TITLE: $X$ is extremely disconnected if and only if every open subspace is $C^{*}$ - embedded.
QUESTION [0 upvotes]: we shall say that a subspace $S$ of $X$ is $C$-embedded ($C^{*}$-embedded) in $X$ if every function in $C(S)$ ($ C^{*}(S) $) can be extended to a function in $C(X)$($C^{*}(X)$).
Urysohn extension theorem: A subspace $S$ of $X$ is  $C^{*}$-embedded in $X$ if only if any two completely separated sets in $S$ are  completely separated in $X$.

a space $X$  is said to be extremely disconnected if every open set has an open closure. $X$ is basically disconnected  if every cozero-set has an open closure. hence any extremely disconnected  space is basically disconnected
we know that$X$  is  extremely disconnected if and only if every pair of disjoint open sets have disjoint closures.

so, my questions are:

1: $X$ is  extremely disconnected if and only if every open subspace is $C^{*}$ - embedded.
2:Every dense subspace $X$ of an  extremely disconnected space $T $ is  extremely disconnected. In fact,
    disjoint open not sets in $X$ have disjoint open closure in $T$.

REPLY [1 votes]: Fact 2. follows from the general fact:

Let $X$ be a space, $U$ open in $X$, $D$ dense in $X$. Then $\overline{D \cap U} = \overline{U}$.

$D \cap U$ is the general form of an open subset of $D$. So by
$$\operatorname{cl}_D(U \cap D) = \operatorname{cl}_X(U \cap D) \cap D = \operatorname{cl}_X(U) \cap D$$
its closure is clopen in $D$.
Fact 1. Let $X$ be extremally disconnected and $O$ be an open subspace of $X$.
Let $A$ and $B$ be completely separated subsets of $O$. Then they are separated by two disjoint open sets (of $O$, and hence of $X$). In an extremally disconnected space disjoint open sets are completely separated (fact 2 in your question here). Apply this to $X$ and we have that $A$ and $B$ are completely separated in $X$. Urysohn then implies that $O$ is $C$-embedded in $X$. The reverse is trivial: if $U$ and $V$ are disjoint open then $O=U \cup V$ is $C$-embedded and $U$ and $V$ are completely separated in $O$ and hence by $C$-embeddedness have disjoint closures in $X$.


TITLE: Proving that the complement of a bipartite graph is perfect
QUESTION [7 upvotes]: In Section 1.6 of Harris et al.'s book Combinatorics and Graph Theory, there is a question asking to show that the chromatic number equals the clique number for the complement of a bipartite graph. I assigned the problem to my students, without thinking much about the solution.
Now that I've given it some thought, I've found what seems to be a very natural proof using Hall's marriage theorem. Unfortunately, my students don't know this result ... it doesn't appear until Section 1.7 of the book!
My question is this: is there a way of showing $\chi(\overline{G})=\omega(\overline{G})$ for bipartite $G$, that avoids Hall's theorem? In particular is there a way that might be found by a student unfamilar with Hall, who has only seen the basics of coloring (definition of $\chi$, greedy algorithm, Brooks' theorem, some basic bounds)?

REPLY [1 votes]: A first principles proof appears on page 129 of Diestel's Graph Theory. It is a bit lengthy, so I will not reproduce it here. Also, it is the original proof and Lovász and is quite clever. I would not expect your students to come up with this on their own.
Edit: I just noticed you want the result only for complements of bipartite graphs. The proof I reference is for general graphs. Perhaps it simplifies greatly if you make the additional assumption.


TITLE: Rank Nullity Theorem application
QUESTION [0 upvotes]: Show that {T $\in$ $L(R^5, R^4)$: dim null T > 2} is not a subspace of $L(R^5, R^4)$
I have no idea how to show this isn't a subspace the farthest I have gotten is to show that
dim range T < 3
But how would I go from there?

REPLY [1 votes]: This question does not lend itself to an application of the rank nullity theorem.
Hint: what does it mean for a set to be a subspace of $L(R^5,R^4)$? How can a subset fail to be a subspace?
Note that
$$
\pmatrix{
1&0&0&0&0\\
0&1&0&0&0\\
0&0&0&0&0\\
0&0&0&0&0
} +
\pmatrix{
0&0&0&0&0\\
0&0&0&0&0\\
0&0&1&0&0\\
0&0&0&1&0
} =
\pmatrix{
1&0&0&0&0\\
0&1&0&0&0\\
0&0&1&0&0\\
0&0&0&1&0
}
$$


TITLE: 2D representation of 3D algebra
QUESTION [0 upvotes]: I'm working with a Lie algebra that has 3 generator, $t_i$, that satisfy the following rules:
$$[t_2, t_3] = it_2, \quad [t_3, t_1] = it_1, \quad [t_1, t_2] = it_3   \tag1$$
A 3D representation is easy if we choose the adjoint representation, but I want a 2D representation. I'm trying computing the commutators (with general $2\times 2$ matrices) but I got non-sense results like parameter equal its negative or a generator with a zero in all entries. Does anyone know a more efficient method to compute this or know a group with this algebra?

REPLY [1 votes]: It deals with SU(2). Take for the $t_k$ a representation by Pauli matrices $\sigma_k$ as well explained in the following document :{http://hepwww.rl.ac.uk/Haywood/Group_Theory_Lectures/Lecture_3.pdf} where $\epsilon_{ijk}$ is the classical alternate tensor ($\epsilon_{ijk}=1$ if $(i,j,k)$ is an even permutation of $(1,2,3)$,  $\epsilon_{ijk}=-1$ otherwise). The presence of this tensor explains why the three identities you have given do not look connected by a circular permutation of indices.


text:
\begin{document}
\begin{center}
{\Large\bf On generalized fractional kinetic equations}\\[1cm]
{\large R.K. Saxena}\\
Department of Mathematics and Statistics,
Jai Narain Vyas University Jodhpur
342001, INDIA\\[0.5cm]
{\large A.M. Mathai}\\
Department of Mathematics and Statistics, McGill University,\\
805 Sherbooke Street West, Montreal, CANADA H3A 2K6\\[0.5cm]
{\large H.J. Haubold}\\
Office for Outer Space Affairs, United Nations,\\
P.O. Box 500, A-1400 Vienna, AUSTRIA\\
\end{center}
\noindent
{\bf Abstract}\par
In a recent paper,   Saxena et al.[1] developed the solutions of three generalized fractional kinetic equations in terms of the Mittag-Leffler functions. The object of the present paper is to further derive the  solution of further generalized fractional kinetic equations. The results are obtained in a compact form in terms of generalized Mittag-Leffler functions. Their relation to fundamental laws of physics is briefly discussed.
\section{Introduction}\par

The fundamental laws of physics are written as equations for the time evolution of a quantity $X(t), dX(t)/dt=-AX$, where this could be Maxwell's equations or Schr\"odinger's equation (if A is limited to linear operators), or it could be Newton's law of motion or Einstein's equations for geodesics (if A may also be a nonlinear operator [2,3,27]). The mathematical solution (for linear operators A) is $X(t) = X(0)exp\{-At\}$.

In thermodynamical or statistical applications one is mostly interested in mean values of the quantity $<X(t)>$. In this case, A is a characteristic time scale $A^{-1}=\tau$
in the evolution equation for $<X(t)>$. It then follows that $<X(t)>$ decays exponentially toward equilibrium $<X(t)> = <(X(0)>exp\{-t/\tau\}.$

In 1988, Tsallis [4] generalized the entropic function of Boltzmann-Gibbs statistical mechanics, $s=-\int dx p(x)lnp(x),$
to nonextensive statistical mechanics with $S_q[p]=\left\{1-\int dx[p(x)]^q\right\}/(q-1)$
that leads to q-exponential distributions $p_q(x)\propto\left[1-(1-q)x^2/kT\right]^{1/(1-q)}.$
Such a distribution reduces to Gaussian distribution for $q = 1$ and for $q = 2$ to a Cauchy-Lorentz distribution, to name two examples. In an attempt to incorporate L\'evy distribution into statistical mechanics, Tsallis has also shown that the above distribution becomes a L\'evy distribution for $q > 5/3$. Recently, Tsallis [4] used the mathematical simplicity of reaction-type equations, $dX/dY = Y^q$, to emphasize the natural outcome of the above distribution function $p_q(x)$ which corresponds exactly to the solution of the reaction equation of nonlinear type. The solution has power-law behavior. In the following we show that the fractional generalization of the linear reaction-type equation also leads to power-law behavior. In both cases, solutions can be expressed in terms of generalized Mittag-Leffler functions.

\section{Generalized Mittag-Leffler function}
 A generalization of  the Mittag-Leffler function [5,6]
\begin{equation}
E_\alpha(z):=\sum^\infty_{n=0}\frac{z^n}{\Gamma(\alpha n+1)},
\end{equation}
and its generalized form
\begin{equation}
E_{\alpha,\beta}(z):=
\sum^\infty_{n=0}\frac{z^n}{\Gamma(\alpha n+\beta)}\;\;\;(\alpha, \beta \in C, Re(\alpha)>0)
\end{equation}
was introduced by Prabhakar [7] in terms of the series representation
\begin{equation}
E^\gamma_{\alpha, \beta}(z):=\sum^\infty_{n=0}\frac{(\gamma)_nz^n}{\Gamma(\alpha n+\beta)(n!)}, \;\;\;(\alpha, \beta, \gamma \in C, Re(\alpha)>0),
\end{equation}
where $(\gamma)_n$  is  Pochammer's symbol defined by
\begin{equation}
(\gamma)_n=\left\{^{1,n=0}_{\gamma(\gamma+1)\ldots(\gamma+n-1), n\in N}
\right.\;\;
\gamma \neq 0
\end{equation}
It is an entire function of order $\rho = [Re(\alpha)]^{-1}$  [7].  This function  has been studied  by Wiman [8,9], Agarwal [10], Humbert [11]  and Humbert and Agarwal [12] and several others. Some special cases of (3) are given below:
\begin{equation}
(i)\;\;\; E_\alpha(z)=E^1_{\alpha, 1}(z),
\end{equation}
\begin{equation}
(ii)\;\; E_{\alpha, \beta}(z)= E^1_{\alpha, \beta}(z),
\end{equation}
\begin{equation}
(iii)\; \Phi(\beta, \gamma; z)=\; _1F_1(\beta;\gamma;z) = \Gamma(\gamma)E_{1,\gamma}^\beta(z),
\end{equation}
where $\Phi(\beta, \gamma;z)$    is  Kummer's confluent hypergeometric function defined in
  Erd\'elyi et al. ([13], p. 248, eq.1]).
        Mellin-Barnes integral representation for the  function defined by (3)  follows from the integral
\begin{equation}
E^\gamma_{\alpha,\beta}(z)=\frac{1}{2\pi\omega\Gamma(\gamma)}\int_\Omega \frac{\Gamma(-s)\Gamma(\gamma+s)(-z)^s}{\Gamma(\beta+s\alpha)}ds,
\end{equation}
where $\omega=(-1)^{1/2}$. The contour $\Omega$ is a straight line parallel to the imaginary axis separating the poles of $\Gamma(-s)$ at the points
$s = \nu\;\;(\nu=0,1,2,\ldots)$ from  those of $\Gamma(\gamma+s)$  at the points  $s = -\gamma-\nu\;\;(\nu=0,1,2,\ldots)$.
The poles of the integrand of (8) are assumed to be simple. (8) can be established by calculating the residues at the poles of $\Gamma(-s)$  at   the points , $s =\nu\;\;(\nu=0,1,2,\ldots,).$
 It follows from (8) that $E^\gamma_{\alpha,\beta}(z)$ can be represented in the form
\begin{equation}
E_{\alpha,\beta}^\gamma(z)=\frac{1}{\Gamma(\gamma)}H_{1,2}^{1,1}\left[-z|^{(1-\gamma,1)}_{(0,1),(1-\beta,\alpha)}\right], (Re(\alpha)>0;\alpha,\beta,\gamma \in C),
\end{equation}
where $H^{1,1}_{1,2}(z)$ is the H-function. A detailed account of the theory and applications of the H-function is available from   Mathai and Saxena [14]. This function can also be represented by
\begin{equation}
E^\gamma_{\alpha,\beta}(z)=\frac{1}{\Gamma(\gamma)}\;_1\Psi_1\left[^{(\gamma,1)}_{\beta,\alpha)};z\right],
\end{equation}
where $_1\Psi_1(z)$   is a special case of  Wright's generalized hypergeometric function $_p\Psi_q(z)$
      [15,16] ; also see,  Erd\'elyi  et al. ([13], Section 4.1 ), defined by

\begin{equation}
_p\Psi_q\left[^{(a_1,A_1),\ldots,(a_p,A_p)}_{(b_1,B_1),\ldots,(b_q,B_q)};z\right]=\sum^\infty_{n=0}\frac{\prod^p_{j=1}\Gamma(a_j+A_jn)}{\prod ^q_{j=1}\Gamma(b_j+B_jn)}\frac{z^n}{(n)!},
\end{equation}
where  $1+\sum_{j=1}^q B_j-\sum^p_{j=1}A_j\geq 0$ (equality only holds for appropriately bounded $z$).
        When $\gamma=1$, (9) and (10) give rise to  (12) and (13) given below:
\begin{eqnarray}
E_{\alpha,\beta} (z) & = & _1\Psi_1\left[^{(1,1)}_{(\beta,\alpha)};z\right],\\
& = & H^{1,1}_{1,2}\left[-z\left[^{(0,1)}_{(0,1),(1-\beta, \alpha)}\right]\right.,
\end{eqnarray}
where $Re(\alpha)>0; \alpha, \beta, \gamma \in C.$

If we further take $\beta=1$ in (12) and (13) we find that
\begin{eqnarray}
E_\alpha(z) & = &_1\Psi_1\left[^{(1,1)}_{(1,\alpha)};z\right],\\
& = & H^{1,1}_{1,2}\left[-z|^{(0,1)}_{(0,1),(0,\alpha)}\right],
\end{eqnarray}
for $Re(\alpha)>0, z\in C.$
The following integral gives the Laplace transform of $E^\gamma_{\alpha,\beta}(z)$.
\begin{equation}
\int^\infty_0 e^{-pt}t^{\beta-1}E^{\gamma}_{\alpha,\beta}(at^\alpha) dt=p^{-\beta}(1-ap^{-\alpha})^{-\gamma},
\end{equation}
where $Re(p) > |a|^{\frac{1}{Re(\alpha)}}, Re(\beta)>0, Re(p)>0,$
which can be established by means of the Laplace integral
\begin{equation}
\int^\infty_0 e^{-pt}t^{\rho-1}dt = \frac{\Gamma(\rho)}{p^\rho}, Re(p)>0, Re(\rho)>0.
\end{equation}
For $\gamma=1,$ (16) reduces to an elegant formula
\begin{equation}
        \int^\infty_0 e^{-pt}t^{\beta-1} E_{\alpha, \beta}(at^\alpha)dt = p^{-\beta}(1-ap^{-\alpha})^{-1},
\end{equation}
where $Re(\beta) > 0, Re(p) > 0 , |p| > |a|^{\frac{1}{Re(\alpha)}}$ .
        In an attempt to investigate the functions which when fractionally differentiated (of any order) reappear,  Hartley and Lorenzo [17]  came across a  special function of  the form
\begin{eqnarray}
F_q[-a,t] &=& t^{q-1}\sum^\infty_{n=0}\frac{(-a)^nt^{nq}}{\Gamma(nq+q)}, Re(q)>0,\\
& = & t^{q-1} E_{q,q}(-at^q).
\end{eqnarray}
This function has been studied  earlier by  Robotnov [18,19] in connection with hereditary integrals for application to continuum mechanics. The Laplace transform of this function is given by
\begin{equation}
L[F_q(a;t)] = \frac{1}{p^q-a}, Re(q)>0.
\end{equation}

A generalization of the F-function is presented by Lorenzo and Hartley [20] by means of the  following series representation:
\begin{eqnarray}
R_{\nu,\mu}[a,c,t]&=& \sum^\infty_{n=0}\frac{a^n(t-c)^{(n+1)\nu-\mu-1}}{\Gamma[(n+1)\nu-\mu]}, t>c>0\\
& = & (t-c)^{\nu-\mu-1} E_{\nu,\nu-\mu}[a(t-c)^\nu], t>c>0.
\end{eqnarray}
The Laplace transform of the R-function is derived  by  Lorenzo and Hartley [20] in the form
 \begin{equation}
L\left\{R_{\nu,\mu}(a,c,t)\right\}= \frac{e^{-cp}p^\mu}{p^\nu-a}, \;\;c\geq 0,
\end{equation}
where $Re(\nu-\mu)>0, Re(p)>0.$
When $c=0$, (23) reduces to
\begin{equation}
L\left\{R_{\nu,\mu}(a,0,t)\right\}=\frac{p^\mu}{p^\nu-a}, Re(\nu-\mu)>0, Re(p)>0.
\end{equation}
Finally we recall the definition of Riemann-Liouville  operator of fractional integration in the form
\begin{equation}
_0D_t^{-\nu}f(t)= \frac{1}{\Gamma(\nu)}\int_0^t(t-u)^{\nu-1}f(u)du,
\end{equation}
with    $_aD_t^0 f(t) = f(t)$ [21,22,23].
The standard kinetic equation, when integrated, yields
\begin{equation}
N_i(t)-N_0=c\;_0D_t^{-1}N(t),
\end{equation}
where $_0D_t^{-1}$  is the standard Riemann-Liouville integral operator. Here it can be mentioned that in the original paper of  Haubold and Mathai [24], the number  density of the species  $i, N_i = N_i(t)$ is a function of time and $N_i(t=0) = N_0$   is the number  density of species  $i$ at time  $t= 0$. If we drop the index $i$ in (27) and replace $c$ by $c^\nu$, then the solution of the generalized equation
 \begin{equation}
N(t)-N_0=-c^\nu\;_0D_t^{-\nu}N(t),
\end{equation}
is obtained, Haubold and Mathai [24]  as
\begin{equation}
N(t)=N_0\sum^\infty_{k=0}\frac{(-1)^k(ct)^{k\nu}}{\Gamma(k\nu+1)}.
\end{equation}
By virtue of (1), (29) can be written in a compact form as
\begin{equation}
N(t)  = N_0E_\nu(-c^\nu t^\nu).
\end{equation}
 In the following, we investigate the  solution of two more  generalized fractional kinetic equations. The results are obtained in a compact form in terms of generalized Mittag-Leffler functions and are suitable for computation. A detailed account of the operators of fractional integration and their applications is available from a recent survey paper of Srivastava and  Saxena [25].

\section{Generalized fractional kinetic equations}
{\bf Theorem 1.} If  $c > 0, \nu>0, \mu>0$, then for the solution of the equation
\begin{equation}
N(t)-N_0t^{\mu-1}E^\gamma_{\nu, \mu}[-c^\nu t^\mu]=-c^\nu _0D_t^{-\nu}N(t),
\end{equation}
there holds the formula
\begin{equation}
N(t)=N_0t^{\mu-1}E_{\nu,\mu}^{\gamma+1}(-c^\nu t^\nu).
\end{equation}
{\bf Proof.} By the application of convolution theorem of Laplace transform (Erd\'elyi et al. [26]) we see that  (26) can be written as
\begin{eqnarray}
L\left\{_0D_t^{-\nu}f(t);p\right\}& = &L\left\{\frac{t^{\nu-1}}{\Gamma(\nu)}\right\}L\left\{f(t)\right\},\\
& = & p^{-\nu}F(p),
\end{eqnarray}
where   $F(p)  = \int^\infty_0 e^{-pu} f(u)du, Re (p)>0$.
Projecting  the equation  (31) to Laplace transform , we obtain
\begin{eqnarray}
N(t)=L[N(t);p] & = & N_0\frac{p^{-\mu}[1+(p/c)^{-\nu}]^{-\gamma}}{[1+(p/c)^{-\nu}]}\nonumber\\
& = & N_0p^{-\mu}[1+(p/c)^{-\nu}]^{-(\gamma+1)}.
\end{eqnarray}
On using the formula (16), we find that
\begin{equation}
L^{-1}[p^{-\mu}\left\{1+(p/c)^{-\nu}\right\}^{-(\gamma+1)}]=t^{\mu-1}E^{\gamma+1}_{\nu, \mu}(-c^\nu t^\nu).
\end{equation}
The result (32) now readily follows from (36).\par
If we set $\gamma=1$ then (n)! is cancelled. Then in view of the formula
\begin{equation}
\beta E^2_{\beta, \gamma}(z) = E_{\beta, \gamma-1}(z)+ (1-\gamma+\beta) E_{\beta, \gamma}(z),
\end{equation}
which follows from the definition  of $E^\gamma_{\alpha, \beta}(z)$ given by  (3) ,we arrive at\\[0.3cm]
{\bf Corollary 1.1.} If $c > 0, \mu>0, \nu>0$, then for the solution of
\begin{equation}
N(t)-N_0 t^{\mu-1} E_{\nu, \mu}[-c^\nu t^\nu] = -c^\nu\; _0D_t^{-\nu}N(t),
\end{equation}
there holds the formula
\begin{equation}
N(t) = \frac{N_0t^{\mu-1}}{\nu}[E_{\nu, \mu-1}(-c^\nu t^\nu)+(1-\mu+\nu)E_{\nu, \mu}(-c^\nu t^\nu)].
\end{equation}
When $\gamma=2$, then by virtue of the following identity
\begin{eqnarray}
E^3_{\beta, \gamma}(z)&=&\frac{1}{2\beta^2}[E_{\beta, \gamma-2}(z)+(3\beta+3-2\gamma)E_{\beta, \gamma-1}(z)\nonumber\\
&+& \left\{2\beta^2+\gamma^2+3\beta-2\gamma-3\beta\gamma+1\right\}E_{\beta,\gamma}(z)],
\end{eqnarray}
which follows as a consequence of the definition  (3) , we obtain\\[0.3cm]
{\bf Corollary 1.2.}  If $c > 0, \nu>0, \mu>0$, then for the solution of
\begin{equation}
N(t)-N_0 t^{\mu-1}E^2_{\nu, \mu}[-c^\nu t^\nu]= -c^\nu _0D_t^{-\nu}N(t),
\end{equation}
there holds the relation
\begin{eqnarray}
N(t)& = & N_0 t^{\mu-1}E^3_{\nu,\mu}(-c^\nu t^\nu),\\
 & = & \frac{N_0 t^{\mu-1}}{2\nu^2}\left[E_{\nu,\mu-2}(-c^\nu t^\nu)+\left\{3(\nu+1)-2\mu\right\}E_{\nu, \mu-1}(-c^\nu t^\nu)\right.\nonumber\\
& + & \left.\left\{2(\nu^2+\mu^2)+3\nu-2\mu-3\nu\mu+1\right\}E_{\nu,\mu}(-c^\nu t^\nu)\right].
\end{eqnarray}
Next, if we set $\gamma=0$, then by virtue of the identity $E^0_{\nu,\mu}(z)=\frac{1}{\Gamma(\mu)}$, we arrive at another result given by Saxena et al. [1].\\[0.3cm]
{\bf Theorem 2.} If  $c > 0 , b\geq 0, Re (p ) > 0 , \nu > \mu+1$, then for the solution of
\begin{equation}
N(t)-N_0R_{\nu, \mu}(-c^\nu, b,t)=-c^\nu\; _0D_t^{-\nu}N(t),
\end{equation}
there holds the formula
\begin{equation}
N(t)=\frac{N_0}{\nu}(t-b)^{\nu-\mu-1}[E_{\nu,\nu-\mu-1}(-c^\nu(t-b)^\nu)+(\mu+1)E_{\nu,\nu-\mu}(-c^\nu(t-b)^\nu)].
\end{equation}
{\bf Proof.} Taking Laplace transform of  both sides of (45), it gives
\begin{eqnarray*}
N(t)& = & L\left\{N(t);p\right\}=N_0 L^{-1}\left[\frac{e^{-bp}p^{\mu-\nu}}{\left\{1+(c^\nu/p^\nu)\right\}^2}\right]\\
& = & N_0 L^{-1}\left[e^{-bp}p^{\mu-\nu}\sum^\infty_{n=0}\frac{(2)_n}{(n)!}(\frac{-c^\nu}{p^\nu})^n\right]\\
& = & N_0\sum^\infty_{n=0}\frac{(2)_n(-c)^{n\nu}}{(n)!}L^{-1}[e^{-bp}p^{\mu-\nu-n\nu}]\\
& = & N_0(t-b)^{\nu-\mu-1}\sum^\infty_{n=0}\frac{(2)_n(-c)^{n\nu}(t-b)^{n\nu}}{(n)!\Gamma(n\nu+\nu-\mu)}\\
& = & \frac{N_0}{\nu}(t-b)^{\nu-\mu-1}\sum^\infty_{n=0}\frac{(-c)^{n\nu}[(n\nu+\nu-\mu-1)+(1+\mu)](t-b)^{n\nu}}{\Gamma(n\nu+\nu-\mu)}\\
& = & \frac{N_0}{\nu}(t-b)^{\nu-\mu-1}\left[\sum^\infty_{n=0}\frac{\left\{-c^\nu(t-b)^\nu\right\}^n}{\Gamma(n\nu+\nu-\mu-1)}+(\mu+1)\sum^\infty_{n=0}\frac{\left\{-c^\nu(t-b)^\nu\right\}^n}{\Gamma(n\nu+\nu-\mu)}\right]\\
& = & \frac{N_0}{\nu}(t-b)^{\nu-\mu-1}[E_{\nu,\nu-\mu-1}\left\{-c^\nu(t-b)^\nu\right\}+(\mu+1)E_{\nu,\nu-\mu}\left\{-c^\nu(t-b)^\nu\right\}],
\end{eqnarray*}
which is same as (45). This completes the proof of  theorem 2.
If we set $\mu=0$, theorem 2 reduces to\\[0.3cm]
{\bf Corollary 2.1.} If $c> 0 , b\geq 0, \nu>1$, then for the solution of
\begin{equation}
N(t)-N_0R_{\nu,0}(-c^\nu, b,t)=-c^\nu\;_0D_t^{-\nu}N(t),
\end{equation}
there holds the formula
\begin{equation}
N(t) = \frac{N_0}{\nu}(t-b)^{\nu-1}[E_{\nu,\nu-1}(-c^\nu(t-b)^\nu)+E_{\nu,\nu}(-c^\nu(t-b)^\nu)].
\end{equation}
For $b = 0$, theorem 2 yields\\[0.3cm]
{\bf Corollary 2.2.} If $c > 0, \nu>\mu+1$, then for the solution of
\begin{equation}
N(t)-N_0R_{\nu,\mu}(-c^\nu,0,t)=-c^\nu\; _0D_t^{-\nu}N(t),
\end{equation}
there holds the formula
\begin{equation}
N(t)=\frac{N_0}{\nu}t^{\nu-\mu-1}[E_{\nu,\nu-\mu-1}(-c^\nu t^\nu)+(\mu+1)E_{\nu, \nu -\mu}(-c^\nu t^\nu)].
\end{equation}
If we further take $\mu=0$ then the above corollary reduces to the following result:\par
        If $c > 0 , \nu>1$, then the solution of
\begin{equation}
N(t)-N_0 F_\nu[-c^\nu,t]=-c^\nu D_t^{-\nu}N(t),
\end{equation}
is given by
\begin{equation}
N(t)=\frac{N_0t^{\nu-1}}{\nu}[E_{\nu,\nu-1}(-c^\nu t^\nu)+E_{\nu,\nu}(-c^\nu t^\nu)].
\end{equation}\par
\bigskip
\noindent
{\bf References}\par
\medskip
\noindent
[1] R.K. Saxena, A.M.Mathai, H.J. Haubold, Astrophys. and \par
Space Sci., 282 (2002) 281. \par
\noindent
[2] J. Jorgenson, S. Lang, in: Mathematics Unlimited - 2001\par
and Beyond, Eds. B. Engquist, W. Schmid, Springer-Verlag,\par
Berlin and Heidelberg 2001, 655.\par
\noindent
[3] R. Hilfer, in : Applications of Fractional Calculus in Physics,\par
Ed. R. Hilfer, World Scientific, Singapore 2000, 1.\par
\noindent
[4] C. Tsallis, in: Nonextensive Entropy: Interdisciplinary Applications,\par
Eds. M. Gell-Mann, C. Tsallis, Oxford University Press,\par
New York 2003,1.\par
\noindent
[5] G.M. Mittag-Leffler, C.R. Acad. Sci. Paris
                    (Ser.II )137 (1903) 554.\par
\noindent
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\end{document}

config: arxiv


TITLE: An example of Risch algorithm for integrating $y$, $F(x,y)=0$.
QUESTION [3 upvotes]: I would like to compute the integral
$$ \int y\,dx, \qquad y=\sqrt{x+\sqrt{x+1}},\\ F(x,y)=y^4-2xy^2+x^2-x-1=0, $$
in closed form, where $F(x,y)$ is a polynomial in $\mathbb{C}[x,y]$.
I am trying to apply the Risch algorithm (from "Algorithms for Computer Algebra" by Geddes, Czapor and Labahn) by hand to integrate this in terms of elementary functions, but I don't understand it. I need to write the integrand $f(x,y)=y$ in the form $a(x,y)/d(x)$ with $d(x)$ square-free, so $a=y$ and $d=1$. Then the zeros of $d$ determine the poles of $f$ and the logarithmic components of the integral, so since $d$ has no zeros, either there is no logarithmic component or the integral is not elementary (I am not sure which).
Am I missing something obvious? The integral has a closed form:
$$\int y\,dx = \frac{-3+8x+2\sqrt{1+x}}{12}\sqrt{x+\sqrt{x+1}} + \frac{5}{8}\log\left( 1+2\sqrt{x+1} + 2\sqrt{x+\sqrt{x+1}} \right). $$

REPLY [1 votes]: You should first find a basis for your function and then apply hermite reduction (p. 564 in your book)


TITLE: Showing a complex function defined as an integral is holomorphic
QUESTION [1 upvotes]: Let $\gamma:[0,1] \to \mathbb{C}$ be any $C^1$ curve. Define $$f(z)=\oint_\gamma \frac{1}{\zeta-z}d\zeta.$$
  Prove that $f$ is holomorphic on $U=\mathbb{C}-\hat{\gamma}$ where $\hat{\gamma}=\{ \gamma(t):0\leq t\leq 1 \} .$ In case $\gamma(t)=t,$ show that there is no way to extend $f$ to a continuous function on all of $\mathbb{C}$.

I know in my head we're really discussing the complex logarithm here, and in the second part in specific we are discussing, say, $Log(\frac{z}{z-1})$ which has $[0,1]$ as its branch cut (not sure if I am using the terminology correctly here). My initial thoughts on the first part are to somehow use Morera's Theorem or simply to use the definition of differentiation, but I can't sort out how the details would work.

REPLY [0 votes]: For the second part of the question, if $\gamma(t)=t$ and $z=x+iy$ is not on the positive real axis, then
$f(z)=\oint_\gamma \frac{1}{\zeta-z}d\zeta=\int_{0}^{1}\frac{1}{(t-x)-iy}dt=\int_{0}^{1}\frac{(t-x)+iy}{(t-x)^2+y^2}dt$
$=\int_{0}^{1}\frac{(t-x)}{(t-x)^2+y^2}dt + iy\int_{0}^{1}\frac{1}{(t-x)^2+y^2}dt$
$=(1/2)\int\frac{ds}{s}ds + (i/y)\int_{0}^{1}\frac{1}{[(t-x)/y]^2+1}dt$ for $s=(t-x)^2+y^2$.
$=(1/2)ln|s|+iarctan((t-x)/y)$ evaluated over the interval $[0,1]$.
Back substituting the expression for $s$, one sees that the final number is
$f(z)=(1/2)ln|1-2x+r^2|-ln|r|+i[arctan((1-x)/y)+arctan(x/y)]$,
where $r^2$ as usual is $x^2 + y^2$.
Now suppose $f$ extends. Taking $z=cos\tau+isin\tau$, which is a parameterization of the unit circle, we have for $\tau \neq n\pi$,
$f(z(\tau))=(1/2)ln|2-2cos\tau|+i[arctan((1-cos\tau)/sin\tau)+arctan(cot\tau)]$,
Now since we believe that $f$ is continuous over all of $\mathbb{C}$, the functions corresponding to its real and imaginary parts also extend continuously over $\mathbb{C}$. In particular, the function
$g(\tau):=Im(f(z(\tau)))= arctan((1-cos\tau)/sin\tau)+arctan(cot\tau)$,
where this equality holds for $\tau \neq n\pi$ extends to a continuous function of $\tau$. But this cannot be since the limit of $g$ as $\tau$ approaches 0 from the left is $arctan(-\infty)+arctan(-\infty)=-\pi$. On the other hand, the limit as $\tau$ approaches 0 from the right is $arctan(\infty)+arctan(\infty)=\pi$, so there's no hope of finding a value of $g$ at $\tau=0$ (hence, of $f$ at $z=1$) that makes this function continuous.
Note that the imaginary parts differ on opposite sides of $\gamma$ by $2\pi$. This is as expected.


TITLE: How could I find an orthogonal basis for $H^k$? - HELP!!
QUESTION [2 upvotes]: Let $P^k=$homogeneous polynomials of degree $k$ in $x$, $y$, $z$, $k=0, 1, 2, \dots, $ i.e. $P^k= \text{span} \{x^{k_x}y^{k_y}z^{k_z} : k_x+k_y+k_z=k\}$ and $H^k= \{f \in P^k : \Delta f = 0\}$, where $\Delta$ is the usual Laplacian operator.
Question : Without using the spherical coordinate, if I give $P^3= \{x^3, y^3, z^3, x^2y, x^2z,xy^2,y^2z, y^2z,xz^2,yz^2, xyz\}$, what could be a basis for $H^3$ and why? I think I can use recursively the fact that $H^0=P^0$ and $H^1=P^1$, but I don't even know how. It seems like as $Δ$ is a linear map, and you have bases of $P^k$ and $P^{k−2}$, it is a linear algebra exercise to find a basis of $\kerΔ$ in $P^k$. I think an excellent article for that should be SPHERICAL HARMONICS AND HOMOGENEOUS HARMONIC POLYNOMIALS page $5$, but I am not quite sure how to do it.
Thanks!

REPLY [0 votes]: Yes, you can consider it a standard (but quite tedious) linear algebra exercise for any fixed $n,k$ (in your example, $n=k=3$).
More generally, the article you linked provides the following answer on page 3. Although it's not really clear what you mean by "finding a basis".
$H^k$ is the orthogonal complement of all polynomials of the form $|x|^2 P(x)$, where $P \in \mathcal{P}^{k-2}$. Here the inner product is
$$ \langle x^\alpha,x^\beta \rangle = \alpha_1! \ldots \alpha_n! \delta_{\alpha,\beta}. $$
A nice proof is given in E.M. Stein's Singular Integrals and Differentiability Properties of Functions. First, any polynomial $P = \sum_\alpha a_\alpha x^\alpha \in \mathcal{P}^k$ corresponds to a differential operator $P(D) = \sum_\alpha a_\alpha D^\alpha$, where $D^\alpha = \frac{\partial}{\partial x_{\alpha_1}} \ldots \frac{\partial}{\partial x_{\alpha_k}}$. This correspodence is a ring homomorphism and in particular $(|x|^2)(D) = \Delta$.
Then the inner product in $\mathcal{P}^k$ can be introduced as
$$ \langle P,Q \rangle = P(D) Q. $$
Check that this coincides with the previous one!
Let $Q \in \mathcal{P}^k$. From the identity
$$ \langle |x|^2P, Q \rangle = P(D) (\Delta Q)  = \langle P, \Delta Q \rangle $$
we see that $Q \perp |x|^2P$ for every $P \in \mathcal{P}^{k-2}$ if and only if $\Delta Q = 0$.


TITLE: Combinations in discrete math help please
QUESTION [1 upvotes]: I have a question that needs some explaining and it has to do with two sentences that are supposedly asking different things, but to me I can't seem to find out why both sentences are different
Question 1: How many ways are there for 12 ice cream cones to have all same flavors out of 31 flavors? Answer to this is obviously 31
VS.
Question 2: How many ways are there if a flavor may be ordered as many as 12 times (31 total diff flavors to choose from)? Answer is (31+12-1,12)<<<< combinations with repetitions
Can someone explain to me how these two questions are different because in my eyes I can't seem to see what the difference is..My english comprehension in math is way below average >.>..I keep reading these two over and over again and I cna't seem to see why these two are asking different things...

REPLY [0 votes]: Q1. How many ways are there for 12 ice cream cones to have all same flavors out of 31 flavors?

This means all cones have same flavour.
And since we have 31 different flavours to choose from, the answer is obviously 31.

Question 2: How many ways are there if a flavor may be ordered as many as 12 times (31 total diff flavors to choose from)?

Here, each flavour can be ordered 0, 1, 2, .... 12 times to fill each cone.
this can be done in (n+m-1, m) ways.
An example with smaller numbers: flavours: 3 (ABC) and cones:2
combinations:
repeating: AA/BB/CC
non repeating: AB/BC/CA

(3+2-1,2)=(4,2) = 6

You can refer to this link.


TITLE: Is there an isomorphism class which is a set?
QUESTION [2 upvotes]: I'm studying Set Theory in my own, with Goldrei's textbook. The chapter I'm reading is on order-isomorphism and well-ordering. One exercise asks (i) to argue that, in general, a collection of well-ordered sets order-isomorphic to a given well-ordered set is a proper class (rather than a set). The proof, IMHO, is fairly straightforward and applicable to any isomorphism class. Then the same exercises asks (ii) if there is a well-ordered set, X say, s.t. the collection of all well-ordered set, order-isomorphic to X, is a set (rather than a proper class). I stumbled on that one.

REPLY [0 votes]: Let’s assume $X\neq\emptyset$ (else the property is trivially true). First note that you are looking at classes of order-isomorphic ordered sets, so pairs $(A,\leq)$ so that all for all such pairs exists a bijection that leaves the order intact. Let’s call this class $Z$. On the other hand if there is a bijection between two sets $A,B$ then this will transform any ordering on $A$ into an ordering on $B$.
Thus by projecting on the first coordinate we get $|X|$ where $|X|$ denotes the class of cardinality.
So if $Z$ is a set, then also $|X|$ is a set.
Now, can this be? Let’s differentiate two cases: If $|X|=n>0\in\mathbb N$ then $B:=\{0,1,\ldots,n-2,|X|\}$ would be a set of that cardinality. On the other hand, if $|X|$ is infinite then $B=X\cup\{|A|\}$ has the same cardinality as $X$.
So in both cases we’d find some $B\in|X|$ so that $|X|\in B$. By ZF this is not possible.
Thus this is not in fact a set.
EDIT: On the ZF part. Set $C=\{B,|X|\}$. Then $C,B$ are not disjoint, for both contain $|X|$, and $C,|X|$ are not disjoint, for both contain $B$. Thus this violates the axiom of regularity.


TITLE: Find n in arithmetic sequence
QUESTION [0 upvotes]: How can solve for n if n∈ℕ:
$$
(4+8+12 +  \ldots +4(n−1)) + (2+4.5+7 + \ldots +(4+5n)/2) = 126.5
$$
Thank you

REPLY [0 votes]: Hint:The first sum is:
$s_1=4(1+2+3+. . . n-1)=4\times\frac{(n-1)(n-1+1)}{2}=2n(n-1)$
The second sum $s_2$ is arithmetic progression $a=4.5$ and $d=2.5$ . We have:
$s_1+2+s_2=126.5$ ⇒ $s_1+s_2=124.5=3\times 41.5$
Let number of terms of $s_1$ be n and number of terms of $s_2$ be k, we can write:
$2n(n+1)+\frac{k}{2}[2\times 4.5 +(k-1)2.5]=3\times 41.5$
This relation deduces that  n and k must be divisible by 3. Now we have to check multiples of 3 for n and k such as (n, k)=(3,3), (6,3), (3, 6) (6, 6) . . etc. We find that $n=k=6 $ is the correct answer, because:
$s_1=2\times 6(6-1)=60$
$s_2=3[2\times 4.5+(6-1)2.5]=64.5$
⇒ $s_1 + s_2=60+64.5=124.5$


TITLE: Geometrical meaning of substituting third variable as a function of other two when doing a surface integral in differential forms
QUESTION [0 upvotes]: Suppose I have to integrate $dy \wedge dz$ over the unit hemisphere centered at the origin. I have:
$$ \int_S dy \wedge dz$$
The equation of hemisphere is given as $z=\sqrt{1-x^2-y^2}$, I have $dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y } dy$. I can substitute this expression of $dz$ in the above to get:
$$ \int_S  \frac{\partial z}{\partial x}dy \wedge dx $$
Conceptually, what exactly happened in the above substitution? I am a bit confused because $dy \wedge dz$ is an area element in the $y-z$ plane and $dy \wedge dx$ is an area element in the $x-y$ plane. Note that I'm using the interpretation of two form as something you project the area spanned by two vectors into.

REPLY [0 votes]: The chosen coordinate chart for the upper hemisphere $S_+$ of the unit sphere $S$ is
$$\tag{1}
\{x^2+y^2\le 1\}\ni (x,y)\mapsto \begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}x\\y\\\sqrt{1-x^2-y^2}\end{pmatrix}\in S_+.
$$
Intuitively and geometrically the $2$-form $dx\wedge dy$ is an oriented surface element that is spanned by the tangent vectors on $S$ in the $x$- and $y$-direction. In $\mathbb R^3$ we know that this can be expressed by a cross product
$$
\begin{pmatrix}\frac{\partial x}{\partial x}\\ \frac{\partial y}{\partial x}\\\frac{\partial z}{\partial x} \end{pmatrix}\times\begin{pmatrix}\frac{\partial x}{\partial y}\\ \frac{\partial y}{\partial y}\\\frac{\partial z}{\partial y} \end{pmatrix} =\begin{pmatrix}1\\0\\\frac{-x}{z}\end{pmatrix}\times
\begin{pmatrix}0\\1\\\frac{-y}{z}\end{pmatrix}=
\frac{1}{z}\begin{pmatrix}x\\y\\z\end{pmatrix}\,.
$$
Then we have a vector valued surface integral
$$\tag{2}
\int_{S_+}dx\wedge dy=\int_{x^2+y^2\le 1}\frac{1}{z}\begin{pmatrix}x\\y\\z\end{pmatrix}\,dx\,dy.
$$
The integration variables $x,y$ on the RHS of this formula are ordinary numbers.
When you want to consider
$$
\int_{S_+}dy\wedge dz
$$
I think you should use the chart
$$\tag{2}
\{y^2+z^2\le 1\}\ni (y,z)\mapsto \begin{pmatrix}x\\y\\z \end{pmatrix}=\begin{pmatrix}\sqrt{1-y^2-z^2}\\y\\z\end{pmatrix}\in S_+.
$$
and repeat the whole exercise. But note that the hemisphere $S_+$ is now a different one.