TITLE: Values of c for which the given quotient ring is a field.
QUESTION [4 upvotes]: I am stuck with the problem :

Find all values of 'c' in $F_{5}=\frac{\mathbb{Z}}{5\mathbb{Z}}$  such
that the quotient ring $\frac{F_{5}}{⟨X^3 + 3X^2 + cX + 3⟩}$ is a
field. Justify your answer.

My approach was, we've got a theorem for commutative ring R that if I is a maximal ideal in R then R/⟨I⟩ is a field. Now to prove $⟨X^3 + 3X^2 + cX + 3⟩$ is a maximal ideal in the given field we need to show that this is irreducible. So, I think for the set of values of 'c' for which this polynomial is irreducible, will be the set for which the above quotient ring is a field.
But I don't know how to find all the values of 'c' for which $⟨X^3 + 3X^2 + cX + 3⟩$ is irreducible except to try each value of 'c' individually and then use some irreducibility test.
Is there a proper and simpler way to find such 'c'. Please help me in finding such values.

REPLY [2 votes]: Actually, there is a theorem by Dickson to decide whether a cubic polynomial is irreducible over a finite field.

For $f(x)=x^3+bx+c$ over $\mathbb F_q$, where $q=p^n$ with $p>3$, its discriminant is $D(f)=-4b^3-27c^2$. Then $f$ is irreducible over $\mathbb F_q$ if and only if $D(f)$ is a square in $\mathbb F_q$, say $D(f)=81d^2$, and $\frac12(-c+dw)$ is a cube in $\mathbb F_q$ if $q\equiv1\pmod3$, or in $\mathbb F_{q^2}$ if $q\equiv2\pmod3$, where $w$ is a root of $x^2+3$ ($w\in\mathbb F_q$ if $q\equiv1\pmod3$ and $w\in\mathbb F_{q^2}$ if $q\equiv2\pmod3$).

In your case, $f(x)=x^3+3x^2+cx+3=x^3=(x+1)^3+(c-3)(x+1)-c+5$, so we can consider $g(x)=x^3+(c-3)x-c$ WLOG. It is kind of complicated, but there is another criterion which is more pratical in this case.

A polynomial of degree $2$ or $3$ is irreducible over a field $F$ if and only if it has no root in $F$.

Therefore we can just let $c$ vary over $\mathbb F_5$. Then $f$ is irreducible, if and only if $f(\alpha)\ne0$ for all $\alpha\in\mathbb F_5$.


TITLE: How prove this limit $\lim\limits_{n\rightarrow \infty} \frac{f_n}{f_{n+1}}=a$ given two other limits related to $f_n$
QUESTION [7 upvotes]: Let $(f_n)$- real sequence such that $$ \lim_{n\rightarrow \infty} \frac{f_{n+1}f_n-f_{n-1}f_{n+2}}{f_{n+1}^2-f_nf_{n+2}}=a+b, $$ and $$ \lim_{n\rightarrow \infty} \frac{f_{n}^2-f_{n-1}f_{n+1}}{f_{n+1}^2-f_nf_{n+2}}=ab   \quad    (|a|<|b|). $$
Prove that:$$\lim_{n\rightarrow \infty} \frac{f_n}{f_{n+1}}=a $$
I think we must prove $\displaystyle\lim_{n\rightarrow \infty} \frac{f_n}{f_{n+1}} $ exists,and we prove this limit is $a$,But I can't prove this limit exists.
My idea: since
$$\lim_{n\to\infty}\dfrac{\dfrac{f_{n}}{f_{n+1}}-\dfrac{f_{n-1}}{f_{n}}\dfrac{f_{n}}{f_{n+1}}\dfrac{f_{n+2}}{f_{n+1}}}{1-\dfrac{f_{n}}{f_{n+1}}\dfrac{f_{n+2}}{f_{n+1}}}=a+b$$
and
$$\lim_{n\to\infty}\dfrac{\left(\dfrac{f_{n}}{f_{n+1}}\right)^2-\dfrac{f_{n-1}}{f_{n}}\dfrac{f_{n}}{f_{n+1}}}{1-\dfrac{f_{n}}{f_{n+1}}\dfrac{f_{n+2}}{f_{n+1}}}=ab$$
But I felt this deal is not useful,
Other idea: I want to take the
Fibonacci sequence  to solve this problem, But I can't,
Thank you

REPLY [5 votes]: This is a supplement to several other answers about existence of the limit.
Let $P_n = \left|\begin{matrix}f_{n+1} & f_{n+2}\\ f_{n-1} & f_n\end{matrix}\right|$
and $Q_n = \left|\begin{matrix}f_{n} & f_{n+1} \\ f_{n-1} & f_n\end{matrix}\right|$,
the given condition can be rewritten as
$$\lim_{n\to\infty} \frac{P_n}{Q_{n+1}} = a + b
\quad\text{ and }\quad\lim_{n\to\infty} \frac{Q_n}{Q_{n+1}} =  ab\tag{*1}$$
Notice
$$f_n P_n - f_{n+1}Q_n = f_{n-1} Q_{n+1}
\quad\iff\quad f_{n+1} = \frac{P_n}{Q_n} f_n - \frac{Q_{n+1}}{Q_n} f_{n-1}
$$
The sequence satisfies a non-linear recurrence relations whose coefficients converge to
some finite limit. The "limit" of the recurrence relation has the form
$$f_{n+1} = \left(\frac{1}{a} + \frac{1}{b}\right) f_{n} - \frac{1}{ab} f_{n-1}$$
with characteristic polynomial $\lambda^2 - \left(\frac{1}{a} + \frac{1}{b}\right)\lambda + \frac{1}{ab} = \left(\lambda - \frac{1}{a}\right)\left(\lambda - \frac{1}{b}\right)$.
By assumption, $|a| < |b|$. This means distinct roots of this characteristic polynomial has distinct modulus. By Poincare-Perron theorem$\color{blue}{^{[1]}}$, we have either

$f_n = 0$ for all sufficiently large $n$ or
$\lim\limits_{n\to\infty} \frac{f_{n+1}}{f_n}$ exists and equal to $\frac{1}{a}$ or $\frac{1}{b}$.

The assumption $(*1)$ is written in such a way that $Q_{n+1}$ is non-zero for sufficient large $n$. So we can ignore the possibility that $f_n$ vanishes for large $n$. This implies the limit $\lim\limits_{n\to\infty} \frac{f_{n}}{f_{n+1}}$ does exist and equal to either $a$ or $b$.
Notes

$\color{blue}{[1]}$ For more details of Poincare-Perron theorem and a proof of it, please consult Chapter 8 of Saber Elaydi's book An Introduction to Difference Equations.


TITLE: Binomial Distribution Confidence Interval
QUESTION [0 upvotes]: Normally when I see confidence intervals it is in attempt to estimate a population parameter (probably poor wording). What I am trying to do is form a confidence interval for some theoretical values, and see if a simulation is behaving properly. For a simple example, consider a dice rolled $3333$ times. I am trying to construct a confidence interval $(90\%, 95\%, 99\%, 99.9\%)$ for the number of times a $6$ is rolled. I know the standard deviation is $\sqrt{np(1-p)}$, but am not sure how to use that to get a confidence interval. Also, is there a name for what I am trying do to. It doesn't seem like I am hypothesis testing. Thanks!

REPLY [1 votes]: You can use the approximate confidence interval for a proportion. In general it is used for normally distributed variables. But you have 3333 identical and independent distributed (iid) random variables. So you can apply the central limit theorem.
The two sided limits are
$\Large{\left[\hat p-z_{1-\frac{\alpha}{2}}\cdot \sqrt{\frac{\hat p\cdot (1-\hat p)}{n}}, \  \hat p+z_{1-\frac{\alpha}{2}}\cdot \sqrt{\frac{\hat p\cdot (1-\hat p)}{n}}\right]}$
You have a sample size of $n=3333$. For a confidence level fo 90% the significance level is $\alpha=1-0.9=0.1$. Therefore  $1-\frac{\alpha}{2}=0.95$
To find $z_{0.95}$ you have to find the value of z, where the value of the standard normal distribution is 0.95. If you look at this table, you will find that $z_{0.95}\approx 1.645$


TITLE: Confusion over term "uniformly distributed" in a probability problem
QUESTION [0 upvotes]: I'm trying to determine a joint PDF for random variables $X$ and $Y$ in this problem:
$(X,Y)$ is uniformly distributed on the subset of $\mathbb R^2$, defined by $0<X<2$ and $0 <Y<X^3$.
I'm not sure I understand what's meant by "uniformly distributed". Are they saying all valid $(X,Y)$ combinations have equal probability? How does one even begin to define a joint PDF for this?
Would it be something like
$f_{X,Y}(x,y) = \begin{cases}k,&0<X<2, 0 <Y<X^3 \\0,& \text{otherwise}   \end{cases}$
Where $k$ is some positive constant

REPLY [2 votes]: It means
$$f_{X,Y}(x,y) = \begin{cases}k,&0<X<2, 0 <Y<X^3 \\0,& \text{otherwise}   \end{cases}$$
where $k$ is a positive constant.
To find $k$, we have to find the area of the support.
we have $$k^{-1}= \int_0^2 x^3 \, dx$$


TITLE: a sequence and which of the following are true??
QUESTION [0 upvotes]: Consider a sequence $\{a_n\}$ of positive numbers satisfying the condition $a_na_{n+2} \le a_{n+1}^2, \forall n\in N$, then $\{a_n\}$ is a
a. convergent sequence if $a_1\ne2a_2$.
b. monotonically increasing sequence if $a_1\ne2a_2$.
c. convergent sequence if $a_1=2a_2$.
d. monotonically increasing sequence if $a_1=2a_2$.
From  $a_na_{n+2} \le a_{n+1}^2$, we have $$\frac{a_{n+2}}{a_{n+1}} \le \frac{a_{n+1}}{a_n}$$. Now if we look at $$\frac{a_n}{a_1}=\frac{a_n}{a_{n-1}}.\frac{a_{n-1}}{a_{n-2}}....\frac{a_2}{a_1} \le \frac{a_2}{a_1}.\frac{a_2}{a_1}....\frac{a_2}{a_1} $$. Since the RHS goes on for $n-1$ times,
$$\frac{a_n}{a_1}\le (\frac{a_2}{a_1})^{n-1}$$.
From (c), we definitely have $\{a_n\}$ as bounded as $$\frac{a_n}{a_1}\le (\frac{1}{2^{n-1}})$$. But that doesn't guarantee that the sequence is convergent. Or does it??

REPLY [1 votes]: $$\dfrac{a_n}{a_1}\le \dfrac{1}{2^{n-1}}\implies0<a_n\le\dfrac{1}{2^{n-1}}a_1\to0~\text{as}~n\to\infty\implies a_n\to0~\text{as}~n\to\infty$$

REPLY [1 votes]: a. Consider $a_{n} = n$;
b. Consider a constant sequence;
c. Consider your own work and get the $a_{1}$ to the other side of the equation. Do you see why it is convergent?
d. Consider $a_{n} = \frac{1}{2^{n}}$.


TITLE: Difference between two definitions for recurrence
QUESTION [1 upvotes]: Let $\{X_n\}$ be a Markov chain on a state-space $E$. A state $i$ is recurrent if $$P(X_n = i\;\text{for some} \;n\geq 1|X_0=i) = 1\tag{Definition 1}$$
$$\text{for some}\; n\geq 1\; P(X_n=i|X_0=i)=1\tag{Definition 2}$$
Definition $2$ is supposed to be incorrect. Could someone provide an example where definition $1$ holds but definition $2$ does not or vice versa? Or otherwise show me why they aren't equivalent. I've been told it might have something to do with when $n$ tends to infinity.

REPLY [0 votes]: If $\{X_n\}$ is any Markov chain with finite state space such that $0<p_{ij} <1$ for all $i,j$ then the first condition holds because such a chain is (positive) recurrent. However condition 2) fails because all elements of the n-th power of the transition matrix are all also less than $1$. 2) implies 1) is obvious.


TITLE: dy/dx implicity
QUESTION [1 upvotes]: Question: Find $\frac{dy}{dx}$ implicity if $x=\tan(xy)$.
So far, I have used implicit differentiation on both sides of the equation and ended up with two different answers. The first one was a bit more work than the second one.
My first answer is $\frac{dy}{dx}=\frac{\cos^2{xy}(1-y\sec^2{xy})}{x}$.
My second answer is $1=y\sec^2{xy}+x\sec^2{xy}\frac{dy}{dx}$
therefore $\frac{dy}{dx}=\frac{1 - y\sec^2{xy}}{x\sec^2{xy}}$.
The wording of the question is what's really throwing me off. Which of these is the correct answer, or are they both wrong?
Your help is always appreciated!

REPLY [3 votes]: Your answers are identical:
$$
\frac{1}{\sec^{2}(xy)} = \cos^{2}(xy).
$$


TITLE: Trivial case of $s \in T$
QUESTION [0 upvotes]: I was reading paper A Formal Basis for the Heuristics Determination of Minimum Cost Paths in section B "Some Definition About Graphs" there is footnote that say "We exclude the trivial case of $s \in T$" I don't get it, is it trivial like empty set? or have different meaning? You can see the paper in here
Sorry, if this question not appropriate.

REPLY [1 votes]: This is what I can gather. We are starting from node $s$, and $T$ is a set of nodes we would like to reach with minimum cost. If $s\in T$, then we are already there, without any cost or search, so that particular case is not interesting in the least and trivial. The authors are therefore excluding such cases for the rest of the paper.


TITLE: how to prove that $\exp(x)\geq x+1$
QUESTION [0 upvotes]: Hey I want the easiest method to prove $\exp(x)\geq x+1$.

The only method I use is to consider a new function $F$, that realizes
$F(x)=\exp(x)-x-1$ then calculate the derivative then use its monotony to prove that  $F(x)<0$
I'm only a high school student, if you could use function study it would be easier for me to understand.
So do you have any better method? this one takes me some time to write down I would love a easier method

REPLY [4 votes]: $$\exp(x) = 1 + \int_0^{x} \exp(t) \ \mathrm dt > 1 + \int_0^x 1 \ \mathrm dt = 1 + x$$

REPLY [0 votes]: Defining $$f(x)=e^x-x-1$$ then $$f'(x)=e^x-1$$ and $$f''(x)=e^x>0$$ Can you proceed?


TITLE: Integer length altitudes of a scalene triangle $ABC$
QUESTION [1 upvotes]: It is given that two of the altitudes of a scalene triangle $ABC$ have length $4$ and $12$. If the length of the third altitude is also an integer, find it's maximum value.

What I Tried: I have no good idea for this. Here is a picture for an example in Geogebra :-

I have to find the maximum value of $AG$ . In this example I suppose, the value of $CD$ can be increased more if we don't take an obtuse triangle, but all I am trying to do here is hit and trial.
Moreover, in this case $AG$ is not an integer (unless you consider it to be $4$) , so how do you make sure that the other altitude is an integer too?
Can anyone help me with this?

REPLY [1 votes]: $$ \triangle = \dfrac{1}{2}a\cdot4 = \dfrac{1}{2}b\cdot12 = \dfrac{1}{2}c\cdot h$$
$$ {a = \dfrac{\triangle}{2}, b = \dfrac{\triangle}{6}, c = \dfrac{2\triangle}{h}} $$
Now use
$$ a-b \lt c \lt a+b$$
I got $$\boxed{h_{\text{max}}=5}$$

REPLY [1 votes]: Consider the area of the triangle.
Let $AB = x$. Then the triangle has area $6x$ and thus $AC$ is equal to $3x$.
By triangle inequality, $BC > 2x$.
Since the area is $6x$, this implies that the remaining altitude cannot exceed $6$.
This argument also leads to the triangle inequality for altitudes:
$$\frac1 {h_a} < \frac1{h_b} + \frac1{h_c} \text{ , etc.}$$
To show the existence of a triangle with altitudes $12, 4, 5$, we take note of the corresponding side lengths $x, 3x, 2.4x$ and use Heron's Formula to obtain the corresponding $x$. The lengths will then form a (constructible) triangle with the desired altitudes.


TITLE: Link between derangement number and Bell polynomial
QUESTION [2 upvotes]: I have empirical evidence to support the following conjectured identity between the derangement numbers and incomplete Bell polynomials,
$!n = \sum_{i=1}^n B_{n,i}(0,1!,2!,\dots, (n-i)!)$,
in that it holds for the first 8 or so derangement numbers. I can find several related results online, such as
$n! = \sum_{i=1}^n B_{n,i}(0!,1!,2!,\dots, (n-i)!)$
(where the factorial is used on the first element also) from Wikipedia, and
$!n = \sum_{i=1}^n i! B_{n,i}(0,1,2,\dots,(n-i))$
from a paper of Qi and Guo.
I wonder if anyone has any ideas as to whether the result I think I have is either surprising or intuitive, whether it actually holds, and if so whether there exists already a proof of it somewhere.

REPLY [1 votes]: $$\sum_{i=1}^n B_{n,i}(0,1!,2!,\dots, (n-i)!)
=B_n(0,1!,\ldots,(n-1)!)$$
where $B_n$ is the complete Bell polynomial.
Also
$$B_n(x_1,x_2,x_3,\ldots,x_n)
=\sum_{j_1,\ldots,j_n\atop j_1+2j_2+\cdots+nj_n=n}
\frac{n!}{j_1!j_2\cdots j_n!}\left(\frac{x_1}{1!}\right)^{j_1}
\left(\frac{x_2}{2!}\right)^{j_2}\cdots\left(\frac{x_n}{n!}\right)^{j_n}.$$
So
$$B_n(0,x_2,x_3,\ldots,x_n)
=\sum_{j_2,\ldots,j_n\atop 2j_2+\cdots+nj_n=n}
\frac{n!}{j_2!\cdots j_n!}\left(\frac{x_2}{2!}\right)^{j_2}\cdots\left(\frac{x_n}{n!}\right)^{j_n}$$
and so
$$B_n(0,1!,2!,\ldots,(n-1)!)
=\sum_{j_2,\ldots,j_n\atop 2j_2+\cdots+nj_n=n}
\frac{n!}{j_2!\cdots j_n!}\left(\frac1{2}\right)^{j_2}\cdots\left(\frac1{n}\right)^{j_n}.$$
In this last sum, the summand is the number of permutations in $S_n$
with cycle structure $2^{j_2}3^{j_3}\cdots n^{j_n}$. Adding them all
up gives the number of derangements.


TITLE: elements of $SL(2,\mathbb{Z})$ which fix roots of Klein's absolute invariant $j(\tau)$
QUESTION [2 upvotes]: As a followup to this question (resulting video here), I'd like to make a video showing elements of $\mathbf{SL}(2,\mathbb{R})$ which fix roots of Klein's absolute invariant $j(\tau)$, stylized before and after frames of such a video below:

Given a root $\alpha$ of Klein's absolute invariant $j(\tau)$, which words (in generator matrices $S$ and $T$) of $\mathbf{SL}(2,\mathbb{Z})$ fix $\alpha$?
edit: it's live: Klein's j(τ) whirling upon SL(2,R) with its root exp(2πi/3)  fixed

REPLY [5 votes]: Because of the modular symmetries of $j$,
the zeros of $j$ are precisely the $\operatorname{SL}(2,\mathbb{Z})$-transforms
of the fundamental-domain zero $\zeta_3=\mathrm{e}^{2\pi\mathrm{i}/3}$.
Let $A\in\operatorname{SL}(2,\mathbb{Z})$ be such that it fixes $\zeta_3$,
and let $B\in\operatorname{SL}(2,\mathbb{Z})$ map $\zeta_3$
to another zero $\alpha$ of $j$.
Then $M=BAB^{-1}$ fixes $\alpha$.
Conversely, if $M\in\operatorname{SL}(2,\mathbb{Z})$ fixes $\alpha$,
then $A=B^{-1}MB$ fixes $\zeta_3$.
Therefore, the key task is to find all $A\in\operatorname{SL}(2,\mathbb{Z})$
that fix $\zeta_3$.
Note that "$M$ fixes $\alpha$" is equivalent to
"$M$ has eigenvector $\begin{pmatrix}\alpha\\1\end{pmatrix}$",
but the following reasoning does not need that.
Let $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\operatorname{SL}(2,\mathbb{Z})$,
then
$$\frac{a\zeta_3+b}{c\zeta_3+d} =
\frac{ac+bd-bc + \zeta_3}{c^2-cd+d^2}$$
This shall equal $\zeta_3$. Comparing imaginary and real parts separately, we find
that we have to seek integer solutions of
$$\begin{aligned}
 c^2-cd+d^2 &= 1
\\ ac+bd-bc &= 0
\\ ad-bc &= 1
\end{aligned}$$
For the first equation, note that $c=0$ implies $|d|=1$ and vice versa. Furthermore,
$c^2-cd+d^2 = (c \pm d)^2 - (1\pm2) cd \geq |cd|$ (choose the $\pm$ such that the right summand becomes nonnegative). These two considerations imply that $c,d$ cannot have
absolute value greater than $1$. With the search space thus narrowed, we find
$(c,d)\in\{\pm(0,1),\pm(1,0),\pm(1,1)\}$.
From that, the remaining two equations can be solved for $a,b$
by writing them as the linear system
$$\begin{pmatrix}c&d-c\\d&-c\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix} =
\begin{pmatrix}0\\1\end{pmatrix}$$
whose matrix is its own inverse, leading to
$$\begin{aligned}a &= d-c & b &= -c\end{aligned}$$
and thus to the solution set
$$A\in\left\{
\pm\begin{pmatrix}1&0\\0&1\end{pmatrix},
\pm\begin{pmatrix}-1&-1\\1&0\end{pmatrix},
\pm\begin{pmatrix}0&-1\\1&1\end{pmatrix}
\right\} = \left\{\pm I, \pm T^{-1}J, \pm JT\right\}$$
where
$$\begin{aligned}
 I &= \begin{pmatrix}1&0\\0&1\end{pmatrix}
& T &= \begin{pmatrix}1&1\\0&1\end{pmatrix}
& J &= \begin{pmatrix}0&-1\\1&0\end{pmatrix}
\end{aligned}$$
For use as transformations on $\mathbb{H}$, the $\pm$ does not matter.
Note that $T^{-1}J = (JT)^2$ and that $(JT)^3=-I$, so the solution set $A$
is the cyclic group generated by $JT$.
(You can easily verify that the solution set for any given fixed point
$\alpha$ is always a group.)
Conjugate the matrices with a suitable $B\in\operatorname{SL}(2,\mathbb{Z})$
to fix zeros of $j$ other than $\zeta_3$.
Edit: Clarified the reasoning a bit.


text:
\begin{document}

\title{PLB-spaces of holomorphic functions with logarithmic growth conditions}
\author{S.-A.~Wegner\,$^{\flat}$}
\date{November 23, 2011}
\maketitle
\renewcommand{\thefootnote}{}
\hspace{-1000pt}\footnote{$^{\flat}$\hspace{1pt}Sven-Ake Wegner, Fachbereich C -- Mathematik und Naturwissenschaften, Arbeitsgruppe Funktionalanalysis, Bergische Universit\"at Wuppertal, Gau\ss{}\-stra\ss{}e 20, D-42097 Wuppertal, Germany, Phone:\spacea{}\spacea{}+49\spacea{}(0)\spacea{}202\spacea{}/\spacea{}439\spacea{}-\spacea{}2531, e-mail: wegner@math.uni-wuppertal.de.\vspace{3pt}}
\hspace{-1000pt}\footnote{\hspace{4.5pt}2010 \emph{Mathematics Subject Classification}: Primary 46E10; Secondary 46A13.}
\hspace{-1000pt}\footnote{\hspace{4.5pt}\emph{Key words and phrases}: PLB-space, derived projective limit functor, weighted space.}
\vspace{-45pt}
{\small
\begin{abstract}\noindent{}Countable projective limits of countable inductive limits, called PLB-spaces, of weighted Banach spaces of continuous functions have recently been investigated by Agethen, Bierstedt and Bonet. In a previous article, the author extended their investigation to the case of holomorphic functions and characterized when spaces over the unit disc w.r.t.~weights whose decay, roughly speaking, is neither faster nor slower than that of a polynomial are ultrabornological or barrelled. In this note, we prove a similar characterization for the case of weights which tend to zero logarithmically.
\end{abstract}}

\section{Introduction}\label{Introduction}
\vspace{-10pt}

PLB-spaces, i.e.~countable projective limits of countable inductive limits of Banach spaces arise naturally in analysis. The space of distributions, the space of real analytic functions and several spaces of ultradistributions and ultradifferentiable functions constitute prominent examples for spaces of this type. In fact, all these spaces turn out to be even PLS-spaces, which means that the linking maps in the inductive spectra of Banach spaces are not only continuous but even compact. During the last years the theory of PLS-spaces has played an important role in the application of abstract functional analytic methods to several classical problems in analysis. We refer to the survey article \cite{Domanski2004} of Doma\'{n}ski for applications, examples and further references.
\smallskip
\\A fundamental tool in the theory of PLS-spaces is the so-called first derived functor of the projective limit functor $\Projeins$. In the late sixties, Palamodov \cite{Palamodov1968} applied this concept from homological algebra to the theory of locally convex spaces. Since the mid eighties, Vogt \cite{VogtLectures} and many others intensified the research on this subject. We refer to the book of Wengenroth \cite{Wengenroth} for a systematic exposition of the theory and a detailed list of references. Among many other results, Vogt \cite{VogtLectures, Vogt1989}, see \cite[3.3.4 and 3.3.6]{Wengenroth}, proved that there is a connection between the vanishing of $\Projeins$ on a countable projective spectrum of LB-spaces and locally convex properties of the corresponding projective limit (e.g.~being ultrabornological or barrelled). In addition, Vogt \cite[Section 4]{Vogt1989} gave characterizations of the vanishing of $\Projeins$ and the forementioned properties in the case of sequence spaces. Recently,
Agethen, Bierstedt, Bonet \cite{ABB2009} extended his results to weighted PLB-spaces of continuous functions. In \cite{WegnerJMAA} the author investigated weighted PLB-spaces of holomorphic functions, first on arbitrary balanced domains in $\mathbb{C}^d$ and then on the unit disc with certain technical assumptions on the weights which, roughly speaking, mean that the weights tend to zero like a polynomial.
\smallskip
\\In this article we restrict ourselves to the unit disc right from the beginning and consider weights which tend to zero logarithmically (see the remarks at the end of Section \ref{Notation:sec}). More formally, we consider weights satisfying the so-called condition (LOG) invented by Bonet, Engli\v{s}, Taskinen \cite{BoEnTa2005}. In Section \ref{Notation:sec} we state its definition and also the formal definition of the PLB-space $AH(\mathbb{D})$ which is the object of our study. In Section \ref{UaB} we recall several weight conditions due to Vogt \cite{Vogt1992} as well as variants introduced in \cite{WegnerJMAA}. Then we present the first result, Theorem \ref{LOG-main-result}, which provides necessary and sufficient conditions for the vanishing of $\Projeins$ and for $AH(\mathbb{D})$ being ultrabornological and barrelled. In Section \ref{Inter} we then treat the question of interchangeability of projective and inductive limit in the definition of $AH(\mathbb{D})$, where we also are able  to prove
necessary and sufficient conditions, see Theorem \ref{Int-result}, using a condition due to Vogt \cite{Vogt1983} and a variant introduced in \cite{WegnerJMAA}. Since in all our results up to this point the necessary and the sufficient conditions differ, we next identify additional assumptions that allow to turn the results of the previous sections into characterizations, see Section \ref{FurtherAss}. In addition, at the end of Section \ref{FurtherAss} we make some remarks on the construction of examples.

\section{Notation}\label{Notation:sec}
\vspace{-10pt}

Let $\mathbb{D}$ denote the unit disc of the complex plane and $H(\mathbb{D})$ the space of all holomorphic functions on $\mathbb{D}$ endowed with the topology co of uniform convergence on the compact subsets. Let $\mathcal{A}=((a_{N,n})_{N\in\mathbb{N}})_{n\in\mathbb{N}}$ be a double sequence of strictly positive and continuous functions (weights) on $\mathbb{D}$ which is decreasing in $n$ and increasing in $N$, i.e.~$a_{N,n+1}\leqslant{}a_{N,n}\leqslant{}a_{N+1,n}$ holds for all $N$ and $n$; this condition will be assumed on the double sequence $\mathcal{A}$ in the remainder of this article. We define
\begin{align*}
Ha_{N,n}(\mathbb{D})&:=\{\,f\in H(\mathbb{D})\: ; \: \|f\|_{N,n}:=\sup_{z\in\mathbb{D}}a_{N,n}(z)|f(z)|<\infty\,\},
\end{align*}
which is a Banach space for the norm $\|\cdot\|_{N,n}$ whose closed unit ball we denote by $B_{N,n}$. By definition, $Ha_{N,n}(\mathbb{D})\subseteq Ha_{N,n+1}(\mathbb{D})$ holds with continuous inclusion for all $N$ and $n$ and we can define for each $N$ the weighted inductive limit
$$
\mathcal{A}_NH(\mathbb{D}):=\ind{n}Ha_{N,n}(\mathbb{D})
$$
which is a complete and hence regular LB-space by Bierstedt, Meise, Summers \cite[end of the remark after 1.13]{BMS1982}. For each $N$ we have $\mathcal{A}_{N+1}H(\mathbb{D})\!\subseteq\!\mathcal{A}_NH(\mathbb{D})$ with continuous inclusion. Hence, $\mathcal{A}H\!:=\!(\mathcal{A}_NH(\mathbb{D}))_{N\in\mathbb{N}}$ is a projective spectrum of LB-spaces with inclusions as linking maps and we can now form the following projective limit, called \textit{weighted PLB-space of holomorphic functions}
$$
AH(\mathbb{D}):=\proj{N}\mathcal{A}_NH(\mathbb{D})=\ind{n}\proj{N}Ha_{N,n}(\mathbb{D})
$$
which is the object of our study in this work. We refer the reader to the book of Wengenroth \cite{Wengenroth} for a detailed exposition of the theory of projective spectra of locally convex spaces $\mathcal{X}=(X_N)_{N\in\mathbb{N}}$, their projective limits $\proj{N}X_N$ and the derived functor $\Projeins$.
\smallskip
\\In this article we consider weights of the following special type; the definition of the so-called condition (LOG) is due to Bonet, Engli\v{s}, Taskinen \cite[4.1]{BoEnTa2005} and was used to prove a projective description for weighted LB-spaces of holomorphic functions. For every $\kappa\in\mathbb{N}$ we put $r_{\kappa}:=1-2^{-2^{\kappa}}$, $r_0:=0$ and $I_{\kappa}:=[r_{\kappa},r_{\kappa+1}]$. We say that the sequence $\mathcal{A}=((a_{N,n})_{N\in\mathbb{N}})_{n\in\mathbb{N}}$ satisfies condition (LOG) if each weight in the sequence is radial and approaches monotonically zero as $r\nearrow1$ and if there exist constants $0<a<1<A$ such that
$$
\text{(LOG 1)}\;\;A\cdot a_{N,n}(r_{\kappa+1})\geqslant a_{N,n}(r_{\kappa})\;\text{ and\; (LOG 2)}\;\;a_{N,n}(r_{\kappa+1})\leqslant a\cdot a_{N,n}(r_{\kappa})
$$
hold for all $N$, $n$ and $\kappa\in\mathbb{N}$. For our investigation we need the following well-known fact; for a proof we refer to \cite[Proof of Remark 1.1]{WegnerBET}: If $v$ is a radial weight which is decreasing on  $[0,1[$ and $(r_n)_{n\in\mathbb{N}}\subseteq[0,1[$ a sequence with $r_n\nearrow1$ as $n\rightarrow\infty$, then for $g\in Hv(\mathbb{D})$ we have $g_n\rightarrow g$ w.r.t.~co, where $g_n(z):=g(r_nz)$ for $z\in\mathbb{D}$.
\smallskip
\\Let us add the following explanatory comment on the meaning of condition (LOG). We assume that $v\colon\mathbb{D}\rightarrow\:]0,\infty[$ satisfies (LOG), put $K:=\{1/(1-r_{\kappa}) \:;\:\kappa\in\mathbb{N}\}$ and consider $w\colon K\rightarrow\,]0,\infty[$ with $w(1/(1-r_{\kappa}))=v(r_{\kappa})$. Then $v(0)(1/A)^{\log\log \nu}\leqslant w(\nu)\leqslant v(0)a^{\log\log\nu}$ holds for all $\nu\in K\backslash\{1\}$, where $\log$ denotes the binary logarithm. Therefore, $w(\nu)=\mathcal{O}(1/(\log \nu)^c)$ and $w(\nu)=\Omega(1/(\log \nu)^C)$ is valid for $\nu\rightarrow\infty$, where $C=-\log1/A>0$ and $c=-\log a>0$. In this sense, the weights satisfying (LOG) tend to zero logarithmically.

\section{Proj$^{\text{\,1}}$ and locally convex properties}\label{UaB}
\vspace{-10pt}

In this section we present necessary and sufficient conditions for the vanishing of $\Projeins\mathcal{A}H$ and $AH(\mathbb{D})$ being ultrabornological and barrelled. To formulate our results, we need the following notation due to Vogt \cite{Vogt1992}. We say that the sequence $\mathcal{A}$ satisfies condition (Q) if
$$
\textstyle\forall \: N \; \exists \: M \geqslant N,\, n \; \forall \: K \geqslant M,\, m,\, \varepsilon > 0 \; \exists \: k, \, S>0 : \frac{1}{a_{M,m}} \leqslant \max\big(\frac{\varepsilon}{a_{N,n}},\frac{S}{a_{K,k}}\big),
$$
we say that it satisfies (wQ) if
$$
\textstyle\forall \: N \; \exists \: M \geqslant N,\, n \; \forall \: K \geqslant M,\, m \; \exists \: k, \, S>0 : \frac{1}{a_{M,m}} \leqslant \max\big(\frac{S}{a_{N,n}},\frac{S}{a_{K,k}}\big).
$$
Condition (Q) implies condition (wQ); the converse is not true, cf.~Bierstedt, Bonet \cite{BB1994}. We define condition $\text{(Q)}^{\sim}$ by the same quantifiers as in (Q) but the estimate replaced with $(a_{M,m}^{-1})^{\sim}\leqslant\max(\varepsilon a_{N,n}^{-1},S a_{K,k}^{-1})^{\sim}$, where for a given weight $a$, $(1/a)^{\sim}\colon \mathbb{D}\rightarrow \mathbb{R}$, $z\mapsto\sup\{|g(z)|\:;\:g\in H(\mathbb{D}),\,a|g|\leqslant 1\}$ is the associated growth condition and $\tilde{a}:=1/(1/a)^\sim$ the associated weight, cf.~Bierstedt, Bonet, Taskinen \cite{BBT}. We define $\text{(wQ)}^{\sim}$ by the same quantifiers as in (wQ) and the estimate replaced with $(a_{M,m}^{-1})^{\sim}\leqslant\max(\varepsilon(a_{N,n}^{-1})^{\sim},S(a_{K,k}^{-1})^{\sim})$.
\smallskip
\\Now we are able to present our first result;  its proof was inspired by the method developed in \cite[Section 4]{BoEnTa2005}, see also \cite{WegnerBET}.

\begin{thm}\label{LOG-main-result} Let $\mathcal{A}$ satisfy condition (LOG). Then we have the implications (i)$\Rightarrow$(ii) $\Rightarrow$(iii)$\Rightarrow$(iv)$\Rightarrow$(v), where
\vspace{5pt}
\\\begin{tabular}{rlrl}
(i)   & $\mathcal{A}$ satisfies condition $\text{(Q)}^{\sim}$,            & (iv) & $AH(\mathbb{D})$ is barrelled, \\
(ii)  & $\Projeins \mathcal{A}H=0$,                                                    & (v)  & $\mathcal{A}$ satisfies condition $\text{(wQ)}^{\sim}$. \\
(iii) & $AH(\mathbb{D})$ is ultrabornological,                                         &      & \\
\end{tabular}
\end{thm}
\begin{proof}\textquotedblleft{}(i)$\Rightarrow$(ii)\textquotedblright{}\; In order to show that $\Projeins \mathcal{A}H=0$ holds, we use Braun, Vogt \cite[Theorem 8]{BraunVogt1997} -- which was obtained independently by Frerick, Wengenroth \cite{FrerickWengenroth1996}. That is, we have to verify condition $\text{(}\overline{\text{P}_{\!2}}\text{)}$
$$
\forall \: N \; \exists \: M,\,n \; \forall \: K,\,m,\,\varepsilon>0 \; \exists \: k,\,S>0\: \colon \: B_{M,m}\subseteq \varepsilon B_{N,n}+SB_{K,k}.
$$
We denote by $0<a<1<A$ the constants of (LOG 1) and (LOG 2) and put $B:=\max\big(\sum_{\kappa=0}^{\infty}a^{\kappa},\,\sup_{\kappa>t+2}2^{-\kappa}A^{\kappa-t}2^{-2^{\kappa-1}}\big)$. In addition, we put $T:=2A^2(B+A^2)+ 4(A^2+2B)$.
\smallskip
\\For given $N$ we select $M$ and $n$ as in $\text{(Q)}^{\sim}$. For given $K,\,m,\,\varepsilon>0$ we put $\varepsilon':=\frac{\varepsilon}{2T}$ and choose $k$ and $S'>0$ according to $\text{(Q)}^{\sim}$ w.r.t.~$\varepsilon'$ and put $S:=2TS'$. Now we fix an arbitrary $f\in B_{M,m}$. We have $|f|\leqslant\frac{1}{a_{M,m}}$, i.e.~with \cite[1.2.(iii)]{BBT} it follows $|f|\leqslant (\frac{1}{a_{M,m}})^{\sim}$. By the estimate in $\text{(Q)}^{\sim}$ we obtain $|f|\leqslant\max(\varepsilon'(\frac{1}{a_{N,n}}),\,S'(\frac{1}{a_{K,k}}))^{\sim}\leqslant\max(\frac{\varepsilon'}{a_{N,n}},\frac{S'}{a_{K,k}})$ where the last estimate follows from \cite[1.2.(i)]{BBT}. We put $u:=\min(\frac{a_{N,n}}{\varepsilon'},\,{\textstyle\frac{a_{K,k}}{S'}})$. Hence $\|f\|_u:=\sup_{z\in\mathbb{D}}u(z)|f(z)|\leqslant1$. By defining $u_0:=a_{N,n}$, $u_1:=a_{K,k}$, $a_0:=1/\varepsilon'$ and $a_1:=1/S'$ we get $u=\min(a_0u_0,\,a_1u_1)$. We put (according to \cite[Proof of 4.5]{BoEnTa2005}) $f_{r_{\kappa}}(z):=f(r_{\kappa}z)$ and obtain
$f_{r_{\kappa}}\rightarrow f$ within the compact open topology (cf.~our remarks after the definition of (LOG) in Section \ref{Notation:sec}).
\medskip
\\Since all the weights in $\mathcal{A}$ are non-increasing, this is also true for $u$. Hence
$$
\text{(1)}\;\;\;\;\;\;\inf_{|z|\in I_{\kappa}}u(z)\:=\:u(r_{\kappa+1})\:\geqslant\:u(r_{\kappa+2})\:\,=\hspace{-5pt}\inf_{|z|\in I_{\kappa+1}}\!\!u(z)\!\!\stackrel{\scriptscriptstyle\text{(LOG 1)}}{\geqslant}\!\!A^{-2}u(r_{\kappa})
$$
holds. For every $\kappa$ in $\mathbb{N}$ we pick $i(\kappa)\in\{0,\,1\}$ such that
$$
\text{(2)}\;\;\;\;\;\;u(r_{\kappa})=a_{i(\kappa)}u_{i(\kappa)}(r_{\kappa})=a_{i(\kappa)}\sup_{|z|\in I_{\kappa}}u_{i(\kappa)}(z)
$$
is valid. For $\nu\in\mathbb{N}$ and $\ell\in\{0,\,1\}$ we define $N_{\ell}:=\{\,\kappa\in\mathbb{N}\:;\:\kappa\leqslant \nu\text{ and } i(\kappa)=\ell\,\}$. For each $\kappa\geqslant1$ we put $g_{\kappa}(z):=f(r_{\kappa+1}z)-f(r_{\kappa}z)$ and $g_0(z):=f(0)$. Finally, we define
$$
\text{(3)}\;\;\;\;\;\;h_{\ell}\,:=\!\Bigsum{\kappa\in N_{\ell}}{}g_{\kappa}
$$
for $\ell\in\{0,1\}$ and compute $f_{r_{\nu+1}}=g_0+h_0+h_1$. For the constant function $g_0$ we have $|g_0(z)|=|f(0)|=|f(r_0)|\leqslant a_{i(0)}^{-1}u_{i(0)}(0)^{-1}$ that is $a_{N,n}(z)|g_0(z)|\leqslant\varepsilon'\leqslant\frac{\varepsilon}{2}$ (if $i(0)=0$) or $a_{K,k}(z)|g_0(z)|\leqslant S'\leqslant{\textstyle\frac{S}{2}}$ (if $i(0)=1$). Now we fix $\ell\in\{0,1\}$, pick $\kappa\in N_{\ell}$ and estimate $|g_{\kappa}(z)|$ for different $z$.\vspace{3pt}
\begin{compactitem}
\item[1.] Assume first $|z|\geqslant r_{\kappa-1}$ (where we put $r_{\kappa-1}:=r_0$ for $\kappa=0$).\vspace{3pt}
\begin{compactitem}
\item[a.] Let $\kappa\geqslant2$. Then we have $|r_{\kappa}z|\geqslant(1-2^{-2^{\kappa}})(1-2^{-2^{\kappa-1}})\geqslant r_{\kappa-2}$ and thus $r_{\kappa-2}\leqslant|r_{\kappa}z|\leqslant |r_{\kappa+1}z|\leqslant r_{\kappa+1}$. Since $\|f\|_u\leqslant1$, we have $|f(z)|\leqslant u(z)^{-1}$ on $\mathbb{D}$. Since $u$ is non-increasing and by (1) we get the estimate $|g_{\kappa}(z)|\leqslant{}|f(r_{\kappa}z)|+|f(r_{\kappa+1}z)|\leqslant{}2\,\max\,(\,\sup_{r\in I_{\kappa-2}}u(r)^{-1},\:\sup_{r\in I_{\kappa-1}}u(r)^{-1},\linebreak{}\:\sup_{r\in I_{\kappa}}u(r)^{-1}) \leqslant 2u(r_{\kappa+1})^{-1}\leqslant 2 A^2 u(r_{\kappa})^{-1}= 2A^2 a_{\ell}^{-1}u_{\ell}(r_{\kappa})^{-1}$ where last equality follows since $u(r_{\kappa})=a_{i(\kappa)}u_{i(\kappa)}(r_{\kappa})$ and $\kappa\in N_{\ell}$ implies $i(\kappa)=\ell$ (cf.~(2)).\vspace{3pt}
\item[b.] Let $\kappa=1$. In this case we have $|g_1(z)|\leqslant |f(r_2z)|+|f(r_1z)|\leqslant2\linebreak{}\sup_{r_0\leqslant r\leqslant r_2}u(r)^{-1}\leqslant 2\max(\sup_{r\in I_0}u(r)^{-1},\,\sup_{r\in I_1}u(r)^{-1})=2u(r_2)^{-1}\leqslant2A^2a_{\ell}^{-1}u_{\ell}(r_1)^{-1}$\! where we use (1) for the last estimate.
\vspace{3pt}
\item[c.] Let $\kappa=0$. We have $|g_{\kappa}(z)|=|f(0)|$ and $\|f\|_u\leqslant1$ implies in particular $u(0)|f(0)|\leqslant1$, i.e.~
$
|g_{\kappa}(z)|=|f(0)|  \leqslant u(0)^{-1}= u(r_0)^{-1}=a_{i(0)}^{-1}u_{i(0)}(r_0)^{-1}\leqslant 2A^2a_{i(\kappa)}^{-1}u_{i(\kappa)}(r_{\kappa})^{-1}=2A^2a_{\ell}^{-1}u_{\ell}(r_{\kappa})^{-1}
$
by (2), since $A>1$ and by our selection $\kappa\in N_{\ell}$.\vspace{3pt}
\end{compactitem}
To sum up, in case 1.~we have
$$
\text{(4)}\;\;\;\;\;\;|g_{\kappa}(z)|\leqslant 2A^2 a_{\ell}^{-1}u_{\ell}(r_{\kappa})^{-1}
$$
for $|z|\geqslant r_{\kappa-1}$ and $\kappa\geqslant0$.\vspace{5pt}
\item[2.] Assume that $\kappa>t+1$ and $|z|\in I_t$, i.e.~$r_t\leqslant|z|\leqslant r_{t+1}$. We have $|g_{\kappa}(z)|=|f(r_{\kappa}z)-f(r_{\kappa+1}z)|$ by definition. By the mean value theorem there exists $\xi$ between $r_{\kappa}z$ and $r_{\kappa+1}z$ with $|f(r_{\kappa}z)-f(r_{\kappa+1})|=|f'(\xi)||r_{\kappa}z-r_{\kappa+1}z|\leqslant|f'(\xi)||r_{\kappa}-r_{\kappa+1}|$. Since $|r_{\kappa+1}-r_{\kappa}|\leqslant 2^{-2^{\kappa}}$ holds, the above yields $|g_{\kappa}(z)|\leqslant\sup_{r_{\kappa}r_t\leqslant|\xi|\leqslant r_{\kappa+1}r_{t+1}}|f'(\xi)|2^{-2^{\kappa}}$. Our assumption $t<\kappa-1$ implies $|\xi|\leqslant r_{\kappa+1}r_{t+1}< r_{t+1}\leqslant r_{\kappa}$ and we thus may use the Cauchy formula
$$
\text{(5)}\;\;\;\;\;\;|f'(\xi)|\leqslant{\textstyle\frac{1}{2\pi}}\int_{|\eta|=r_{\kappa}}{\textstyle\frac{|f(\eta)|}{|\eta-\xi|^2}}d\eta
$$
to estimate $|f'(\xi)|$. We have $|f(\eta)|\leqslant u(\eta)^{-1}=u(r_{\kappa})^{-1}$, since $\|f\|_u\leqslant1$ and $u$ is radial. For $|\eta|=r_{\kappa}$ and $\kappa>t+2$ straightforward computations show $\frac{1}{|\eta-\xi|^2}\leqslant4\cdot2^{2^{\kappa-1}}$, i.e.~$|f'(\xi)|\leqslant{\textstyle\frac{2\pi r_{\kappa}}{2\pi}}\cdot4\cdot2^{2^{\kappa-1}}u(r_{\kappa})^{-1}\leqslant 4\cdot2^{2^{\kappa-1}}u(r_{\kappa})^{-1}$ holds by (5) and we get
$|g_{\kappa}(z)|\leqslant4\cdot2^{2^{\kappa-1}}a_{\ell}^{-1}u_{\ell}(r_{\kappa})^{-1}$. If $\kappa=t+2$, similar computations show $|g_{\kappa}(z)|\leqslant 4a_{\ell}^{-1}u_{\ell}(r_{\kappa})^{-1}$. Now we use (LOG 1) $(\kappa-t)-$times to obtain $u_{\ell}(r_t)\leqslant Au_{\ell}(r_{\kappa+1})\leqslant A^2u_{\ell}(r_{t+2})\leqslant\cdots\leqslant A^{\kappa-t}u_{\ell}(r_{t+\kappa-t})\leqslant A^{\kappa-t}u_{\ell}(r_{\kappa})$. Since $|z|\geqslant r_t$ and since $u_{\ell}$ is radial and decreasing for $r\nearrow 1$ we have $u_{\ell}(r_t)\geqslant u_{\ell}(z)$ and thus we get $u_{\ell}(z)\leqslant u_{\ell}(r_{t})\leqslant A^{\kappa-t}u_{\ell}(r_{\kappa})$, which finally yields $u_{\ell}(r_{\kappa})^{-1}\leqslant A^{\kappa-t}u_{\ell}(z)^{-1}$. For $\kappa>t+2$ we get $|g_{\kappa}(z)|\leqslant 4a_{\ell}^{-1}u_{\ell}(z)^{-1}A^{\kappa-t}2^{-2^{\kappa-1}}$ from the latter and thus $|g_{\kappa}(z)|\leqslant4\cdot2^{-\kappa}Ba_{\ell}^{-1}u_{\ell}(z)^{-1}$ by our selection of $B$. If $\kappa=t+2$ we get $|g_{\kappa}(z)
|\leqslant 4a_{\ell}^{-1}u_{\ell}(z)^{-1}A^{2}$.
\smallskip
To sum up, in case 2.~we have
$$
\text{(6)}\;\;\;\;\;\;|g_{\kappa}(z)|\leqslant 4a_{\ell}^{-1}u_{\ell}(z)^{-1}\,
\begin{cases}
\;2^{-\kappa}B & \text{if }\;\kappa>t+2\\
\;\;\,A^{2} & \text{if }\;\kappa=t+2
\end{cases}
$$
for $|z|\in I_t$ and $\kappa$ as indicated above.
\end{compactitem}
To complete the proof, let now $z\in\mathbb{D}$ be arbitrary. We select $t\in\mathbb{N}$ such that $|z|\in I_t=[r_t,r_{t+1}]$. Then
$$
\text{(7)}\;\;\;\;|h_{\ell}(z)|\,\stackrel{\text{\tiny dfn}}{=}\,\big|\!\!\Bigsum{\kappa\in N_{\ell}}{}\!g_{\kappa}(z)\hspace{0.5pt}\big|\,\leqslant\!\Bigsum{\stackrel{\kappa\in N_{\ell}}{\scriptscriptstyle\kappa\leqslant t+1}}{}\hspace{-2pt}|g_{\kappa}(z)|\:+\!\!\Bigsum{\stackrel{\kappa\in N_{\ell}}{\scriptscriptstyle\kappa>t+1}}{}\hspace{-2pt}|g_{\kappa}(z)|\:=:\:G_{\ell}(z)+H_{\ell}(z).
$$
\begin{compactitem}
\item[(i)] Consider $G_{\ell}(z)$, that is all occurring $\kappa$ satisfy $0\leqslant\kappa\leqslant t+1$ and $\kappa\in N_{\ell}$. Thus we have $\kappa-1\leqslant t$, hence $|z|\geqslant r_t\geqslant r_{\kappa-1}$ (remember that we defined $r_{-1}:=r_0=0$). By (4) we therefore have
$$
\text{(8)}\;\;\;\;\;\;G_{\ell}(z)\,\stackrel{\text{\tiny dfn}}{=}\!\!\Bigsum{\stackrel{\kappa\in N_{\ell}}{\scriptscriptstyle\kappa\leqslant t+1}}{}\hspace{-2pt}|g_{\kappa}(z)|\:\leqslant\!\Bigsum{\stackrel{\kappa\in N_{\ell}}{\scriptscriptstyle\kappa\leqslant t+1}}{}\hspace{-1.5pt}2A^2a_{\ell}^{-1}u_{\ell}(r_{\kappa})^{-1}.
$$
(LOG 2) implies $u_{\ell}(r_{\kappa+1})\leqslant a u_{\ell}(r_{\kappa})$, i.e.~$u_{\ell}(r_{\kappa})^{-1}\leqslant a u_{\ell}(r_{\kappa+1})^{-1}$ for arbitrary $\kappa$. Iterating this estimate $t-\kappa$ times for a fixed $\kappa\leqslant t$ we get $u_{\ell}(r_{\kappa})^{-1}\leqslant a u_{\ell}(r_{\kappa+1})^{-1}\leqslant\cdots\leqslant a^{t-\kappa}u_{\ell}(r_{\kappa+t-\kappa})^{-1}=a^{t-\kappa}u_{\ell}(r_{t})^{-1}$.
With the latter we get from (8)
\begin{align*}
G_{\ell}(z)
& \leqslant 2A^2a_{\ell}^{-1}\big(\Bigsum{\kappa=0}{t}u_{\ell}(r_{\kappa})^{-1} + u_{\ell}(r_{t+1})^{-1}\big) \\
&\leqslant 2A^2a_{\ell}^{-1}\big(\Bigsum{\kappa=0}{t}a^{t-\kappa}u_{\ell}(r_t)^{-1} + A^2u_{\ell}(r_{t})^{-1}\big) \\
&= 2A^2a_{\ell}^{-1}u_{\ell}(r_t)^{-1}\big(\Bigsum{\sigma=0}{t}a^{\sigma} + A^2\big) \\
&\leqslant 2A^2(B+A^2)a_{\ell}^{-1}u_{\ell}(z)^{-1}
\end{align*}
where we used that $B>\Bigsum{\kappa\in\mathbb{N}}{}a^{\kappa}$, that $u_{\ell}$ is radial and decreasing for $r\nearrow1$ and $|z|\geqslant r_t$, whence $u_{\ell}(r_t)^{-1}\leqslant u_{\ell}(z)^{-1}$.\vspace{3pt}
\item[(ii)] Consider $H_{\ell}(z)$. Then all the occurring $\kappa$ satisfy $\kappa>t+1$ and $\kappa\in N_{\ell}$. By (6) we obtain
\begin{align*}
\;\;\;\;\;\;\;H_{\ell}(z)\,\stackrel{\text{\tiny dfn}}{=}\hspace{-1.8pt}\Bigsum{\stackrel{\kappa\in N_{\ell}}{\scriptscriptstyle\kappa>t+1}}{}\hspace{-2pt}|g_{\kappa}(z)|\;&=\;\delta_{i(t+2),\ell}\,|g_{t+2}|\:+\hspace{-2pt}\Bigsum{\stackrel{\kappa\in N_{\ell}}{\scriptscriptstyle\kappa>t+2}}{}\hspace{-1.5pt}|g_{\kappa}(z)|\\
&\leqslant\;4a_{\ell}^{-1}u_{\ell}(z)^{-1}A^2\,+\hspace{-2pt}\Bigsum{\stackrel{\kappa\in N_{\ell}}{\scriptscriptstyle\kappa>t+2}}{}\!4\cdot2^{-\kappa}Ba_{\ell}^{-1}u_{\ell}(z)^{-1}\\
&\leqslant\;\big(4A^2+4B\Bigsum{\kappa=0}{\infty}2^{-\kappa}\big)a_{\ell}^{-1}u_{\ell}(z)^{-1}\\
&=\;4(A^2+2B)a_{\ell}^{-1}u_{\ell}(z)^{-1},
\end{align*}
where $\delta$ denotes the Kronecker symbol.\vspace{3pt}
\end{compactitem}
Combining the estimates in (i) and (ii) we obtain $|h_{\ell}(z)|\leqslant G_{\ell}(z)+H_{\ell}(z)\leqslant(2A^2(B+A^2)+ 4(A^2+2B))a_{\ell}^{-1}u_{\ell}(z)^{-1}=Ta_{\ell}^{-1}u_{\ell}(z)^{-1}$ from (7), that means $u_{\ell}(z)|h_{\ell}(z)|\leqslant Ta_{\ell}^{-1}$ for each $z\in\mathbb{D}$ and $\ell=0,\,1$. By the definition of $u_{\ell}$ and $a_{\ell}$ this means $a_{N,n}(z)|h_0(z)|=u_0(z)|h_0(z)|\leqslant T\varepsilon'\leqslant{\textstyle\frac{\varepsilon}{2}}$ and $a_{K,k}(z)|h_1(z)|=u_1(z)|h_1(z)|\leqslant TS'\leqslant{\textstyle\frac{S}{2}}$ for each $z\in\mathbb{D}$. Hence $h_0\in \frac{\varepsilon}{2}B_{N,n}$ by the first estimate and $h_1\in \frac{S}{2}B_{K,k}$ by the second estimate, i.e.~$f_{r_{\nu+1}}=g_0+h_0+h_1\in{\textstyle\frac{\varepsilon}{2}}B_{N,n}+{\textstyle\frac{S}{2}}B_{K,k}+{\textstyle\frac{\varepsilon}{2}}B_{N,n}+{\textstyle\frac{S}{2}}B_{K,k}\subseteq\varepsilon B_{N,n} + S B_{K,k}$. Since $\varepsilon B_{N,n} + S B_{K,k}$ is co-compact, $f\in \varepsilon B_{N,n} + S B_{K,k}$ follows
from our remark after the definition of (LOG) and we are done.
\smallskip
\\\textquotedblleft{}(ii)$\Rightarrow$(iii)$\Rightarrow$(iv)\textquotedblright{}\; The first implication holds in general, see Wengenroth \cite[3.3.4]{Wengenroth} (cf.~Vogt \cite[5.7]{VogtLectures}). The second also holds in general, see e.g.~\cite{MeiseVogtEnglisch}.
\smallskip
\\\textquotedblleft{}(iv)$\Rightarrow$(v)\textquotedblright{}\; The assumptions of (LOG) imply that the topology of $Ha_{N,n}(\mathbb{D})$ is stronger than co and that the polynomials are contained in the latter space for any $N$ and $n$; by Bierstedt, Bonet, Galbis \cite[remark previous to 1.2]{BBG} the second statement is equivalent to requiring that each weight $a_{N,n}$ extends continuously to $\bar{\mathbb{D}}$ with $a_{N,n}|_{\partial\mathbb{D}}=0$. Therefore, the statement follows from \cite[Section 3]{WegnerJMAA}.
\end{proof}

\section{Interchangeability of projective and inductive limit}\label{Inter}
\vspace{-10pt}

Given a sequence of weights $\mathcal{A}=((a_{N,n})_{N\in\mathbb{N}})_{n\in\mathbb{N}}$ on the unit disc we can -- in addition to the PLB-space investigated in the preceding section -- also associate a \textit{weighted LF-space of holomorphic functions} by defining
\begin{align*}
\mathcal{V}H(\mathbb{D})&:=\ind{n}\proj{N}Ha_{N,n}(\mathbb{D}).
\end{align*}
This space constitutes the holomorphic version of the weighted LF-space of continuous functions investigated by Bierstedt, Bonet \cite{BB1994} and has been studied by several authors in different contexts, cf.~the survey \cite{Bierstedt2001} of Bierstedt for detailed references. In \cite[Section 3]{WegnerJMAA} the author showed that $\mathcal{V}H(\mathbb{D})\subseteq AH(\mathbb{D})$ holds in general with continuous inclusion. In order to investigate when $\mathcal{V}H(\mathbb{D})= AH(\mathbb{D})$ holds algebraically, in \cite{WegnerJMAA} the following condition based on research of Vogt \cite[1.1]{Vogt1983} was used: We say that a sequence $\mathcal{A}$ as above satisfies condition (B) if
$$
\forall \: (n(N))_{N\in\mathbb{N}}\subseteq \mathbb{N} \; \exists \: m \; \forall \: M \; \exists \: L,\, c>0 \colon a_{M,m} \, \leqslant \, c \max_{\scriptscriptstyle N=1,\dots,L} a_{N,n(N)}.
$$
Condition $\text{(B)}^{\sim}$ is defined by the same quantifiers and the estimate replaced by $\tilde{a}_{M,m} \, \leqslant \, c (\max_{\scriptscriptstyle N=1,\dots,L} a_{N,n(N)})^{\sim}$. Now we are ready to state the result on interchangeability of projective and inductive limit.

\begin{thm}\label{Int-result}Let $\mathcal{A}$ satisfy condition (LOG).
\begin{compactitem}\vspace{2pt}
\item[(i)] $AH(\mathbb{D})=\mathcal{V}H(\mathbb{D})$ holds algebraically if and only if $\mathcal{A}$ satisfies $\text{(B)}^{\sim}$.\vspace{3pt}
\item[(ii)] If $AH(\mathbb{D})=\mathcal{V}H(\mathbb{D})$ holds algebraically and topologically then $\mathcal{A}$ satisfies the conditions $\text{(B)}^{\sim}$ and $\text{(wQ)}^{\sim}$.\vspace{3pt}
\item[(iii)] If $\mathcal{A}$ satisfies the conditions $\text{(B)}^{\sim}$ and $\text{(Q)}^{\sim}$ then $AH(\mathbb{D})=\mathcal{V}H(\mathbb{D})$ holds algebraically and topologically.
\end{compactitem}
\end{thm}
\begin{proof}(i) As in the proof of the implication \textquotedblleft{}(iv)$\Rightarrow$(v)\textquotedblright{} of Theorem \ref{LOG-main-result} the statement is even true in a more general setting, see \cite[Section 5]{WegnerJMAA}.
\smallskip
\\(ii) This follows directly from (i) and Theorem \ref{LOG-main-result}.
\smallskip
\\(iii) Let $\text{(B)}^{\sim}$ and $\text{(Q)}^{\sim}$ be satisfied. By (i) the identity $\mathcal{V}H(\mathbb{D})\rightarrow AH(\mathbb{D})$ is one-to-one and as we noted above it is continuous. Since $AH(\mathbb{D})$ is ultrabornological by Theorem \ref{LOG-main-result} and $\mathcal{V}H(\mathbb{D})$ is webbed, we can apply the open mapping theorem (cf.~Meise, Vogt \cite[24.30]{MeiseVogtEnglisch}) and are done.
\end{proof}

\section{Further assumptions on the defining double sequence and some remarks on examples}\label{FurtherAss}
\vspace{-10pt}

In the previous sections we established necessary and sufficient conditions for the vanishing of $\Projeins$, for $AH(\mathbb{D})$ being ultrabornological and barrelled and also for the interchangeability of projective and inductive limit. Unfortunately, in almost all results the necessary and the sufficient conditions are distinct. Therefore it is desirable to identify additional general assumptions on the sequence $\mathcal{A}$ which allow to prove characterizations. In the sequel we present a setting in which this is possible and in which we in addition get by on associated weights which is -- in view of the complexity of our conditions -- clearly also desirable. In order to do so, we need the following condition $(\Sigma)$, which was introduced by Bierstedt, Bonet \cite[Section 5]{BB1994} and constitutes a generalization of condition (V) invented by Bierstedt, Meise, Summers \cite{BMS1982}. We say that a double sequence $\mathcal{A}=((a_{N,n})_{N\in\mathbb{N}})_{n\in\mathbb{N}}$ on $\mathbb{D}$ satisfies
condition $(\Sigma)$ if
$$
\textstyle\forall\:N\;\exists\: K\geqslant N\:\forall\:k\;\exists\:n\geqslant k\colon \frac{a_{N,n}}{a_{K,k}} \text{ vanishes at } \infty \text{ on } \mathbb{D}.
$$
Moreover, let us -- according to Taskinen \cite{Taskinen2001} -- call a weight $a$ essential, if there exists $C>0$ such that $(1/a)^{\sim}\leqslant 1/a \leqslant C(1/a)^{\sim}$ holds. Following the lines of \cite{WegnerJMAA} we obtain the last result of this note which should be compared with the results explained in \cite{ABB2009}.

\begin{thm}\label{TopologicalResultsSigma}Let $\mathcal{A}$ satisfy the conditions (LOG) and $\text{(}{\mathit{\Sigma}}\text{)}$.\vspace{2pt}
\begin{compactitem}
\item[(1)]If all weights in $\mathcal{A}$ are essential, then the following are equivalent.\vspace{3pt}
\\\begin{tabular}{rlrl}\vspace{1pt}
(i)  & $\mathcal{A}$ satisfies (Q).                              &(iv)  & $AH(\mathbb{D})$ is barrelled. \\\vspace{1pt}
(ii) & $\Projeins \mathcal{A}H=0$.                               &(v)  & $\mathcal{A}$ satisfies (wQ). \\\vspace{1pt}
(iii)  & $AH(\mathbb{D})$ is ultrabornological.                    &      &
\end{tabular}\vspace{2pt}
\item[(2)]If $\mathcal{A}$ is contained in a set of essential weights which is closed under finite maxima, then the following are equivalent.\vspace{3pt}
\\\begin{tabular}{rlrl}\vspace{1pt}
(i)  & $\mathcal{A}$ satisfies (Q) and (B).\hspace{18.5pt} & (iii)   & $\mathcal{A}$ satisfies (wQ) and (B).\\\vspace{1pt}
\end{tabular}\vspace{-13pt}
\\\begin{tabular}{rl}
\hspace{-3pt}(ii) & $AH(\mathbb{D})=\mathcal{V}H(\mathbb{D})$ holds algebraically and topologically.
\end{tabular}
\end{compactitem}
\end{thm}
\noindent{}To conclude our investigation let us remark that examples for PLB-spaces can be constructed (according to the result \cite[7.2]{WegnerJMAA} of Bonet) as follows: Put $a_{N,n}(z)=a(|z|)^{\alpha_n}v(|z|)^{\beta_N}$ with $a$, $v\colon[0,1]\rightarrow\;]0,1]$ continuous, decreasing with $\lim_{r\nearrow1}a(r)=\lim_{r\nearrow1}v(r)=0$, $\alpha_n\nearrow\alpha\in\;]0,\infty]$ and $\beta_N\searrow\beta\in[0,\infty[$. If now the conditions (df) and (m) of \cite[7.2]{WegnerJMAA}
are both \textit{not} satisfied, then $AH(\mathbb{D})$ is a (proper) PLB-space. The latter is valid if for instance $\limsup_{r\nearrow1}\log v(r)/\log a(r)=\limsup_{r\nearrow1}\log a(r)/\log v(r)$ $=\infty$ holds. Sequences with the forementioned property can be constructed following the lines of Bierstedt, Bonet \cite[Claim on p.~765]{BB2006}.

\vspace{-5pt}

\subsection*{Acknowledgment}
\vspace{-10pt}

This article arises from a part of the author's doctoral thesis, which was started at the University of Paderborn under the direction of Klaus D.~Bierstedt and after his sudden death  finished under the direction of Jos\'{e} Bonet at the Universidad Polit\'{e}cnica de Valencia. The author is greatly indebted to both of his supervisors; in addition he likes to thank Thomas Sauerwald for some clarifying discussions on the asymptotic interpretation of condition (LOG). Finally, the author likes to thank the referee for pointing out some mistakes in the submitted manuscript.

\setlength{\parskip}{0cm}

\small

config: arxiv


TITLE: Metrizable space and homeomorphism
QUESTION [3 upvotes]: Show that a topological space $X$ is metrizable if and only if there exists a homeomorphism of $X$ onto a subspace of some metric  space $Y$.
First, assume that there exists an homeomorphism $f$ between a topological space $X$, and a subspace of a metric space $(Y,d)$, define for $a,b$ in $X$
$$d'(a,b)= d(f(a),f(b)) $$
$d'$ is a metric on $X$, How can I prove that  it generates  the topology on $X$?
Also, for the converse, can I just take a metric space $Y$ such that $X\subset Y$ ? or how can I prove that? .

REPLY [1 votes]: For your first question see the answer of Ahmed.
The converse can easily cause some confusion. This answer shows a route.
Let $X=\langle S,\tau\rangle$ where $\tau$ denotes the original topology and $S$ denotes the underlying set.
The fact that $X$ is metrizable tells us that there exists a metric $d:S\times S\to[0,\infty)$ such that the topology $\tau_d$ induced by metric $d$ coincides with $\tau$.
So if $Y:=\langle S,d\rangle$ then the map $S\to S$ prescribed by $s\mapsto s$ is a homeomorphism $X\to Y$.
$\langle S,d\rangle$ is nothing more than a notation of $\langle S,\tau_d\rangle$ emphasizing that the topology is generated by metric $d$.
Finally observe that metric space $Y$  can be looked at as a subspace of itself.


TITLE: Evaluating an integral with Laplace
QUESTION [3 upvotes]: We need to evaluate the following integral:
$$\int_{0}^{\infty}\frac{\cos(tx)}{x^2+a^2}dx$$
There is the following note: "You may interchange taking the Laplace transform and integrating."
I have no idea how to solve this, I need help. Thank you!

REPLY [2 votes]: Let $$f(t)=\int_0^{\infty} \frac{cos(tx)}{x^2+a^2}\, \mathrm{d}x$$
Then, $$\mathcal{L} (f(t)) = \int_0^{\infty} e^{-st}\left( \int_0^{\infty} \frac{cos(tx)}{x^2+a^2}\, \mathrm{d}x\right) \, \mathrm{d}t$$
On interchanging, $$\mathcal{L} (f(t)) = \int_0^{\infty} \frac{1}{x^2+a^2} \left( \int_0^{\infty} e^{-st} cos(tx) \, \mathrm{d}t \right) \, \mathrm{d}x$$
Now utilizing the Laplace Transform of $cos(tx)$ to solve the inner integral, we have
$$\mathcal{L} (f(t)) = \int_0^{\infty} \frac{s}{(x^2+a^2)(x^2+s^2)} \, \mathrm{d} x$$
On solving and simplifying this integral using partial fractions and the basic rules of definite integration, we have
$$\mathcal{L} (f(t)) = \frac{\pi}{2a(a+s)}$$
Applying the inverse Laplace Transform on both sides, we have
$$f(t)=\mathcal{L}^{-1}\left(\frac{\pi}{2a(a+s)}\right)$$
$$f(t)=\frac{\pi}{2a} \mathcal{L}^{-1}\left(\frac{1}{a+s}\right)$$
Thus, $$f(t)=\frac{\pi}{2a} e^{-at}$$
and we are done.


TITLE: Why is the feasible set of a LP a polyhedron?
QUESTION [0 upvotes]: I still cannot understand why the topic is true.
The following is from Convex Optimization, S. Boyd

My claim is the following:

The feasible region for $Gx \preceq h$ is a polyhedron, which is easy to understand and can be easily drawn on the paper.
The feasible region for $Ax=b$ is the intersection of all affine set defined by $a_i^Tx = b_i$ for each $i$, where $a_i, \forall i$ are rows of $A$.  So there are three possibilities:  1. No solution. Then the feasible set of LP is empty. 2. One solution. Then the feasible set of LP is "at most" a point. 3. $\infty$ solution. Then the feasible set of LP is "at most" a line.

Why is the feasible set a polyhedron?

REPLY [1 votes]: The feasible region for $Ax=b$ is an intersection of affine sets, therefore also an affine set. However there are no further restrictions on the dimension of the solution space for $Ax=b$. It may very well be two- or higher-dimensional.


TITLE: function equation $f(x)+f(y)=f(\frac {x+y} {1+xy})$
QUESTION [0 upvotes]: Is there exist $f:(-1,1)\to R$ such that $f(x)+f(y)=f(\frac {x+y} {1+xy})$?
how to find $f$?

REPLY [4 votes]: The functional equation for hyperbolic tangent is
$$ g(x):=\tanh(x),\quad g(x+y) = \frac{g(x)+g(y)}{1+g(x)g(y)}. $$ The Wikipedia article
Hyperbolic functions is one source for this.
The functional equation for its inverse function is
$$ f(x):=\tanh^{-1}(x),\quad f(x)+f(y) = f\Big(\frac{x+y}{1+xy}\Big). $$
Note that, just like with Cauchy's functional equation, without any
other assumptions such as continuity, and so on, there
could be many strange solutions. Consult the Wikipedia
article for details.


TITLE: How do I interpret Euler's formula?
QUESTION [34 upvotes]: I don't understand the formula at all:
$$e^{ix} = \cos(x) + i \sin(x)$$
I've tried reading all sorts of webpages and answers on the subject but it's just not clicking with me. I don't understand how we can define things when these are already known quantities. We've already got an exact definition for $e$ as $\lim_{n \to 0} (1+\frac{1}{n})^n$, and we've got $i = \sqrt{-1}$, and we've got $\cos$ and $\sin$ as the $x$ and $y$ coordinates of where right triangles meet the unit circle.
So I don't understand how we get from those figures to Euler's formula. Then again I also don't understand why complex numbers are of the form $a + bi$ so that might have something to do with it. I have a hard time separating what identities are "emergent" from the previous mathematics and which pieces are "by definition" and why they are defined that way. I don't understand why we start talking about "rotations" as if it's obvious that's what's happening. I don't even know why $e$ is involved in any of this to begin with.
Is $e^{ix}$ simply being written in the slightly rearranged complex form $a + ib$ where $a = \cos(x)$ and $b = \sin(x)$? Is this to be interpreted as $\cos(x)$ being the $x$-coordinate and somehow $i \sin(x)$ is the $y$ coordinate?

REPLY [2 votes]: I'll try to address these pieces

I don't understand how we can define things when these are already known quantities. We've already got an exact definition for $e$ as $\lim_{n \to 0} (1+\frac{1}{n})^n$, and we've got $i = \sqrt{-1}$, and we've got $\cos$ and $\sin$ as the $x$ and $y$ coordinates of where right triangles meet the unit circle.

I have a hard time separating what identities are "emergent" from the previous mathematics and which pieces are "by definition" and why they are defined that way.

You're right, we do have those definitions. And yes, someone can't come up to you and define $e$ as $3.7$, because $e$ already has a definition.
We've had a definition of what $e^x$ is when $x$ is a real number. What we didn't have was a definition of $e^x$ when $x$ is an imaginary number. Before we give the definition, $e^{bi}$ is some meaningless writing. You could define it to to be whatever you like, ex. $b$, or $5$, or $\log 10b$, etc. But people chose to define it as $\cos(x) + i \sin(x)$ because it would have some really cool propreties (ex. defining it this way preserves some of the propreties of
 the real $e^x$ we are used to).
Another common example is infinite sums. We have defined $\sum_{i=1}^n a_i$ as $a_1+a_2+ \cdots + a_n$. But again this does not tell us anything about the meaning of the symbol $\sum_{i=1}^{\infty} a_i$. We could again assign anything we want to it, such as $7$, or $a_4$, or $\pi$. But we have decided to assign it $\lim_{n \to \infty} \sum_{i=1}^n a_i$


TITLE: Inequality involving sums and products (related to multi-stage rockets)
QUESTION [3 upvotes]: Let $a_i$ and $b_i$ be strictly positive real numbers for $i=1,\ldots, n$. I wonder whether the following inequality holds in general or does not:$$\frac{\sum_{i=1}^n a_i+\sum_{i=1}^n b_i}{\sum_{i=1}^n a_i}<\prod_{k=1}^n\frac{\sum_{i=1}^k a_i+\sum_{i=1}^k b_i}{\sum_{i=1}^k a_i+\sum_{i=1}^{k-1} b_i}.$$
I have not been able to prove it in general, but I am not finding a countexample [EDIT: the non-strict inequality holds for $n=1$, as pointed out by Oskar, whom I thank], either, therefore I am inclined to suspect that it is true. How can it be proved, if it is? I thank you so much for any answer!

Background: the question has arisen in my mind because I read in a physics text that the efficience of a one stage rocket, whose velocity variation can be expressed, in absence of external forces and for constant speed $\mathbf{v}_{gr}$ and rate of expulsion $\frac{dm}{dt}$ of the gas, by the Ciolkovskij rocket equation
$$\mathbf{v}_f-\mathbf{v}_i=\mathbf{v}_{gr}\ln\Big(\frac{m_f}{m_i}\Big)$$
where $m_i$ and $m_f$ respectively are the initial (rocket and gas) and final (rocket only) masses, can be increased by using multi-stage rockets.
I would say that, if we call $a_i$ the mass of stage $i$ and $b_i$ the mass of the gas contained in it, the above speed variation, considering a one stage rocket made of $n$ "stage-like" parts, which expels the gas at once, can be written as $$\Delta v_{\text{one stage rocket}}=-\|\mathbf{v}_{gr}\|\ln\frac{\sum_{i=1}^n a_i}{\sum_{i=1}^n a_i+\sum_{i=1}^n b_i}=\|\mathbf{v}_{gr}\|\ln\frac{\sum_{i=1}^n a_i+\sum_{i=1}^n b_i}{\sum_{i=1}^n a_i}$$ while, if we discharge the gas contained in each stage $i$ at speed $\mathbf{v}_{gr}$ and then stage $i$ is detached, I would say that the variation is (letting $\sum_{i=1}^0 b_i:=0$) $$\Delta v_{\text{multi-stage rocket}}=\sum_{k=1}^n \|\mathbf{v}_{gr}\| \ln\frac{\sum_{i=1}^k a_i+\sum_{i=1}^k b_i}{\sum_{i=1}^k a_i+\sum_{i=1}^{k-1} b_i}$$and $\Delta v_{\text{one stage rocket}}<\Delta v_{\text{multi-stage rocket}}$ is equivalent to the inequality that is the object of the question.

REPLY [1 votes]: Let denote:
$$ s_k=\sum_{i=1}^k a_i\ \ \ \ \ t_k=\sum_{i=1}^kb_i$$
hence the inequality to prove is :

$$P(n)\ \ \ \ \ \ \ \ \ \ 1+\frac{t_n}{s_n}\leq \prod_{k=1}^{n} \frac{s_k+t_k}{s_k+t_{k-1}}=\prod_{k=1}^{n} \left(1+\frac{b_k}{s_k+t_{k-1}}\right)$$

Proof (by induction on $n$)
Basis step $P(1)$ is obviously true because:
$$(1+\frac{b}{a})\leq \prod_{i=1}^1 \left(1+\frac{b_i}{s_i+t_{i-1}}\right)=(1+\frac{b}{a})$$
Induction step Let's assume that $P(n)$ is true, given $a_1,\cdots,a_{n+1},b_1,\cdots,b_{n+1}$ strictly positive numbers we have:$$s_nt_n<s_{n+1}t_{n}$$
because $0<a_{n+1}$and this is equivalent to (by adding $s_ns_{n+1}$ in both sides of the inequality and multiplying by $(t_{n+1}+s_{n+1})$) $$(s_{n+1}+t_{n+1})(s_{n+1}+t_n)s_n<s_{n+1}(s_n+t_n)(s_{n+1}+t_{n+1}) $$
which  implies that:
$$1+\frac{t_{n+1}}{s_{n+1}}<\left(1+\frac{t_n}{s_n}\right)\left(1+\frac{b_{n+1}}{s_{n+1}+t_n}\right)  $$
hence using induction hypothesis:
$$ 1+\frac{t_{n+1}}{s_{n+1}}< \left(1+\frac{t_n}{s_n}\right)\cdot \left(1+\frac{b_{n+1}}{s_{n+1}+t_n}\right) \leq \prod_{k=1}^{n+1} \left(1+\frac{b_k}{s_k+t_{k-1}}\right)$$
and the induction terminates.
The inequality is strict for $n\geq 2$ because in the induction step even if $P(n)$ is not strict, it implies that $P(n+1)$ strict, so the only non strict inequality is $P(n)$


TITLE: confusion in David Burton number theory
QUESTION [0 upvotes]: From elementary number theory by  David Burton books
Theorem $2.4$: Let $a$ and $b$ be integers, not both zero. Then $a$ and $b$ are relatively prime
if and only if there exist integers $x$ and $y$ such that $1 = ax + by.$
I think this theorem is not true:  take $a=2, b=3 , x= 1, y=2$;
then $1 \neq 2.1 + 3.2= 2+6=8$ but here gcd$( 2,3)=1$.
My confusion is that  Is this theorem is true/false ?

REPLY [2 votes]: The theorem is true.
Given $a=2$ and $b=3$, there exist integers $x$ and $y$ (such as $-1$ and $1$, respectively)
such that $ax+by=1$.


TITLE: basis extension intuition
QUESTION [0 upvotes]: If we have vectors $v_1, v_2, v_3\in \mathbb R^4$ and we want to extend them to a basis of $\mathbb R^4$, why can we take the vectors as rows and then doing rows reductions with Gauß algorithm and then choose the canonical vector $e_k$ where index $k$ is one of the components that can be set as a free variable? I want to find a reasoning for doing this.

REPLY [0 votes]: Call the matrix whose rows are $v_1, v_2, v_3$, say, $A$. Then what we are doing in picking the $e_k$ corresponding to a free variable is finding a basis for the null space of $A$ (which suggests that this works in general, not just when we're missing one vector).
The null space of $A$ completes the basis of $\mathbb R^4$ because null space of $A$ is equal to the orthogonal complement of the row space of $A$ (this is some version of the "four subspaces theorem", relating row and/or column spaces to null spaces).


TITLE: Are $X$ and $Y$ independent random variables?
QUESTION [0 upvotes]: A box has six balls numbered from one to six. The balls numbered $1$ and $2$ are red while the remaining balls are white. Two balls are drawn at random from the box (without replacement and without order), let $X$ and $Y$ be the random variables representing the number of red balls and the number of even balls in the sample, respectively. Are $X$ and $Y$ independent random variables?
My intuition tells me that they are not independent. But I don't know how to progress with this problem.

REPLY [2 votes]: The total number of pairs can be "6 choose 2", $\binom{6}{2}$ and the number of pairs when one of the balls is red and the other is even are 5:
$$ {(1,2), (1,4), (1,6), (2,4), (2,6)}.$$ Note $(2,2)$ is not included in the set since you are choosing without replacement. The probability of the event then is $5/15=1/3$ which is not the same as the probability of choosing red and even: $$P(X)P(Y)=(2/6)(3/6).$$ Therefore, $X$ and $Y$ are not independent.


TITLE: Equation of plane -- point/vector pedagogy
QUESTION [1 upvotes]: Suppose we have a point $\mathbf P$ and a vector $\mathbf n$ in plain ordinary 3D space. Here I am deliberately using upper-case letters for points, and lower-case points for vectors, since they are two different things. Then the set of points $\mathbf X$ that satisfy
$(\mathbf X - \mathbf P) \cdot \mathbf n = 0$ is a plane, as we all know. The quantity $\mathbf X - \mathbf P$ is the difference of two points, so it's a vector, and taking its dot product with $\mathbf n$ is legitimate. So far, so good.
But, it's very tempting to rewrite this as $\mathbf X \cdot \mathbf n  = \mathbf P \cdot \mathbf n$. Then, if I let $\mathbf P \cdot \mathbf n = d$, the plane equation becomes $\mathbf X \cdot \mathbf n  = d$. From a programming point of view, this is nice -- I can now represent the plane by four numbers ($\mathbf n$ and $d$) rather than six ($\mathbf n$ and $\mathbf P$). Also, if I write $\mathbf P = (a,b,c)$, then the equation becomes $ax + by + cz = d$, which is the plane representation that we all know from high school geometry. Nice, comfy, familiar. Good.
The problem is that expressions like $\mathbf X \cdot \mathbf n$ and $\mathbf P \cdot \mathbf n$ don't make sense -- everyone knows that you can't take dot products of points and vectors. Computationally, everything works fine, but pedagogically, it feels like something is wrong.
Can anyone make some sense out of this, please? I'd like to write plane equations in the form $\mathbf X \cdot \mathbf n  = d$ and still be able to sleep peacefully at night..

REPLY [1 votes]: I guess this depends on the way things are defined. Some guys say that the space is $\mathbb{R}^3$ from the start, and for them points and vectors are the same. They sleep at night very peacefully I guess.
Then again, others appreciate a distinction between affine spaces and vector spaces. This distinction is there throughout school geometry. In this case, you'll probably have to randomly select an origin $O$ and instead of $X \cdot n$ write $(X-O) \cdot n$.
But, there's a third option! You can sort of make sense of the notation $X \cdot n$ and still respect the fact that affine spaces don't have a starting point. Problem is, you probably shouldn't show this to children, otherwise they might become the ones losing their sleep. Also, you won't be able to compare values $X_1 \cdot n_1$ and $X_2 \cdot n_2$ if $n_1 \neq n_2$.
You have probably already guessed where I'm driving at. Suppose $n$ is a vector. Define an equivalence relation $\sim_n$ like this: $X \sim_n Y$ iff $(Y-X)\cdot n = 0$. Then, if $A$ is the whole affine space and $V$ is the set of its vectors, then $\sim_n$ is an equivalence relation on $A$ and $A/\sim_n$ has a natural structure of an affine space (whose space of vectors is $\langle n \rangle$). Then you can simply denote by $X \cdot n$ the class of $X$ with respect to equivalence $\sim_n$.
With this notation, $X \cdot n = const$ really defines a plane, with the only drawback that the constant on the right is not a number, but rather just an element of a one-dimensional something without any point of reference. Now that I think about it, this constant is the plane itself ))
This is mostly useless I guess, except maybe for the sleeping problem.


TITLE: Calculate a location on a square, on square's projection on a plane
QUESTION [1 upvotes]: I have the projection of a square on a plane. I know it's four corners' coordinates on said plane. (This is from a picture of a square taken by a camera at an unknown relation to the square.)
I now need to be able to locate any point on the square (in square coordinates) on the plane. (I assume that the center of the square (point 0.5/0.5) is "located" on the plane at the intersection of the diagonals, but I need to be able to locate any arbitrary point.)
I first thought of calculating the percent of the distance on the sides and then finding their intersection. For example, to find point 0.2,0.6 I would find the 0.2 point on the top and bottom sides, create a line between them, and do the same for point 0.6 on the left and right sides. Then find the intersection of these 2 lines. But this would only be an approximation, because the closer side of the square to the camera would have more square-distance per projection-distance.
So how do I achieve the correct calculation?

REPLY [1 votes]: The scene plane in which the square lies and the image plane are related by a planar projective transformation. This answer explains a way to compute the transformation. To use that method, you need to choose a coordinate system for the square’s plane. Your idea of treating it as a unit square by computing the fractions of the distances along both dimensions does that, and makes the matrix $A$ for the source part of the map particularly easy to compute: solving $$\begin{pmatrix}0&1&0\\0&0&1\\1&1&1\end{pmatrix} \begin{pmatrix}\lambda\\\mu\\\tau\end{pmatrix} = \begin{pmatrix}1\\1\\1\end{pmatrix}$$ gives $(-1,1,1)^T$, and so $$A^{-1} = \begin{pmatrix}1&1&-1\\1&0&0\\0&1&0\end{pmatrix}.$$ Take care to put the corresponding vertices of the image quadrilateral in the correct order when computing $B$. Note that this doesn’t require knowing anything about the perspective projection itself: you just need the vertex coordinates in the image.


text:
\section{Fusion rules for free products and the quantum group $\Akn$}

In this section we describe the fusion rules of the free complexification $\Akn \cong \widetilde{\Ahn}$. Instead of referring to $\Akn$ explicitly, we will work in a more general setting and deduce its fusion rules as a corollary. Roughly the main statement of this section is given by the following theorem. See theorem \ref{theorem_fusion_rules_of_complexifications} for a precise statement.

\begin{theorem}\label{theorem_main_statement_simplified}
  Let $(A,U)$ be an orthogonal compact matrix quantum group, i.e. $\overline{U} = U$, such that its fusion rules are free. Assume further that $1 \notin U^{\boxtimes 2k + 1}$ for any $k \in \NN$. Then the fusion rules of $(\widetilde{A}, \widetilde{U})$ are free and can be described in terms of the fusion rules of $(A,U)$.
\end{theorem}

The following theorem is due to Wang \cite{wang98}.
\begin{theorem}\label{theorem_irr_rep_of_free_products}
  Let $(A,\Delta_A)$ and $(B, \Delta_B)$ be compact quantum groups. Let $(U^\alpha)_{\alpha \in \mathscr{A}}$ and $(U^\beta)_ {\beta \in \mathscr{B}}$ be complete sets of representatives of irreducible corepresentations of $A$ and $B$, respectively. Then the corepresentations $(W^{\gamma_1} \boxtimes \dotsm \boxtimes W^{\gamma_n})$ with $n \in \NN$, all $W^{\gamma_i}$ in $\{U^\alpha \, | \, \alpha \in \mathscr{A}\}$ and $\{U^\beta \, | \, \beta \in \mathscr{B}\}$ and neighbours not from the same set, form a complete set of irreducible representations of the free product $(A, \Delta_A) * (B, \Delta_B)$.
\end{theorem}

The following observation will be useful when studying the fusion rules of a free complexification.

\begin{remark}\label{remark_fusion_rules_in_free_products}
  Let $A * B$ be a free product of compact quantum groups with irreducible corepresentations $W^{\gamma_1} \boxtimes \dotsm \boxtimes W^{\gamma_n}$ and $W^{\delta_1} \boxtimes \dotsm \boxtimes W^{\delta_m}$ as in the last theorem. Then
  \begin{enumerate}
    \item If $W^{\gamma_n}$ and $W^{\delta_1}$ are not corepresentations of the same factor of the free product, then $W^{\gamma_1} \boxtimes \dotsm \boxtimes W^{\gamma_n} \boxtimes W^{\delta_1} \boxtimes \dotsm \boxtimes W^{\delta_m}$ is an irreducible corepresentation of $A * B$.
    \item If $W^{\gamma_n}$ and $W^{\delta_1}$ are corepresentations of the same factor and $W^{\gamma_n} \boxtimes W^{\delta_1} = \sum_{i = 1}^k W^{\epsilon_i} + \delta_{\overline{W^{\gamma_n}}, W^{\delta_1}} \cdot 1$ is the decomposition into irreducible corepresentations, then
    \begin{multline*}
      W^{\gamma_1} \boxtimes \dotsm \boxtimes W^{\gamma_n} \boxtimes W^{\delta_1} \boxtimes \dotsm \boxtimes W^{\delta_m} \\
      = \sum_{i = 1}^k{(W^{\gamma_1} \boxtimes \dotsm \boxtimes W^{\gamma_{n-1}} \boxtimes W^{\epsilon_i} \boxtimes W^{\delta_2} \boxtimes \dotsm \boxtimes W^{\delta_m})} \\
      + \delta_{\overline{W^{\gamma_n}}, W^{\delta_1}} \cdot W^{\gamma_1} \boxtimes \dotsm \boxtimes W^{\gamma_{n-1}} \boxtimes W^{\delta_2} \boxtimes \dotsm \boxtimes W^{\delta_m}
    \end{multline*}
    and the first $k$ summands of this decomposition are irreducible.
  \end{enumerate}
\end{remark}

For the rest of this section fix an orthogonal compact matrix quantum group $(A,U)$ such that its fusion rules are described by a free fusion ring over the fusion set $S$. Assume further that $1 \notin U^{\boxtimes 2k + 1}$ for any $k \in \NN$. \\
Note that the fusion ring of $\widetilde{A}$ is the fusion subring of $\Rep(A * \cont(\textup{S}^1))$ that is generated by $U \boxtimes z$, where $z$ denotes the identity on the circle.\\
We will construct the free complexification $\widetilde{S}$ of $S$ and prove that the fusion rules of $(\widetilde{A}, \widetilde{U})$ are described by $\widetilde{S}$. We begin by constructing $\widetilde{S}$. \\
Let $\Repirreven$ (respectively $\Repirrodd$) be the set of classes of irreducible corepresentations of $A$ that appear as subrepresentations of an even (respectively odd) tensor power of $U$. We have $\Repirreven \cap \Repirrodd = \emptyset$ due to Frobenius duality and the requirement $1 \notin U^{2k + 1}$ for all $k \in \NN$. Let $\Seven \subset S$ (resp. $\Sodd \subset S$) be the set of elements corresponding to corepresentations from $\Repirreven$ (resp. $\Repirrodd$). The set $\widetilde{S}$ is then by definition the disjoint union $\Seven \sqcup \Seven \sqcup \Sodd \sqcup \Sodd$. Denote the first copy of $\Seven$ (resp. $\Sodd$) by $\Sevenone$ (resp. $\Soddone$) and the second one by $\Seventwo$ (resp. $\Soddtwo$). \\

What follows is motivated by the following point of view:
\begin{remark}\label{remark_elements_of_S_tilde}
  We consider element of $\Sevenone$ as a plain copy of those in $\Seven$. The elements of $\Seventwo$ are of the form $z^*\cdot s \cdot z$ for some $s \in \Seven$. Similarly we consider elements of $\Soddone$ as $s \cdot z$ and elements of $\Soddtwo$ as $z^* \cdot s$ for $s \in \Sodd$.
\end{remark}

Define a conjugation on $\widetilde{S}$ by the conjugation on $S$ leaving $\Sevenone$ and $\Seventwo$ globally invariant and exchanging $\Soddone$ and $\Soddtwo$. Note that $\overline{\Seven} = \Seven$ and $\overline{\Sodd} = \Sodd$, i.e. the conjugation on $\widetilde{S}$ is well defined. A fusion on $\widetilde{S}$ can be defined according to the following table.
\begin{center}
\begin{tabular}{c||c|c|c|c}
  $\cdot$	&\Sevenone	&\Seventwo	&\Soddone	&\Soddtwo	\\
  \hline
  \Sevenone	&$\Sevenone \cup \{\emptyset\}$	&$\emptyset$	&$\Soddone \cup \{\emptyset\}$	&$\emptyset$	\\
  \hline
  \Seventwo	&$\emptyset$	&$\Seventwo \cup \{\emptyset\}$	&$\emptyset$	&$\Soddtwo \cup \{\emptyset\}$	\\
  \hline
  \Soddone	&$\emptyset$	&$\Soddtwo \cup \{\emptyset\}$	&$\emptyset$	&$\Sevenone \cup \{\emptyset\}$	\\
  \hline
  \Soddtwo	&$\Soddtwo \cup \{\emptyset\}$	&$\emptyset$	&$\Seventwo \cup \{\emptyset\}$	&$\emptyset$
\end{tabular}
\end{center}
The row gives the element which is fused from the right with an element coming from the set indicated by the column. The fusion is empty if this is indicated by the table and is otherwise the usual fusion of two elements of $S$ lying in the part of $\widetilde{S}$ indicated by the table. Note that this definition makes sense, since $\Seven \cdot \Seven, \Sodd \cdot \Sodd \subset \Seven \cup \{\emptyset\}$ and $\Seven \cdot \Sodd, \Sodd \cdot \Seven \subset \Sodd \cup \{\emptyset\}$. It is easy to see that $\widetilde{S}$ with this structure is a fusion set. \\
Now we can state a precise version of \ref{theorem_main_statement_simplified}.

\begin{theorem}\label{theorem_fusion_rules_of_complexifications}
  Let $(A,U)$ be an orthogonal compact matrix quantum group such that its fusion rules are described by a free fusion ring over the fusion set $S$. Assume further that $1 \notin U^{\boxtimes 2k + 1}$ for any $k \in \NN$. Then the fusion rules of $(\widetilde{A}, \widetilde{U})$ are given by the free complexification $\widetilde{S}$ of $S$.
\end{theorem}

We construct a complete set of corepresentations of $\widetilde{A}$. In order to do so we associate an irreducible corepresentations of $(\widetilde{A}, \widetilde{U})$ to any element of $\widetilde{R} := \Repirreven \sqcup \Repirreven \sqcup \Repirrodd \sqcup \Repirrodd$. We denote the $i$-th copy of $\Repirreven$ ($\Repirrodd$) by $\textup{Rep}_\textup{even}^{\textup{irr}, (i)}$ ($\textup{Rep}_\textup{odd}^{\textup{irr}, (i)}$). Let $V$ be a irreducible corepresentation in $\Repirreven$. Then $V$ and $z^* \cdot V \cdot z$ are corepresentations of $\widetilde{A}$. Actually, if $V$ is an irreducible subrepresentation of $U^{\boxtimes 2k}$ then $V$ is an irreducible subrepresentation of $(\widetilde{U} \boxtimes \overline{\widetilde{U}})^{\boxtimes k}$ and $z^* \cdot V \cdot z$ is an irreducible subrepresentation of $(\overline{\widetilde{U}} \boxtimes \widetilde{U})^{\boxtimes k}$. We consider $V$ as an element of $\Repirrevenone$ and $z^* \cdot V \cdot z$ as an element of $\Repirreventwo$. Similarly we see that if $V \in \Repirrodd$ then we can associate with it corepresentations $V \cdot z \in \Repirroddone$ and $z^* \cdot V \in \Repirroddtwo$. Note that elements $s$ from $\widetilde{S}$ give corepresentations $\widetilde{U}_s$ by this identification.
Consider a word $w = w_1 \dotsc w_k$  with letters in $\widetilde{R}$. We say that $w$ is reduced if in the sequence $\widetilde{U}_{w_1}, \dotsc, \widetilde{U}_{w_n}$ a $z$ is never followed by $z^*$ and $U_x$ is always followed by $z$ or $z^*$. In formal terms:
\begin{align*}
  \forall 1 \leq i \leq k - 1: \: & (w_i \in \Repirrevenone \cup \Repirroddtwo \Rightarrow w_{i + 1} \in \Repirreventwo \cup \Repirroddtwo) \wedge \\
    & (w_i \in \Repirreventwo \cup \Repirroddone \Rightarrow w_{i + 1} \in \Repirrevenone \cup \Repirroddone)
\end{align*}
Any such reduced word $w = w_1 \dotsc w_k$ gives rise to an irreducible corepresentation of $\widetilde{A}$ by $\widetilde{U}_w := \widetilde{U}_{w_1} \boxtimes \dotsc \boxtimes \widetilde{U}_{w_k}$ and different reduced words give rise to inequivalent corepresentations by \ref{theorem_irr_rep_of_free_products}. Since any iterated tensor product of $\widetilde{U}$ and $\overline{\widetilde{U}}$ decomposes as a sum of irreducible corepresentations of the type $\widetilde{U}_w$, where $w$ is a reduced word with letters in $\widetilde{R}$, any irreducible corepresentation of $\widetilde{A}$ is equivalent to some $\widetilde{U}_w$.

\begin{definition}
Consider now a word $w = w_1 \dotsc w_k$ with letters in $\widetilde{S}$. It is called connected if every $z$ is followed by a $z^*$. Formally:
\begin{align*}
  \forall 1 \leq i \leq k-1: \:	& (w_i \in \Sevenone \cup \Soddtwo \Rightarrow w_{i + 1} \in \Sevenone \cup \Soddone) \wedge \\
    & (w_i \in \Seventwo \cup \Soddone \Rightarrow w_{i + 1} \in \Seventwo \cup \Soddtwo)
\end{align*}
\end{definition}

The following definition says how we can associate irreducible corepresentations of $\widetilde{A}$ to words with letters in $\widetilde{S}$.

\begin{definition}
  If $w$ is an arbitrary word with letters in $\widetilde{S}$ then it has a unique decomposition $w = x_1 \dotsc x_l$ into maximal connected words. This gives rise to a unique reduced word $w'$ with letters in $\widetilde{R}$. We set $\widetilde{U}_w := \widetilde{U}_{w'}$
\end{definition}

Next we have to do some preparations in order to prove theorem  \ref{theorem_fusion_rules_of_complexifications}.

\begin{definition}
  Let $x=x_1 \dotsc x_m$ be a word in $\widetilde{S}$. Then $\check{x_i}$ is the letter in $S$ corresponding to $x_i$ and $\check{x} := \check{x_1}\check{x_2} \dotsc \check{x_m}$.
\end{definition}

\begin{remark}\label{remark_irreps_from_connected_words}
  Note that if $x$ is a connected word with letters in $S$ then according to remark \ref{remark_elements_of_S_tilde} it can be written as $z^{i_0} \cdot \check{x} \cdot z^{i_1}$, $i_0,i_1 \in \{0,1,-1\}$ and we have $\widetilde{U}_x = z^{i_0} \boxtimes U_{\check{x}} \boxtimes z^{i_1}$.
\end{remark}

\begin{definition}
Let $x,y$ be connected words with letters in $\widetilde{S}$. We say that $(x,y)$ fits together if $xy$ is a connected word.
\end{definition}

\begin{lemma}\label{lemma_fusion_of_connected_words}
  Let $x = x_1 \dotsc x_m$ and $y=y_1 \dotsc y_n$ be connected words with letters in $\widetilde{S}$ such that $(x_m,y_1)$ fits together. Write $\widetilde{U}_x = z^{i_0} \boxtimes U_{\check{x}} \boxtimes z^{i_1}$ and $\widetilde{U}_y = z^{j_0} \boxtimes U_{\check{y}} \boxtimes z^{j_1}$. Then
  \[\widetilde{U}_x \boxtimes \widetilde{U}_y = z^{i_0} \boxtimes \left (\sum_{x = ac, y = \overline{c}b}{U_{\check{a}\check{b}} \boxplus U_{\check{a} \cdot \check{b}}} \right) \boxtimes z^{j_1} = \sum_{x = ac, y = \overline{c}b}{\widetilde{U}_{ab} \boxplus \widetilde{U}_{a \cdot b}} .\]
\end{lemma}
\begin{proof}
  Since $(x,y)$ fits together, we have $z^{i_1} \boxtimes z^{j_0} = 1$. So by remark \ref{remark_fusion_rules_in_free_products} the first equation follows. We have to prove that for all $x = ac$, $y= \overline{c}b$
  \begin{enumerate}
    \item $z^{i_0} \boxtimes U_{\check{a}\check{b}} \boxtimes z^{j_1} = \widetilde{U}_{ab}$
    \item $z^{i_0} \boxtimes U_{\check{a} \cdot \check{b}} \boxtimes z^{j_1} = \widetilde{U}_{a\cdot b}$.
  \end{enumerate}
  In order to prove (1), note that $ab$ is connected, since $a,b$ are connected and $(a,b)$ fits together. So (1) follows from the way irreducible corepresentations are associated to connected words remarked in \ref{remark_irreps_from_connected_words}. \\
  For (2) note that, since $(a,b)$ fits together, $\check{a}\cdot \check{b} = \emptyset$ if and only if $a\cdot b = \emptyset$. If $a\cdot b \neq \emptyset$ then it is connected and (2) follows by remark \ref{remark_irreps_from_connected_words} again.
\end{proof}

Now we can give the proof of Theorem \ref{theorem_fusion_rules_of_complexifications}
\begin{proof}[Proof of Theorem \ref{theorem_fusion_rules_of_complexifications}]
  Let $x = x_1 \dotsc x_k$ and $y=y_1 \dotsc y_l$ be words with letters in $\widetilde{S}$. We have to show that
  \[\widetilde{U}_x \boxtimes \widetilde{U}_y = \sum_{x = ac, y=\overline{c}b}{\widetilde{U}_{ab} \boxplus \widetilde{U}_{a \cdot b}}\]
  Let $x=u_1 \dotsc u_m$ and $y=v_1 \dotsc v_n$ be the decomposition in maximal connected words. We identify them with letters in $\widetilde{R}$.  Then
  \begin{align*}
    \widetilde{U}_x	& = z^{i_0} \boxtimes U_{\check{u_1}} \boxtimes z^{i_1} \boxtimes U_{\check{u_2}} \boxtimes z^{i_2} \boxtimes \dotsm \boxtimes U_{\check{u_{m-1}}} \boxtimes z^{i_{m-1}} \boxtimes \underbrace{z^{i_m} \boxtimes U_{\check{u_m}} \boxtimes z^{i_{m+1}}}_{=\widetilde{U}_{u_m}}, \\
    \widetilde{U}_y	& = \underbrace{z^{j_0} \boxtimes U_{\check{v_1}} \boxtimes z^{j_1}}_{= \widetilde{U}_{v_1}} \boxtimes z^{j_2} \boxtimes U_{\check{v_2}} \boxtimes z^{j_3} \boxtimes \dotsm \boxtimes z^{j_{n-1}} \boxtimes U_{\check{v_{n-1}}} \boxtimes z^{j_n} \boxtimes U_{\check{v_n}} \boxtimes z^{j_{n+1}}
  \end{align*}
  with $i_1,...,i_{m-2},j_3,...,j_n \in \{1,*\}$, $i_0, i_m, j_0, j_2 \in \{0,*\}$ and $i_{m-1}, i_{m+1},j_1, j_{n+1} \in \{0, 1\}$. \\
  We are going to consider the two cases $(x_k,y_1)$ do or do not fit together. Assume that $(x_k, y_1)$ do not fit together. This means $z^{i_{m+1}} \cdot z^{j_0} \neq 1$. Then $\widetilde{U}_x \boxtimes \widetilde{U}_y$ is irreducible by Theorem \ref{theorem_irr_rep_of_free_products}. Moreover, $xy = u_1 \dotsc u_mv_1 \dotsc v_n$ is a decomposition in maximal connected words. So $\widetilde{U}_x \boxtimes \widetilde{U}_y = \widetilde{U}_{xy}$. On the other hand $(x_k, y_1)$ not fitting together implies $x_k \neq \overline{y_1}$ and $x_k \cdot y_1 = \emptyset$. So $\sum_{x = ac, y=\overline{c}b}{\widetilde{U}_{ab} \boxplus \widetilde{U}_{a \cdot b}} = \widetilde{U}_{xy}$. This completes the proof for the first case. \\
  Assume now that $(x_k,y_1)$ fits together. This means $z^{i_{m+1}} \cdot z^{j_0} = 1$. By Lemma \ref{lemma_fusion_of_connected_words}
  \begin{align*}
  \begin{split}
    \widetilde{U}_x \boxtimes \widetilde{U}_y & = z^{i_0} \boxtimes U_{\check{u_1}} \boxtimes z^{i_1} \boxtimes \dotsm \boxtimes U_{\check{u_{m-1}}} \boxtimes z^{i_{m-1}} \boxtimes \\
    & (\sum_{u_m = ac, v_1 = \overline{c}b}{\widetilde{U}_{ab} \boxplus \widetilde{U}_{a \cdot b}}) \boxtimes z^{j_2} \boxtimes U_{\check{v_2}} \boxtimes z^{j_3} \boxtimes \dotsm \boxtimes U_{\check{v_n}} \boxtimes z^{j_{n+1}} \\
    & = z^{i_0} \boxtimes U_{\check{u_1}} \boxtimes z^{i_1} \boxtimes \dotsm \boxtimes U_{\check{u_{m-1}}} \boxtimes z^{i_{m-1}} \boxtimes \\
    & ((\sum_{u_m = ac, v_1 = \overline{c}b, |a| \geq 1 \text{ or } |b| \geq 1}{\widetilde{U}_{ab} \boxplus \widetilde{U}_{a \cdot b}}) \boxplus \delta_{u_m,\overline{v_1}}\cdot 1) \boxtimes \\
    & z^{j_2} \boxtimes U_{\check{v_2}} \boxtimes z^{j_3} \boxtimes \dotsm \boxtimes U_{\check{v_n}} \boxtimes z^{j_{n+1}}.
  \end{split}
  \end{align*}
  By applying the induction hypothesis to the term
  \begin{gather*}
  z^{i_0} \boxtimes U_{\check{u_1}} \boxtimes \dotsm \boxtimes z^{i_{m-1}} \boxtimes \delta_{u_m,\overline{v_1}}\cdot 1 \boxtimes z^{j_2} \boxtimes U_{\check{v_2}} \boxtimes \dotsm \boxtimes z^{j_{n+1}} \\
  = \delta_{u_m,\overline{v_1}} \cdot \widetilde{U}_{u_1u_2 \dotsc u_{m-1}} \boxtimes \widetilde{U}_{v_2v_3 \dotsc v_n}
  \end{gather*}
  we obtain
  \[\widetilde{U}_x \boxtimes \widetilde{U}_y = \sum_{x = ac, y=\overline{c}b}{\widetilde{U}_{ab} \boxplus \widetilde{U}_{a \cdot b}}.\]
\end{proof}

We are now going to deduce the fusion rules of $\Akn$. The following result is proven in \cite{banicavergnioux09} and describes the fusion rules of $\Ahn$.

\begin{theorem}
  Let $S_\textup{h} := \{u, p\}$ with fusion $u \cdot u = p \cdot p = p$, $u \cdot p = p \cdot u = u$ and trivial conjugation. The fusion rules of $(\Ahn, U_\textup{h})$ are given by the free fusion ring over $S_\textup{h}$ in such a way that $U_u \simeq U_\textup{h}$ and $U_p \boxplus 1 \simeq (u_{ij}^2)$.
\end{theorem}

Using this theorem we obtain the following corollary in the case $A = \Akn$.

\begin{corollary}
The irreducible corepresentations of $\Akn$ are described by the fusion set $S_{\textup{k}} := \{u,v,p,q\}$ with fusion given by
\begin{center}
\begin{tabular}{r||ccccc}
  $\cdot$	& $u$		& $v$		& $p$		& $q$		\\
  \hline \hline
  $u$		& $\emptyset$	& $q$		& $u$		& $\emptyset$	\\
  $v$		& $p$		& $\emptyset$	& $\emptyset$	& $v$		\\
  $p$		& $\emptyset$	& $v$		& $p$		& $\emptyset$	\\
  $q$		& $u$		& $\emptyset$	& $\emptyset$	& $q$		\\
\end{tabular}
\end{center}
and conjugation $\overline{u} = v$, $\overline{p} = p$, $\overline{q} = q$. \\
The elements of $S_{\textup{k}}$ correspond to the following corepresentations.
\begin{itemize}
  \item The class of the fundamental corepresentation $U$ is $U_u$.
  \item The class of $\overline{U}$ is $U_v$.
  \item The class of the corepresentation $(u_{ij}^* \cdot u_{ij})$ is $U_p \boxplus 1$
  \item The class of the corepresentation $(u_{ij} \cdot u_{ij}^*)$ is $U_q \boxplus 1$
\end{itemize}
\begin{proof}
  We only have to prove the part about the concrete description of $U_u$, $U_v$, $U_p$ and $U_q$. The fact that $U_u$ is the class of the fundamental corepresentation is obvious from the construction. $U_v \simeq \overline{U}$ follows directly. \\
  It is easy to check that $(u_{ij}^* \cdot u_{ij})$ and $(u_{ij} \cdot u_{ij}^*)$ are corepresentation of $\Akn$. We have the decomposition $U \boxtimes \overline{U} \simeq U_{uv} \boxplus U_p \boxplus 1$. Moreover the construction in this section shows that $U_{uv}$ is $n^2 - n$ dimensional and $U_p$ is $n-1$ dimensional. Since $(u_{ij} \cdot u_{ij}^*)$ is non trivial, it suffices to give at least two linearly independent intertwiners from the $n$ dimensional corepresentation $(u_{ij} \cdot u_{ij}^*)$ to $U \boxtimes \overline{U}$. Two such intertwiners are $\CC^n \rightarrow (\CC^n)^{\otimes 2}: \: e_i \mapsto e_i \otimes e_i$ and $\CC^n \rightarrow (\CC^n)^{\otimes 2}: \: e_i \mapsto \sum_j{e_j \otimes e_j}$. \\
  The proof for $(u_{ij} \cdot u_{ij}^*)$ works similarly.
\end{proof}
\end{corollary}