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).
Related
I have a Matlab function G(x,y,z). At each given (x,y,z), G(x,y,z) is a scalar. x=(x1,x2,...,xK) is a Kx1 vector.
Let us fix y,z at some given values. I would like your help to understand how to compute the derivative of G with respect to xk evaluated at a certain x.
For example, suppose K=3
function f= G(x1,x2,x3,y,z)
f=3*x1*sin(z)*cos(y)+3*x2*sin(z)*cos(y)+3*x3*sin(z)*cos(y);
end
How do I compute the derivative of G(x1,x2,x3,4,3) wrto x2 and then evaluate it at x=(1,2,6)?
You're looking for the partial derivative of dG/dx2
So the first thing would be getting rid of your fixed variables
G2 = #(x2) G(1,x2,6,4,3);
The numerical derivatives are finite differences, you need to choose an step h for your finite difference, and an appropriate method
The simplest one is
(G2(x2+h)-G2(x2))/h
You can make h as small as your numeric precision allows you to. At the limit h -> 0 the finite difference is the partial derivative
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 have a function X_t which I have defined anonymously to take as variables t which is a scalar time and z which is a vector of potentially arbitrary dimension.
That is,
X_t = #(t,z) fun(t,z).
I want to find the integral L2 norm of this object over the space of z's. That is, I want to find
X_t_norm = #(t) integral(#(z) abs(X_t(t,z))^2, -infinity,infinity).
Now clearly two things are a problem here.
Firstly, taking the limits as infinite isn't going to work, but I think I should be fine to take the limits as just large numbers (100 would certainly be enough for my purposes).
However, the real problem comes with taking this integral. My variable z is a vector of somewhat arbitrary (even) dimension and as a result I've gotten stuck figuring out how to pass z into the integral in such a way that I can compute this n-dimensional integral. (Basically I want to vary the length of z for different cases and compute the L2 norm for each of these cases).
In particular, I'm not sure how to tell the integral function to compute the integral over every component of z between the limits discussed above.
Any help would be greatly appreciated!
I'm trying to compute a rather ugly integral using MATLAB. What I'm having problem with though is a part where I multiply a very big number (>10^300) with a very small number (<10^-300). MATLAB returns 'inf' for this even though it should be in the range of 0-0.0005. This is what I have
besselFunction = #(u)besseli(qb,2*sqrt(lambda*(theta + mu)).*u);
exponentFuncion = #(u)exp(-u.*(lambda + theta + mu));
where qb = 5, lambda = 12, theta = 10, mu = 3. And what I want to find is
besselFunction(u)*exponentFunction(u)
for all real values of u. The problem is that whenever u>28 it will be evaluated as 'inf'. I've heared, and tried, to use MATLAB function 'vpa' but it doesn't seem to work well when I want to use functions...
Any tips will be appreciated at this point!
I'd use logarithms.
Let x = Bessel function of u and y = x*exp(-u) (simpler than your equation, but similar).
Since log(v*w) = log(v) + log(w), then log(y) = log(x) + log(exp(-u))
This simplifies to
log(y) = log(x) - u
This will be better behaved numerically.
The other key will be to not evaluate that Bessel function that turns into a large number and passing it to a math function to get the log. Better to write your own that returns the logarithm of the Bessel function directly. Look at a reference like Abramowitz and Stegun to try and find one.
If you are doing an integration, consider using Gauss–Laguerre quadrature instead. The basic idea is that for equations of the form exp(-x)*f(x), the integral from 0 to inf can be approximated as sum(w(X).*f(X)) where the values of X are the zeros of a Laguerre polynomial and W(X) are specific weights (see the Wikipedia article). Sort of like a very advanced Simpson's rule. Since your equation already has an exp(-x) part, it is particularly suited.
To find the roots of the polynomial, there is a function on MATLAB Central called LaguerrePoly, and from there it is pretty straightforward to compute the weights.
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.