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.
Related
Can someone please explain why the following symmetric function cannot pass a certain limit of negative values?
D = 0.1; l = 4;
c = #(x,v) (v/D).*exp(-v*x/D)./(1-exp(-v*l/D));
v_vec = -25:0.01:25;
figure(2)
hold on
plot(v_vec,c(l,v_vec),'b')
plot(v_vec,c(0,v_vec),'r')
Notice at the figure where the blue line chops, this is where I get inf/nan values.
It seems that Matlab is trying to compute a result that is too large, outputs +inf, and then operates on that, which yields +/- inf and NaNs.
For instance, at v=-25, part of the function computes exp(-(-25)*4/0.1), which is exp(1000), and that outputs +inf. (larger than the largest representable double precision float).
You can potentially solve that problem by rewriting your function to avoid operating of such very large (or very small) numbers, say by reorganising the fraction containing exp() functions.
I did encounter the same hurdle using exp() with arguments triggering overflow. Sometimes it is difficult to trace back numeric imprecision or convergence errors. In principle the function definition using exp() only create intermediate issues as your purpose as a transition function. The intention I guess was to provide a continuous function.
My solution to this problem is to divide the argument into regions and provide in each region an approximation function. In your case zero for negative x and proportional to x for positive x. In between you can use the orginal function. Care should be taken to match the approximation at the borders of the regions and the number of continuous differentiations which is important for convergence in loops.
I've written some Matlab procedures that evaluate orthogonal polynomials, and as a sanity check I was trying to ensure that their dot product would be zero.
But, while I'm fairly sure there's not much that can go wrong, I'm finding myself with a slightly curious behaviour. My test is quite simple:
x = -1:.01:1;
for i0=0:9
v1 = poly(x, i0);
for i1=0:i0
v2 = poly(x,i1);
fprintf('%d, %d: %g\n', i0, i1, v1*v2');
end
end
(Note the dot product v1*v2' needs to be this way round because x is a horizontal vector.)
Now, to cut to the end of the story, I end up with values close to 0 (order of magnitude about 1e-15) for pairs of degrees that add up to an odd number (i.e., i0+i1=2k+1). When i0==i1 I expect the dot product not to be 0, but this also happens when i0+i1=2k, which I didn't expect.
To give you some more details, I initially did this test with Chebyshev polynomials of first kind. Now, they are orthogonal with respect to the weight
1 ./ sqrt(1-x.^2)
which goes to infinity when x goes to 1. So I thought that leaving this term out could be the cause of non-zero dot products.
But then, I did the same test with Legendre polynomials, and I get exactly the same result: when the sum of the degrees is even, the dot product is definitely far from 0 (order of magnitude 1e2).
One last detail, I used the trigonometric formula cos(n*acos(x)) to evaluate the Chebyshev polynomials, and I tried the recursive formula as well as one of the formulas involving the binomial coefficient to evaluate the Legendre polynomials.
Can anyone explain this odd (pun intended) behaviour?
You're being misled by symmetry. Both Chebyshev and Legendre polynomials are eigenfunctions of the parity operator, which means that they can all be classified as either odd or even functions. I guess the same goes for your custom orthogonal polynomials.
Due to this symmetry, if you multiply a polynomial P_n(x) by P_m(x), then the result will be an odd function if n+m is odd, and it will be even otherwise. You're computing sum_k P_n(x_k)*P_m(x_k) for a symmetric set of x_k values around the origin. This implies that for odd n+m you will always get zero. Try computing sum_k P_n(x_k)*Q_m(x_k) with P a Legendre, and Q a Chebyshev polynomial. My point is that for n+m=odd, the result doesn't tell you anything about orthogonality or the accuracy of your integration.
The problem is that probably you're not integrating accurately enough. These orthogonal polynomials defined on [-1,1] vary quite rapidly on their domain, especially close to the boundaries (x==+-1). Try increasing the points of your integration, using a non-equidistant mesh, or a proper integration using integral.
Final note: I'd advise you against calling your functions poly, since that's a MATLAB built-in. (And so is legendre.)
I am processing a set of data using ridge regression. I found a very interesting phenomenon when apply the learned function to data. Namely, when the ridge parameter increases from zero, the test error keeps increasing. But if we penalize small coefficients(set the parameter <0), the test error can even be smaller.
This is my matlab code:
for i = 1:100
beta = ridgePolyRegression(ty_train,tX_train,lambda(i));
sqridge_train_cost(i) = computePolyCostMSE(ty_train,tX_train,beta);
sqridge_test_cost(i) = computePolyCostMSE(ty_valid,tX_valid,beta);
end
plot(lambda,sqridge_test_cost,'color','b');
lambda is the ridge parameter. ty_train is the output of the training data, tX_train is the input of training data. Also, we use a quadratic function regression here.
function [ beta ] = ridgePolyRegression( y,tX,lambda )
X = tX(:,2:size(tX,2));
tX2 = [tX,X.^2];
beta = (tX2'*tX2 + lambda * eye(size(tX2,2))) \ (tX2'*y);
end
The plotted picture is:
Why the error is minimal when lambda is negative? Is it a sign of under-fitting?
You should not use negative lambdas.
From (probabilistic) theoretic point of view, lambda relates to the inverse of variance of parameter prior distribution, and variance can't be negative.
From computational point of view, it can (given it's less that the smallest eigenvalue of the covariance matrix) turn your positive-definite form into an indefinite form, which means you'll have not a maximum, but a saddle point. It also means there are points where your target function is as small (or as big) as you want, so you can reduce loss indefinitely and no minimum / maximum exists at all.
Your optimization algorithm gives you just a stationary point, which will be a global maximum if and only if the form is positive definite.
Short Answer: When lambda is negative, you're actually overfitting your data. Hence, it's reasonable to get much less error.
Long Answer:
The regularization term (or the penalty term as described by many statisticians) aims to penalize the weights (or the betas as written in the coming Eq.) for going too high (overfitting) and going too low (underfitting). Giving you the power to control how your model behaves, and you usually aim the "right fitting" model.
For mathematical intuition, you can check the following Eq. (P. S. Equation is screenshotted from Elements of Statistical Learning by Trevor Hastie et. al)
When you decide to make your lambda negative, the penalty term is indeed turned into a utility term that helps to increase the weights (i.e., overfitting).
Overfitting is, simply, understanding your data along with the features more than you should, because you do not have the whole population yet; therefore, what you understood so far is possibly wrong on a different dataset.
So, you should never be using negative values of lambdas.
I have simple question. I'm trying to evaluate improper integral of 0th order Bessel function using Matlab R2012a:
v = integral(#(x)(besselj(0, x), 0, Inf)
which gives me v = 3.7573e+09. However this should be v = 1 in theory. When I'm trying to do
v = integral(#(l)besselj(0,l), 0, 1000)
it results to v = 1.0047. Could you briefly explain me, what is going wrong with integration? And how to properly integrate Bessel-type functions?
From the docs to do an improper integral on an oscillatory function:
q = integral(fun,0,Inf,'RelTol',1e-8,'AbsTol',1e-13)
in the docs the example is
fun = #(x)x.^5.*exp(-x).*sin(x);
but I guess in your case try:
q = integral(#(x)(besselj(0, x),0,Inf,'RelTol',1e-8,'AbsTol',1e-13)
At first I was sceptical that taking an integral over a Bessel function would produce finite results. Mathematica/Wofram Alpha however showed that the result is finite, but it is not for the faint of heart.
However, then I was pointed to this site where it is explained how to do it properly, and that the value of the integral should be 1.
I experimented a bit to verify the correctness of their statements:
F = #(z) arrayfun(#(y) quadgk(#(x)besselj(0,x), 0, y), z);
z = 10:100:1e4;
plot(z, F(z))
which gave:
so clearly, the integral indeed seems to converges to 1. Shame on Wolfram Alpha!
(Note that this is kind of a misleading plot; try to do it with z = 10:1e4; and you'll see why. But oh well, the principle is the same anyway).
This figure also shows precicely what the problem is you're experiencing in Matlab; the value of the integral is like a damped oscillation around 1 for increasing x. Problem is, the dampening is very weak -- as you can see, my z needed to go all the way to 10,000 just to produce this plot, whereas the oscillation amplitude was only decreased by ~0.5.
When you try to do the improper integral by messing around with the 'MaxIntervalCount' setting, you get this:
>> quadgk(#(x)besselj(0,x), 0, inf, 'maxintervalcount', 1e4)
Warning: Reached the limit on the maximum number of intervals in use.
Approximate bound on error is 1.2e+009. The integral may not exist, or
it may be difficult to approximate numerically.
Increase MaxIntervalCount to 10396 to enable QUADGK to continue for
another iteration.
> In quadgk>vadapt at 317
In quadgk at 216
It doesn't matter how high you set the MaxIntervalCount; you'll keep running into this error. Similar things also happen when using quad, quadl, or similar (these underly the R2012 integral function).
As this warning and plot show, the integral is just not suited to approximate accurately by any quadrature method implemented in standard MATLAB (at least, that I know of).
I believe the proper analytical derivation, as done on the physics forum, is really the only way to get to the result without having to resort to specialized quadrature methods.
This is my first post to stackoverflow, so if this isn't the correct area I apologize. I am working on minimizing a L1-Regularized System.
This weekend is my first dive into optimization, I have a basic linear system Y = X*B, X is an n-by-p matrix, B is a p-by-1 vector of model coefficients and Y is a n-by-1 output vector.
I am trying to find the model coefficients, I have implemented both gradient descent and coordinate descent algorithms to minimize the L1 Regularized system. To find my step size I am using the backtracking algorithm, I terminate the algorithm by looking at the norm-2 of the gradient and terminating if it is 'close enough' to zero(for now I'm using 0.001).
The function I am trying to minimize is the following (0.5)*(norm((Y - X*B),2)^2) + lambda*norm(B,1). (Note: By norm(Y,2) I mean the norm-2 value of the vector Y) My X matrix is 150-by-5 and is not sparse.
If I set the regularization parameter lambda to zero I should converge on the least squares solution, I can verify that both my algorithms do this pretty well and fairly quickly.
If I start to increase lambda my model coefficients all tend towards zero, this is what I expect, my algorithms never terminate though because the norm-2 of the gradient is always positive number. For example, a lambda of 1000 will give me coefficients in the 10^(-19) range but the norm2 of my gradient is ~1.5, this is after several thousand iterations, While my gradient values all converge to something in the 0 to 1 range, my step size becomes extremely small (10^(-37) range). If I let the algorithm run for longer the situation does not improve, it appears to have gotten stuck somehow.
Both my gradient and coordinate descent algorithms converge on the same point and give the same norm2(gradient) number for the termination condition. They also work quite well with lambda of 0. If I use a very small lambda(say 0.001) I get convergence, a lambda of 0.1 looks like it would converge if I ran it for an hour or two, a lambda any greater and the convergence rate is so small it's useless.
I had a few questions that I think might relate to the problem?
In calculating the gradient I am using a finite difference method (f(x+h) - f(x-h))/(2h)) with an h of 10^(-5). Any thoughts on this value of h?
Another thought was that at these very tiny steps it is traveling back and forth in a direction nearly orthogonal to the minimum, making the convergence rate so slow it is useless.
My last thought was that perhaps I should be using a different termination method, perhaps looking at the rate of convergence, if the convergence rate is extremely slow then terminate. Is this a common termination method?
The 1-norm isn't differentiable. This will cause fundamental problems with a lot of things, notably the termination test you chose; the gradient will change drastically around your minimum and fail to exist on a set of measure zero.
The termination test you really want will be along the lines of "there is a very short vector in the subgradient."
It is fairly easy to find the shortest vector in the subgradient of ||Ax-b||_2^2 + lambda ||x||_1. Choose, wisely, a tolerance eps and do the following steps:
Compute v = grad(||Ax-b||_2^2).
If x[i] < -eps, then subtract lambda from v[i]. If x[i] > eps, then add lambda to v[i]. If -eps <= x[i] <= eps, then add the number in [-lambda, lambda] to v[i] that minimises v[i].
You can do your termination test here, treating v as the gradient. I'd also recommend using v for the gradient when choosing where your next iterate should be.