How can I numerically integrate Exp[-Abs[x1]-Abs[x2]]/Abs[x1-x2] for x1 and x2 from -Infinity to Infinity?
What strategy would be adapted, Monte-Carlo? The Vegas algorithm (I tried implementations in Python and Mathematica) might work, but I cannot find a package which enables me to exclude a set of points, which Mathematica's NIntegrate does (here: Exclusions->x1 == x2). The latter however gives different answers depending on the strategy (without error messages and with integral errors only going up to 1.1, using https://mathematica.stackexchange.com/questions/75426/obtaining-an-nintegrate-error-estimate). I also go to tangent space using x->Tan[alpha] to compress the integration limits to +-pi/2.
Thank you.
Density Plot of function along x1 and x2 (in tangent space)
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I need a robust integration algorithm for f(x)exp(-x) between x=0 and infinity, with f(x) a positive, differentiable function.
I do not know the array x a priori (it's an intermediate output of my routine). The x array is typically ~log-equispaced, but highly irregular.
Currently, I'm using the Simpson algorithm, buy my problem is that often the domain is highly undersampled by the x array, which produces unrealistic values for the integral.
On each run of my code I need to do this integration thousands of times (each with a different set of x values), so I need to find an efficient and robust way to integrate this function.
More details:
The x array can have between 2 and N points (N known). The first value is always x[0] = 0.0. The last point is always a value greater than a tunable threshold x_max (such that exp(x_max) approx 0). I only know the values of f at the points x[i] (though the function is a smooth function).
My first idea was to do a Laguerre-Gauss quadrature integration. However, this algorithm seems to be highly unreliable when one does not use the optimal quadrature points.
My current idea is to add a set of auxiliary points, interpolating f, such that the Simpson algorithm becomes more stable. If I do this, is there an optimal selection of auxiliary points?
I'd appreciate any advice,
Thanks.
Set t=1-exp(-x), then dt = exp(-x) dx and the integral value is equal to
integral[ f(-log(1-t)) , t=0..1 ]
which you can evaluate with the standard Simpson formula and hopefully get good results.
Note that piecewise linear interpolation will always result in an order 2 error for the integral, as the result amounts to a trapezoid formula even if the method was Simpson. For better errors in the Simpson method you will need higher interpolation degrees, ideally cubic splines. Cubic Bezier polynomials with estimated derivatives to compute the control points could be a fast compromise.
I am developing a 2D plotting program for functions of 1 variable. It is designed to be very simple so that the user should not have to select the initial plot limits (or "range").
Are there known algorithms that can find the most interesting plot limits, knowing only the function f(x) ?
Notes:
The definition of interesting plot limits is not well defined here. This is part of the question: what is the most interesting part of the plot?
I already have an algorithm to determine the range of x values where the function f has finite values.
I am using Javascript, but any language is ok.
I don't want to use existing libraries.
The function f is restricted to expressions that the user can write with the basic math operators + - * / ^ and functions exp log abs sqrt sin cos tan acos asin atan ceil floor.
Using the Google graphs, you can get some examples of automatic limits. Typing graph sin(x) works pretty well, but graph exp(x) and graph log(x) don't really give the best results. Also, graph sin(x*100)*exp(-x^2) does not choose the limits I would qualify as the most informative. But it would be good enough for me.
UPDATE:
I found that PlotRange in Mathematica does that automatically very well (see here). Is the source code available, or a reference explaining the algorithm? I could not find it anywhere.
UPDATE:
I started using an adaptative refinement algorithm to find informative plot ranges, inspired from this site. It is not working perfectly yet, but the current progress is implemented in my project here. You can try plotting a few functions and see how it works. When I have a fully working version I can post an answer.
I don't have a complete answer, but I might have some useful ideas to start with.
For me, interesting parts of a graph include:
All the roots of the function, except where there is an infinite number of roots (where we might be interested in no more than 8 of each).
All the roots of the first and second derivatives of the function, again except where there is an infinite number of roots.
The behaviour of the function around x = 0.
Locations of asymptotes, though I wouldn't want the graph to plot all the way to infinity.
To see the features of the graph, I'd like it to take up a "reasonable" amount of the rectangular graphing window. I think that might be achieved by aiming to have the integral of the absolute value of the function over the plotted range equal be in the range of, say, 20-80% of the graphing window.
Thus, a sketch of an heuristic for setting plot limits might be something like:
Find the range that includes all the roots of the function, its first and second derivatives, or (for functions with infinite numbers of roots) the (say) 8 roots closest to x=0.
If the range does not already include x=0, expand the range to include x=0.
Expand the x range by, say, 10% in each direction to ensure that all "interesting" points are well within the window.
Set the y range so that the integral of the absolute value of the function is (say) 30% of the area of the graphing window.
I have a problem involving 3d positioning - sort of like GPS. Given a set of known 3d coordinates (x,y,z) and their distances d from an unknown point, I want to find the unknown point. There can be any number of reference points, however there will be at least four.
So, for example, points are in the format (x,y,z,d). I might have:
(1,0,0,1)
(0,2,0,2)
(0,0,3,3)
(0,3,4,5)
And here the unknown point would be (0,0,0,0).
What would be the best way to go about this? Is there an existing library that supports 3d multilateration? (I have been unable to find one). Since it's unlikely that my data will have an exact solution (all of the 4+ spheres probably won't have a single perfect intersect point), the algorithm would need to be capable of approximating it.
So far, I was thinking of taking each subset of three points, triangulating the unknown based on those three, and then averaging all of the results. Is there a better way to do this?
You could take a non-linear optimisation approach, by defining a "cost" function that incorporates the distance error from each of your observation points.
Setting the unknown point at (x,y,z), and considering a set of N observation points (xi,yi,zi,di) the following function could be used to characterise the total distance error:
C(x,y,z) = sum( ((x-xi)^2 + (y-yi)^2 + (z-zi)^2 - di^2)^2 )
^^^
^^^ for all observation points i = 1 to N
This is the sum of the squared distance errors for all points in the set. (It's actually based on the error in the squared distance, so that there are no square roots!)
When this function is at a minimum the target point (x,y,z) would be at an optimal position. If the solution gives C(x,y,z) = 0 all observations would be exactly satisfied.
One apporach to minimise this type of equation would be Newton's method. You'd have to provide an initial starting point for the iteration - possibly a mean value of the observation points (if they en-circle (x,y,z)) or possibly an initial triangulated value from any three observations.
Edit: Newton's method is an iterative algorithm that can be used for optimisation. A simple version would work along these lines:
H(X(k)) * dX = G(X(k)); // solve a system of linear equations for the
// increment dX in the solution vector X
X(k+1) = X(k) - dX; // update the solution vector by dX
The G(X(k)) denotes the gradient vector evaluated at X(k), in this case:
G(X(k)) = [dC/dx
dC/dy
dC/dz]
The H(X(k)) denotes the Hessian matrix evaluated at X(k), in this case the symmetric 3x3 matrix:
H(X(k)) = [d^2C/dx^2 d^2C/dxdy d^2C/dxdz
d^2C/dydx d^2C/dy^2 d^2C/dydz
d^2C/dzdx d^2C/dzdy d^2C/dz^2]
You should be able to differentiate the cost function analytically, and therefore end up with analytical expressions for G,H.
Another approach - if you don't like derivatives - is to approximate G,H numerically using finite differences.
Hope this helps.
Non-linear solution procedures are not required. You can easily linearise the system. If you take pair-wise differences
$(x-x_i)^2-(x-x_j)^2+(y-y_i)^2-(y-y_j)^2+(z-z_i)^2-(z-z_j)^2=d_i^2-d_j^2$
then a bit of algebra yields the linear equations
$(x_i-x_j) x +(y_i-y_j) y +(zi-zj) z=-1/2(d_i^2-d_j^2+ds_i^2-ds_j^2)$,
where $ds_i$ is the distance from the $i^{th}$ sensor to the origin. These are the equations of the planes defined by intersecting the $i^{th}$ and the $j^{th}$ spheres.
For four sensors you obtain an over-determined linear system with $4 choose 2 = 6$ equations. If $A$ is the resulting matrix and $b$ the corresponding vector of RHS, then you can solve the normal equations
$A^T A r = A^T b$
for the position vector $r$. This will work as long as your sensors are not coplanar.
If you can spend the time, an iterative solution should approach the correct solution pretty quickly. So pick any point the correct distance from site A, then go round the set working out the distance to the point then adjusting the point so that it's in the same direction from the site but the correct distance. Continue until your required precision is met (or until the point is no longer moving far enough in each iteration that it can meet your precision, as per the possible effects of approximate input data).
For an analytic approach, I can't think of anything better than what you already propose.
The kernel trick maps a non-linear problem into a linear problem.
My questions are:
1. What is the main difference between a linear and a non-linear problem? What is the intuition behind the difference of these two classes of problem? And How does kernel trick helps use the linear classifiers on a non-linear problem?
2. Why is the dot product so important in the two cases?
Thanks.
When people say linear problem with respect to a classification problem, they usually mean linearly separable problem. Linearly separable means that there is some function that can separate the two classes that is a linear combination of the input variable. For example, if you have two input variables, x1 and x2, there are some numbers theta1 and theta2 such that the function theta1.x1 + theta2.x2 will be sufficient to predict the output. In two dimensions this corresponds to a straight line, in 3D it becomes a plane and in higher dimensional spaces it becomes a hyperplane.
You can get some kind of intuition about these concepts by thinking about points and lines in 2D/3D. Here's a very contrived pair of examples...
This is a plot of a linearly inseparable problem. There is no straight line that can separate the red and blue points.
However, if we give each point an extra coordinate (specifically 1 - sqrt(x*x + y*y)... I told you it was contrived), then the problem becomes linearly separable since the red and blue points can be separated by a 2-dimensional plane going through z=0.
Hopefully, these examples demonstrate part of the idea behind the kernel trick:
Mapping a problem into a space with a larger number of dimensions makes it more likely that the problem will become linearly separable.
The second idea behind the kernel trick (and the reason why it is so tricky) is that it is usually very awkward and computationally expensive to work in a very high-dimensional space. However, if an algorithm only uses the dot products between points (which you can think of as distances), then you only have to work with a matrix of scalars. You can implicitly perform the calculations in the higher-dimensional space without ever actually having to do the mapping or handle the higher-dimensional data.
Many classifiers, among them the linear Support Vector Machine (SVM), can only solve problems that are linearly separable, i.e. where the points belonging to class 1 can be separated from the points belonging to class 2 by a hyperplane.
In many cases, a problem that is not linearly separable can be solved by applying a transform phi() to the data points; this transform is said to transform the points to feature space. The hope is that, in feature space, the points will be linearly separable. (Note: This is not the kernel trick yet... stay tuned.)
It can be shown that, the higher the dimension of the feature space, the greater the number of problems that are linearly separable in that space. Therefore, one would ideally want the feature space to be as high-dimensional as possible.
Unfortunately, as the dimension of feature space increases, so does the amount of computation required. This is where the kernel trick comes in. Many machine learning algorithms (among them the SVM) can be formulated in such a way that the only operation they perform on the data points is a scalar product between two data points. (I will denote a scalar product between x1 and x2 by <x1, x2>.)
If we transform our points to feature space, the scalar product now looks like this:
<phi(x1), phi(x2)>
The key insight is that there exists a class of functions called kernels that can be used to optimize the computation of this scalar product. A kernel is a function K(x1, x2) that has the property that
K(x1, x2) = <phi(x1), phi(x2)>
for some function phi(). In other words: We can evaluate the scalar product in the low-dimensional data space (where x1 and x2 "live") without having to transform to the high-dimensional feature space (where phi(x1) and phi(x2) "live") -- but we still get the benefits of transforming to the high-dimensional feature space. This is called the kernel trick.
Many popular kernels, such as the Gaussian kernel, actually correspond to a transform phi() that transforms into an infinte-dimensional feature space. The kernel trick allows us to compute scalar products in this space without having to represent points in this space explicitly (which, obviously, is impossible on computers with finite amounts of memory).
The main difference (for practical purposes) is: A linear problem either does have a solution (and then it's easily found), or you get a definite answer that there is no solution at all. You do know this much, before you even know the problem at all. As long as it's linear, you'll get an answer; quickly.
The intuition beheind this is the fact that if you have two straight lines in some space, it's pretty easy to see whether they intersect or not, and if they do, it's easy to know where.
If the problem is not linear -- well, it can be anything, and you know just about nothing.
The dot product of two vectors just means the following: The sum of the products of the corresponding elements. So if your problem is
c1 * x1 + c2 * x2 + c3 * x3 = 0
(where you usually know the coefficients c, and you're looking for the variables x), the left hand side is the dot product of the vectors (c1,c2,c3) and (x1,x2,x3).
The above equation is (pretty much) the very defintion of a linear problem, so there's your connection between the dot product and linear problems.
Linear equations are homogeneous, and superposition applies. You can create solutions using combinations of other known solutions; this is one reason why Fourier transforms work so well. Non-linear equations are not homogeneous, and superposition does not apply. Non-linear equations usually have to be solved numerically using iterative, incremental techniques.
I'm not sure how to express the importance of the dot product, but it does take two vectors and returns a scalar. Certainly a solution to a scalar equation is less work than solving a vector or higher-order tensor equation, simply because there are fewer components to deal with.
My intuition in this matter is based more on physics, so I'm having a hard time translating to AI.
I think following link also useful ...
http://www.simafore.com/blog/bid/113227/How-support-vector-machines-use-kernel-functions-to-classify-data
Does anyone have experience with algorithms for evaluating hypergeometric functions? I would be interested in general references, but I'll describe my particular problem in case someone has dealt with it.
My specific problem is evaluating a function of the form 3F2(a, b, 1; c, d; 1) where a, b, c, and d are all positive reals and c+d > a+b+1. There are many special cases that have a closed-form formula, but as far as I know there are no such formulas in general. The power series centered at zero converges at 1, but very slowly; the ratio of consecutive coefficients goes to 1 in the limit. Maybe something like Aitken acceleration would help?
I tested Aitken acceleration and it does not seem to help for this problem (nor does Richardson extrapolation). This probably means Pade approximation doesn't work either. I might have done something wrong though, so by all means try it for yourself.
I can think of two approaches.
One is to evaluate the series at some point such as z = 0.5 where convergence is rapid to get an initial value and then step forward to z = 1 by plugging the hypergeometric differential equation into an ODE solver. I don't know how well this works in practice; it might not, due to z = 1 being a singularity (if I recall correctly).
The second is to use the definition of 3F2 in terms of the Meijer G-function. The contour integral defining the Meijer G-function can be evaluated numerically by applying Gaussian or doubly-exponential quadrature to segments of the contour. This is not terribly efficient, but it should work, and it should scale to relatively high precision.
Is it correct that you want to sum a series where you know the ratio of successive terms and it is a rational function?
I think Gosper's algorithm and the rest of the tools for proving hypergeometric identities (and finding them) do exactly this, right? (See Wilf and Zielberger's A=B book online.)