I try to use wolfram Alpa to compute a complex integral,
integrate( 0.0016*v^(-0.5)*(1-exp(- 0.0112*v^0.5))+ 0.0036*v^(-0.5)*(1-exp(-0.0090*v^0.5)))
I get the following result
0.285714 e^(-0.0112 sqrt(v))+0.8 e^(-0.009 sqrt(v))+0.0104 sqrt(v)+constant
I am wondering what this constant means? The reason I am asking is because I need to evaluate the result at certain points of v to obtain a definite integral.
Cheers
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how are you?.
I'm trying to create a function that calculates the z-transform of a transfer function using the residues method but for that, I need the factors of the characteristic equation and the powers of the factors, so, in order to do that I tried to factorize polynomials with non-integer coefficients but after trying everything that I read I couldn't factorize make maxima to factorize those polynomials the way I need it.
For giving an example, I have this characteristic equation: "s·(s^2+0.1·s)", the factors should be "s^2" and "s + 0.1" but maxima allways gives me "(s^2·(10·s + 1))/10".
Why I'm signalling this?, well, as I learned that maxima treates the outputs equation as list so I can have its dimension and separate the factos by its positions in the list to measure the powers of the factors and do what I need, but like maxima gives me the result that is shown above then the dimension of the list is different and it will make my function to work differently and possibly have errors.
The result that is shown is given by maxima no matter if I use factor, gfactor, or expand or whatever other function that I found and I know that result is because maxima are rationalizing the polynomial before working with it but I don't need that behavior, I only need the pure factors, so, how can I have the result that I want?.
Thanks in advance for the help.
I have two - likely simple - questions that are bothering me, both related to quadratic programming:
1). There are two "standard" forms of the objective function I have found, differing by multiplication of negative 1.
In the R package quadprog, the objective function to be minimized is given as −dTb+12bTDb and in Matlab the objective is given as dTb+12bTDb. How can these be the same? It seems that one has been multiplied through by a negative 1 (which as I understand it would change from a min problem to a max problem.
2). Related to the first question, in the case of using quadprog for minimizing least squares, in order to get the objective function to match the standard form, it is necessary to multiply the objective by a positive 2. Does multiplication by a positive number not change the solution?
EDIT: I had the wrong sign for the Matlab objective function.
Function f(b)=dTb is a linear function thus it is both convex and concave. From optimization standpoint it means you can maximize or minimize it. Nevertheless minimizer of −dTb+12bTDb will be different from dTb+12bTDb, because there is additional quadratic term. Matlab implementation will find the one with plus sign. So if you are using different optimization software you will need to change d→−d to get the same result.
The function −dTb+12bTDb where D is symmetric and convex and thus has unique minimum. In general that is called standard quadratic programming form, but that doesn't really matter. The other function dTb−12bTDb is concave function which has unique maximum. It is easy to show that for, say, bounded function f(x) from above the following holds:
argmaxxf=argminx−f
Using the identity above value b∗1 where −dTb+12bTDb achieves minimum is the same as the value b∗2 which achieves maximum at dTb−12bTDb, that is b∗1=b∗2.
Programmatically it doesn't matter if you are minimizing −dTb+12bTDb or maximizing the other one. These are implementation-dependent details.
No it does not. ∀α>0 if x∗=argmaxxf(x), then x∗=argmaxxαf(x). This can be showed by contradiction.
I use the scipy.optimize.minimize ( https://docs.scipy.org/doc/scipy/reference/tutorial/optimize.html ) function with method='L-BFGS-B.
An example of what it returns is here above:
fun: 32.372210618549758
hess_inv: <6x6 LbfgsInvHessProduct with dtype=float64>
jac: array([ -2.14583906e-04, 4.09272616e-04, -2.55795385e-05,
3.76587650e-05, 1.49213975e-04, -8.38440428e-05])
message: 'CONVERGENCE: REL_REDUCTION_OF_F_<=_FACTR*EPSMCH'
nfev: 420
nit: 51
status: 0
success: True
x: array([ 0.75739412, -0.0927572 , 0.11986434, 1.19911266, 0.27866406,
-0.03825225])
The x value correctly contains the fitted parameters. How do I compute the errors associated to those parameters?
TL;DR: You can actually place an upper bound on how precisely the minimization routine has found the optimal values of your parameters. See the snippet at the end of this answer that shows how to do it directly, without resorting to calling additional minimization routines.
The documentation for this method says
The iteration stops when (f^k - f^{k+1})/max{|f^k|,|f^{k+1}|,1} <= ftol.
Roughly speaking, the minimization stops when the value of the function f that you're minimizing is minimized to within ftol of the optimum. (This is a relative error if f is greater than 1, and absolute otherwise; for simplicity I'll assume it's an absolute error.) In more standard language, you'll probably think of your function f as a chi-squared value. So this roughly suggests that you would expect
Of course, just the fact that you're applying a minimization routine like this assumes that your function is well behaved, in the sense that it's reasonably smooth and the optimum being found is well approximated near the optimum by a quadratic function of the parameters xi:
where Δxi is the difference between the found value of parameter xi and its optimal value, and Hij is the Hessian matrix. A little (surprisingly nontrivial) linear algebra gets you to a pretty standard result for an estimate of the uncertainty in any quantity X that's a function of your parameters xi:
which lets us write
That's the most useful formula in general, but for the specific question here, we just have X = xi, so this simplifies to
Finally, to be totally explicit, let's say you've stored the optimization result in a variable called res. The inverse Hessian is available as res.hess_inv, which is a function that takes a vector and returns the product of the inverse Hessian with that vector. So, for example, we can display the optimized parameters along with the uncertainty estimates with a snippet like this:
ftol = 2.220446049250313e-09
tmp_i = np.zeros(len(res.x))
for i in range(len(res.x)):
tmp_i[i] = 1.0
hess_inv_i = res.hess_inv(tmp_i)[i]
uncertainty_i = np.sqrt(max(1, abs(res.fun)) * ftol * hess_inv_i)
tmp_i[i] = 0.0
print('x^{0} = {1:12.4e} ± {2:.1e}'.format(i, res.x[i], uncertainty_i))
Note that I've incorporated the max behavior from the documentation, assuming that f^k and f^{k+1} are basically just the same as the final output value, res.fun, which really ought to be a good approximation. Also, for small problems, you can just use np.diag(res.hess_inv.todense()) to get the full inverse and extract the diagonal all at once. But for large numbers of variables, I've found that to be a much slower option. Finally, I've added the default value of ftol, but if you change it in an argument to minimize, you would obviously need to change it here.
One approach to this common problem is to use scipy.optimize.leastsq after using minimize with 'L-BFGS-B' starting from the solution found with 'L-BFGS-B'. That is, leastsq will (normally) include and estimate of the 1-sigma errors as well as the solution.
Of course, that approach makes several assumption, including that leastsq can be used and may be appropriate for solving the problem. From a practical view, this requires the objective function return an array of residual values with at least as many elements as variables, not a cost function.
You may find lmfit (https://lmfit.github.io/lmfit-py/) useful here: It supports both 'L-BFGS-B' and 'leastsq' and gives a uniform wrapper around these and other minimization methods, so that you can use the same objective function for both methods (and specify how to convert the residual array into the cost function). In addition, parameter bounds can be used for both methods. This makes it very easy to first do a fit with 'L-BFGS-B' and then with 'leastsq', using the values from 'L-BFGS-B' as starting values.
Lmfit also provides methods to more explicitly explore confidence limits on parameter values in more detail, in case you suspect the simple but fast approach used by leastsq might be insufficient.
It really depends what you mean by "errors". There is no general answer to your question, because it depends on what you're fitting and what assumptions you're making.
The easiest case is one of the most common: when the function you are minimizing is a negative log-likelihood. In that case the inverse of the hessian matrix returned by the fit (hess_inv) is the covariance matrix describing the Gaussian approximation to the maximum likelihood.The parameter errors are the square root of the diagonal elements of the covariance matrix.
Beware that if you are fitting a different kind of function or are making different assumptions, then that doesn't apply.
I am new on Semantic Web Rules Language and I am writing some rules in order to calculate the probability of - discrete and continuous - distributions.
I know that with SWRL I can do subtractions, addition, multiplication and divisions.
But what about exponentiation, summation, calculation of mathematical functions? Is there a way to do this in SWRL?
Just an example to place my question :
You know, for example, for Triangular distribution, we need basic mathematical calculus (subtractions and divisions), but for Beta Distribution we need exponentiation and calculus of the beta function..
Is there a way to do this in SWRL?
Thanks
The standard describes what math functions should be available, and these include exponentiation:
8.2. Math Built-Ins
…
swrlb:pow
Satisfied iff the first argument is equal to the result of the second argument raised to the third argument power.
There's no built in for the Beta function, though. You'd need to look into the reasoner that you're using and see whether you can implement additional mathematical builtins.
summation, calculation of mathematical functions
For summations, you may find the aggregate functions in SPARQL useful, but only if the terms you need to sum are available individually. You won't easily be able to express arbitrary sums like ∑i=1…n i2. You might find support for extension functions in SPARQL implementations, too.
Basically, I have a set of up to 100 co-ordinates, along with the desired tangents to the curve at the first and last point.
I have looked into various methods of curve-fitting, by which I mean an algorithm with takes the inputted data points and tangents, and outputs the equation of the cure, such as the gaussian method and interpolation, but I really struggled understanding them.
I am not asking for code (If you choose to give it, thats acceptable though :) ), I am simply looking for help into this algorithm. It will eventually be converted to Objective-C for an iPhone app, if that changes anything..
EDIT:
I know the order of all of the points. They are not too close together, so passing through all points is necessary - aka interpolation (unless anyone can suggest something else). And as far as I know, an algebraic curve is what I'm looking for. This is all being done on a 2D plane by the way
I'd recommend to consider cubic splines. There is some explanation and code to calculate them in plain C in Numerical Recipes book (chapter 3.3)
Most interpolation methods originally work with functions: given a set of x and y values, they compute a function which computes a y value for every x value, meeting the specified constraints. As a function can only ever compute a single y value for every x value, such an curve cannot loop back on itself.
To turn this into a real 2D setup, you want two functions which compute x resp. y values based on some parameter that is conventionally called t. So the first step is computing t values for your input data. You can usually get a good approximation by summing over euclidean distances: think about a polyline connecting all your points with straight segments. Then the parameter would be the distance along this line for every input pair.
So now you have two interpolation problem: one to compute x from t and the other y from t. You can formulate this as a spline interpolation, e.g. using cubic splines. That gives you a large system of linear equations which you can solve iteratively up to the desired precision.
The result of a spline interpolation will be a piecewise description of a suitable curve. If you wanted a single equation, then a lagrange interpolation would fit that bill, but the result might have odd twists and turns for many sets of input data.