## Uniqueness of the gamma function with the Euler reflection formula

Is the gamma function unique? $$f (x)$$ so that
$$f (x) f (1-x) = dfrac { pi} { sin pi x}?$$
I looked up the Bohr Mollerup theorem, although I wonder if it is possible to uniquely characterize the gamma function with less than 3 conditions.

## Is the broker a design pattern and is it used in design patterns by Gamma et al. Mentioned?

Software architecture says Bass

The broker pattern defines a runtime component called a broker
mediates communication between multiple clients and servers.

The original version of the broker pattern as documented by Gamma, Helm, Johnson and Vlissides
[Gamma 94], is shown in Figure 13.6.

I do not find Gamma's book with a pattern called a broker. If the broker in Design Patterns by Gamma et al. Mentioned?

I can see that the broker is listed as an architectural pattern in https://en.wikipedia.org/wiki/Template:Design_Patterns_patterns.
Is the broker a design pattern or architectural pattern?

Many Thanks.

## problem

The negative log probability of the Dirichlet multinominal distribution is given by
$$f ( alpha) = – L ( alpha) = sum_ {i = 1} ^ n sum_ {j = 1} ^ d log frac { gamma ( alpha_j)} { gamma ( alpha_j + mathbf {x} _ {ij})} + text {constant}$$
from where $$mathbf {x} _ {ij} in {0, 1 }$$ and $$alpha in mathbb {R} ^ d, alpha_j> 0$$

Then determined whether $$f ( alpha)$$ is convex or not.

## What have I done

It is natural to calculate the Hessian $$nabla_ alpha ^ 2 f ( alpha)$$, However, when I looked in Wikipedia, the derivative of the gamma function looks way too complicated.

Then I tried to find some counterexamples $$d = 1$$ Case. After several simulations performed by me, this function actually seems to be convex.

I wonder if we could analytically Determine the convexity of this function.

## Photo Editing – What does the gamma histogram mean?

A good way to analyze an image is to measure different locations (densities or intensities) and then draw these measurement points using graph paper. When done, the diagram resembles a half bell curve. We divide this curve into regions. The lower part is called "toe". This graph graphically indicates that the image is starting to form slowly. Next, the area of ​​the straight line will be referred to. This part of the graph graphically shows a proportional response to the exposure light. We can measure the angle of the straight line with a protractor. When the image contrast is low, the angle of the straight line is pressed. If the image has a high contrast, the straight line shows that fact. Its angle will be 45 ° or larger.
Historically, we have transformed this angle into a tangent (TAN) using trigonometry. Most images with sufficient contrast have a straight line with an angle of about 40 °. The tan of 40 = about 0.8. This is traditionally the target contrast for images that show a respectable contrast. We're talking about the science of sensitometry – taking test shots and the science of densitometry – measuring photographic images. It is common to call the TAN of the angle of the straight line "gamma".

The following graphic refers to the film – digital images are a subset of the film when it comes to the determination. we also draw these.

35 = gamma 0.7 = flat 45 = gamma 1.0 = contrast

## Number theory – Stirling's approximation for normalized \$ Gamma \$

To let
$$H (s) = frac {1} {2} s (1-s) pi ^ {- s / 2} gamma left ( frac {s} {2} right).$$
With the approach of Stirling for the gamma function I would like to prove that
$$frac {H (1/2 + it) overline {H} (1/2 + it + iu)} { left | H (1/2 + it) overline {H} (1/2 + it +) iu) right |} = left ( frac {2 pi} {t} right) ^ {iu / 2} left (1+ mathcal {O} left ( frac {u ^ 2 + 1)} {T} right) right)$$
from where $$T and $$| u | leq Delta$$, Do you have any idea how to show it?
I think I should use the approach of the Stirling
$$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 might work, as we normalize our estimation function, we can use the fact
$$z = | z | e ^ {i cdot arg (z)}$$
So what we want to appreciate is basically
$$e ^ {i cdot arg (H (1 / + it) overline {H} (1/2 + it + iu))}.$$
We use that for that $$arg (z) = Im ( log z)$$so $$arg ( Gamma (s)) = Im ( ln Gamma (s))$$ for which we use the Stirling approximation.
The contribution of the non-gamma factor is easier to estimate and should be
$$( pi) ^ {iu / 2}$$
Therefore, only an estimate of the gamma contribution remains.
For this purpose we use the approach of Stirling (in response to this question there are many useful approaches)
$$arg left ( gamma ( frac {1} {4} + i frac {t} {2}) right) = In left[left(frac{1}{4}+ifrac{t}{2}-frac{1}{2}right)ln(frac{1}{4}+ifrac{t}{2})-frac{1}{4}-ifrac{t}{2}+frac{1}{2}ln(2pi)+mathcal{O}left(frac{1}{t}right)right]$$
thus
$$arg left ( gamma ( frac {1} {4} + i frac {t} {2}) right) = left[frac{tln(1/16+t^2/4)}{4}-frac{1}{4}arctan(2t)-frac{t}{2}+mathcal{O}(1/t)right]$$
If I use only the first expression of such an extension, 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 similar
$$e ^ {i cdot arg left ( gamma ( frac {1} {4} -i frac {t} {2} -i frac {u} {2}) right)} sim left ( frac {1} {16} + left ( frac {t} {2} + frac {u} {2} right) ^ 2 right) ^ {- i (t + u) / 4}$$
So I think it remains to be proved:
$$left ( frac {1} {16} + frac {t ^ 2} {4} right) ^ {it / 4} cdot left ( frac {1} {16} + left ( frac .) {t} {2} + frac {u} {2} right) ^ 2 right) ^ {- i (t + u) / 4} = left ( frac {2} {t} right) ^ {iu / 2} left (1+ mathcal {O} left ( frac {u ^ 2 + 1} {T} right) right)$$
and that all additional conditions in the serial extension of $$ln gamma (s)$$ go also in the error expression.
Thanks in advance for any help!

## Real analysis – derivation of the gamma function at 3/2

It is known that the derivative is the gamma function $$– gamma$$ at 1, i. $$gamma & # 39; (1) = – gamma.$$
However, I wonder if the value or a good rapprochement is known
$$Gamma & # 39; (3/2) = int_0 ^ infty sqrt {x} cdot e ^ {- x} cdot ln (x) dx.$$

## multivariable calculus – Gradient of \$ -y ^ T log left (f left (W x; gamma, beta right) right) \$ w.r.t. \$ left {W, gamma, beta right } \$?

How to calculate the gradient of
begin {align} L left (W, gamma, beta right): = -y ^ T log left (f left (W x; gamma, beta right) right) end
in memory of $$left {W, gamma, beta right }$$, from where $$x in mathbb {R} ^ n$$, $$W in mathbb {R} ^ {m times n}$$, and $$y in mathbb {R} ^ m$$, but $$y_i in {0,1 }$$, $$f (z; gamma, beta)$$ is parameterized with $$gamma$$ and $$beta$$?

The definition of
eqalign { f (z; gamma, beta) & = gamma left (z- mu (z) right) \ left ( sigma (z) + epsilon right) ^ {- 1/2} + beta cr mu (z) & = alpha 1 ^ Tz cr sigma (z) & = alpha sum_ {k = 1} ^ m left (ex[k] – mu (z) right) ^ 2 equiv alpha 1 ^ T left[ left( z- mu(z) right) odot left(z – mu(z) right) right] cr from where $$1 ^ T$$ is a row vector with all, $$odot$$ is an elementwise multiplication and $$alpha$$ and $$epsilon$$ are known scalars.

## Calculus – Understand how to integrate parts into the gamma function

The gamma function is defined as …

$S = \int_{0}^{n}t^{s-1}(1-\frac{t}{n})^{n}dt$
,

I'll see how Gaussian representation The gamma function is derived and the first step is the integration of parts. No steps are shown and the following is the result of applying the integration of parts …

$S = \frac{t^{2}}{s}(1-\frac{t}{n})^{n}\binom{n}{0} + \frac{n}{ns}\int_{0}^{n}t^{s}(1-\frac{t}{n})^{n-1}dt$
,

I am confused how these values ​​were derived. That's my attitude …

$u = (1-\frac{t}{n})^{n}$
,

$dv = t^{s-1}$
,

I imagine they chose it that way u and dv, which does ______________ mean…

$v = \frac{t^{s}}{s}$
,

$du = ?$
,

I'm not sure how they came you, I tried to derive u and ended at …

$$e[n ln(1-t/n)]$$

and then got another answer after trying to derive it. Can someone show me how you is derived or shows me where I am wrong so that I can complete the parts through integration?

## Color Correction – How can gamma be applied correctly to linear raw files from an image processing camera?

Visually, this looks absolutely correct to me. In your unmatched image on my color-calibrated monitor, the steps between the gray spots appear at irregular intervals. In the adjusted image, they are more noticeable.

Is color saturation a natural consequence of applying a gamma value, and if so, what can be done to compensate for this effect?

No. This is a consequence if correct black levels and white points are not set. You should do this before you apply the gamma curve.

The gamma curve that you used has the following form:

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

This is basically a boring, "flat" curve – that is, even though it has no linear value, it is perceptively basically like that. You can see, however, that the values ​​are all in the middle. This is very functional and maybe that's what you want for image processing, but in general it's not what we want visually Ones. You may just want to increase the contrast by dragging the black and white dots as follows:

What gives a picture like this:

… much less washed out.

Your camera may have an adjustable black level, and you may want to raise it slightly. You must also set a suitable white level for your camera for conversion. You may want to check the Dcraw code to see how it normally does. (Well, spoiler: dcraw sets the white level to the 99th percentile of the histogram.)

Incidentally, this result has such a histogram:

what you can see extends to the extremes of the histogram. Since I am working with an 8-bit image in an 8-bit area, you can see that the colors are getting sparse. For "real work" you want to work with a higher bit depth (and probably only apply a transformation instead of a series).

This is definitely a bit boring. In the visual world, we may want to apply an s-curve to increase the power:

yielding

… or something like that.

## raw – How to apply gamma correctly to the image processing camera

I'm using a machine vision camera that produces linear pixels, according to the manufacturer. As a result, I need to apply a gamma correction to the image before I can see the result. However, after applying a gamma correction to the image, the colors appear washed out.

Before each gamma correction:

After applying 2.2 gamma:

Overall, the brightness in the picture looks much better. The colors in the Color Checker, however, seem very oversaturated. Is color saturation a natural consequence of applying a gamma value, and if so, what can be done to compensate for this effect?