#trigamma

[yonatanpandukristanto]

Mesin Pembelah & Pelubang Kelapa Muda #trigammateknikmandiri #trigammateknik #trigamma #mesinbelahkelapamuda #mesinbelahkelapa #kelapamuda #coconut #coconutmachine #ypk (at CV Trigamma Teknik) https://www.instagram.com/p/CDklmBLHGOX/?igshid=1htdiylqgxx8e

[bubbloquacious]

Fake numbers, but instead it’s fake variables and we have MEGA XI

[kerlonask]

Julia gama

Julia gama series#

In 2010, a friend encouraged her to enter beauty pageants and she entered A Mais Bela Gaúcha – which was to end up participating two years later, in 2012. Later on, he practiced sports such as volleyball, judo and boxing. In his youth, he even participated in children’s figure skating championships. De 28 años, la joven fue la primera finalista del Miss Universo de esa época y, aunque no ganó, actualmente se encuentra viviendo en un apartamento en la ciudad de Sao Paulo, en su natal Brasil. Julia Gamma says she was a chemical engineering student and worked as an actress in China, where she moved in 2016 and lived for three years. En el certamen del Miss Universo 2020, Andrea Meza tuvo varias contrincantes, entre ellas la modelo Julia Gama. I hope she does her best with this title to help people and give voice to the issues. Even if Some were disappointed in some way, I want to ask for a lot of respect. I just want to wish Andrea good luck, have a wonderful tenure. Julia Gama in childhood After completing her schooling, Julia enrolled herself in the Brazil’s Federal University of Rio Grande do Sul to pursue Chemical Engineering. On the winner Andrea Meza, Julia asked for respect: “I know that not everyone is satisfied with the result, and not everything has gone as we had planned, but this result really means a lot to me. Julia Gama was born on Tuesday, ( age 28 years as of 2021 ), in Porto Alegre, Brazil.

Julia gama series#

Brazil won the competition only in 1963 with Ida Maria Vargas, and in 1968 with Martha Vasconcelos. Mets and Yankees collide again in the Subway Series Julia Gama and Rafael Nieves show how much they know each other in Game Time Julia Gama and Rafael. She has featured her mother on her social media channels. She has gained popularity there for her original vocal and acoustic guitar covers of popular internationally-renowned singles. Brazilian singer who is famous for her eponymous YouTube channel. The second place has not been won by the Brazilian since 2007, when Natalia Guimarães of Minas Gerais took the place. Julia Gama is best known as a YouTube Star. The winner of the competition was Mexican Andrea Meza. inverse of digamma function at x using fixed-point iteration algorithm) trigamma (x) trigamma function (i.e the logarithmic second derivative of gamma at x) polygamma (m,x) polygamma function (i.e the (m+1)-th derivative of the loggamma function at x) gamma (a,z) upper incomplete gamma function. Local News Miss Brazil on fellow Latina and newly crowned Miss Universe 2020 Andrea Meza: I feel represented. “What an intense and unforgettable night, a mixture of feelings that I will never be able to translate! I want you to feel my love on stage! Representing Brazil was one of the greatest tributes of my life,” declared the Brazilian. Gaucho Julia Gama, Miss Brazil, came second. Julia Gama was born in in November 14, 1995. Brazil won second place in the 69th edition of the Miss Universe contest held in Miami, USA. Julia Gama net worth and salary: Julia Gama is a YouTube Star who has a net worth of 23 million.

[qifeng618618]

Some papers published in Applied and Computational Mathematics by Dr. Prof. Feng Qi

Some papers published in Applied and Computational Mathematics by Dr. Prof. Feng Qi

Some papers published in Applied and Computational Mathematics Feng Qi, Necessary and sufficient conditions for a difference defined by four derivatives of a function containing trigamma function to be completely monotonic, Applied and Computational Mathematics 21 (2022), no. 1, accepted on 30 December 2021. Related…

[craigbrownphd-blog-blog]

A Beautiful Result in Probability Theory

#ICYDK: This is another spectacular property of the exponential distribution, and also the first time an explicit formula is obtained for the variance of the range, besides the uniform distribution. It has important consequences, and the result is also useful in applications. Theorem The range R(n) associated with n independent random variables with an exponential distribution of parameter l satisfies Before proving the theorem, note that the first formula is well known, only the second one is new. The standard proof for the expectation is not considered simple: it is based on computing the expectation for the maximum (see here) and the fact that the minimum also has an exponential distribution with known expectation (see here). Our proof is simpler and also covers the variance. Proof The general distribution of the range is known for any distribution, see here. The range is defined as In the case of the exponential distribution, the range computed on n random variables has the following density (see here page 3):  With a simple change of variable, the k-th moment of the range is equal to Using WolframAlpha (see here and here) one obtains Thus, The two symbols H(n-1) and ψ1(n) represent the harmonic numbers and the Trigamma function, respectively. To complete the proof, use the fact that ∎  There are a number of interesting consequences to this result. First, the expectation of the range grows indefinitely and is asymptotically equal to log(n). Also, the variance of the range grows slowly and eventually converges to p^2 / 6 This is in contrast to the uniform distribution: its range is bounded, and its variance tends to zero as fast as 1 / n^2, see section 2.3 in this article.   This result is pretty deep. It is almost like the range, for the exponential distribution, is made up of a weighted sum of independent exponential variables with same parameter λ, with the k-th term added into the sum contributing with a weight equal to 1 / k.  But perhaps most importantly, we found the two extreme cases to a new statistical theorem (see here, section 1) stating that the length of any confidence interval attached to an estimator is asymptotically equal to A / n^B, with B between 0 and 1. This length is usually proportional to the standard deviation of the estimator in question. In practice, in almost all cases, B = 1/2. However, here we have: * For the range, if the variables are independently and uniformly distributed, then B = 1. * For the range, if the variables are independent with an exponential distribution, then B = 0. http://bit.ly/2WcU6S6

[mathematicianadda]

Integrate $\operatorname{PV}\int_0^{\infty}\frac{x\tan(\pi x)}{(1+x^2)^2}dx$ https://ift.tt/eA8V8J

A friend of mine send me the problem to integrate

$$\operatorname{PV}\int_0^{\infty}\frac{x\operatorname{tan}(\pi x)}{(1+x^2)^2}dx$$ where $\operatorname{PV}$ is Cauchy principle value.

I'm getting $\frac{1}{2}\psi^1\left(\frac{1}{2}+i\right)$ which is complex trigamma argument however, he has got answer in real closed form $\frac{\pi^2}{(e^{-\pi}+e^{\pi})^2.}$.

My work

I evaluted integral as follows

Recalling the result to due Weiestrass factorization theorem $$ \operatorname{cos}(\pi x)=\prod_{n\geq 0}\left(1-\frac{4 x}{(2n+1)^2}\right)$$ taking $\log$ and differentiating with rest to $x$ we have $$\operatorname{tan}(\pi x)=\frac{8}{\pi}\sum_{n\geq 0}\frac{x}{(2n+1)^2-4x^2}$$ and thus subbing this result the integral, we have $$\frac{8}{\pi}\sum_{n\geq 0}\int_0^{\infty}\frac{x^2dx}{((2n+1)^2-4x^2)(1+x^2)^2}$$

Making partial fraction and integration gives us $$\sum_{n\geq 0}\frac{2}{(2n+2i+1)^2}=\frac{1}{2}\psi^{1}\left(\frac{1}{2}+i\right)$$

As the per WA $$\Re\frac{1}{2}\left(\psi^1\left(\frac{1}{2}+i\right)\right)=\frac{\pi^2}{(e^{-\pi }+e^{\pi})^2}\tag{1}\label{mainfm}$$

My question is, How do i prove the relation \eqref{mainfm}?

I tried for the references of relationship between hyperbolic function and trigamma function however, I cannot get any such relationship.

Any sort of help/ reference or different approches will be appreciated. Thank you.

Interestingly, making change of $\operatorname{tan}(\pi x)$ as $\operatorname{tanh}(\pi x)$ I came up with the following closed form

$$\int_0^{\infty}\frac{x\tanh(\pi x)}{(1+x^2)^2}dx=\frac{\pi^2}{4}-2$$

from Hot Weekly Questions - Mathematics Stack Exchange Naren Naruto from Blogger https://ift.tt/30s7aT8

[kjunichi]

GitHub - JuliaMath/SpecialFunctions.jl: Julia implementations of Bessel and Airy functions [はてなブックマーク]

GitHub - JuliaMath/SpecialFunctions.jl: Julia implementations of Bessel and Airy functions

Special mathematical functions in Julia, including Bessel, Hankel, Airy, error, Dawson, eta, zeta, digamma, inverse digamma, trigamma, and polygamma functions. These functions were formerly part of Ba...

kjw_junichi プログラミング, julia

from kjw_junichiのブックマーク http://ift.tt/2jXlRem

[hannahlettington-blog]

Finally a Tri Gamma sister

[nanabear-blog]

One of the best people I was blessed to meet at USF & get to call my sister/baby bear got married today. Congratulations Lexie + Chris! Thank you for inviting us to celebrate your beautiful union 😘🙌🏽🎉👰🏽🍾 #trigamma #canyonview #duenaspartyoftwo (at Canyon View)

[simplyjanel]

Couldn't have asked for a better Big💕💕 #trigamma #biglittlereveal (at University of San Francisco)

[yonatanpandukristanto]

THUMBLING CAP TESTER MACHINE for AQUA 5 GALLON - made by CV TRIGAMMA TEKNIK MANDIRI #cvtrigammateknikmandiri #cvttm #trigamma #trigammateknik #trigammateknikmandiri #ypk #aqua #aquagoldenmississipi #tirtainvestama #thumblingcaptester (at CV Trigamma Teknik) https://www.instagram.com/p/CCmvtkmnUmD/?igshid=1tmreqkzalucf

[mark-see]

Congratulations to @kallmekimi and Chad!! I never got to go to Princess Diana & Prince Charles's wedding, buuuut def felt like it after this weekend lol! #SF #xyz #trigamma #royalwedding #juliamorgan #si #usf #cyclone #bringingitbackto2007 (at The Julia Morgan Ballroom)

[monikaroxanne]

well lots of planning, lots of passive aggressive talks, lots of help kinda sorta not rly, lots of flowers, lots of ppl, lots of money, lots of drinking after but never ending support. that's a wrap, fashion show 2k14.

[qifeng618618]

Some papers published in Arab Journal of Basic and Applied Sciences by Dr. Prof. Feng Qi

Some papers published in Arab Journal of Basic and Applied Sciences Feng Qi, Bounds for completely monotonic degree of a remainder for an asymptotic expansion of the trigamma function, Arab Journal of Basic and Applied Sciences 28 (2021), in press.

[craigbrownphd-blog-blog]

A Beautiful Result in Probability Theory

#ICYMI: This is another spectacular property of the exponential distribution, and also the first time an explicit formula is obtained for the variance of the range, besides the uniform distribution. It has important consequences, and the result is also useful in applications. Theorem The range R(n) associated with n independent random variables with an exponential distribution of parameter l satisfies Before proving the theorem, note that the first formula is well known, only the second one is new. The standard proof for the expectation is not considered simple: it is based on computing the expectation for the maximum (see here) and the fact that the minimum also has an exponential distribution with known expectation (see here). Our proof is simpler and also covers the variance. Proof The general distribution of the range is known for any distribution, see here. The range is defined as In the case of the exponential distribution, the range computed on n random variables has the following density (see here page 3):  With a simple change of variable, the k-th moment of the range is equal to Using WolframAlpha (see here and here) one obtains Thus, The two symbols H(n-1) and ψ1(n) represent the harmonic numbers and the Trigamma function, respectively. To complete the proof, use the fact that ∎  There are a number of interesting consequences to this result. First, the expectation of the range grows indefinitely and is asymptotically equal to log(n). Also, the variance of the range grows slowly and eventually converges to p^2 / 6 This is in contrast to the uniform distribution: its range is bounded, and its variance tends to zero as fast as 1 / n^2, see section 2.3 in this article.   This result is pretty deep. It is almost like the range, for the exponential distribution, is made up of a weighted sum of independent exponential variables with same parameter λ, with the k-th term added into the sum contributing with a weight equal to 1 / k.  But perhaps most importantly, we found the two extreme cases to a new statistical theorem (see here, section 1) stating that the length of any confidence interval attached to an estimator is asymptotically equal to A / n^B, with B between 0 and 1. This length is usually proportional to the standard deviation of the estimator in question. In practice, in almost all cases, B = 1/2. However, here we have: * For the range, if the variables are independently and uniformly distributed, then B = 1. * For the range, if the variables are independent with an exponential distribution, then B = 0. http://bit.ly/2YNwn7O

[mathematicianadda]

Evaluate $\int _0^{2 \pi }\int _0^{2 \pi }\log (3-\cos (x+y)-\cos (x)-\cos (y))dxdy$

How can we prove that: $$\int _0^{2 \pi }\int _0^{2 \pi }\log (3-\cos (x+y)-\cos (x)-\cos (y))dxdy= -4 \pi ^2 \left(\frac{\pi }{\sqrt{3}}+\log (2)-\frac{\psi ^{(1)}\left(\frac{1}{6}\right)}{2 \sqrt{3} \pi }\right)$$ Where $\psi^{(1)}$ denotes trigamma function. It's J. Borwein's review on experimental mathematics that offers this interesting identity (I've verified it numerically). The literature refer this formula to V.Adamchik, but I haven't find any related source dealing with this kind of integrals. In fact, I've asked this question on this site a year before but no answer was given. Since I have still no solution, I'd like you to give some suggestions again. Any help will be appreciated.

from Hot Weekly Questions - Mathematics Stack Exchange from Blogger https://ift.tt/2Vy5430