complex analysis – In a $ Gamma $ network the sum $ sum_ {0 neq w in Gamma} frac {1} {| w | ^ 3} $ converge

Today, I read about the famous Weierstrass $ wp $ operate in this document. To prove the absolute convergence of $ wp $ they use the fact that $ sum_ {w in Gamma setminus {0 }} frac {1} {| w | ^ 3} $ converge (see page 9 of the document).

How do they prove it? Well, the argument is exactly as follows:

Yes $ Gamma: = langle w_1, w_2 row_ { mathbb {Z}} $ or $ w_1 / w_2 notin mathbb {R} $ then there is a natural number $ k such as $ | mw_1 + nw_2 | geq (| m | + | n |) k $. For each $ N $ there is exactley $ 4N $ choice of $ (m, n) $ such as $ | m | + | n | = N $. Using this, we have:

$$ sum_ {w in Gamma setminus {0 }} frac {1} {| w | ^ 3} = sum _ {(m, n) in mathbb {Z} ^ 2 setminus {(0,0) }} frac {1} {| mw_1 + nw_2 | ^ 3} leq frac {1} {k} sum _ {(m, n) in mathbb {Z} ^ 2 setminus {(0,0) }} frac {1} { | m | + | n |) ^ 3} $$
$$ = sum_ {N = 1} ^ { infty} frac {4} {N ^ 2} < infty $$

Nice is not it? But I'm not sure of the first line of the previous argument. Do not they assume the absolute convergence of the thing that they are trying to prove that converges? Can we avoid doing this?

Thanks to all the guys in advance

Note

I do not think it's important to know what is the function of Weierstrass to understand my question, but just in case: given $ Gamma $

$$ wp (z): = sum_ {w in Gamma setminus {0 }} left ( frac {1} {| z + w | ^ 2} – frac {1} {| w | ^ 2} right) $$

what is the inverse transformation of Laplace of some form of gamma function?

$ F (s) = frac { Gamma (1 + b) Gamma (1 + as)} {s ( Gamma (1 + b + as) – Gamma (1 + b) Gamma (1 + as) ))} $

What is Laplace's inverse transformation of this function?

a and b are positive real numbers. Even if it can be solved, b is an integer will be useful.

Thank you so much!

Complex Analysis – Prove the integral of $ f (z) = frac {1-e ^ {2iz}} {4z ^ 2} $ tends to $ 0 $ on the curve $ gamma = Re ^ {it} $ as $ R rightarrow { infty} $

I am asked to prove that the integral of $ f (z) = frac {1-e ^ {2iz}} {4z ^ 2} $ on the curve $ gamma = Re ^ {it} $ $ 0 leq t leq pi $, when $ R rightarrow { infty} $, go to zero. To do this I've tried $$ int_ gamma f (z) dz = int_0 ^ { pi} f ( gamma) gamma dt = int_0 ^ { pi} frac {1-e ^ {2iRe ^ {it }}} {4 (Re ^ {it}) ^ 2} Rie ^ {it} dt = int_0 ^ { pi} frac {1-e ^ {2iRe ^ {it}} {4R ^ 2e ^ { it}} idt $$
And now I need to see that $$ lim_ {R to infty} int_0 ^ { pi} frac {1-e ^ {2iRe ^ {}}} {4} ^ {it}} idt = 0 $$
However, I do not know how to solve this integral, so I can not continue.

photo editing – What does the gamma histogram represent?

A proven method of image analysis involves measuring various locations (densities or intensities), and then using a grid paper, trace these measured points. Once completed, the plot looks like half a bell curve. We divide this curve into regions. The lower part is called the "toe". This graph shows graphically that the image slowly begins to form. Next is called the region of the straight line. This part of the graph graphically represents a response proportional to the exposed light. We can measure with a protractor the angle of the line. If the contrast of the image is weak, the angle of the straight line is reduced. If the image displays high contrast, the straight line indicates this fact; its angle will measure 45⁰ or more.
Historically, we convert this angle using tangent trigonometry (TAN). Most images with adequate contrast have a straight line with an angle of about 40 °. Tan of 40 = about 0.8. Traditionally, it is the target contrast for images displaying a respectable contrast. We are talking about the science of sensitometry – the manufacture of test exposures, and the science of densitometry – the measurement of photographic images. It is customary to name the TAN from the angle of the right "gamma".

The graph below relates to the film – digital images are a subset of the film when it comes to making decisions; we represent them as well.

35⁰ = gamma 0.7 = flat 45 = gamma 1.0 = contrast

enter the description of the image here

nt.number theory – Stirling approximation for normalization $ Gamma $

Let
$$
H (s) = frac {1} {2} s (1-s) pi ^ {- s / 2} Gamma left ( frac {s} {2} right).
$$

Using Stirling's approximation for Gamma function, I would like to prove that
$$
frac {H (1/2 + it) overline {H} (1/2 + it + iu)} { left | H (1/2 + it) overline {H} (1/2 + it + iu) right |} = left ( frac {2 pi} {t} right) ^ {iu / 2} left (1 + mathcal {O} left ( frac {u ^ 2 + 1} {T} right) right)
$$

or $ T <t <2T $ and $ | u | leq Delta $. Do you have an idea of ​​how to show it?
I guess I should use Stirling's approximation
$$
ln Gamma (s) = (s-1/2) ln ss + frac {1} {2} ln 2 pi + sum_ {m = 1} ^ { infty} frac {B_ { 2m}} {2m (2m-1) s ^ {2m-1}}
$$

What I thought I could work is: since we normalize our function to estimate, we can use the fact that
$$
z = | z | e ^ {i cdot arg (z)}
$$

so what we want to estimate is basically
$$
e ^ {i cdot arg (H (1 / + it) overline {H} (1/2 + it + iu))}}.
$$

To do this, we use it $ arg (z) = Im ( log z) $, So $ arg ( Gamma (s)) = Im ( ln Gamma (s)) $ for which we use Stirling's approximation.
The contribution of the non-gamma factor is easier to estimate and should be
$$
( pi) ^ {iu / 2}
$$

All that remains is to estimate the gamma contribution.
To this end, we use Stirling's approximation (in an answer to this question, there are many useful approximations) to get
$$
arg left ( Gamma ( frac {1} {4} + i frac {t} {2})) right) = Im left[left(frac{1}{4}+ifrac{t}{2}-frac{1}{2}right)ln(frac{1}{4}+ifrac{t}{2})-frac{1}{4}-ifrac{t}{2}+frac{1}{2}ln(2pi)+mathcal{O}left(frac{1}{t}right)right]
$$

So
$$
arg left ( Gamma ( frac {1} {4} + i frac {t} {2})) right) = left[frac{tln(1/16+t^2/4)}{4}-frac{1}{4}arctan(2t)-frac{t}{2}+mathcal{O}(1/t)right]
$$

If I only use the first term of such an expansion, I get
$$
e ^ {i cdot arg left ( Gamma ( frac {1} {4} + i frac {t} {2})) right)} sim left ( frac {1} {16} + frac {t ^ 2} {4} right) ^ {it / 4}
$$

and similarly
$$
e ^ {i cdot arg left ( Gamma ( frac {1} {4} -i frac {t} {2} -i frac {u} {2} {}} right)} sim left ( frac {1} {16} + left ( frac {t} {2} + frac {u} {2} right) ^ 2 right) ^ {- i (t + u) / 4}
$$

So, I think that it remains to be proven, that is that
$$
left ( frac {1} {16} + frac {t ^ 2} {4} right) ^ {it / 4} cdot left ( frac {1} {16} {16} + left ( frac {t} {2} + frac {u} {2} right) ^ 2 right) ^ {- i (t + u) / 4} = left ( frac {2} {t} right) ^ {iu / 2} left (1 + mathcal {O} left ( frac {u ^ 2 + 1} {T} right) right)
$$

and that all the additional terms in the expansion series of $ ln Gamma (s) $ also go into the error term.
Thank you in advance for any help!

real analysis – Derivative of gamma function at 3/2

It is well known that the derivative of the gamma function is $ – gamma $ at 1, that is to say $$ Gamma (1) = – gamma. $$
I wonder though whether the value or a decent approximation is known to
$$ Gamma (3/2) = int_0 ^ infty sqrt {x} cdot e ^ {- x} cdot ln (x) dx. $$

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?