## Mathematica erroneously says this gamma integral does not converge

Mathematica does not recognize this case:

``````Integrate((T^a/(Gamma(a))) *  (E^(-Tx)) (x^(a - 1)), {x, 0, Infinity})
Integrate::idiv: Integral of (E^-Tx T^a x^(-1+a))/Gamma(a) does not converge on {0,(Infinity)}.
``````

This should converge because

## coding theory – Variance Gamma parameter estimation in R studio

I am using the vgFit function in R Studio to estimate parameters for stock prices. I converted the stock prices to a vector and I get this error. My code was:

``````param<-c(0,1,0,1)
fit<-vgFit(c(x), param=param)
``````

The error is:

``````Error in optim(paramStart, llsklp, NULL, method = startMethodSL, hessian = FALSE,  :
function cannot be evaluated at initial parameters
``````

## analytic number theory – Is every modular function on \$Gamma\$ univalent?

Suppose $$f$$ is a modular function on $$Gamma$$ then it has a fundamental region $$R_L$$. Since $$f$$ is modular, it can be expressed as a rational function of $$J(tau)=frac{g_2^3(tau)}{Delta(tau)}$$. To prove this we actually construct a rational function
$$g(tau)=frac{prod(J(a_i)-J(tau))}{prod(J(b_i)-J(tau))}$$
where the $$a_i,b_i$$ are zeros and poles of $$f$$. It seems to me that this construction implies the possibility of having multiple poles in the fundamental region. I know for sure that $$J$$ has only one pole since it attains every value exactly once.

Since a modular function is univalent if and only if its fundamental region has genus $$1$$ and if we identify the edges of $$R_Gamma$$ it would become homeomorphic to $$mathbb{P}^1$$, any modular function that has the fundamental region $$R_Gamma$$ should be univalent, that is, any modular function is univalent in the closure of its fundamental region.

This really confused me and I am not sure if this is correct.

## Fitting data to left skewed gamma distribution

How can I fit the following two set of data to a left skewed gamma function, which I what I think should fit the data best?:

data 1 is here: https://pastebin.com/X2HTjTP7
data 2 is here: https://pastebin.com/8Rh4BHDT

Is there any other suggestion of what would be the best distribution or equation to fit the data?

A picture of how data 1 looks is here:

A picture of how data 2 looks is here:

## simplifying expressions – Is there any way to prioritize Gamma function?

In Mathematica, if you type

``````Gamma[1, 0, -a]
``````

It automatically simplifies to `1 - E^a`. I wonder though if there is a way to stop Mathematica from simplify Gamma function automatically.

I understand that I can use `Inactive` or `HoldFrom`. But I actually want having to remember using Inactive every time.

## How to calculate gamma function values ​​for quaternions with Mathematica?

How can we calculate the values ​​for that? gamma Function with Quaternions on Mathematica?

For example:

``````Gamma(5) = 24

N(Gamma(I)) = -0.15495 - 0.498016 I

N(Gamma(1 + 2 I)) = 0.151904 + 0.0198049 I

N(Gamma(Quaternion(5, 0, 0, 0))) = Gamma(Quaternion(5., 0., 0., 0.)) (* which is wrong *)
``````

How can we calculate specifically?

``````Gamma(Quaternion(1, 2, 3, 4))?
``````

Tries:

``````N(Gamma(FromQuaternion(Quaternion(1, 2, 0, 0)))) (* works *)
N(Gamma(FromQuaternion(Quaternion(1, 2, 3, 4)))) (* doesn't work *)

QuaternionPower(x_, y_) := E^(y ** Log(x))
QuaternionGamma(z_) := Integrate(QuaternionPower(E, -x + Log(x)*(z - 1)), {x, 0, Infinity})
N(QuaternionGamma(FromQuaternion(Quaternion(1, 2, 0, 0)))) = 0.151904 + 0.0198049 I (* works *)
N(QuaternionGamma(FromQuaternion(Quaternion(1, 2, 3, 0)))) = ConditionalExpression(Gamma((1.  + 2. I) + 3. J), Re(J) > -0.333333) (* doesn't work *)
``````

## Logic – If a set of sentences \$ Gamma \$ has an infinite model, does it also have a denumerable model?

I think that's wrong, since we can only conclude from the Downward-Lowenheim-Skolem theorem $$Gamma$$ has an enumeration model, it is not shown whether this enumeration model is infinite. Certainly ours $$Gamma$$ already has an infinite model, but nothing says that its enumerable model must be identical to its infinite model.

I can't see how this claim is true, but I think it should be true?

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## 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 1-2 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
$$a geq 0, c geq 0, x geq 0, y geq 0,$$
$$a + c + x + y = 1,$$
$$(a + c) ^ 2 + (x + y) ^ 2 + 2xc leq 1-2 gamma,$$
Where $$gamma$$ is a fixed constant so that $$0 leq gamma leq 1/4$$.

Maple showed me that the maximum value is reached at $$frac { log (6)} {2} + frac { sqrt {1-4 gamma}} {2} log (3/2)$$that is given by $$a = frac {1+ sqrt {1-4 gamma}} {2}$$, $$b = frac {1- sqrt {1-4 gamma}} {2}$$, $$c = y = 0$$.

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 $$log (4) c + log (3) a + log (2) x leq frac { log (6)} {2} + frac { sqrt {1-4 gamma}} { 2} log (3/2)$$ 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 {5-1} {2} x + frac {5- 3} {2} y right) – log (9/8) \ & = log (3/2) left ( frac {5+ sqrt {1-4 gamma}} {2} right) – log (9/8) + log (3/2) left ( frac {ax + 3 (cy) – sqrt {1-4 gamma}} {2} right) \ & = frac { log (6)} {2} + frac { sqrt {1-4 gamma}} {2} + log (3/2) left ( frac {ax + 3) ( cy) – sqrt {1-4 gamma}} {2} right). \ end {align *}
That is, to get the inequality I want, it would be enough to show it $$a-x leq 3 (y-c) + sqrt {1-4 gamma}$$. 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