probability distributions – What generative model produces a scale-free network with a specific gamma?

So, what I'm trying to do is to rewire a randomly directed graph (especially a Boolean lattice) so that the lower degree distribution is scaleless. However, I need a generative model that allows me to specify $ gamma $ in $ P (k) sim k ^ {- gamma} $.

I imagine that there is a certain ratio of rewiring $ n / m $ (rewiring $ n $ out of $ m $ total edges) with preferential attachment (proportional to the existing outer degree) that will produce this result on average. But I do not know how to convert $ gamma $ in this report.

I would like to know if such a generative model exists. What would be even better is to learn how to arrive at such a model, whether it exists or not.

color correction – How to properly apply gamma to linear raw files from a machine vision camera?

Visually, that seems to me quite correct. On your unadjusted image on my color-calibrated monitor, the steps between the gray areas appear irregularly spaced. In the adjusted picture, they are more perceptual.

You ask:

Is the desaturation of colors a natural consequence of the application of a gamma and, if so, what can be done to compensate for this effect?

Short answer:

No, it's a consequence of not setting the black levels and the white point correctly. You should do it before applying the gamma curve.

Long answer:

The gamma curve you applied has the following form:

original with histogram of linear version with gamma curve to apply

The light line applies to the values ​​displayed in the histogram. Once you have applied this curve, you get a histogram like this:

applied curve

This is basically a boring "flat" curve – that is, even if its value is not linear, that is perceptually basically yes. However, you can see that the values ​​are all grouped in the middle. C & # 39; very functional and may be what you want for the purpose of artificial vision, but is not usually what we want for visual each. We may want to increase the contrast by simply pulling the black and white dots, like this:

increase the contrast

Which gives an image like this:

much less washed out

… much less washed out.

Your camera may have an adjustable black level, and you may want to increase it slightly. You will also need to set an appropriate white level for your camera to use for conversion. You may want to look at the dcraw code to see how this is usually done. (Well, spoiler: dcraw sets the white level on the 99th percentile of the histogram.)

This result has a histogram like this, by the way:

increased contrast

that you can see extends further to the extremes of the histogram. Since I'm working with an 8-bit image in an 8-bit space, you can see that colors are becoming rare. for "real work", you would like to work in a higher bit depth (and probably apply a single transformation rather than a series).

In any case, it's still a bit boring. In the visual world, we may want to apply a curve in s, like this one, to increase the punch:

S curve

resulting in

dynamic!

… or something like that.

raw – How to properly apply gamma to the machine vision camera

I use a machine vision camera that, according to the manufacturer, produces linear pixels. Therefore, I have to apply a gamma correction to the image before I can visualize the result. However, after applying a gamma correction to the image, the colors appear faded.

Before any gamma correction:

Example of raw output

After applying a gamma 2.2:

Example of port 2.2 gamma application

Overall, the brightness of the image is significantly better. However, the colors of the color checker look very desaturated. Is the desaturation of colors a natural consequence of the application of a gamma and, if so, what can be done to compensate for this effect?

numerics – Calculation between gamma functions

I calculated gamma functions in Mathematica while that does not give me an agreed answer.

By definition, $ Gamma[alpha]= int_0 ^ infty t ^ { alpha-1} e ^ {- t} dt $, $ Gamma[alpha,z]= int_z ^ infty t ^ { alpha-1} e ^ {- t} dt $ and $ Gamma[alpha,z_1,z_2]= int_ {z_1} ^ {z_2} t ^ { alpha-1} e ^ {- t} dt $. So we should have $ Gamma[alpha]-Gamma[alpha,z]$ should be equal to $ Gamma[alpha,0,z]$. But this is not the case. I simply leave $ alpha = $ 200 and $ z $ from 1 to 5, the numerical values ​​do not match; see below.

    α = 200.
Table[{Gamma[α] - Gamma[α, z]Gamma[α, 0, z]}, {z, 1, 5}

{{0. * 10 ^ 359, 0.00184859}, {0. * 10 ^ 359, 1.0983 * 10 ^ 57}, {0. * 10 ^ 359.6.71225 * 10 ^ 91},
{0. * 10 ^ 359, 2.41279 * 10 ^ 116}, {0. * 10 ^ 359, 2.14999 * 10 ^ 135}}

Why is that?

$ bigcap I_ gamma $ unlimited if $ I_0 supset cdots supset I _ { gamma} cdots $ are unlimited

Let be $ kappa $ a regular cardinal and $ I_0 supset cdots supset I _ { gamma} cdots $ are unbound subsets of $ kappa $ for $ gamma < lambda < kappa $ or $ lambda $ is limited. I want to show $ bigcap I_ gamma $ is limitless. Is it true? It's easy to show that this is not empty for the regularity of $ kappa $.

calculation and analysis – Can this problem of integral equation $ int_ Gamma frac {e ^ {ik | xy |}} {4 pi | xy |}} x) $ be solved?

I do not know if Mathematica is able to solve 2D / 3D integral equations. I found this page in the documentation, but this only applies to 1D.

The following is what I would like to solve, this can be considered a problem of electromagnetism but it is an important point. Leave the field of the incident $ u_ {x_0} ^ {in} (x) $ by given by a point source to $ x_0 $:$$
u_ {x_0} ^ {in} (x) = frac {e ^ {ik | x-x_0 |}} {4 pi | x-x_0 |},
$$

Then I have to find $ varphi in Gamma $ such as$$
S_ Gamma ^ k[varphi](x) = u_ {x_0} ^ {in} (x), quad quad forall x in Gamma,
$$

or
$$
S_ Gamma ^ k[varphi](x): = int_ Gamma frac {e ^ {ik | x-y |}} {4 pi | x-y |} varphi (y) dy,
$$

and
$ Gamma in mathbb {R} ^ $ 3 is the triangle defined by its vertices as $$ Gamma: = {v_1, v_2, v_3 },
$$

with begin {align} v_1 & = (4,0,0), \ v_2 & = (8,0,0), \ v_1 & = (6,2,0). end {align}

The problem is in 3D, but the domain of integration is a 2D triangle on the $ x $$ y $ plane, with a singular integral when $ x = y $.

Is it possible to solve this problem with Mathematica?

photoshop – How to know if TIF images have a gamma correction or not?

I have several DNG image files, saved with a Nexus 5X LGE under Open Camera for Android as a camera app. For further processing of the images, I converted the DNG files into TIF files using Camera Raw 8 and Photoshop CC 2015. I converted the images without changing the color settings or d & # 39; exposure.
As far as I know, DNG files must be linear (without gamma correction). My question is whether the converted TIF file is always linear or not. Does Photoshop / Camera Raw make a gamma correction for TIF files?

I would be happy to receive help for this matter.
Better,
Hans

calculation and analysis – Infinite complex of hypergeometric + gamma function

I am trying to solve an integral that gives a hypergeometric function + gamma function. The fact is that my values ​​of not (see the code below) are integers, so n = 1,2,3,4 .... But that gives a complex singularity when I want to introduce the value of not in the expression once the integral has already been calculated.

F[r_, k_, kl_, n_] : = Sin[k*kl*r]/ (k ^ 2 * r) * k ^ -n
L & # 39; Integration[r_, kl_, n_, m_] : =
Supposing[{kl > 0, n > 0, r > 0, m > 0}, 
   Integrate[f[r, k, kl, n], {k, 1, m}]]// FullSimplify

a function[r_] : = Integration[r, kl, n, Infinity] // FullSimplify
a function[r]

The result is:

-kl (kl r) ^ n Cos[(n [Pi]) / 2]Gamma[-1 - n] + (
kl HypergeometricPFQ[{-(n/2)}, {3/2, 1 - n/2}, -(1/4) kl^2 r^2]) / not

So Mathematica can solve the integral for any value of not. But when I substitute in this expression the value of not that I want then I get a singularity (of course, because of the gamma function and the hypergeometric function)

-kl (kl r) ^ n Cos[(n [Pi]) / 2]Gamma[-1 - n] + (
kl HypergeometricPFQ[{-(n/2)}, {3/2, 1 - n/2}, -(1/4) kl^2 r^2]) /
not /. n -> 2

I receive

Infinity :: indet: Undetermined expression ComplexInfinity + ComplexInfinity encountered.

But I do not underestimate because if I realize the integral given the value of not already, so I get an analityical expression for that

F[r_, k_, kl_, n_] : = Sin[k*kl*r]/ (k ^ 2 * r) * k ^ -n
L & # 39; Integration[r_, kl_, n_, m_] : =
Supposing[{kl > 0, n > 0, r > 0, m > 0}, 
   Integrate[f[r, k, kl, n], {k, 1, m}]]// FullSimplify

a function[r_] : = Integration[r, kl, 2, Infinity] // FullSimplify
a function[r]

I receive:

(kl r Cos[kl r] +
kl ^ 3 r ^ 3 CosIntegral[kl r] + (2 - kl ^ 2 r ^ 2) Sin[kl r]) / (6 r)

And there is no kind of singularity in this expression. My question is this: there is a way to get a general expression in terms of variable not, without having a singular behavior as it is due to the gamma and the hypergeometric function? Mathematica can simplify this expression in order to cancel the divergences?

Ag.algebraic geometry – Is there any manual proof that $ Gamma ( mathbb {C} P ^ n, E) $ is a finite dimension for a holomorphic vector bundle $ E $?

Please let me know if this question is appropriate for Mathoverflow.

Let $ E $ to be a finite holomorphic vector beam (or more generally a coherent analytic sheaf) on a compact and complex manifold $ X $. By the theorem of finitude of Cartan-Serre, cohomology $ H ^ q (X, E) $ is a finite dimensional vector space for any $ q $, and in particular the space of the global sections $ Gamma (X, E) = H ^ 0 (X, E) $ is of finite dimension.

The proof is based on Hodge's theory or on the properties of Frechet's spaces.

My question is, if we only consider $ X = mathbb {C} P ^ n $, could we prove that $ Gamma ( mathbb {C} P ^ n, E) $ is the finite dimension by a direct, elementary computation?

statistics – Moment method with gamma distribution

I am more confused as to a specific step in getting the MOM than to completely get the MOM:

Given a random sample of $ Y_1, Y_2, …, Y_i $ ~ $ Gamma ( alpha, beta) $ find the mother

So I found the population and examples of moments

$ u_1 ^ {$ = alpha beta $
$ u_2 ^ {+ # 39;} = sigma ^ 2 + mu ^ 2 = alpha ^ 2 beta ^ 2 + alpha beta ^ 2 $
$ m_1 ^ { #} = highlight Y $

$ m_2 ^ {} = {frac {1} {n} sum_ {i = 1} ^ {n} Y_i ^ 2 $

solve for $ hat alpha_ {MOM} $ I receive $ hat alpha_ {MOM} = frac { overline Y} { beta} $

The solutions say that it ends up being equal to:

$ hat alpha_ {MOM} = frac { frac {1} {n} ( sum_ {i = 1}

I think I forget a property because I do not know how they transformed the initial $ hat alpha_ {MOM} $ equation in the 2nd $ hat alpha_ {MOM} $ equation.