TITLE: Punctured Ball not homeomorphic to Euclidean Space
QUESTION [2 upvotes]: Just another ordinary day with another (great) ordinary mathy conversation. A friend and I got to ask this:
Problem. Prove that $B(0,r) \setminus \{0 \} \subseteq \mathbb{R}^n$ is not homeomorphic to open balls for $r > 0$.
I have not yet had a topology course and my friend has just started a topology course, so we tried to find a solution as elementary as possible.
It seems easy enough, though we struggled...

It suffices to prove that it's not homeomorphic to $\mathbb{R}^n$.
Most elementary topological invariants don't work.

The best we came up with, is to compute the fundamental group and take generalizations of the fundamental group for $n > 2$. That should probably work.
So here's my question: Is there a different way to prove this result?

REPLY [2 votes]: As done here you can use the following topological invariant:

For every compact subset $K$ of $X$ there is a compact $K^\prime\subset X$ containing $K$ such that $X-K^\prime$ is connected.

$\mathbb{R}^n$ has this property, but $\mathbb{R}^n-\{0\}$ (or $B(0,r)-\{0\}$) doesn't.


TITLE: Changing Momentum
QUESTION [1 upvotes]: Okay, so I am having a hard time wrapping my head around this. Before looking at the answer, I worked the example and got the answer only because I used dimensional analysis to get me to the right units for force. However, to be honest, I don't think I even understand the concept here. I know mass is changing and velocity is constant, but what is throwing me off is where it says "The water strikes the window at $32\, m/s$ so each kilogram of water loses $32\, kgm/s$ of momentum. Water strikes the window at the rate of $45\,kg/s$, so the rate at which it loses momentum to the window is $1400\, kgm/s^2$" How is it losing $32\, kgm/s$ of momentum and then at the same time also losing it at a rate of $1400\,kg m/s^2$?
Why does it even mention "The water strikes the window at $32\,m/s$ so each kilogram of water loses $32\,kgm/s$ of momentum" what is the purpose of that sentence. That is really confusing me.
EDIT: I added this simple diagram to show how I see this system. Where is $32\,kgm/s$ of momentum come in? (I am aware that the window is now showing)

REPLY [2 votes]: Let's take the picture you drew to visualize this and run with it; I think it's a pretty good way to think about the problem.
Every second, a $45\,\mathrm{kg}$ chunk of water strikes the window horizontally after which its horizontal velocity drops to zero instantly.  Since the momentum of the water is its mass times its velocity, and since the velocity of the water before it strikes the window is $32\,\mathrm m/\mathrm s$, we find that before it strikes the window, each $45\,\mathrm{kg}$ chunk of water has a total momentum equal to $45\,\mathrm{kg}\cdot 32\,\mathrm m/\mathrm s = 1440 \,\mathrm{kg}\, \mathrm m/\mathrm s$.
On the other hand, right after the water strikes the window, its velocity is zero, so it has zero momentum.  Therefore, the magnitude of the change in momentum of each such chunk of water is $1440 \,\mathrm{kg}\, \mathrm m/\mathrm s$, and this change in momentum occurs every second, namely a full chunk hits the window and stops every second.
It follows that the rate of change of the momentum of the stream of water has magnitude $1440 \,\mathrm{kg}\, \mathrm m/\mathrm s^2$; this is the rate at which the momentum of water stream decreases due to the successive chunks hitting the wall and stopping second after second.


TITLE: Physical intuition on vortex sound generation
QUESTION [3 upvotes]: Whenever a fluid has a nonzero vorticity, it looses some of its energy in a form of sound wave. Formally is this mechanism described by Lighthill's equation or some related model (like e.g. Curle's theory). But I have a lack of physical intuition on this process. Derivation of the Lighthill's model is based on disturbances of momentum flux tensor in a stokesian fluid, so nothing straightforward. I am fully capable of understanding this derivation with all the maths, but I can't "explain that to my grandmother in plain english".
Could anybody provide that sort of explanation?
Note 1: I am mainly interested in generating noise in turbulent fluid states. Tone-like sounds generated by periodical emerging of a macroscopic vortex eddies (Strouhal's Aeolian sound) are clear to me.
Note 2: I am aware of brilliant M. C. Howe's publication on the Theory of Vortex Sound. I have read it. But it is mainly on applied maths - that's why I am asking this question.

REPLY [2 votes]: Good question. My Problem is that I can't give you a "Mainstream answer".
The reason is that Turbulence is considered happening In three space dimensions and time, given an initial velocity field, there exists a vector velocity and a scalar pressure field, which are both smooth and globally defined, that solve the Navier–Stokes equations.
But with these premise you are not able to produce sound, which needs vibration. Why would a  single object start to ever vibrate through internal stresses and tensions?
There simply must be something else, as the vibration allways needs some kind of a shock, or sudden release.
My explanation is, that the Turbulence is collision and friction, instead of viscous forces. And these collisions are able to produce vibrations, aka Sound. A Turbulent flow is a fluid cutted in peaces, internal surface tensions inside the fluid.
But as said, this is not "Mainstream physics", the mainstream physics just haven't explained this yet. This forces me to make another approach; Vortex tube The German language version of this wiki explains how the whistling sound is crucial for the Functionality. If the sound is removed, the temperature differences which can be produced drops from 40 Kelvin to just few Kelvin's. This supports the idea of the separating surfaces inside the fluid. The whistling sound comes from these surfaces, and when this sound disappears the fluids are contacted and the temperature is transferred.
Answer; The noise is caused by the collisions of the separated Fluid components.
I hope this helps. But also feel free to vote this down. Our thoughts or votes doesn't really change anything in nature.


TITLE: Unitarity in numerical simulations of Schrodinger equation
QUESTION [0 upvotes]: I would like to simulate directly time evolution of a quantum state in basic Schrodinger equation
$$
i\hbar\frac{\text{d} \psi}{\text{d} t}=\hat{H}\psi,
$$
where $\psi$ has high dimensionality. I would like to model the evolution by discretizing the time. Unfortunately, most numerical schemes violate the unitarity of evolution. Is there any numerical scheme that preserves the unitarity without global rescaling?

REPLY [4 votes]: Check out Visscher's method, which is described in Wally's answer here, at scicomp.stackexchange.
I have used it to create some fun animations of the Schrodinger equation.


TITLE: Stability of a system of ODEs
QUESTION [1 upvotes]: It is well known that for a system of ODEs, $\dot{\boldsymbol{y}} = \boldsymbol{Ay}$, the global stable equlibrium point is given by the eigenvector that correponds to the largest eigenvalue of $\boldsymbol{A}$, provided that the matrix $\boldsymbol{A}$ is primitive. This is according to the Perron-Frobenius theorem.
Then my question is that for a system $\dot{\boldsymbol{y}} = \boldsymbol{Ay} + \boldsymbol{B}\boldsymbol{y}\boldsymbol{y}^{T}$, do we have a similar theory for the global stability as that for $\dot{\boldsymbol{y}} = \boldsymbol{Ay}$?
We know that if we have $\boldsymbol{y}^{\ast}$ such that $\boldsymbol{0} = \boldsymbol{Ay}^{\ast} + \boldsymbol{B}\boldsymbol{y}^{\ast}\boldsymbol{y}^{\ast T}$, and the eigenvalues of the Jacobian matrix $J(\boldsymbol{y}^{\ast})$ are negative, we can conclude $\boldsymbol{y}^{\ast}$ is locally stable, but how about the global stablibility?
If all the entries of $\boldsymbol{y}(0)$ are between $0$ and $1$, do things change?
Thank you guys for helpful discussions!

REPLY [1 votes]: In the case of $1\times 1$ matrices, and $b\neq 0$, your equation is
$y'=ay+by^2$,
 which is never globally stable.


TITLE: The set of diffeomorphisms preserving some metric.
QUESTION [5 upvotes]: Let $M$ be a finite-dimensional, smooth manifold. Call a diffeomorphism $f : M \rightarrow M$ diagonalizable if there exists a Riemannian metric $g$ on $M$ such that $f : (M, g) \rightarrow (M, g)$ is an isometry. I have some questions regarding such objects.
a) Is the set Diag(M) of all diagonalizable diffeomorphisms a group under composition?
b) Note that, in order to be diagonalizable, a diffeomorphism must possess the following well-known property of isometries:
$$\text{If}~f(p) = p~\text{and}~df(p) = \mathrm{Id}_{T_pM}, \text{for some $p \in M$, then}~f = \mathrm{Id}_M. (*)$$
Is $(*)$ also a sufficient condition? In other words, given a diffeomorphism $f \in \mathrm{Diff}(M)$ satisfying $(*)$, is there a Riemannian metric for which $f$ is an isometry? Maybe this is too general, because any diffeomorphism not fixing any point satisfies $(*)$, but I don't know the answer.
My motivation here is to know how large is the set of diffeomorphisms that could be isometries within the set of all diffeomorphisms. I apologize if I'm missing some standard notation and/or vocabulary here. I'd appreciate, as always, any references.
Thanks in advance.

REPLY [0 votes]: Regarding your question (a), the set $\text{Diag(M)}$ of all diagonalizable diffeomorphisms is not always closed under composition. (So in particular it is not a group).
For example take $M=\mathbb{R}^n$ , and consider linear diffeomorphims.
It is proved here that a linear automorphism $T:V \rightarrow V$ preserves some inner product on $V$ if and only if the matrix of $T$ w.r.t an arbitrary basis is similar to an orthogonal matrix. The point is that composition of two transformation of this type is not necessarily also of that type. (Which amounts to proving that the union of conjugacy classes of $O(n)$ is not a subgroup, see here).
In particular, if you take $A = \pmatrix{0&2\\-1/2&0}$, then $A,A^T$ both preserves some inner products on $\mathbb{R}^2$ , but their product $AA^T = \pmatrix{4&0\\0&\frac14}$ does not.
While the result above holds pointwise (it is an algebraic result, true for a single vector space, which can be thought of as a tangent space), note that for linear transformations, we can identify them with their differentials (after identifying the tangent spaces canonically with $\mathbb{R}^n$).
Now arise the natural question: what about other manifolds?


text:
\begin{document}

\title[{Analytic computations of digamma functions }]{Analytic computations of digamma function\\ using some new identities}

\author[M.I. Qureshi and M. Shadab]{Mohammad Idris Qureshi and Mohd Shadab$^{*}$}

\address{Mohammad Idris Qureshi:   Department of Applied Sciences and Humanities,
 Faculty of Engineering and Technology,
 Jamia Millia Islamia (A Central University),
 New Delhi 110025, India}
\email{miqureshi\_delhi@yahoo.co.in}

\address{Mohd Shadab: Department of Applied Sciences and Humanities,
 Faculty of Engineering and Technology,
 Jamia Millia Islamia (A Central University),
 New Delhi 110025, India}
\email{shadabmohd786@gmail.com}

\keywords{ Digamma (Psi) function; Generalized hypergeometric series; Euler's constant; Gamma function. }

\subjclass[2010]{11J81, 33B15, 11J86, 33C05.}

\thanks{*Corresponding author}

\begin{abstract}
Motivated by rigorous development in the theory of digamma functions, we have first derived some new identities for the digamma function, and then computed the values of digamma function for the fractional orders using these identities conveniently.
 \end{abstract}

\maketitle

\section{Introduction and preliminaries}\label{Intro}
A natural property of digamma (Psi function) function to be used as application in the theory of beta distributions-probability models for the domain [0,1]. It is mainly used in the theory of special functions with a wide range of the applications. Digamma functions are directly connected with many special functions such as Riemann's zeta function and Clausen's function etc.\\

 Many authors have contributed to develop the theory of polygamma function with respect to properties \cite{Qi, Gauss, Jensen, Lehmer, Mahler}, inequalities \cite{Alzer, Batir, Clark}, monotonicity \cite{Qi1, Qi2, Qi3, Qi4}, series \cite{Bor, de, Lewin, Wu, Gosper, Grossman}, and fractional calculus \cite{Adam, Al-saqabi, Sri1, Sri2}.\\

The Gamma function, $\Gamma{(z)}$, was introduced by Leonard Euler as a generalization of the factorial function on the sets, $\mathbb{R}$ of all real numbers, and $\mathbb{C}$ of all complex numbers. It (or, Euler's integral of second kind) is defined by

\begin{eqnarray}\label{eq(1.1)}
\Gamma{(z)}&=&\int_{0}^{\infty}\exp{(-t)}t^{z-1}dt,\qquad \Re(z)>0\nonumber\\
&=&\lim_{n\rightarrow\infty}\int_{0}^{n}{\left(1-\frac{t}{n}\right)^n}t^{z-1}dt.
\end{eqnarray}
\vskip.2cm
In 1856, Karl Weierstrass gave a novel definition of gamma function
\begin{eqnarray}\label{eq(1.2)}
\frac{1}{\Gamma{(z)}}&=&z \exp{(\gamma z)} \prod_{n=1}^{\infty}\left[\left(1+\frac{z}{n}\right)\exp{\left(-\frac{z}{n}\right)}\right],
\end{eqnarray}
where $\gamma= 0.577215664901532860606512090082402431042\dots$ is called Euler-Mascheroni constant, and
$\frac{1}{\Gamma{(z)}}$ is an entire function of $z$, and
\begin{eqnarray}
\gamma&=&\lim_{n\rightarrow\infty}\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{n}-\ell n{(n)}\right)\nonumber.
\end{eqnarray}

The function

\begin{eqnarray}
\psi(z)=\frac{d}{dz}\{\ell n \,{\Gamma(z)}\}=\frac{\Gamma ^{\prime}(z)}{\Gamma(z)},
\end{eqnarray}
or, equivalently
\begin{eqnarray}
\ell n \,{\Gamma(z)}=\int_{1}^{z}\psi(\zeta)d\zeta,
\end{eqnarray}

is the logarithmic derivative of the gamma function (Psi function or digamma function).\\

$\psi^{(i)}(z)$ for $i\in \mathbb{N}$ are called the polygamma functions, and $\psi$ has the presentation as
\begin{eqnarray}
\psi(z)=\frac{\Gamma ^{\prime}(z)}{\Gamma(z)}
=-\gamma+\int_{0}^{\infty}\frac{e^{-t}-e^{-zt}}{1-e^{-t}}dt  \quad (\gamma= \text{ Euler's constant}).
\end{eqnarray}

The Psi function has following series representation
\begin{eqnarray}
\psi(z)=-\gamma-\frac{1}{z}+\sum_{n=1}^{\infty}\frac{z}{n(z+n)}, \quad z \ne 0,-1,-2,-3,\dots
\end{eqnarray}

The {\it generalized hypergeometric function} ${_p}F_q$, is defined by

\begin{eqnarray}\label{eq.(1.11)}
{_p}F_{q} \left[ \begin{array}{r}(a_p); \\ (b_q); \end{array} z \right] = \sum_{m=0}^{\infty} \frac{[(a_p)]_m}{[(b_q)]_m} \frac{z^m}{m!} ,
\end{eqnarray}

\begin{itemize}
  \item p and q are positive integers or zero,
  \item $z$ is a complex variable,
  \item $(a_p)$ designates the set ${a_1,a_2, . . . , a_p}$,
  \item the numerator parameters $a_1, . . . , a_p \in \mathbb{C}$ and the denominator parameters $b_1, . . . , b_q \in \mathbb{C}\setminus {\mathbb{Z}^{-}_{0}}, $
  \item $[(a_r)]_k = \displaystyle\prod_{i=1}^{r} (a_i)_k$. By convention, a product over the empty set is 1,
  \item $(a)_k$ is the Pochhammer's symbol.
\end{itemize}

The widely used Pochhammer symbol $(\lambda)_{\nu}$ ~$(\lambda, \nu \in\mathbb{C})$ is defined by
\begin{equation}
\left(\lambda\right)_{\nu}:=\frac{\Gamma\left(\lambda+\nu\right)}{\Gamma\left(\lambda\right)}=\begin{cases}
\begin{array}{c}
1\\
~\\
\lambda\left(\lambda+1\right)\ldots\left(\lambda+n-1\right)
\end{array} & \begin{array}{c}
\left(\nu=0;\lambda\in\mathbb{C}\setminus\left\{ 0\right\} \right)\\
~\\
\left(\nu=n\in\mathbb{N};\lambda\in\mathbb{C}\right),
\end{array}\end{cases}
\end{equation}
it being understood $conventionally$ that $\left(0\right)_{0}=1$, and assumed $tacitly$ that the $\Gamma$ quotient exists.
\vskip.2cm
Thus, if a numerator parameter is a negative integer or zero, the $_pF_q$ series terminates, and then we are led to a generalized hypergeometric polynomial.
\vskip.2cm

In 1813, Gauss \cite{Gauss} (see also, Jensen \cite[p.146, eq.(32)]{Jensen};  \cite[p.19, (1.7.3) eq.(29)]{Erde}; B\"{o}hmer \cite[p.77]{Boh} ) discovered an interesting formula for digamma (Psi) function as follows

\begin{eqnarray}\label{eq(Gauss)}
\psi(p/q) = -\gamma - \ell n\,{(q)}-\frac{\pi}{2}\cot{\left(\frac{\pi p}{q}\right)}+\sum_{j=1}^{[\frac{q}{2}]}{'}\left\{\cos\left({\frac{2\pi j p}{q}}\right)\ell n\, {\left(2-2\cos{\frac{2\pi j}{q}}\right)}\right\},
 \end{eqnarray}

where $1 \le p < q$ and $p, q$ are positive integers, and accent(prime) to right of the summation sign indicates the term corresponding to (last term) $j = \frac{q}{2}$ (when $q$ is positive even integer) should be divided by 2.\\

A different form of Gauss formula is also given in N. Nielsen \cite[p. 22, an equation between equations (7) and (8)]{Nielsen} as follows

\begin{eqnarray}\label{eq(Gauss1)}
\psi(p/q) = -\gamma - \ell n\,{(q)}-\frac{\pi}{2}\cot{\left(\frac{\pi p}{q}\right)}+\sum_{j=1}^{q-1}\left\{\cos\left({\frac{2\pi p j }{q}}\right)\ell n\, \left(2\sin\left(\frac{\pi j}{q}\right)\right)\right\},
 \end{eqnarray}

where $1 \le p < q$ and $p, q$ are positive integers.

\vskip.2cm

Afterwards, in 2007, a simplified treatment of the above formula was made by Murty and Saradha \cite[p. 300, after eq.(4)]{Murty} (see also, Lehmer \cite[p. 135, after eq.(20)]{Lehmer}) ) such that

\begin{eqnarray}\label{eq(Murty)}
\psi(p/q)=-\gamma-\ell n\,{(2q)}-\frac{\pi}{2}\cot{\left(\frac{\pi p}{q}\right)}+2\sum_{j=1}^{[\frac{q}{2}]}\left\{\cos{\left(\frac{2\pi pj}{q}\right)}\ell n\,{ \sin{\left(\frac{\pi j}{q} \right)}}\right\},
\end{eqnarray}
where $p = 1, 2, 3, \dots ,(q-1) ,\,  q = 2, 3, 4, \dots ; (p,q)= 1$.\\

Also, we have verified the results \eqref{eq(Gauss)}, \eqref{eq(Gauss1)} and \eqref{eq(Murty)} by taking different values of $p$ and $q$.

An erroneous formula of digamma function is also recorded in Gradshteyn and Ryzhik \cite[p. 904, eq 8.363(6)]{Gradshteyn} such that

\begin{eqnarray}\label{eq(Err)}
\psi(p/q) \circeq -\gamma-\ell n\,{(2q)}-\frac{\pi}{2}\cot{\left(\frac{\pi p}{q}\right)}+2\sum_{j=1}^{[\frac{q+1}{2}]-1
}\left\{\cos{\left(\frac{2\pi pj}{q}\right)}\ell n\,{ \sin{\left(\frac{\pi j}{q} \right)}}\right\},
\end{eqnarray}
where $p = 1, 2, 3, \dots ,(q-1) ,\,  q = 2, 3, 4, \dots ; (p,q)= 1$ and the symbol $\circeq$ exhibits the fact that equation \eqref{eq(Err)} does not hold true as stated.

{\bf Some important facts:}
\begin{itemize}
\item We can not compute the value of digamma function when $p > q$ or ( and ) $\frac{p}{q}$ is negative fraction using Gauss formula \cite{Gauss}.

\item We can not compute the value of digamma function when $p > q$
using Jensen formula \cite{Jensen}.

\item We can not compute the value of digamma function when $\frac{p}{q}$ is negative
using Jensen \cite{Jensen}.

\item  Murty and Saradha \cite[p. 300]{Murty} corrected a formula of Lehmer \cite[p. 135]{Lehmer} for $\psi(\frac{p}{q})$ .

\item The value of digamma function has been proved transcendental with the help of Gauss formula in \cite{Murty}.
\end{itemize}

\vskip.2cm
\section{Some new identities for digamma function}
Here, we derive some interesting  and new identities for the computation of digamma function of fractional order.
\vskip.2cm

Some functional relations for digamma function that are easily derivable from the properties of the gamma function. Indeed, from the formula,

\begin{eqnarray}\label{eq(main-2.1)}
\Gamma(z+1)=z\Gamma(z),\quad
\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin{(\pi z)}},  \quad z\ne0,\pm1,\pm2,\pm3,\dots
\end{eqnarray}
taking $\ell n$ both sides and differentiating the above equation with respect to z, we derive the some basic identities for digamma function as follows
\begin{eqnarray}\label{eq(main-2.2)}
\psi(z+1) = \psi(z)+\frac{1}{z}, \quad
\psi(1-z) = \psi(z)+\pi \cot{(\pi z)},  \quad z\ne0,\pm1,\pm2,\pm3,\dots
\end{eqnarray}

\begin{eqnarray}\label{eq(main-2.3)}
\psi(z+n) = \frac{1}{z} + \frac{1}{z+1} + \dots  +\frac{1}{z+n-1} + \psi(z).
\end{eqnarray}

On setting $z=1-z$ in equation \eqref{eq(main-2.2)}, we get
\begin{eqnarray}\label{eq(main-2.4)}
\psi(-z) =  \frac{1}{z} + \psi(1-z).
\end{eqnarray}

On comparing the values of $\psi(1-z)$ from the equations \eqref{eq(main-2.2)} and \eqref{eq(main-2.4)}, we get a new identity

\begin{equation}\label{eq(main-2.5)}
\psi(z)+\pi \cot{(\pi z)} =  \psi(-z) - \frac{1}{z}.
\end{equation}

By setting $z=\frac{p}{q}, \,\, 1\leq p < q$ in equations \eqref{eq(main-2.2)} and \eqref{eq(main-2.4)}, we get more identities, which would be used to derive our main identities to compute the values of digamma function

\begin{eqnarray}\label{eq(dig-2.6)}
\psi\left(\frac{p+q}{q}\right) =  \frac{q}{p} + \psi\left(\frac{p}{q}\right), \text{  and  }
\psi\left(\frac{-p}{q}\right) =  \frac{q}{p} + \psi\left(\frac{q-p}{q}\right), \quad 1\leq p < q.
\end{eqnarray}

\vskip.2cm

For the sake of convenient computation of digamma function, we derive some more identities, which are simple but more applicable in the computation of digamma function for $\frac{p}{q} > 1$. For this concern, we connect the Murty and Saradha's corrected formula for digamma function \eqref{eq(Murty)} with our above derived identity \eqref{eq(dig-2.6)}, and get the result as follows
\begin{eqnarray}\label{eq(main-2.7)}
\psi\left(\frac{q-p}{q}\right) = -\gamma-\ell n{(2q)}+\frac{\pi}{2}\cot{\left(\frac{\pi p}{q}\right)}+2\sum_{j=1}^{[\frac{q}{2}]}\left\{\cos{\left(\frac{2\pi pj}{q}\right)}\ell n{ \sin{\left(\frac{\pi j}{q} \right)}}\right\},
\end{eqnarray}
$(p,q)=1$; $1\leq p < q$.

\vskip.2cm

Now, we derive the identity for the computation of digamma function for negative fractions ($-\frac{p}{q}$). For this motive, we derive the identity in the similar manner as used in above identity, and get the result as follows
\begin{eqnarray}\label{eq(main-2.8)}
\psi\left(\frac{-p}{q}\right) = \frac{q}{p} - \gamma-\ell n{(2q)} - \frac{\pi}{2}\cot{\left(\frac{\pi (q-p)}{q}\right)}+2\sum_{j=1}^{[\frac{q}{2}]}\left\{\cos{\left(\frac{2\pi (q-p)j}{q}\right)}\ell n{ \sin{\left(\frac{\pi j}{q} \right)}}\right\},\nonumber\\
\end{eqnarray}
$(p,q)=1$; $1\leq p < q$.
\section{Numeric computations of digamma function}

\begin{table}[h!]
\centering

\caption{$\psi$- Function(Negative Fractional Valued)}
\vskip.1cm
\label{table-1.2}
\begin{tabular}{|c  |c  |c  |c|c|c|c|c|c|}
\hline
Ser. No. &$z=\frac{p}{q}$ & $\psi(z)= \frac{\Gamma^{'}{(z)}}{\Gamma{(z)}}$  \\ [5pt]\hline

1 &$-\frac{2}{3}$ & $ -\gamma+\frac{3}{2} - \frac{\pi \sqrt{3}}{6} - \frac{3\ell n \,3}{2}$  \\[5pt] \hline

2 &$-\frac{3}{4}$ & $-\gamma+\frac{4}{3} - \frac{\pi}{2} -3\ell n \,2 $  \\[5pt] \hline

3 &$-\frac{1}{2}$ & $-\gamma +2 - 2\ell n \,2 $   \\ [5pt] \hline

4 &$-\frac{1}{3}$ & $-\gamma +3 + \frac{\sqrt{3}\pi}{6} - \frac{3}{2}\ell n \,3 $   \\ [5pt] \hline

5 &$-\frac{1}{4}$ & $-\gamma +4 + \frac{\pi}{2} - 3\ell n \,2 $   \\ [5pt] \hline

6 &$-\frac{5}{8}$ & $-\gamma +\frac{8}{5} - \frac{(\sqrt{2} -1)\pi}{2} -4\ell n \,2 + \sqrt{2}\ell n \,(1+\sqrt{2}) $  \\[5pt] \hline

7 &$-\frac{3}{8}$ & $-\gamma+\frac{8}{3} + \frac{(\sqrt{2}-1)\pi}{2} -4\ell n \,2 + \sqrt{2}\ell n \,(1+\sqrt{2}) $  \\[5pt] \hline

8 &$-\frac{1}{8}$ & $-\gamma +8 + \frac{(1+\sqrt{2})\pi}{2} - 4\ell n \,2 -\sqrt{2}\ell n \,(1+\sqrt{2})$   \\ [5pt] \hline

9 &$-\frac{5}{6}$ & $-\gamma +\frac{6}{5} - \frac{\pi \sqrt{3}}{2} - \frac{3}{2}\ell n \,3 - 2 \ell n \,2$   \\ [5pt] \hline

10 &$-\frac{3}{2}$ & $-\gamma+\frac{8}{3} -2\ell n \,2 $  \\[5pt] \hline

11 &$-\frac{7}{3}$ & $-\gamma+\frac{117}{28} +\frac{\pi\sqrt{3}}{6}-\frac{3}{2}\ell n \,3 $  \\[5pt] \hline

\end{tabular}
\end{table}

\newpage

\begin{table}[h!]
\centering

\caption{$\psi$- Function(Positive Fractional Valued)}
\vskip.1cm
\label{table-1.2}
\begin{tabular}{|c  |c  |c  |c|c|c|c|c|c|}
\hline
Ser. No. & $z=\frac{p}{q}$ & $\psi(z)= \frac{\Gamma^{'}{(z)}}{\Gamma{(z)}}$  \\ [5pt]\hline

1 &$\frac{1}{2}$ & $-\gamma - 2\ell n \, 2 $   \\ [5pt] \hline

2 &$\frac{1}{3}$ & $-\gamma -\frac{\sqrt{3} \pi}{6}- \frac{3}{2}\ell n \,3 $   \\ [5pt] \hline

3 &$\frac{1}{4}$ & $-\gamma - \frac{\pi}{2} - 3\ell n \,2$   \\ [5pt]\hline

4 &$\frac{1}{5}$ & $-\gamma -\ell n 10 - \left(\frac{1+\sqrt{5}}{\sqrt{(10-2\sqrt{5})}} \right) \frac{\pi}{2} +  \frac{1}{2}\{ \sqrt{5} \ell n \,\left(\frac{\sqrt{5} - 1}{2}\right) - \ell n \,\frac{\sqrt{5}}{4} \}$  \\[5pt] \hline

5 &$\frac{1}{6}$ & $-\gamma -\ell n \, 12 - \frac{\pi \sqrt{3}}{2} - \ell n \, \sqrt{3} $  \\[5pt] \hline

6 &$\frac{1}{8}$ & $-\gamma - \frac{(1+\sqrt{2})\pi}{2} -4\ell n \,2 - \sqrt{2}\ell n \,(1+\sqrt{2})$   \\[5pt] \hline

7 &$\frac{1}{10}$ & $-\gamma -\ell n \, 20 - \left(\frac{\sqrt{(10+2\sqrt{5})}}{\sqrt{5} -1} \right) \frac{\pi}{2} +  \frac{1}{2}\{ \sqrt{5} \ell n \, (\sqrt{5} - 2) - \ell n \, \sqrt{5} \}$  \\[5pt] \hline

8 &$\frac{1}{12}$ & $-\gamma -\ell n \, 24 - \left(2+\sqrt{3} \right) \frac{\pi}{2} +  \{ \sqrt{3} \ell n \, (2-\sqrt{3}) - \ell n \, \sqrt{3} \}$  \\[5pt] \hline

9 &$\frac{2}{3}$ & $ -\gamma +\frac{\sqrt{3} \pi}{6} - \frac{3}{2} \ell n \, 3$   \\ [5pt] \hline

10 &$\frac{2}{5}$ & $-\gamma -\ell n 10 - \left(\frac{\sqrt{5} -1}{\sqrt{(10+2\sqrt{5})}} \right) \frac{\pi}{2} +  \frac{1}{2}\{ \sqrt{5} \ell n \, \left(\frac{\sqrt{5} + 1}{2}\right) - \ell n \, \frac {\sqrt{5}}{4} \}$  \\[5pt] \hline

11 &$\frac{3}{4}$ & $-\gamma + \frac{\pi}{2} -3\ell n \,2 $  \\[5pt] \hline

12 &$\frac{3}{5}$ & $-\gamma -\ell n 10 + \left(\frac{\sqrt{5} -1}{\sqrt{(10+2\sqrt{5})}} \right) \frac{\pi}{2} +  \frac{1}{2}\{ \sqrt{5} \ell n \, \left(\frac{\sqrt{5} + 1}{2}\right) - \ell n \, \frac {\sqrt{5}}{4} \}$  \\[5pt] \hline

13 &$\frac{3}{8}$ & $-\gamma - \frac{(\sqrt{2}-1)\pi}{2} - 4\ell n \,2 + \sqrt{2}\ell n \,(1+\sqrt{2}) $  \\[5pt] \hline

14 &$\frac{3}{10}$ & $-\gamma -\ell n \, 20 - \left(\frac{\sqrt{(10-2\sqrt{5})}}{1+\sqrt{5}} \right) \frac{\pi}{2} +  \frac{1}{2}\{ \sqrt{5} \ell n \, (2+ \sqrt{5}) - \ell n \, \sqrt{5} \}$  \\[5pt] \hline

15 &$\frac{4}{5}$ & $-\gamma -\ell n 10 + \left(\frac{1+\sqrt{5}}{\sqrt{(10-2\sqrt{5})}} \right) \frac{\pi}{2} +  \frac{1}{2}\{ \sqrt{5} \ell n \,\left(\frac{\sqrt{5} - 1}{2}\right) - \ell n \,\frac{\sqrt{5}}{4} \}$  \\[5pt] \hline

16 &$\frac{5}{6}$ & $-\gamma -\ell n \, 12 + \frac{\pi \sqrt{3}}{2} - \ell n \, \sqrt{3} $  \\[5pt] \hline

17 &$\frac{5}{8}$ & $-\gamma + \frac{(\sqrt{2}-1)\pi}{2} - 4\ell n \,2 + \sqrt{2}\ell n \,(1+\sqrt{2}) $  \\[5pt] \hline

18 &$\frac{5}{12}$ & $-\gamma -\ell n \, 24 - \left(2-\sqrt{3} \right) \frac{\pi}{2} +  \{ \sqrt{3} \ell n \, (2+\sqrt{3}) - \ell n \, \sqrt{3} \}$  \\[5pt] \hline

19 &$\frac{7}{8}$ & $-\gamma + \frac{(1+\sqrt{2})\pi}{2} - 4\ell n \,2 - \sqrt{2}\ell n \,(1+\sqrt{2}) $  \\[5pt] \hline

20 &$\frac{7}{10}$ & $-\gamma -\ell n \, 20 + \left(\frac{\sqrt{(10-2\sqrt{5})}}{1+\sqrt{5}} \right) \frac{\pi}{2} +  \frac{1}{2}\{ \sqrt{5} \ell n \, (2+ \sqrt{5}) - \ell n \, \sqrt{5} \}$  \\[5pt] \hline

21 &$\frac{7}{12}$ & $-\gamma -\ell n \, 24 + \left(2-\sqrt{3} \right) \frac{\pi}{2} +  \{ \sqrt{3} \ell n \, (2+\sqrt{3}) - \ell n \, \sqrt{3} \}$  \\[5pt] \hline

22 &$\frac{9}{10}$ & $-\gamma -\ell n \, 20 + \left(\frac{\sqrt{(10+2\sqrt{5})}}{\sqrt{5} -1} \right) \frac{\pi}{2} +  \frac{1}{2}\{ \sqrt{5} \ell n \, (\sqrt{5} - 2) - \ell n \, \sqrt{5} \}$  \\[5pt] \hline

23 &$\frac{11}{12}$ & $-\gamma -\ell n \, 24 + \left(2+\sqrt{3} \right) \frac{\pi}{2} +  \{ \sqrt{3} \ell n \, (2-\sqrt{3}) - \ell n \, \sqrt{3} \}$  \\[5pt] \hline

\end{tabular}
\end{table}

\newpage

\begin{table}[h!]
\centering

\caption{$\psi$- Function(Fractional Valued, $p>q$ )}
\vskip.1cm
\label{table-1.2}
\begin{tabular}{|c  |c  |c  |c|c|c|c|c|c|}
\hline
Ser. No. &$z=\frac{p}{q}$ & $\psi(z)= \frac{\Gamma^{'}{(z)}}{\Gamma{(z)}}$  \\ [5pt]\hline

1 &$\frac{7}{3}$ & $-\gamma+\frac{15}{4} -\frac{ \pi \sqrt{3}}{6}- \frac{3}{2}\ell n \,3 $   \\ [5pt] \hline

2 &$\frac{3}{2}$ & $ -\gamma +2 - 2 \ell n \, 2$   \\ [5pt] \hline

3 &$\frac{5}{2}$ & $-\gamma+\frac{8}{3} -2\ell n \,2 $  \\ [5pt] \hline

\end{tabular}
\end{table}

The following errata are found in a paper of Jensen \cite[p. 147]{Jensen} such that

\begin{eqnarray}\label{eq(Err1)}
\psi(3/5) \circeq -\gamma  + \frac{\pi}{2} \sqrt{\left(1-\frac{2}{\sqrt{5}} \right)}- \frac{5}{4}  \ell n \, 5 +\frac{\sqrt{5}}{4}  \ell n\, \left(\frac{3+2\sqrt{5}}{2} \right),
 \end{eqnarray}

\begin{eqnarray}\label{eq(Err2)}
\psi(4/5) \circeq -\gamma  + \frac{\pi}{2} \sqrt{\left(1+\frac{2}{\sqrt{5}} \right)}- \frac{5}{4}  \ell n \, 5 -\frac{\sqrt{5}}{4}  \ell n\, \left(\frac{3+2\sqrt{5}}{2} \right),
 \end{eqnarray}

where the symbol $\circeq$ exhibits the fact that each of the equations \eqref{eq(Err1)} and \eqref{eq(Err2)} does not hold true as stated.

The following are the corrected forms of above equations
\begin{eqnarray}
\psi(3/5) = -\gamma  + \frac{\pi}{2} \sqrt{\left(1-\frac{2}{\sqrt{5}} \right)}- \frac{5}{4}  \ell n \, 5 +\frac{\sqrt{5}}{4}  \ell n\, \left(\frac{3+\sqrt{5}}{2} \right),
 \end{eqnarray}

\begin{eqnarray}
\psi(4/5) = -\gamma  + \frac{\pi}{2} \sqrt{\left(1+\frac{2}{\sqrt{5}} \right)}- \frac{5}{4}  \ell n \, 5 -\frac{\sqrt{5}}{4}  \ell n\, \left(\frac{3+\sqrt{5}}{2} \right).
 \end{eqnarray}

\vskip.2cm
{\bf Concluding Remark :}
We conclude our present investigation by observing that several digamma functions for positive and negative fractional values have been deduced using our new identities \eqref{eq(dig-2.6)}, \eqref{eq(main-2.7)} and \eqref{eq(main-2.8)} in an analogous manner.

config: arxiv


TITLE: Where does the term "marginal" in "marginal probability" or "marginal distribution" come from?
QUESTION [4 upvotes]: Where does the term "marginal" in "marginal probability" or "marginal distribution" come from?

REPLY [5 votes]: According to the Wikipedia article about marginal distribution:

These terms are dubbed "marginal" because they used to be found by summing values in a table along rows or columns, and writing the sum in the margins of the table.


TITLE: Finding a base of a matrix.
QUESTION [1 upvotes]: I'm struggling with a question where I know an application but I need to find the base of a given matrix.
set $u : E \to E$ with E an R-vector space in n dimension. (sorry for my english)
it is given that $$u²+4Id=0$$
a few questions before this one, I had to prove that u didn't have any eigenvalue and wasn't bijective.I could prove it and hope I could use $ker(u-\lambda Id)\neq0$ and $det(u)\neq0$
then, here is the Question :
 $n = 2$ I'm asked to demonstrate if there is an existing base of E where the matrix of $u$ is:
$$A=\begin{pmatrix}
0&&-4 \\
1&&0 \\
\end{pmatrix}$$
so when I saw that question I thought I had to prove that this matrix is similar to another matrix of $u$ in a different base so I tried :
set $\forall x \in E$, let's take $(u(x),x)$ (I don't know what other base I can take)
$\forall x \in E$, $\lambda_1 u(x)+\lambda_2 x =0$ we know that $ker(u-\lambda Id)\neq0$ which means that $\forall x \in E , u(x)\neq \lambda x$ if $x\neq0$
but the problem is that I don't know if I'm allowed to suppose my $x\neq 0$ to say my family is linearly independant.
Besides, even though it may be a base of E, I'm not convinced of having the base of my matrix given upthere.
thank you in advance !

REPLY [2 votes]: Notice that $u\ne0$ so $\ker u\ne \Bbb R^2$ so let $x\not\in\ker u$ hence $u(x)\ne0$ and then the set $(u(x),x)$ is linearly independent. In fact, if we have $\alpha,\beta\in\Bbb R$ such that
$$\alpha u(x)+\beta x=0\tag1$$
then we apply $u$ to $(1)$ to get
$$-4\alpha x+\beta u(x)=0\tag2$$
and from $(1)$ and $(2)$ we find $\beta^2+4\alpha^2=0$ and then $\alpha=\beta=0$. Conclude.


TITLE: Locally Compact Quantum Groups application
QUESTION [12 upvotes]: Recently Mathematicians in harmonic analysis become more and more interested in Locally compact quantum groups and try to transfer concepts from abstract harmonic analysis to the setting of locally compact quantum groups.
I know that the first goal of defining the category of LCQGs was extending the Pontryagin duality for locally compact abelian groups. In fact this leads me to study LCQGs and do research in this area.
But after a while I wonder whether there exists any application for locally compact quantum groups in the real world for example physics.

REPLY [6 votes]: Locally compact quantum groups can be used to construct quantum channels, see the recent preprint http://arxiv.org/abs/1210.2738 by Jason Crann and Matthias Neufang.


TITLE: In a proof by contradiction, how do we know the assumption is the cause of the contradiction?
QUESTION [47 upvotes]: In a proof by contradiction, how do we know the assumption is the cause of the contradiction?  And not just the result of some other property more fundamental to numbers?
In other words, how can we be sure we arrived at the contradiction because of the proposition $P$ we assumed, and not because of some other proposition $Q$ we are not even aware of?

The answer that comes in mind is "Well, if there were some proposition $Q$ causing the contradiction, then it should also manifest in the event of $ \neg P$."
Is it this straightforward or there's more to it?  What if the proposition $Q$ can only manifest in the event of $P$, and not at all in the event of $\neg P$?
Sorry if this is nonsense - just trying to clear out some confusions.

REPLY [0 votes]: It is not that by $P\Rightarrow \bot$ (the symbol $\bot$ denotes contradiction) we assume $\neg P$ just because it seems there is something "wrong" with $P$. It is more that we cannot avoid it anyway.
Either the contradiction arises from $P$ and it would be reasonable to assume $\neg P$, or the contradiction remains when removing $P$, but then the already existing inconsistent axiom system suffices to prove $\neg P$ (as well as $P$) using the principle of explosion. So in either way, adding $\neg P$ is reasonable $-$ or at least not more inconsistent as the axiom
system itself.


TITLE: Analyticity of this multivariate Laplace transform
QUESTION [0 upvotes]: Consider a $d$-dimensional random vector $\mathbf X$ with laplace transfrom $$L(\mathbf t) = \left(1 + \langle \mathbf \theta, \mathbf t \rangle\right)^{-\alpha}$$
Where $\mathbf \theta$ is a $d$-dimensional vector of non-negatives reals (some can be zero, not all), and $\alpha$ is a strictly positive real (that cannot be zero). The coresponding random vector writes as $\mathbf X = \left(\theta_1 Y,...,\theta_d Y\right)$ for $Y$ a random variable following a gamma distribution with shape $\alpha$ and unit scale, that is with laplace tranform $\left(1 + t\right)^{-\alpha}$. Therefore, $\mathbf X$ is a comonotonous vector, living on a straigth half-line in $\mathbb R^d$.
Question: Considering $L$ as a function from $\mathbb C^d$ to $\mathbb C$, what is the region of convergence of $L$ ? Where are it's singularities ?

REPLY [1 votes]: $\alpha >0$ and
$$\int_{\Bbb{R}^n} e^{-\langle t, x\rangle} d\mu_X(x)= \int_{ \Bbb{R}}e^{-\langle t, \theta y\rangle} d\mu_Y(y)=
\int_{[0,\infty)}e^{-\langle t, \theta \rangle y} \frac{y^{\alpha-1} e^{-y}}{\Gamma(\alpha)} dy
 $$ converges for $\Re(\langle \theta,t\rangle) > -1$ and diverges for $\Re(\langle \theta,t\rangle) \le -1$.
It extends analytically to $\{ t\in \Bbb{C}^n,\langle \theta,t\rangle\in U\}$ where $U$ is any simply connected open of $\Bbb{C}-\{-1\}$.
If $\alpha$ is an integer then it extends to a meromorphic function on $\Bbb{C}^n$ with only one pole of order $\alpha$ at the hypersurface $\langle \theta,t\rangle=-1$.


TITLE: Changing base of an inequality
QUESTION [0 upvotes]: I was considering,
$$-1 \leq \cos\left(\theta - \frac{\pi}{4} \right) \leq1 $$
Can I do this?
$$2^{-1}\leq 2^{\cos\left(\theta - \frac{\pi}{4} \right)} \leq 2^1 $$
And further,
Can I multiply with $\frac{1}{\sqrt{2}}$ in the powers of this inequality, that is
$$2^{\frac{-1}{\sqrt{2}}}\leq 2^{\frac{\cos\left(\theta - \frac{\pi}{4} \right)}{\sqrt{2}}} \leq 2^{\frac{1}{\sqrt{2}}} $$
I found that the above result is correct.
But, I am wondering, Am I allowed to do this in any case, like changing bases to "ANYTHING" and multiplying with "ANYTHING" in the powers of an inequality?

REPLY [0 votes]: Note:
$$1) \quad a^x<a^y,a>0 \iff \begin{cases}x<y,a>1\\ x>y,0<a<1\end{cases} \\
2) \quad ax<ay \iff \begin{cases}x<y,a>0\\ x>y,a<0\end{cases}.$$


TITLE: Sieve of Eratosthenes Time Complexity Clarification
QUESTION [1 upvotes]: I've found plenty of sources claiming that the time complexity of the prime sieving algorithm Sieve of Eratosthenes is $O(n\log(\log n))$ where $n$ is the input. However, is this $\log_{10}$ or $\ln$? I assume it's $\ln$ but according to some notational conventions, just $\log$ refers to $\log_{10}$ and I can't find a source that clarifies this problem.
Does anyone know?
EDIT: I know that in a case where you have only one logarithm, you can scale between different bases using constants. However, this is not true when you have several nested logarithms. I.e. $\log_{10}(\log_{10} n)/\ln(\ln n))$ does not equal $\log_{10}(\log_{10} (n+1))/\ln(\ln (n+1)))$. Because of this, the logarithmic base does matter here. (I think, correct me if I'm wrong).

REPLY [2 votes]: In this particular case, the base doesn't matter.
$$n\log_b(\log_b(n))=n\frac{\ln\left(\dfrac{\ln(n)}{\ln(b)}\right)}{\ln(b)}=n\frac{\ln(\ln(n))-\ln(\ln(b))}{\ln(b)}=O(n\ln(\ln(n)))$$ as the second term is of a lower order.


TITLE: Diophantine equations of the form $z^2=x^4+m^2 y^4$?
QUESTION [1 upvotes]: What is known about Diophantine equations of the form $z^2=x^4+m^2 y^4$ for various values of $m$?
 Of course we can use infinite descent to show that when $m=1$ there are no solutions.  Also when $m=2$ there are no nontrivial solutions, since this would imply that there exists solutions to $a^2=b^4-c^4$.
However when $m=3$ there are infinitely many solutions. For example $(x,y,z)=(2,1,5)$ is a primitive solution from which infinitely many non-primitive solutions can be produced. Is there any way to determine which values of $m$ allow for solutions?

REPLY [1 votes]: Standard methods show that the quartic $z^2=x^4+m^2y^4$ is birationally equivalent to the elliptic curve
\begin{equation*}
V^2=U^3-4m^2U = U(U-2m)(U+2m)
\end{equation*}
with
\begin{equation*}
\frac{x}{y}=\frac{V}{2U}
\end{equation*}
This is the congruent number elliptic curve for triangle area $N=2m$. Thus the quartic has a non-trivial solution when $2m$ is a congruent number.
As an example, there will be a solution when $m=7$ since $14$ is a congruent number. The generator of the elliptic curve is $(U,V)=-(7/4,147/8)$ which gives $x=21$, $y=4$ and $z=455$.