text:
\begin{document}
\title{Colored Jones polynomials without tails}

\author[C. Lee]{Christine Ruey Shan Lee}
\author[R. van der Veen]{Roland van der Veen}

\address[]{Department of Mathematics, University of Texas at Austin, Austin TX 78712}
\email[]{clee@math.utexas.edu}

\address[]{Mathematisch Instituut, Leiden University, Leiden, Netherlands}
\email[]{r.i.van.der.veen@math.leidenuniv.nl}

\thanks{Lee was partially supported by NSF Grant MSPRF-DMS 1502860. Van der Veen was supported by the Netherlands foundation for scientific research (NWO)}

\begin{abstract}We exhibit an infinite family of knots with the property that the first coefficient of the $n$-colored Jones polynomial grows linearly with $n$.
This shows that the concept of stability and tail seen in the colored Jones polynomials of alternating knots does not generalize naively.
\end{abstract}

\maketitle

\section{Introduction}

The colored Jones polynomials are a well known knot invariant with many connections to other parts of mathematics \cite{Jo,Tu}.
It is the Reshetikhin-Turaev knot invariant corresponding to the $n$-dimensional representation of quantum $sl_2$.
Given a knot $K$ and a natural number $n\geq 2$ the $n$-colored Jones polynomial is a Laurent polynomial $J_{K,n}(v)$. Dasbach-Lin \cite{DL06} observed that for any fixed alternating knot the first coefficients seem to stabilize as $n$ grows. The sequence of stabilized coefficients was called the tail of the knot. Tails are defined and studied in greater detail in \cite{Arm13,CK13,GL15, AD17, Haj17, L18}.

For general knots similar behaviour was expected\footnote{See for example Remark 1.7 of \cite{GaVu17}.} provided one allowed a certain periodicity. For example the leading coefficient of most torus knots is $(-1)^{n}$, and they are shown to have one or more tails \cite{AD18, GL15}; for an index $b$ closed braid consisting of a positive full twist on $b$ strands and $\ell\leq b$ negative crossings, the work of \cite{CK13} determines that up to $b-\ell+1$ first coefficients of its colored Jones polynomial are stable, though it might still admit multiple tails.

In this note we introduce knots where the first coefficient of the colored Jones polynomial does not stabilize at all. Instead it grows without bound, proportional to $n$. Such knots can never have tails, so we propose to name such knots Manx knots for a breed of cats without tails.

Possibly the simplest Manx knot is the pretzel knot $P(5,-3,5)$. The infinite family of Manx knots $C(5,-3,5,u)$ introduced in this paper are the Montesinos knots shown in Figure \ref{fig.ManxKnot}.

\begin{figure}
\begin{center}
\includegraphics[width=5cm]{ManxKnot.png}
\end{center}
\caption{The family of knots $C(r,s,t,u)$ where $r,s,t,u$ denote the number of crossings in each twist region. We have drawn the Manx knot $C(5,-3,5,-3)$.
To make the diagram zero framed we drew $16=r-s+t-u$ additional negative curls.}
\label{fig.ManxKnot}
\end{figure}

We will first give a quick, self-contained definition of the colored Jones polynomial that allows us to compute a formula for the polynomial of our examples of Manx knots. Next we will analyse the degree of the terms in our formula and observe the lack of tail in the main theorem, Theorem \ref{t.main}.

\section{Colored Jones polynomial}

The colored Jones polynomial is usually considered as a knot invariant but from our point of view it is more natural to consider its extension to knotted framed graphs.
A good thing about this definition of the colored Jones polynomial is that it is less dependent on knot diagrams and a little more efficient for computations. The technical reason behind its success is that we diagonalize the R-matrix.

\begin{defn}Knotted trivalent graphs (KTG)
\begin{enumerate}
\item A framed graph is a 1-dimensional simplicial complex $\Gamma$ together with an embedding $\Gamma\to \Sigma$ of $\Gamma$ into a surface with boundary $\Sigma$ as a spine.
\item A coloring of $\Gamma$ is a map $\sigma:E(\Gamma)\to \mathbb{N}$.
\item A Knotted Trivalent Graph (KTG) is a trivalent framed graph embedded (as a surface) into $\mathbb{R}^3$, considered up to isotopy.
\end{enumerate}
\end{defn}

Framed knots and links are special cases of KTGs. We prefer the more general set of KTGs here because there are natural 3-dimensional operations
that can be used to generate the whole theory from the tetrahedron.

\begin{figure}[htp]
\begin{center}
\def \svgwidth{.8\columnwidth}
\input{KTGMoves.pdf_tex}
\end{center}
\caption{Operations on Knotted Trivalent Graphs: framing change $F^e_{\pm}$, unzip $U^e$ and triangle move $A^w$ applied to an edge $e$ and vertex $w$ shown in the middle.}
\label{fig.KTGMoves}
\end{figure}

More precisely it was shown by D. Thurston \cite{ThuKTG} and \cite{Vdv} that the following
three moves suffice to generate any KTG from the planar theta graph. The first operation $F^e_{\pm}$ cuts the surface $\Sigma$ transversal to an edge, rotates one side by $\pi$
and reglues. This is called framing change or connected sum with a M\"{o}bius band. The second operation is called unzip, $U^e$. It doubles a chosen edge along its framing, deletes its end-vertices and joins the result as shown. The final move is called $A^w$ and expands a vertex $w$ into a triangle. All three moves are as shown in Figure \ref{fig.KTGMoves}.

Define $[k] =\frac{v^{2k}-v^{-2k}}{v^2-v^{-2}}$ and $[k]! = [1][2]\dots[k]$ for $k\in \mathbb{N}$ and $[k]! = 0$ if $k\notin \mathbb{N}$. Let
\[\qb{a_1+a_2+\dots+a_r}{a_1,a_2,\dots, a_r} = \frac{[a_1+a_2+\dots+a_r]!}{[a_1]!\dots[a_r]!}\]

The building blocks of the Jones polynomial are the following Laurent polynomials. For each we explicitly state the term of maximal degree for later use.
\[
f(a) = i^{-a}v^{\frac{-a(a+2)}{2}} \qquad O^k = (-1)^k[k+1] = (-1)^kv^{2k}+lot
\]
\[
\theta(a,b,c) = O^{\frac{a+b+c}{2}}\qb{\frac{a+b+c}{2}}{\frac{-a+b+c}{2},\frac{a-b+c}{2},\frac{a+b-c}{2}} = (-1)^{\frac{a+b+c}{2}}
v^{a+b+c+ab+bc+ca-\frac{a^2+b^2+c^2}{2}} +lot
\]
\[\Delta(a,b,c,\alpha,\beta,\gamma) = \sum_{z}\frac{O^z}{O^{\frac{a+b+c}{2}}} \qb{z}{\frac{a+b+c}{2}}\qb{\frac{-a+b+c}{2}}{z-\frac{a+\beta+\gamma}{2}}
\qb{\frac{a-b+c}{2}}{z-\frac{\alpha+b+\gamma}{2}}\qb{\frac{a+b-c}{2}}{z-\frac{\alpha+\beta+c}{2}} = \]
\[
(-1)^{m-\frac{a+b+c}{2}}v^{g(m+1,\frac{a+b+c}{2}+1)+g(\frac{-a+b+c}{2},m-\frac{a+\beta+\gamma}{2})+g(\frac{a-b+c}{2},m-\frac{\alpha+b+\gamma}{2})+g(\frac{a+b-c}{2},m-\frac{\alpha+\beta+c}{2})}+lot
\]
where $g(n,k) = 2k(n-k)$ and $2m = a+b+c+\alpha+\beta+\gamma-\max(a+\alpha,b+\beta,c+\gamma)$

Our formulas agree with the integer normalization of \cite{Cost}. The formula $\Delta$ is the quotient of the $6j$-symbol and a theta; it is in fact a Laurent polynomial. The summation range for $\Delta$ is finite as dictated by the binomials.

The colored Jones polynomial of any KTG can now be defined by the following three equations.
\begin{equation}
\langle F^e_{\pm}(\Ga),\s \rangle = f(\s(e))^{\pm1}\langle \Ga,\s \rangle
\end{equation}
Here $F_{\pm}^e(\Ga)$ means the result after applying framing change to edge $e$ of the KTG $\Ga$.
\begin{equation}
\langle U^e(\Ga),\s' \rangle = \langle \Gamma,\s \rangle\sum_{\s(e)}\frac{O^{\s(e)}}{\theta(\s(e),\s(b),\s(d))}
\end{equation}
The two colorings $\s$ and $\s'$ agree on all edges except that $U^e(\Ga)$ does not have edge $e$. The four edges in $\Ga$ adjacent to $e$ must have the same colors as $b$ and $d$ in $U^e(\Ga)$.
Also, the summation is over all possible values of color of the missing edge $\s(e)$. Most values yield zero because the factorials vanish. The only values that may be non-zero are when $\s(e)$ is between $|\s'(b)-\s'(d)|$ and $\s'(b)+\s'(d)$ and has the same parity. Finally,
\begin{equation}
\langle A^w(\Ga),\s' \rangle = \langle \Ga,\s \rangle\Delta(\s'(a),\s'(b),\s'(c),\s'(\alpha),\s'(\beta),\s'(\gamma))
\end{equation}
In this equation the colorings $\s$ and $\s'$ again agree on all edges where it makes sense. The colors of the six new edges are denoted $a,\alpha,b,\beta,c,\gamma$ as shown in Figure \ref{fig.KTGMoves} and $\Delta$
is defined above.
\begin{defn}
As noted above a $0$-framed knot $K$ is a special case of a $KTG$ with a single edge $e$. In this case we denote its colored Jones polynomial by
$J_{K,n} = \langle K, \sigma \rangle$, where $\sigma(e)=n-1$.
\end{defn}

We study the family of Montesinos knots $C(r,s,t,u)$ shown in Figure \ref{fig.ManxKnot}. For technical reasons we restrict ourselves to $s\leq -3 \leq 3 r,t,u$ all odd. And $u<0$. A similar family was studied in \cite{LYL17}.

\begin{figure}[ht]
\tiny{
\begin{center}
\def \svgwidth{\columnwidth}
\input{KTGComp.pdf_tex}
\end{center}}
\caption{\label{fig.KTGComputation} Starting from a theta graph (left), we first apply the $A$-move to both vertices, and again on one of the newly created vertices. Next change the framing on many edges. The numbers on each edge indicate the number of half twists present. Finally unzip the four edges (shown thickened) marked with $r, s, t, u$, respectively, to obtain the $0$-framed diagram for the knot $C(r,s,t,u)$.}
\end{figure}

To compute the colored Jones polynomial of $K = C(r,s,t,u)$ we take the following steps (see also \cite{LV16}). The first step is to generate our knot $K$ from the theta graph by KTG moves. One way to achieve this is shown in Figure \ref{fig.KTGComputation}. All edges should be thought of as bands with blackboard framing. The numbers indicate half twists (not colors). Note that the unzip applied to a twisted edge produces two twisted bands that form a crossing as shown in the figure on the right. This is natural considering that the black lines stand for actual strips.
Reading backwards and applying the above equations we may compute $J_{K,n+1}$ as follows. The unzips yield four summations, the framing change multiplies everything by the factors $f$, the $A$ moves produce the $\Delta$ terms and finally we are left with a single $\theta(2a,2b,2c)$. Note we have scaled the summation variables $a,b,c,d$ to make them integers instead of even integers.

\[
J_{C(r,s,t,u),n+1} = f(n)^{-2\mathrm{Wr}(r,s,t)-2r-2s-2t+2u}
\sum_{a,b,c,d \in nP\cap \mathbb{Z}}\Delta(2a,2b,2c,n,n,n)^2\times  \]
\[
\Delta(2b,n,n,2d,n,n)
\frac{O^{2a}O^{2b}O^{2c}O^{2d}
f(2a)^r f(2b)^s f(2c)^t f(2d)^{-u}
\theta(2a,2b,2c)}{\theta(2a,n,n)\theta(2b,n,n)\theta(2c,n,n)\theta(2d,n,n)}
\]
Here the writhe is given by $\mathrm{Wr}(r,s,t,u)=r-s+t-u$ and \\$P = \{(a,b,c,d)\in [0,1]^4: |a-b|\leq c \leq a+b \}$.

\section{The lack of tail}

Using the formulas for the building blocks we may determine the degree $\delta(a,b,c,d)$ of each term in the state sum.
It is an explicit piecewise quadratic function on the polytope $nP$. From our assumptions it follows that:

\[\delta(a+1,b,c,d)<\delta(a,b,c,d) \text{ and } \delta(a,b,c+1,d)<\delta(a,b,c,d) < \delta(a,b,c,d-1).\]
Therefore all terms of highest degree satisfy $b=a+c$ and $d = 0$ with $0\leq a,c \leq a+c\leq n$. Explicitly the degree is
$\delta(a,a+c,c,0) = $
\[b (r + 2 c (1 + r) + s-1)-2 (b^2 (1 + r + s)  + c (r + t-2) +
    c^2 (r + t)) - 2 n (-2 r - u) - 2 n^2 (-r - u)\]
Setting $r,s,t = 5,-3,5$ we find $\delta(a,b,c,d) = -4 (a - c)^2 + 2n(18 + 10 n - (2  + n) u)$
It follows that the maxima are on the line $a=c$ and there are $\lfloor n/2 \rfloor$ such maxima.
Looking at the building blocks, each of the corresponding maximal terms in the state sum has the same leading coefficient $(-1)^{8 a + 2 n (-11 + u)}=1$.
This completes the proof of our main theorem:

\begin{theorem} \label{t.main}
If $u\leq -3$ then $C(5,-3,5,u)$ is a Manx knot.\\
 More precisely, the leading coefficient of
$J_{C(5,-3,5,u),n+1}(v)$ is $\lfloor n/2 \rfloor$.
\end{theorem}

The same techniques will also show that the pretzel knot $P(5,-3,5)$ is Manx but this appears to be the only Manx pretzel knot \cite{GLV18}.
We expect that many more Manx knots exist but not in the family $C(r,s,t,u)$ we considered. All Montesinos Manx knots seem to derive from the unique Manx pretzel $P(5,-3,5)$.

Although our Manx knots do not stabilize in the ordinary sense they may still be $c$-stable. This more general notion of stability was introduced in Garoufalidis-Vuong \cite{GaVu17} and allows growing leading coefficients. Briefly, a sequence of Laurent power series $f_n(v)$ with $v$-degree $\delta(n)$ is $c$-stable if there exists a series $F(n,x,v) = \sum_{k=0}^\infty \Phi_k(n,v)x^k$. Here $\Phi_k(n,v)$ are formal Laurent power series in $v$ with coefficients that are integer-valued quasi-polynomials in $n$.
We require that for all $k\in \mathbb{N}$ we have
\[
\lim_{n\to\infty}v^{-k(n+1)}\left(v^{-\delta(n)}f_n(v)-\sum_{j=0}^k\Phi_j(n,v)v^{j(n+1)} \right)=0
\]

In their paper they conjecture that $c$-stability holds for all Reshetikhin-Turaev knot invariants as one scales a dominant weight.

{\bf Acknowledgement.} We thank Stavros Garoufalidis for informing us about the notion of $c$-stability and the organisers of the 2018 Banff workshop on Modular forms and Quantum invariants for providing the right atmosphere for writing this paper. We thank Abhijit Champanerkar for interesting conversations about tails.
\bibliographystyle{plain}

\bibliography{References}

\end{document}


TITLE: Find $\sum_{k=1}^n P(z_k)$ where P is a polynomial and $z_k$ the $k$th root of $z^{13}=1$
QUESTION [2 upvotes]: Question: Given a complex number z such that $z^{13}=1$, find the sum of all possible values of $z+z^3+z^4+z^9+z^{10}+z^{12}$.
I know we have to use roots of unity and try to manipulate the polynomial we want to evaluate, but can't find a pattern.

REPLY [0 votes]: If $z^{13} = 1$ and $z \not= 1$, then $z = e^{2\pi i \frac{k}{13}}$ where $(k, 13) = 1$.
Since $z^{12} = z^{-1}$, $z^{10} = z^{-3}$ and $z^{9} = z^{-4}$ we have
$z+z^3 + z^4 + z^9 + z^{10}+z^{12} = 2(\cos(2 \pi \frac{k}{13}) + \cos(2 \pi \frac{3k}{13}) + \cos(2 \pi\frac{4k}{13}))$ where $z = e^{2\pi i \frac{k}{13}}$.
The total is $2\sum_{k=1}^{12} \cos(2 \pi \frac{k}{13}) + \cos(2 \pi \frac{3k}{13}) + \cos(2 \pi\frac{4k}{13}) = -6$.
If $z = 1$ is included, the total is $0$.


TITLE: Probability that two random numbers have a Sørensen-Dice coefficient over a given threshold
QUESTION [3 upvotes]: The following problem came up at work and my probability knowledge isn't up to the task. Let $a, b \in \mathbb{Z}_{2^n}$ be two $n$ bit integers. Their Sørensen-Dice coefficient is the quantity
$$
DS(a, b) = \frac{2|a \wedge b|}{|a| + |b|}
$$
where $|a|$ denotes the population count of $a$ (number of 1 bits) and $a\wedge b$ is bitwise AND. I would like to know:

Given $t \in [0, 1]$, what is the probability that $DS(a, b) \ge t$ for $a, b$ drawn uniformly at random from $\mathbb{Z}_{2^n}$?

Call this probability $P_n(t)$.  Experimentally this probability appears to have the form $$P_n(t) = \frac{1}{1 + 2^{\alpha(t - 1/2)}}$$ where $\alpha$ is some function of $n$. I expect this to be the form of the final answer, albeit with a more precise description of $\alpha$.
Edit: In practice $n$ is 1024 or, more generally, some even power of 512 (I expect the result to hold for general $n$, but maybe that case is easier to handle as a first step).
I would like to be better at solving these kinds of problems, so I will gratefully accept any recommendations for further reading too.

REPLY [2 votes]: In this question, the Sørensen-Dice coefficient is the ratio $$R_n = \frac {\displaystyle 2\sum_{i=1}^n X_iY_i} {\displaystyle \sum_{i=1}^n X_i + \sum_{i=1}^n Y_i}$$ where all $X_i$ and $Y_i$ are iid $\text{Bernoullli}\left({1\over 2}\right)$.
Now the products $X_iY_i$ are iid $\text{Bernoulli}\left({1\over 4}\right)$, so
$$R_n={U_n\over V_n},$$ where
$$U_n:=2\displaystyle \sum_{i=1}^n X_iY_i\overset{d}{=}2\,\text{Binomial}\left(n,{1\over 4}\right)\overset{d}{\approx} \text{Normal}\left({n\over 2},{3n\over 4}\right)$$ and
$$V_n:=\displaystyle \sum_{i=1}^n X_i + \sum_{i=1}^n Y_i\overset{d}{=} \text{Binomial}\left(2n,{1\over 2}\right)\overset{d}{\approx} \text{Normal}\left({n},{n\over 2}\right)$$
and the correlation coefficient is (after some tedious arithmetic) $$\rho(U_n,V_n)={\text{Cov}(U_n,V_n)\over \sqrt{\text{Var}(U_n)\,\text{Var}(V_n)}}=\sqrt{2\over 3}\quad\text{(independent of $n$ !)}.$$
Thus, for large enough $n$, the ratio $R_n$ is approximately distributed like the ratio of two correlated Normal random variables. An approximation due to Hinkley then gives
$$P[R_n\le t]\approx \Phi\left({\theta_2 t-\theta_1\over \sigma_1\,\sigma_2\,a(t)}\right)$$
where $\Phi$ is the Standard Normal CDF, and
$$\begin{align}
\theta_1&=E(U_n)={n\over 2}\\
\theta_2&=E(V_n)=n\\
\sigma_1^2&=Var(U_n)={3n\over 4}\\
\sigma_2^2&=Var(V_n)={n\over 2}\\[2ex]
a(t)&=\sqrt{{t^2\over \sigma_1^2}-{2\,\rho\,t\over \sigma_1\sigma_2}+{1\over \sigma_2^2}}=\sqrt{{1\over n}\left({4\over 3}t^2-{8\over 3}t+2\right)}.
\end{align}$$
The approximation assumes that $0<{\sigma_2\over\theta_2}\ll 1;$ so, if $n>512$ then ${\sigma_2\over\theta_2}=\sqrt{1\over 2n}\lt 0.04,$ in which case the approximation may be quite acceptable.

For small $n$ it's feasible to directly compute the exact distribution of $R_n$ combinatorially, which I've done for $n\le 12$. The following plot for $n=12$ shows the exact CDF of $R_{12}$ (the step function), over which I've superposed Hinkley's approximation (the smooth curve):


TITLE: Finding the number of frogs in a pond
QUESTION [2 upvotes]: A pond has an unknown number $N$ of frogs. To estimate $N$ Bob collects 40 frogs and marks
them. The next day, he collects 60 frogs, and finds that 13 of them are marked from the previous day.
Assume in both cases that the frogs are a random sample from the $N$.
Find the probability that the second sample has 13 marked frogs.
I can't think of what to do here, so a hint would be helpful. As in, what theorem, etc should I try to use.

REPLY [0 votes]: You should use the hypergeometric distribution. Specifically, the number $X$ of marked frogs in the second sample is a random variable with hypergeometric distribution with paremeters

$N=$ population size,
$k=40$ successes, where success $=$ marked frog (there are $40$ from the previous day).
$n=60$ sample size.

Of course a reasonable assumption is that $N>n$. Hence, for any $$
$$P(X=x)=\frac{\dbinom{k}{x}\dbinom{N-k}{n-x}}{\dbinom{N}{n}}$$ For $x=13$ and for the above values for the parameters, you get that
$$P(X=13)=\frac{\dbinom{40}{13}\dbinom{N-40}{60-13}}{\dbinom{N}{60}}$$


TITLE: Show that a given harmonic function has a specific form
QUESTION [1 upvotes]: Suppose $u$ is a harmonic function in $\mathbb{R}^n$ and satisfies
$$|u(x)| \leq C |x|~\forall x \in \mathbb{R}^n.$$
Show that $u(x)=q \cdot x$ where $q$ is a constant vector.
I'm struggling with this problem. I tried defining a new function, $v(x)=u(x)/{x}$ for $x \neq 0$ and using the given bound $C,$ that leads to a constant vector.  Being dividing by a vector, that couldn't be correct. My gut feeling says that this could be done using Maximum principle. Any help is much appreciated. Thank you.

REPLY [0 votes]: Hint: Because $u$ is harmonic, it is real analytic, and so it can be uniformly approximated in some large open ball by a high-degree polynomial. Deduce that most of the polynomial's coefficients are zero.


TITLE: Averaging decibels
QUESTION [12 upvotes]: Wikipedia:

The decibel (dB) is a logarithmic unit that indicates the ratio of a
  physical quantity (usually power or intensity) relative to a specified
  or implied reference level.

If I measure some physical quantity in decibels, then what is the preferred way to calculate the mean of the measured values? Is it enough to simply average them, or should I convert them back to linear scale, calculate the average, and convert it back to decibels (example)? When should I use which approach, and why?

REPLY [0 votes]: To Pierre Ghislain's answer: AVG(a,b) is not 5kg but 5dB, which converts back to 3.16kg – and he's right, that's not 3.5kg.
In my opinion it depends what the 'average' should express.
(a) If you have two loudspeakers, operating in phase-sync and you want to compute the average output in pascal (for sound pressure), you need to convert the dB back to pascal (or whatever your original unit was) and average the values.
(b) If you have made several measurements of e.g. loudness in dB at different times of the day and you want to know the average 'loudness' during a day (in dB) you will average the dB values as if they where linear numbers, to get an average dB value.
This is nicely illustrated by the two lines that user313951 showed


TITLE: Homology group of $X$
QUESTION [0 upvotes]: Can I say that the homology group  $ H_n(X)$  it gives us the information of how many holes the topological space $X$ has ? And for an example to say that the $H_n(\bigtriangleup^{n} )=0$ because n simplexis dont have holes and compare it to $H_n(X )=1$ where $X$ here is a torus which has one hole? Meaning if $ H_n(X)\neq H_n(Y) $ then $X$ can't be isomorphic to $Y$ ?

REPLY [0 votes]: Intuitively, you can think of $H_n(X)$ as measuring the $n$-dimensional holes of your space $X$. This observation is motivated by the homology groups we encounter on simple spaces such as spheres and tori etc.
Saying $H_n(\Delta^n) = 0 $ would mean there are no $n$-dimensional holes in your space $\Delta^n$. If $\Delta^n$ is supposed to be the $n$-simplex, note that $\Delta^n$ is homeomorphic to the $n$-dimensional ball $D^n$ which is homotopy equivalent to a point $*$ and homotopy equivalent spaces have isomorphic homology groups, therefore
$$H_n(\Delta^n) \cong H_n(D^n) \cong H_n(*) = 0\ \ \text{for}\ \ n > 0$$
A bit more algebraic background: What homology does actually measure is the failure of exactness of a given chain complex $(C_*,\partial_*)$:
$$\cdots \to C_n \xrightarrow{\partial_n}  C_{n-1} \xrightarrow{\partial_{n-1}} C_{n-2} \xrightarrow{\partial_{n-2}}  \cdots $$
Usually one consideres the chain complex of the (free abelian) chain groups $C_k(X)$ that are generated by the (singular) $k$-chains in $X$.
Now by definition, the $n$th homology group $H_n(X)$ is defined as the quotient $$H_n(X) := \ker \partial_{n}/\operatorname{im}\partial_{n+1}$$
Therefore, the "bigger" the homology group $H_n(X)$, the further your chain complex is away from being exact at $C_n(X)$, the "smaller" $H_n(X)$, the closer you are to exactness at $C_n(X)$. Consequently, if $H_n(X) = 0$, your chain complex is exact at degree $n$, i.e.
$$\ker \partial_{n} = \operatorname{im}\partial_{n+1}.$$
That's what homology groups measure algebraically.


TITLE: about possion gamma and exponential distribution
QUESTION [1 upvotes]: about possion gamma and exponential distribution

can someone explain how to sub in the numbers?
i tried subbing in numbers and the outcome its not the same as the answer.
maybe I'm not doing it correct .
I'm getting 1/2 e^-1.5 for the first question .

REPLY [3 votes]: Exponential distributions deal with the amount of time between events.  The first problem is exponential with $\lambda = .5$. Since we are dealing with more than 3 hours between calls, we have
$$P(X\ge{3})=\int_{3}^{\infty}.5e^{-.5x}dx=-e^{-.5x}|_{3}^{\infty}=0-(-e^{-.5(3)})=e^{-1.5}=.2231$$
THe second problem, you need to know that the mean and standard deviation of an exponential distribution is $\frac1{\lambda}$.  Since this number is $2$, $(.5^{-1})$ then the we can find
$$P(X\ge\mu+\sigma)=P(X\ge{4})=\int_{4}^{\infty}.5e^{-.5x}dx=-e^{-.5x}|_{4}^{\infty}=0-(-e^{-.5(4)})=e^{-2}=.1353$$
EDIT:  Continued answer.
For questions 3 and 4, we are talking about Poisson, since we are wanting the number of events given a mean and a particular interval.  Since we know the mean in one hour is .5, the mean in a 3 hour interval is 1.5.  Thus we will use $\lambda=1.5$.  Now we are looking for exactly 2 calls in three hours, thus, using Poisson,
$$P(X=k)=\frac{\lambda^ke^{-\lambda}}{k!}=\frac{1.5^2e^{-1.5}}{2!}=.2510$$
Now for 4, we have our $\lambda=1$ since it is in a 2 hour interval.  The mean and variance for Poisson are both $\lambda$, so $\mu+2\sigma=1+2\sqrt1=3$.  Since we need fewer than 3 we have
$$P(X\le2)=\sum_{k=0}^{2}\frac{1^ke^{-1}}{k!}=e^{-1}\left(1+1+\frac1{2}\right)=.9197$$


TITLE: Set of all ordinals injectable into a well-order S is of strictly greater cardinality than S
QUESTION [0 upvotes]: I searched the site as thoroughly as I could, but couldn't find anything that's close enough to what I'm asking.
Let S be some well-ordered set. Let M(S) be the set of all ordinals α, such that there is an injection from α into S. It's easily provable that M(S) is an ordinal itself, and that |S| ≤ |M(S)|. How can one show, without Choice, that |S| < |M(S)|? That is, in the sense that there is no injection from M(S) into S.
I tried to assume that there exists an injection from M(S) into α, the unique ordinal that's isomorphic to S, aiming to somehow reach the conclusion that α = M(S), in contradiction to the fact that α ∈ M(S). But I've been stuck on this for too long without making progress.
Will appreciate any guidance. Thank you!

REPLY [1 votes]: Suppose there were an injection $M(S)\to S$. Since $M(S)$ is the set of all ordinals injecting into $S$ and $M(S)$ itself is an ordinal we have $M(S)\in M(S)$, a contradiction


TITLE: Equivalent arrangements of arrows
QUESTION [1 upvotes]: Given $n$ arrows arranged so that every arrow starts from the base of one of the arrows and ends on the base of one of the arrows, what should be meant by that two such arrangements are essentially the same? E.g. how many essentially different arrangements exists for $n=2$?
The diagram below shows only a fragment of a complete arrangement.

An arrow could also start or end on its own base.

I regard all arrangements for $n=1$ to be equivalent.

I had the feeling that there was a canonical answer, but that I just couldn't see it. Thinking about it a little more I found:

A morphism is a function between the arrows of two arrangements which
  in the obvious way preserves the structure, and equivalence is
  "isomorphic to".

The accepted answer:

a={ <1,1,1> <2,1,1>}
b={ <1,1,1> <2,1,2>}
c={ <1,1,1> <2,2,1>}
d={ <1,1,1> <2,2,2>}
e={ <1,1,2> <2,1,1>}
f={ <1,1,2> <2,1,2>}
g={ <1,1,2> <2,2,1>}
h={ <1,2,1> <2,1,1>}
i={ <1,2,1> <2,1,2>}
Obviously this nine arrangements are non isomorphic due to definition.

REPLY [1 votes]: Each arrangement of arrows can be represented as a set of triples $ < name, head, tail> $ selected from an alphabet $\Sigma$ with the restriction that each triple has a unique name. In this representation $ \{ <0,0,0> \} $ would be your $ n = 1 $ case where an arrow starts and ends on its own base.
We then can define two arrangements to be equivalent if there is a permutation $ \Sigma \to \Sigma $ remaps one arrangement to the other.
With this definition I was able to iterate 9 examples for n=2.
{ <1,1,1> <2,1,1>}
{ <1,1,1> <2,1,2>}
{ <1,1,1> <2,2,1>}
{ <1,1,1> <2,2,2>}
{ <1,1,2> <2,1,1>}
{ <1,1,2> <2,1,2>}
{ <1,1,2> <2,2,1>}
{ <1,2,1> <2,1,1>}
{ <1,2,1> <2,1,2>}


TITLE: Solve the integral $\int_{(0,\infty)\times(0,\infty)}\frac{1}{\alpha}\left ( \frac{x}{x+y} \right )^{\alpha-1}e^{-\frac{x+y}{\alpha}}dm_2(x,y)$
QUESTION [1 upvotes]: Can I have a hint to solve this integral?
$$
\int_{(0,\infty)\times(0,\infty)}\frac{1}{\alpha}\left ( \frac{x}{x+y} \right )^{\alpha-1}e^{-\frac{x+y}{\alpha}}dm_2(x,y)
$$

REPLY [0 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\ds{\int_{0}^{\infty}\int_{0}^{\infty}{1 \over \alpha}
\pars{x \over x + y}^{\alpha - 1}\expo{-\pars{x + y}/\alpha}\dd y\,\dd x}} =
{1 \over \alpha}\int_{0}^{\infty}\int_{x}^{\infty}
\pars{x \over y}^{\alpha - 1}\expo{-y/\alpha}\dd y\,\dd x
\\[5mm] = &\
{1 \over \alpha}\int_{0}^{\infty}y^{1 - \alpha}
\expo{-y/\alpha}\int_{0}^{y}x^{\alpha - 1}\,\dd x\,\dd y =
{1 \over \alpha}\int_{0}^{\infty}y^{1 - \alpha}
\expo{-y/\alpha}\pars{y^{\alpha} \over \alpha}\dd y =
{1 \over \alpha^{2}}\int_{0}^{\infty}y\expo{-y/\alpha}\dd y
\\[5mm] = &\
\int_{0}^{\infty}y\expo{-y}\dd y = \bbx{1}
\end{align}


TITLE: How many ways of splitting objects equally into groups?
QUESTION [1 upvotes]: How many ways in which $m\cdot n$ distinct objects can be divided equally into $n$ groups?
The answer is $$\frac{(mn)!}{(m!)^n n!}$$
Can someone please supply the intuition behind this answer?
Thanks in advance.

REPLY [2 votes]: First, let us consider an easy example.
If you want to divide $9$ distinct balls in the $A, B, C$ boxes, the answer is
$$\binom{9}{3}\times\binom{6}{3}.$$
First choose three balls for $A$, then choose three balls for $B$.
On the other hand, if you want to divide $9$ distinct balls into three name-less boxes, the answer is
$$\frac{\binom{9}{3}\times\binom{6}{3}}{3!}.$$
Now, let us come back to your question.
Your group has no names, so the answer is
$$\frac{\binom{mn}{m}\times\binom{mn-m}{m}\cdots\times\binom{2m}{m}}{n!}.$$
This is equal to what you wrote.