually an assumption made by other iterative methods such as the Jacobi method or Gauss-Seidel. Actually the convergence behavior of BiCGStab is not very well understood at all, and it is usually only used when memory is a very severe problem but conjugate gradients is not applicable.
If you are really interested in using a Krylov method (such as BiCGStab) to solve your problem, then you generally need to have more understanding of where your matrix is coming from so that you can choose a sensible preconditioner.
So this calls for a bit more information. Do you know more about this matrix? Is it arising from some kind of physical problem? Do you know for example if it is symmetric or positive definite (I will assume not both because you are not using CG).
Let me end with some actionable advice which is very generic, and so not necessarily optimal:
If memory is not an issue, consider using restarted GMRES instead of BiCGStab. My experience is that GMRES has much more robust convergence.
Try an approximate factorization preconditioner such as ILU. MATLAB has a function for this built in.
I have an integration function which does not have indefinite integral expression.
Specifically, the function is f(y)=h(y)+integral(#(x) exp(-x-1/x),0,y) where h(y) is a simple function.
Matlab numerically computes f(y) well, but I want to compute the following function.
g(w)=w*integral(1-f(y).^(1/w),0,inf) where w is a real number in [0,1].
The problem for computing g(w) is handling f(y).^(1/w) numerically.
How can I calculate g(w) with MATLAB? Is it impossible?
Expressions containing e^(-1/x) are generally difficult to compute near x = 0. Actually, I am surprised that Matlab computes f(y) well in the first place. I'd suggest trying to compute g(w)=w*integral(1-f(y).^(1/w),epsilon,inf) for epsilon greater than zero, then gradually decreasing epsilon toward 0 to check if you can get numerical convergence at all. Convergence is certainly not guaranteed!
You can calculate g(w) using the functions you have, but you need to add the (ArrayValued,true) name-value pair.
The option allows you to specify a vector-valued w and allows the nested integral call to receive a vector of y values, which is how integral naturally works.
f = #(y) h(y)+integral(#(x) exp(-x-1/x),0,y,'ArrayValued',true);
g = #(w) w .* integral(1-f(y).^(1./w),0,Inf,'ArrayValued',true);
At least, that works on my R2014b installation.
Note: While h(y) may be simple, if it's integral over the positive real line does not converge, g(w) will more than likely not converge (I don't think I need to qualify that, but I'll hedge my bets).
I have a vector z that I want to approximate by a linear combination of two other vectors (x,y) such that the residual of a*x+b*y and z is minimized. Also I want to keep one coefficient (a) positive for the fitting.
Any suggestions which command may help?
Thanks!
If you didn't have a bound on one of the coefficients, your problem could have been viewed as multiple regression (solved in matlab by regress). Since one of the coefficients is bounded, you should use lsqlin. This function solves least squares problems with bounds or inequalities on the coefficients. Don't forget to include an all-ones intercept predictor if your signals are not centered.
I think that fminsearch would be an overshoot in this case, since lsqlin does exactly what you want.
You have to define a function, which desribes the cost. The lower the cost is, the better the solution. The output must be a single scalar, e.g. the norm of the difference.
To avoid negative values for x, add something like (x<0)*inf. This rejects every solution with a negative x.
If done so, use fminsearch for a numeric solution.
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
I want to numerically integrate the following:
where
and a, b and β are constants which for simplicity, can all be set to 1.
Neither Matlab using dblquad, nor Mathematica using NIntegrate can deal with the singularity created by the denominator. Since it's a double integral, I can't specify where the singularity is in Mathematica.
I'm sure that it is not infinite since this integral is based in perturbation theory and without the
has been found before (just not by me so I don't know how it's done).
Any ideas?
(1) It would be helpful if you provide the explicit code you use. That way others (read: me) need not code it up separately.
(2) If the integral exists, it has to be zero. This is because you negate the n(y)-n(x) factor when you swap x and y but keep the rest the same. Yet the integration range symmetry means that amounts to just renaming your variables, hence it must stay the same.
(3) Here is some code that shows it will be zero, at least if we zero out the singular part and a small band around it.
a = 1;
b = 1;
beta = 1;
eps[x_] := 2*(a-b*Cos[x])
n[x_] := 1/(1+Exp[beta*eps[x]])
delta = .001;
pw[x_,y_] := Piecewise[{{1,Abs[Abs[x]-Abs[y]]>delta}}, 0]
We add 1 to the integrand just to avoid accuracy issues with results that are near zero.
NIntegrate[1+Cos[(x+y)/2]^2*(n[x]-n[y])/(eps[x]-eps[y])^2*pw[Cos[x],Cos[y]],
{x,-Pi,Pi}, {y,-Pi,Pi}] / (4*Pi^2)
I get the result below.
NIntegrate::slwcon:
Numerical integration converging too slowly; suspect one of the following:
singularity, value of the integration is 0, highly oscillatory integrand,
or WorkingPrecision too small.
NIntegrate::eincr:
The global error of the strategy GlobalAdaptive has increased more than
2000 times. The global error is expected to decrease monotonically after a
number of integrand evaluations. Suspect one of the following: the
working precision is insufficient for the specified precision goal; the
integrand is highly oscillatory or it is not a (piecewise) smooth
function; or the true value of the integral is 0. Increasing the value of
the GlobalAdaptive option MaxErrorIncreases might lead to a convergent
numerical integration. NIntegrate obtained 39.4791 and 0.459541
for the integral and error estimates.
Out[24]= 1.00002
This is a good indication that the unadulterated result will be zero.
(4) Substituting cx for cos(x) and cy for cos(y), and removing extraneous factors for purposes of convergence assessment, gives the expression below.
((1 + E^(2*(1 - cx)))^(-1) - (1 + E^(2*(1 - cy)))^(-1))/
(2*(1 - cx) - 2*(1 - cy))^2
A series expansion in cy, centered at cx, indicates a pole of order 1. So it does appear to be a singular integral.
Daniel Lichtblau
The integral looks like a Cauchy Principal Value type integral (i.e. it has a strong singularity). That's why you can't apply standard quadrature techniques.
Have you tried PrincipalValue->True in Mathematica's Integrate?
In addition to Daniel's observation about integrating an odd integrand over a symmetric range (so that symmetry indicates the result should be zero), you can also do this to understand its convergence better (I'll use latex, writing this out with pen and paper should make it easier to read; it took a lot longer to write than to do, it's not that complicated):
First, epsilon(x)-\epsilon(y)\propto\cos(y)-\cos(x)=2\sin(\xi_+)\sin(\xi_-) where I have defined \xi_\pm=(x\pm y)/2 (so I've rotated the axes by pi/4). The region of integration then is \xi_+ between \pi/\sqrt{2} and -\pi/\sqrt{2} and \xi_- between \pm(\pi/\sqrt{2}-\xi_-). Then the integrand takes the form \frac{1}{\sin^2(\xi_-)\sin^2(\xi_+)} times terms with no divergences. So, evidently, there are second-order poles, and this isn't convergent as presented.
Perhaps you could email the persons who obtained an answer with the cos term and ask what precisely it is they did. Perhaps there's a physical regularisation procedure being employed. Or you could have given more information on the physical origin of this (some sort of second order perturbation theory for some sort of bosonic system?), had that not been off-topic here...
May be I am missing something here, but the integrand
f[x,y]=Cos^2[(x+y)/2]*(n[x]-n[y])/(eps[x]-eps[y]) with n[x]=1/(1+Exp[Beta*eps[x]]) and eps[x]=2(a-b*Cos[x]) is indeed a symmetric function in x and y: f[x,-y]= f[-x,y]=f[x,y].
Therefore its integral over any domain [-u,u]x[-v,v] is zero. No numerical integration seems to be needed here. The result is just zero.