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Tag: gamma
Optimization – Maximize $ log (4) c + log (3) a + log (2) x $ if $ a + c + x + y = 1 $, $ (a + c) ^ 2 + (x + y) ^ 2 + 2xc leq 12 gamma $, $ 0 leq gamma leq 1/4 $
I want to find out how to show by hand that the maximum of
$$ log (4) c + log (3) a + log (2) x $$
when
$$ a geq 0, c geq 0, x geq 0, y geq 0, $$
$$ a + c + x + y = 1, $$
$$ (a + c) ^ 2 + (x + y) ^ 2 + 2xc leq 12 gamma, $$
Where $ gamma $ is a fixed constant so that $ 0 leq gamma leq 1/4 $.
Maple showed me that the maximum value is reached at $ frac { log (6)} {2} + frac { sqrt {14 gamma}} {2} log (3/2) $that is given by $ a = frac {1+ sqrt {14 gamma}} {2} $, $ b = frac {1 sqrt {14 gamma}} {2} $, $ c = y = 0 $.
I've tried using Lagrangian multipliers by changing the last constraint to an equality, but it gets very messy and I'm stuck.
I would also be happy if I could just show it $ log (4) c + log (3) a + log (2) x leq frac { log (6)} {2} + frac { sqrt {14 gamma}} { 2} log (3/2) $ for all such $ a, c, x, y $. One method I've tried is to use the concavity of $ log (x) $. The equation of the line that goes through $ (2, log (2)) $ and $ (3, log (3)) $ is $ L (x) = log (3/2) x log (9/8) $. Then by concavity of $ log (x) $ we have $ log (x) leq L (x) $ for all integers $ x $. Then
begin {align *} & log (4) c + log (3) a + log (2) x \
& leq L (4) c + L (3) a + L (2) x \
& = log (3/2) (4c + 3a + 2x + y) – log (9/8) \
& = log (3/2) left ( frac {5 + 3} {2} c + frac {5 + 1} {2} a + frac {51} {2} x + frac {5 3} {2} y right) – log (9/8) \
& = log (3/2) left ( frac {5+ sqrt {14 gamma}} {2} right) – log (9/8) + log (3/2) left ( frac {ax + 3 (cy) – sqrt {14 gamma}} {2} right) \
& = frac { log (6)} {2} + frac { sqrt {14 gamma}} {2} + log (3/2) left ( frac {ax + 3) ( cy) – sqrt {14 gamma}} {2} right). \
end {align *}
That is, to get the inequality I want, it would be enough to show it $ ax leq 3 (yc) + sqrt {14 gamma} $. However, I have not been able to prove this inequality. Any suggestions or help will be greatly appreciated.
Another option would be to use jamming techniques, but I have no experience with them. Thank you for your help.
Statistics – probability between two values of a gamma distribution
I'm trying to solve a problem where I have a random variable X that follows a gamma distribution of $ Gamma ( theta = 5, alpha = 4) $.
I'm trying to find the probability of $ P (10 leq X leq 30) $. Of course, if this were a normal distribution, it would be easy, but I currently don't know how to deal with this problem. I am confused as to whether to convert it to one $ chi ^ 2 $ and how that helps me to solve this problem. Thank you very much
Distinguish the special case of a lower incomplete gamma function
Accept $ P (s, sx) = frac { gamma (s, sx)} { Gamma (s)} $ is the normalized lower incomplete gamma function, where $ x in mathbb {R} ^ {+} $, and $ gamma ( cdot) $ and $ Gamma ( cdot) $ are lower incomplete or complete gamma functions.
I try to evaluate the following discrete differentiation, i. H.
begin {equation}
P left (s + 1, (s + 1) x right) – P (s, sx)
end {equation}
I used the basic property of the incomplete gamma function below, but stayed at this level:
begin {align *}
& = frac { gamma (s + 1, (s + 1) x) – s gamma (s, sx)} {s Gamma (s)} \ & = frac {s gamma (s, (s + 1) x) – x ^ {s} e ^ { x} – s gamma (s, sx)} {s Gamma (s)}
end {align *}
Thank you in advance!
Real analysis – According to convergence or divergence criteria of the given series $ S (x, s) = sum_1 ^ infty frac { sin ^ 2 ( frac {x Gamma (n)} {n})} {n ^ s} $
Consider the following sum:
$$ S (p, s) = sum_1 ^ infty frac { sin ^ 2 ( frac {p Gamma (n)} {n})} {n ^ s} $$
Here , $ p $ is a variable w.r.t with which we will analyze the sum
$ s in (0.1) $
I tried to use the Abel Plana summation formula (ABSF) for the subtotal:
We can get some rough estimates for the integral:
$$ I_1 (q) = int_1 ^ q frac { sin ^ 2 ( frac { Gamma (t)} {t})} {t ^ s} dt $$
How $ q rightarrow infty $
(Integrand swings very wildly in the right halfplane)
But it seems that the second integral in ABSF is impregnable.
Question: Can we get divergence criteria for the sum of the parameters w.r.t? $ p $ and $ s $?
Note: Second integral in ABSF:
$$ I_2 (x) = int_0 ^ infty frac {F (x + iy, s) – F (x – iy, s)} {e ^ {2πy} 1} dy $$
Where , $ F (z) = frac { sin ^ 2 ( frac { Gamma (z)} {z})} {z ^ s} $
I was trying to get an estimate as $ x rightarrow infty $ but to no avail.
Shader – implementation of the gamma adjustment slider
Various online sources talk sufficiently about gamma correction. By following them, I reached a rendering pipeline that looks something like this:
// Both are set to 2.2
uniform float gammaIn;
uniform float gammaOut;
void main()
{
vec3 color = pow(texture(material.albedo, texCoord).rgb, vec3(gammaIn));
// Do lighting and then HDR tone mapping here.
// color = ...
FragColor = pow(color, vec3(1.0 / gammaOut));
}
That is the easy part.
In many games, the slider for adjusting the gamma correction is now implemented, which should take into account the display differences on different monitors that the players may have. This contains some questions that I couldn't find a definitive answer to:

Which value should the setting controller influence? I assume that it will be so
gammaOut
theregammaIn
deals with the decoding of the sRGB image and assuming that all textures are sRGB, this should always be the constant2.2
, With this approach a lower onegammaOut
means darker picture. 
What is the appropriate range of values for the slider? Should start at
1.0
or somewhere higher? Where should it end? 
If I would display the comparison combination image "barely visible / invisible" to help the user with the correct setting, which color values should be "barely visible" for the dark image and its background and also for the "should" be "invisible" bright Image and its background?
Probability – How to show that if $ X sim text {Gamma} (n / 2, 1/2) $ and $ Y sim text {Gamma} (1/2, 1/2) $, then $ X + Y sim text {gamma} ((n + 1) / 2, 1/2) $?
Since $$ f_X (x) = frac {(1/2) ^ {n / 2} x ^ {n / 2 – 1} e ^ { x / 2}} { Gamma (n / 2)} $$ and $$ f_Y (y) = frac {x ^ { 1/2} e ^ { x / 2}} { sqrt {2} Gamma (1/2)} $$
We get that with the convolution formula $$ f_ {X} * f_Y (z) = e ^ { z / 2} frac {(1/2) ^ {(n + 1) / 2}} { Gamma ( frac {n} {2 }) Gamma ( frac {1} {2})} int _ { infty} ^ { infty} (zw) ^ {(n / 2) – 1} w ^ { 1/2} dw $$
I am not sure how to evaluate the integral on the right.
real analysis – properties that are fulfilled by functions with gamma function
Consider the following features:
$$ F_1 (z, s) = frac { sin ^ 2 ( frac {c Gamma (z)} {z})} {z ^ s} $$
$$ F_2 (z, s) = sin ^ 2 ( frac {c Gamma (z)} {z}) sin ^ 2 ( frac {c Gamma (z + d)} {z + d} ) { frac {1} {z ^ s}} + { frac {1} {(z + d) ^ s}} $$
Here, $ c, d $ are constants and $ s in (0.1) $ ,
Now we have to prove whether the following property applies $ F_i $::
$$ int_0 ^ infty  F_i (x + iy) – F_i (x – iy)  e ^ { 2πy} dy $$
exists for everyone $ x≥1 $ and tends to zero if $ x → ∞ $
Also how to prove whether $$ int_1 ^ infty F_i (z, s) ds $$ is convergent or divergent?
Statistics – approximation of the gamma function, decomposition
Is there an approximation of the gamma function of a sum so that the gamma function is broken down into functions of each element in the sum? Example:
$$ Gamma (n_1 + n_2 + … + n_N) = f_1 (n_1), f_2 (n_2), …, f_N (n_N) $$ where that, on the left, can be replaced by some mathematical operation and the elements, i.e. $ n_1 … n_N $ are positive integers and only one element, e.g. $ n_1 $must be a positive real.
nt.number Theory – Does finding positive integer solutions of $ zeta (a / b) = c $ correspond to the decision about the rationality of $ gamma $?
This question requires a little background explanation and is therefore a bit lengthy. Note: The question was originally posted in MSE but received no replies, so it was posted in MO.
For every integer $ n ge 2 $ there is a unique real $ 1 <x_n <$ 1.73 so the RiemannNzeta function $ zeta (x_n) = n $, example
$$ zeta (1.72865 ..) = 2, text {} zeta (1.41785 ..) = 3, text {} zeta (1.29396 ..) = 4, …
$$
It is likely that everyone $ x_n $ are irrational, but that's an open problem. I have developed a methodology to achieve some partial results in this direction.
If a reasonable one $ x_n> 1 $ then exists, WLOG, let $ x_n = 1 + dfrac {a_n} {b_n} = 1 + dfrac {1} {i_n + f_n} $ Where $ i_n $ is the integer part of the irreducible part $ b_n / a_n $ and $ f_n $ is its fraction. We get from the expansion of the Riemann zeta function in the Stieltjes series
$$
f_2 approx. 0.37241 le f_n le f_ {n + 1}
<1 ? about 0.422785
$$
Where $ gamma $ is the EulerMascheroni constant. $ f_n $ Strictly increasing it follows that $ zeta (x) $ decreases sharply in our region of interest $ x ge x_2 $,
So if $ zeta (1 + a_n / b_n) = n $ then we can write $ b_n = ka_n + r_n $ for some integers $ k $ and $ r_n $ so that $ gcd (r_n, a_n) = 1 $ and $ 0.37241 < dfrac {r_n} {a_n} <0.422789 $,
Now for everyone $ a_n ge 1 $ we can eliminate the possible values of $ r_n $ through computer verification to increase the lower limit $ a_n $ As shown below.
Example: Prove $ a_n ge $ 6 to the $ n ge 2 $,
To the $ a_n = $ 1,2,3,4, there is no $ r_n $ satisfying $ 0.37241 < dfrac {r_n} {a_n} <0.422789 $ So these values of $ a_n $ be eliminated. To the $ a_n = 5 $ there is a possible value $ r_n = 2 $ which satisfies both $ gcd (r_n, a_n) = 1 $ and $ 0.37241 < dfrac {r_n} {a_n} <0.422789 $, So we take $ r_n = 2 $ for further testing. We watch that
$$ f_3 approx.0.3932265 < frac {2} {5} <f_4 approx.0.4018059 $$
Since $ dfrac {2} {5} $ lies between two successive values of $ f_3 $ and $ f_4 $ is getting bigger, we'll never have a situation where $ f_n = dfrac {2} {5} $ for each $ n $, Therefore $ r_n = 2 $ is excluded for $ a_n = 5 $, The whole numbers $ 3 $ and $ 4 $ are Witnesses for the elimination of $ dfrac {2} {5} $, With this we have exhausted all possibilities $ a_n = 5 $ and so we can conclude if $ zeta (x_n) in N $ then $ a_n ge $ 6, We can also eliminate them $ a_n = 6.7, ldots $ and so on.
Note: How $ a_n $ increases the number of possible $ r_n $ increases. Everyone $ r_n $ must be removed individually with a witness to completely eliminate $ a_n $, So after a point $ a_n $ will have several witnesses.
definition: A witness for a particular integer pair $ (a, r), a> r $ is an integer $ w_ {a, r} $ so that $ f_ {w_ {a, r}} < dfrac {r} {a} <f_ {1 + w_ {a, r}} $
The existence of a witness $ w_ {a, r} $ is therefore sufficient to prove this $ zeta Big (1 + dfrac {a} {x + r} Big) $ is not an integer for an integer $ x> a> r $, Through actual calculation, I found the witness for every allowed pair $ (a, r), a le 10 ^ 4 $ what that implies
If $ x_n> 1 $ is rational and $ zeta (x_n) in N $ then the difference between
the numerator and denominator of $ x_n $ is bigger than $ 10 ^ 4 $,
Size of the witness
I noticed that most integers have small witnesses, but some of them have big witnesses. I looked at the list of integers $ a $ whose greatest witness for some $ r $ is greater than all witnesses of all integers less than $ a $, Let us call them the maximum witnesses. The first values of the maximum witnesses are given below:
a Max of w_{a,r}
5 3
12 12
19 42
45 130
30 292
97 701
123 3621
518 16503
913 31689
1308 49858
1703 71984
2098 99521
2493 134724
2888 181317
3283 245888
3678 341249
4073 496221
4468 790559
4863 1598102
5258 11274164
Some of the values $ a = $ 5.19,123.5258 are the nonconsecutive denominators in the further group expansion of $ gamma $ but generally each of these values of $ a $ are the denominators in ascending order for the best rational approximation $ gamma $, In other words, the closer it is $ dfrac {r} {a} $ to $ 1 gamma $, the bigger his testimony $ w_ {a, r} $ that eliminates mathematically $ a $ more difficult.
I finally come to my question.
question: The problem is finding positive integer solutions $ zeta (p / q) = u $ synonymous with the decision about the rationality of $ gamma $,
Related question