TITLE: "Nice" Solution to repeated integral
QUESTION [3 upvotes]: I have a problem wherein I have defined a function $I_r(t) = \int e^{(2r-1)at} \int e^{(2r-3)at} \cdots \int e^{at} dt\cdots dt$, and $I_r(0) = 0$, for $r = 1,2,3,\ldots$.
I find that $e^{-ar^2t} I_r(t) = \left(1-e^{-at}\right)^r q(t)$, where $q(exp(-at))$ is a polynomial of $e^{-at}$. Is there a general technique for evaluating repeated integrals of this type that allows me to write $q$ in a nice clean way?
If I took $I^*_r(t) = \int e^{at} \int e^{at}\cdots \int e^{at} dt\cdots dt$ with $I^*_r(0) = 0$, and multiplied by $e^{-art}$, I would get $e^{-art}I^*_r(t) = (1-e^{-at})^r$. I am looking for a nice closed-form solution where I have a quadratic in $r$.
This is related to the derivation of a discrete probability distribution where the transition rate function is quadratic w/r.t. the number of events per cell.

REPLY [2 votes]: I don't think you'll get a closed form but you get get an expression that involves products and sums that isn't too horrible and is easy to evaluate for particular cases. It'll take me forever to type it up here with nice formatting but I'll describe the approach in a way that's easy to reproduce.
We're starting with the number 1 and repeatedly performing these two operations: integration and multiplication by $e^{nat}$ for various $n$. Switch to working with Laplace transforms. In Laplace transform space, integration (starting from zero) is simply multiplication by $1/s$ and multiplication by the exponential takes $f(s)$ to $f(s-na)$.
So the Laplace transform of $I_1$ is $1/(s(s-a))$, the Laplace transform of $I_2$ is $1/(s(s-3a)(s-4a))$ and so on. Notice how when we keep going, the denominator is made up of factors of the form $s-(r^2-i^2)a$. So it's easy to write the Laplace transform of $I_r$ using product notation. You now simply need to invert the Laplace transform. To do this, we need to write our rational function in $s$ as partial fractions. The factors in the denominator are all distinct so this is the easy case. After the inverse transform you should end up with something like
$e^{r^2t}\sum_{i=0}^r\exp(-i^2t)/\prod_{j\ne i}(i^2-j^2)$
I expect that I've made a slip somewhere as I only have a few moments spare. That's why I said 'something like'. But the method is sound and if you put in more work than me you'll get a definite result.
Update: Did some testing in Mathematica. I think the expression above is correct.


text:
\begin{document}

\maketitle

\begin{abstract}

 In this paper, we study the dependence of the weak Lefschetz property
 of

  algebras defined by a special class of monomials ideals in a polynomial ring with coefficient in a field,
    to the characteristic of the base field.
\end{abstract}

 \section{Introduction}
Let $\mathbb{K}$ be an arbitrary infinite field, and let
$R=\mathbb{K}[x_1, \dots, x_r]$ be the polynomial ring  with
standard grading. Let $I$ be a homogeneous ideal in $R$ such that
$A=R/I$ is an Artinian algebra. This is equivalent to say that the
radical of $I$ is equal to $(x_1, \dots, x_r)$, or,  $A$ can be
written as $\bigoplus_{i=0}^e A_i$.

A significant property which a standard graded Artinian
$\mathbb{K}$-algebra may pose is the weak Lefschetz property (WLP
for short). A standard graded $\mathbb{K}$-algebra has the WLP, if
there exists a linear form in $R$ such that for each $0\le i \le
e-1$, the multiplication map $\times \ell:A_i \longrightarrow
A_{i+1}$ has maximal rank, i.e., it is injective or surjective. This
property, not only depends to algebraic structure of the algebra
$A$, but also depends on the characteristic of the base field
$\mathbb{K}$.

In addition to  intrinsic significance of the WLP for a standard
graded Artinian $\mathbb{K}$-algebra, this property is closely
related to some problems in other disciplines of mathematics. For
example the presence of this property is related to existence of
finite projective planes \cite{CMNZ, Maeno}, has connection with
some special family of curves in Algebraic Geometry \cite{Miro}, it
is associated with the problem of enumerating  the plane partitions
in combinatorics \cite{DU, LZ}.

 Even though, this property has
a simple definition, but establishing it for a general standard
graded Artinian $\mathbb{K}$-algebra, is not an easy task. As a
consequences of this fact, classifying all standard graded
$\mathbb{K}$-algebra which pose this property would be a hard
problem. This forces to look for this property in special classes of
standard graded Artinian $\mathbb{K}$-algebras. Among  such
algebras, those for which $I$ is a  monomial ideal, are the most
accessible and excellent  ones.

When the characteristic of the base field $\mathbb{K}$ is zero,
Stanley \cite{Stan1} showed that the algebra $\mathbb{K}[x_1, \dots,
x_r]/(x_1^{d_1}, \dots, x_r^{d_r})$, where $d_is$ are greater than
1, enjoys the SLP, and hence the WLP, but when the characteristic of
the base field is positive, this result is no longer hold. It can be
easily shown that whenever $\mathrm{char\ } \mathbb{K }=p$, and
$r\ge 3$, the algebra $\mathbb{K}[x_1,\dots, x_r]/(x_1^p, \dots,
x_r^p)$, does not have the WLP, while it poses this property for
$r\le 2$. In \cite{MI_MI}, it is shown that if the ideal
$I=(x_1^{d_1}, \dots, x_r^{d_r})$, where $d_1\ge \dots \ge d_r\ge
2$, satisfies the condition $d_1
>[t/2]$, where $t=d_1+d_2+\dots+d_r-n$, then $R/I$ has the WLP,
regardless of the characteristic of $\mathbb{K}$. Moreover, in
\cite[Proposition 3.5]{D}, it is shown that if the $\mathrm{char\
}\mathbb{K}=p>0$ and $d_1\le \lceil t/2 \rceil$, and if $d_2 \le p
\le d_1$ or for some positive integer $m$, the condition $d_1 \le
p^m\le \lceil t/2 \rceil$ holds, then $R/I$ fails to have the WLP.
In \cite{BK}, by a geometric method, the WLP of the algebra
$\mathbb{K}[x,y,z]/(x^d,y^d,z^d)$, in terms of integer $d$ and the
characteristics of the field $\mathbb{K}$ is determined. In
\cite{LZ}, it is proved that  if  $p=\mathrm{char\ }\mathbb{K}$ is a
prime divisor of the number of plane partitions $M(a,b,c)$,  then
the $\mathbb{K}$-algebra
$A=\mathbb{K}[x,y,z]/(x^{a+b},y^{a+c},z^{b+c})$

does not have the WLP. In \cite[Theorem 3.8]{LN}, a complete
classification of the SLP for all monomial ideals $I$, for which $I$
is a complete intersection and
 $\mathbb{K}>0$, is given. In \cite{DU}, by tools
which have been developed in \cite{JJ},  the WLP of the monomial
ideals of the form
$I=(x^{t+\alpha}y^{t+\beta}z^{t+\gamma},x^{\alpha}y^{\beta}z^{\gamma})$,
known as almost complete intersection, are investigated. In
particular, those characteristics that these type of ideals may fail
to have the WLP are determined in terms of the exponents $t,\alpha,
\beta, \gamma$.

In \cite{JM}, the weak Lefschetz property of algebras defined by
another class of monomial ideals which are in the form
\begin{align*}
I_{r,k,d}=(x_1^k, \dots, x_r^k) +(\mathrm{all \  squarefree
monomilas \ of \ degree\ } d)
\end{align*}
are studied and the WLP of these type of ideals, whenever $d=2$ and
$d=3$, are established  (\cite[Theorem 3.3]{JM}), and for general
case is stated as the following conjecture (see \cite[Conjecture
3.4.]{JM})
\begin{con}\label{conj}
Consider the  algebra $R/I_{r,k,d}$, where the ideal $I_{r,k,d}$, is
defined as
\begin{center}
$\left(  x_1^{k}, \dots, x_r^{k}\right) +\left( \mathrm{ all \
square \ free \ monomials \ of \ degree} \ d\right)$.
\end{center}
Then
\begin{flushleft}
(a) If $d=4$, then it has the WLP if and only if $k \mod 4$  is 2 or 3.\\
(b) If $d=5$, then  the WLP fails.\\
(c) If  $d=6$, then the WLP fails.
\end{flushleft}
\end{con}\label{JU_MI_NA}
The proof of part $(b)$ of \cite [Theorem 3.3]{JM}, and the above
conjecture has motivated us to investigate the presence of the WLP
for a class of monomial  ideals which their generators are nearly
similar to the generators of $I_{r,k,d}$. I.e., we consider the
ideals of type
 \begin{align*}
 I=\left(  x_1^{\alpha_1}, \dots, x_r^{\alpha_r}\right) +\left(\textrm{ all\ square \ free\ monomials\ of \ degree}\ d
 \right),
\end{align*}
where $2 \le \alpha_i \le 4$ for $1 \le i \le r$, and study the WLP
behavior of $R/I$ with respect to small prime numbers.

 Our main results
are:

\noindent{\bf Theorem A.} Let $I$ be as the above ideal, where
 $ 2 \le \alpha_i \le 3$ for all $d\geq 4$ and $1 \le i \le r$. Then
 $A=R/I$ does not have the WLP, whenever $\mathrm{char}\ \mathbb{K}=3$.

In particular, if all $\alpha_i$s, are equal to 2,  or are equal to
3, then Theorem A, implies the part $(a)$ of  the conjecture
\ref{JU_MI_NA} can not be  true, while it confirms what is claimed
in  parts $(b),(c)$ of \ref{conj} for a specific value.

 \noindent{\bf Theorem B.} Let $I$ be as the ideal, where
 $2 \le \alpha_i \le 4$ for all $d \geq 5$ and $1 \le i \le r$. Then
 $A=R/I$ does not have the WLP,  whenever $\mathrm{char}\ \mathbb{K}=2$.

Another result of this paper is a little bit different from the
other results. In fact, we determined  all characteristics of the
base field which a special type of ideals  define an algebra without
the WLP.

\begin{prop}
Let
\begin{center}
$I^{'}=\left( x_1^2, \dots, x_r^2\right) +\left( \mathrm{all \
squarefree \ monomials \ of \ degree} \ d\right)$
\end{center}
be an ideal in $R$. Then $R/I^{'}$ is a level algebra and
$e=\mathrm{ Socle Degree }(I^{'})=d-1$. Moreover, $R/I^{'}$ doesn't
have the WLP in $\mathrm{char}\ \mathbb{K}= \mathit{p}$ whenever
$\mathit{p}$ is a prime number less than  $i+2$, where $1 \leq i
\leq \lceil r/2 \rceil$.
\end{prop}

The method of proof of  Theorem A and Theorem B, can be applied to
prove  the failure of the WLP for the following class of monomial
ideals:
\begin{align*}
J=\left( x_1^{\alpha}, \dots, x_r^{\alpha},
x_1^{\alpha-2}x_2^{2},x_1^2x_2^{\alpha -2},

\dots,x_{r-1}^{\alpha-2}{x_{r}}^{2},x_{r-1}^{2}{x_{r}}^{\alpha-2}
\right),
\end{align*}
provided $r\ge 4, \alpha \ge 5.$

 \noindent{\bf Theorem C.} Let $J$ be as the above ideal  in $R$.
Let $\mathrm{char}\ \mathbb{K}=2$. Then  $R/J$ does not have the
WLP.

\section{Preliminaries}

Let $A=R/I=\bigoplus_{i=0}^e A_i$ be a standard graded Artinian
$\mathbb{K}$ algebra. Then the function $h_i=\dim A_i$, for $0\le i
\le e$ is called the Hilbert function of $A$. Since $\dim A_i=0$ for
$i>e$, this function can be represented as an array
$\textbf{h}=(h_0,h_1, \dots, h_e)$, which is called the
$\textbf{h}$-vector of $A$. If we denote the maximal ideal of $A$ by
$\mathfrak{m}$, then the ideal
$$(0:_A \mathfrak{m})=\{ a\in \mid
a\mathfrak{m}=0\}=\mathcal{U}_0\oplus \mathcal{U}_1\oplus \dots
\oplus \mathcal{U}_e,$$ is called the socle of $A$. Since this ideal
gathers the annihilators of $\mathfrak{m}$, its structure is closely
related to the WLP of the algebra $A$. Since $A_{e+1}=0$, it is
clear that $A_e\subset (0:_A \mathfrak{m})$. The integer $e$ is
called the socle degree of $A$. Moreover, if $\mathcal{U}_i=0$ for
all $i <e$, then $A$ is called a level algebra.

\begin{defn}
Let $\ell$ be a general linear form in $R$. We say that the Artinian
ring $A$ has the {\it weak Lefschetz property} (WLP for short) if
the homomorphism induced by multiplication by $\ell$,
\begin{center}
$\times \ell: A_{i}\longrightarrow A_{i+1}$,
\end{center}
has the maximal rank for every $i$, $0\le i \le e-1$ (i.e., it is
injective or
surjective). In this case, the linear form $\ell$ is called the Lefschetz element for $A$.\\
We say that $A$ has the {\it strong Lefschetz property} the (SLP) if
\begin{center}
$\times \ell^{d}: A_{i}\longrightarrow A_{i+d}$
\end{center}
has the maximal rank for every $i$ and $d$, with $0\le i \le e-2$
and $1 \le d \le e-1$.
\end{defn}
If a standard graded $\mathbb{K}$-algebra $A$, has a Lefschetz
element $\ell$, then it can be shown that there is a Zariski open
set in $\mathbb{P}^{r-1}=\mathbb{P}(\mathbb{K}[x_1,\dots,x_r]_1)$
which parameterizes all Lefschetz elements of $A$.

In \cite[Proposition 2.2]{JJ}, it is shown if the field $\mathbb{K}$
is infinite, and the monomial Artinian $I$  satisfies the WLP, then
the linear form $\ell=x_1+\dots+x_r$ would be a Lefschetz element
too. This simplifies the checking for posing or failure of the WLP.
Moreover, in \cite[Proposition 4.3]{LN}, the assumption of being
infinite for $\mathbb{K}$ has been weaken. This will allows us to
use finite fields to construct examples or counterexamples for
posing of failure of the WLP.

\begin{prop}(\cite[Proposition 4.3]{LN})
Let $\mathbb{K}$ be a field and let $\mathbb{K}^{'}$ be an extension
field of $\mathbb{K}$. Let $I\subset \mathbb{K}[x_1, \dots, x_r]$ be
a monomial ideal. Then the following are equivalent.
\begin{itemize}
\item[(a)] $A=\mathbb{K}[x_1, \dots, x_r]/I$ has the WLP.
\item[(b)] $A^{'}=A\otimes_{\mathbb{K}} \mathbb{K}^{'}$ has the WLP.
\item[(c)] $x_1+\dots+x_r$ is a weak Lefschetz element of $A$.
\item[(d)] $x_1+\dots+x_r$ is a weak Lefschetz element of $A^{'}$
\end{itemize}

\end{prop}

\section{Main Results}

In this section, we prove the main results of this paper. The
Artinian algebras that we consider are defined by the ideals of the
following form.
\begin{align*}
I=\left(  x_1^{\alpha_1}, \dots, x_r^{\alpha_r}\right) +\left( all \
square \ free \ monomials \ of \ degree \ d\right),
\end{align*}
where $2\le \alpha_i \le 4$ and $2 \le d \le r$.

 The main idea of the proof of the main
results, is to construct a homogenous polynomial $f \in A=R/I$, such
that for suitable indices $i$, the map $\times \ell: A_i
\longrightarrow A_{i+1}$ fails to be injective.

\begin{thm}({\bf Theorem A.})\label{M}
Let $I$ be as above ideal, where  $d\geq 4$ and $ 2 \le \alpha_i \le
3$ for all $1 \le i \le r$. If  $\mathrm{char}\ \mathbb{K}=3$, then
 $A$ does not have the WLP.
\end{thm}
\begin{proof}
According to Remark \ref{r1}, $h_2\le h_3$.

We show that the map $\times \ell: A_2 \longrightarrow A_3$ can not
be injective. To prove our claim, let
\begin{align*}
f =\sum_{ 1 \le i < m \le r } (-1)^jx_i^jx_m^{2-j}, \mathrm{ \ \
where \ } j=1,2.
\end{align*}
It is clear that $f$ is a nonzero element of $ A_2$. Moreover, in
$A_3$, the element $f \times \ell $, consists of the following
terms:
$$
\begin{array}{llllllllll}

& \mathrm{ for }& j=1,&  - x_i^2x_m,& -x_ix_m^2,& -x_ix_mx_{t},&
\mathrm { where }& t \neq i, m,\ 1 \le t \le r; & \\
& \mathrm{ for  }& j=2,& \ \ x_i^2x_m,&\ \ x_i^2x_k,&\ \
x_i^3,&\mathrm{ where }& 1 \le k < i < m \le r. & &

\end{array}
$$
Notice that for those $\alpha_i s$ which are equal to two, the above
terms can not be appeared in $f \times \ell$. Moreover, as can be
seen from the above expressions, for $j=1$ and $j=2$, the first two
terms  of the first row are respectively, additive inverses of the
first two terms of the second row. Hence, their sum in $f \times
\ell$ cancel each other.

Moreover,  each term $-x_ix_m^2$ in the first row, is the additive
inverse of a monomial $x_i^2x_k$, in the second row, and vice versa.
Hence, their sum in $f \times \ell$ is zero.

On the other hand, by our assumption, $x_i^3$ modulo $I$ is zero.
Hence, the only terms which will be remained in $f \times \ell$, are
in the form $-x_ix_mx_{t}$, where $t \neq i,m$ and $ 1 \le t \le r$.
We  count the occurrences of these monomials in $f\times \ell$ via
two methods.

 {\it First method.} We know that
the number of squarefree monomials of degree 3 in  $r$ variables  is
equal to ${r \choose 3}$. We label these  monomials as $a_{1},
a_{2}, \dots, a_{{r \choose 3}}$ and set $M= a_{1}+ a_{2}+ \dots+
a_{{r \choose 3}}$.

 {\it Second method.} By definition of $f$, the terms
of $f \times \ell$ are products of all $x_{i}x_{m}$ and the
different variables with $x_{i}$ and $x_{m}$ that vary  in $\{x_1,
\dots, x_r\}$. The possible number of  such monomials is equal to
\begin{align*}
(r-2) {r \choose 2}=\frac{r(r-1)(r-2)}{2}.
\end{align*}
We label these  monomials as $b_{1}, b_{2}, \dots, b_{(r-2) {r
\choose 2}}$.

 By comparing the results of these two methods, we
observe that the monomials $x_ix_mx_t$ of  $f \times \ell$, have a
coefficient equal to 3 in $A_{3}$, since
 $b_{1}+ b_{2}+ \dots+ b_{(r-2) {r \choose 2}}=3( a_{1}+ a_{2}+ \dots+ a_{{r \choose 3}})=3M$.
Therefore, $f \times \ell =0$ modulo $I$, while $f \neq 0$. This
means the kernel of the map $\times \ell: A_2\longrightarrow A_3$ is
nontrivial. Hence, $A$ does not have the WLP.
\end{proof}
 \begin{rem}\label{r1}
 Let $I$ be as the above ideal, where  $d\geq 4$ and $ 2 \le \alpha_i \le 3$ for all
$1 \le i \le r$. The number of monomials of degree $i$ in $R_i$, is
equal to $i+(r-1) \choose r-1$. Hence $h_2= {2+(r-1) \choose r-1}
\leq h_3= {3+(r-1) \choose r-1}$, but we should remove those
monomials which are multiples of the generators of $I$. Let $m$ be
the number of these  degree 2 terms  and $n=r-m$ be the number of
these degree 3 terms. Then we have
\begin{center}
$h_2= {2+(r-1) \choose r-1}-m$ and $h_3= {3+(r-1) \choose
r-1}-(m.r)-n$
\end{center}
If $h_2 \leq h_3$, we are nothing to do and it is the case that  we
need for our argument, but  if $h_3 < h_2$, then this would be the
exceptional cases which should be avoided. Hence, if we evaluate
both sides of this inequality, then
\begin{center}
$h_2= {2+(r-1) \choose r-1}-m>h_3= {3+(r-1) \choose
r-1}-(m.r)-n={3+(r-1) \choose r-1}-(m.r)-(r-m),$
\end{center}
which implies  $7r+6rm> r^3+12m$. Note that $m$ and $ n $ varies in
$\{0,1,\dots,r\}$.

Notice that whenever $r$ becomes larger, the above inequality is no
longer hold. Therefore, these exceptional cases happen whenever
 $m=r=4$.
\end{rem}
In \cite[Theorem 4.3]{JM}, it is proved that the Artinian algebra
defined by the ideal in the general form
\begin{align*}
I_{r,r}=(x_1^r, \dots, x_r^r, x_1x_2\cdots x_r),
\end{align*}
fails to have the WLP. As a special case of the above theorem, we
can state the following  result.
\begin{cor}
Let the ideal $I$ be as in Theorem \ref{M}. If $d=r$, then the
algebra $R/I$ does not have the WLP.
\end{cor}

Now we can state a similar result for the case $\mathrm{char}\
\mathbb{K}=2$.

\begin{thm}({\bf Theorem B.})\label{M2}
Let $I$ be  the ideal as in Theorem \ref{M}. Let $d \geq 5$ and $2
\le \alpha_i \le 4$ for all $1 \le i \le r$. Then
 $A$ does not have the WLP,  whenever $\mathrm{char}\ \mathbb{K}=2$.
\end{thm}
\begin{proof}
According to Remark \ref{r2}, $h_3\le h_4$. We show that the map
$\times \ell: A_3 \longrightarrow A_4$ can not be injective. To
prove this, let $f\in A_3$ be as follow
\begin{align*}
f=\sum_{1\le i<m\le r}x_i^jx_m^{3-j} \ \ \mathrm{ where \ } j=1,2,3.
\end{align*}
Then terms of the polynomial $f \times \ell$ can be grouped together
in terms of the value of $j$ and consists of

$$

\begin{array}{llllll}
\mathrm{for \ } j=1, & x_i^2x_m^2,& x_ix_m^3,&  x_ix_m^2x_t; &
 t \neq i, m, 1 \le  t  \le r, & \\
\mathrm{for \ } j=2, & x_i^3x_m,&  x_i^2x_m^2,&  x_i^2x_mx_t; &
 t \neq i, m, 1 \le  t  \le r, &
\\ \mathrm{for \ } j=3, &  x_i^4,&  x_i^3x_m,& x_i^3x_k;&
 1\le k < i<m   \le r.
\end{array}
$$

As the above list shows, the terms $x_i^2x_m^2$  appear  in
 $f \times \ell$ for $j=1$ and  $j=2$.
 Hence, the sum of these terms in  $f \times \ell$ has
coefficient 2, which by our assumption, this sum would become zero.
Moreover, the terms $x_i^3x_m$ exist in the rows $j=2$ and $j=3$.
Hence, the coefficient of their sum is 2. Therefore, these terms
would be killed in $f \times \ell$. As well as,  each monomial
$x_ix_m^3$, in the first row of the above list, is equal to a term
$x_i^3x_k$, in the third row and vice versa. Hence, their sum would
be zero in $f \times \ell$. On the other hand,
 $x_i^4$ is zero   modulo $I$. Hence, it
remains the terms $x_ix_m^2x_t$ and $x_i^2x_mx_t$ in $f \times
\ell$. We  count the number of occurrence of these terms in $f
\times \ell$ via two ways.

{\it First method.} The number of distinct squarefree monomials of
degree 3 in terms of $\{x_1, \dots, x_r\}$ is equal to ${r \choose
3}$, and since the exponent of one of the variables in each terms is
equal to 2, the total number of these type of terms is equal to
\begin{align*}
3 {r \choose 3}= \frac{r(r-1)(r-2)}{2}.
\end{align*}
We label these terms as $a_{1}, a_{2}, \dots, a_{3 {r \choose 3}}$,
and set $M=a_{1}+ a_{2}+ \dots+ a_{3 {r \choose 3}}$.

{\it Second method.} Each of one of the monomials  $x_ix_m^2x_t$ and
$x_i^2x_mx_t$ is obtained by multiplying $x_t \in \{x_1, \dots,
x_r\}$, where $t \neq i, m$, by $x_i x_m^{2}$ and $x_i^2 x_m$ of
$f$. Hence, the total number of these summands is equal to $2(r-2){r
\choose 2}=r(r-1)(r-2)$.
We label these monomials as $b_{1}, b_{2}, \dots, b_{2(r-2){r \choose 2}}$.\\
By comparing the results of these two methods of counting, one can
observe that all monomials in the $f \times \ell$, are multiplied by
two, since $b_{1}+ b_{2}+ \dots+ b_{2(r-2){r \choose 2}}=2(a_{1}+
a_{2}+ \dots+ a_{3 {r \choose 3}})=2M$. Hence, by our assumption,
the sum is equal to 0, and the proof completes.
\end{proof}
 \begin{rem}\label{r2}
 Let $I$ be as the above ideal, where  $d\geq 5$ and $ 2 \le \alpha_i \le 4$ for all
$1 \le i \le r$. Then an argument similar to the one used in the
Remark \ref{r1} shows that $h_3 \leq h_4$.
\end{rem}

Due to its importance, we state a special case of the above theorem
as a corollary.
\begin{cor}
Let the ideal $I$ be as in Theorem \ref{M2}. If $d=r$, then the
algebra $R/I$ does not have the WLP.
\end{cor}
Contrary to the previous results which in characteristics 2 and 3,
we showed that the algebras that we considered do not pose the WLP,
in the next result, we determine all primes that the Artinian
algebra defined a specific ideal may not pose the WLP.
\begin{prop} \label{1}
Let
\begin{center}
$I^{'}=\left( x_1^2, \dots, x_r^2\right) +\left( \mathrm{all \
squarefree \ monomials \ of \ degree} \ d\right)$
\end{center}
be an ideal in $R$. Then $R/I^{'}$ is a level algebra and
$e=\mathrm{ Socle Degree }(I^{'})=d-1$. Moreover, $R/I^{'}$ doesn't
have the WLP in $\mathrm{char}\ \mathbb{K}= \mathit{p}$ whenever
$\mathit{p}$ is a prime number less than  $i+2$, where $1 \leq i
\leq \lceil r/2 \rceil$.
\end{prop}
\begin{proof}

By our assumption on $I$, each homogeneous component $A_i$ of $A$ is
generated by squarefree monomials of degree $i$. Hence $h_i={r
\choose i}$. Therefore,
\begin{center}
$h_0 \leq h_1 \leq \dots \leq h_{\lceil r/2 \rceil}$.
\end{center}

We show that the map $\times \ell: A_i \longrightarrow A_{i+1}$ can
not be injective for any $1 \leq i \leq \lceil r/2 \rceil$. Let
\begin{center}
$f =\sum x_{j_1}x_{j_2} \cdots x_{j_i}$,
\end{center}
where $1\leq j_i \leq r-1$.\\
According to the definition of $f$, the nonzero terms of $f \times
\ell$ are in the form:
\begin{center}
$x_{j_1}x_{j_2} \cdots x_{j_i} x_m$
\end{center}
where $1\leq m \leq r$ and $m$ is not in $\{j_1,\dots, j_i\}$.

Now we want to count the occurrence of these terms in $A_{i+1}$.\\
{\it First method.} We know that the number of squarefree monomials
of degree $i+1$ in $R$
is equal to ${r \choose i+1}$. We label these terms as $a_{1},
a_{2}, \dots, a_{{r \choose i+1}}$,
and set $M= a_{1}+ a_{2}+ \dots+ a_{{r \choose i+1}}$.\\
{\it Second method.} By definition of $f$, these terms of $f \times
\ell$ are products of all $x_{j_1}x_{j_2} \cdots x_{j_i} $ and the
different variables with $x_m$ in $\{x_1,\dots, x_r\}\setminus
\{x_{j_1}, \dots, x_{j_i} \}$. The number of possible such terms is
equal to
\begin{center}
$(r-i) {r \choose i}=\frac{(r-i)r(r-1)\cdots(r-i+1)}{i!}$.
\end{center}
We label these terms as $b_{1}, b_{2}, \dots, b_{(r-i) {r \choose i}}$. \\
By comparing the results of these two methods of counting, we
observe that the terms $x_{j_1}x_{j_2} \cdots x_{j_i}x_m $ of $f
\times \ell$, have coefficient equal to $i+1$ in $A_{i+1}$, because
$b_{1}+ b_{2}+ \dots+ b_{(r-i) {r \choose i}}=(i+1)( a_{1}+ a_{2}+
\dots+ a_{{r \choose i+1}})=(i+1)M$. Therefore, while $f \neq 0$, $f
\times \ell =0$ modulo $I$,  whenever $\mathit{p}$ is a prime
divisor of $ i+1$. Hence, $A$ does not have the WLP in these cases.
\end{proof}
The following corollary, not only shows that the WLP may fail for
only finitely many prime numbers, but also it determines these
primes exactly.
\begin{cor}\label{Cor38}
 Let $I^{'}$ be an ideal as  in
 Proposition \ref{1}.

 Then $R/I^{'}$ has the WLP
 whenever $\mathrm{char}\ \mathbb{K}=\mathit{p}$ is not a prime number less than $ i+2$, where
  $1 \leq i \leq \lceil r/2 \rceil$.
\end{cor}
\begin{proof}

Two cases may arise. In the first case, which $i$ varies in the
range $1 \leq i \leq d-1 < \lceil r/2 \rceil $, we only need to
prove the injectivity of the map $\times \ell:A_i \longrightarrow
A_{i+1}$ for every $i$. But this follows from Proposition
\ref{1}, since all  maps $\times \ell:A_i \longrightarrow A_{i+1}$,

are injective.

In the other case,  $i$ varies in the range  $\lceil r/2 \rceil \leq
i \leq d-1$, and it is enough to prove that all maps $\times \ell:
A_{ i} \longrightarrow A_{i+1}$,
 are surjective. By \cite [Proposition 2.1]{JM} it is enough to do it for $i=\lceil r/2 \rceil $. Let
$x_{j_1}x_{j_2} \cdots x_{j_{i+1}} $ be an element of $A_{i+1}$.
Then it is clear that, it is the image of $f=\sum_{m=1}^i
x_{j_1}\cdots x_{j_{m-1}}\widehat{x_{j_m}}x_{j_{m+1}}\cdots
x_{j_{i+1}}$ under the multiplication map by $\ell$.
\end{proof}

With a method similar to the proofs of Theorems \ref{M} and
\ref{M2}, we can prove the failure of the  WLP for another special
class of monomial ideals.

\begin{thm}({\bf Theorem C.})\label{C}
Let $J$ be the following ideal  of $R$ for which $r\ge 4$ and
$\alpha \geq 5$.
\begin{align*}
J=\left( x_1^{\alpha}, \dots, x_r^{\alpha},
x_1^{\alpha-2}x_2^{2},x_1^2x_2^{\alpha -2},
\dots,

x_{r-1}^{\alpha-2}{x_{r}}^{2},x_{r-1}^{2}{x_{r}}^{\alpha-2}
\right).
\end{align*}
 If $\mathrm{char\ }
 \mathbb{K}=2$, then
 $R/J$ does not have the WLP.
\end{thm}
\begin{proof}

According to Remark \ref{special}, $h_{\alpha-1} \le h_{\alpha}$. We
show that the map $\times \ell: A_{\alpha-1} \longrightarrow
A_\alpha$ is not injective. Let
 $f\in A_{\alpha-1}$ be as follow
\begin{align*}
f=\sum_{1\le i<m\le r}x_i^{\alpha-j}x_m^{j-1} \ \ \mathrm{ where \ }
j=1,2, \alpha-1.
\end{align*}
Then terms of  $f \times \ell$, with respect to different values of
$j$, can be grouped together as follows.
$$
\begin{array}{llllll}
 \mathrm{ for  \ } j=1, & x_i^\alpha, & x_i^{\alpha-1}x_m,&
x_kx_i^{\alpha-1},&\mathrm { where \ }  1 \le  k<  i < m \le r;  \\
\mathrm{ for \  } j=2,& x_i^{\alpha-1}x_m,&   x_i^{\alpha-2}x_m^2,&  x_i^{\alpha-2}x_mx_t,&  \mathrm{where \ }  t\neq i, m, 1 \le  t \le r;   \\
 \mathrm{ for \ } j=\alpha-1,& x_ix_m^{\alpha-1},&
x_i^2x_m^{\alpha-2},& x_ix_m^{\alpha-2}x_t,&   \mathrm{where \ }
t\neq i, m, 1 \le  t \le r.
\end{array}
$$
 The monomials $x_i^\alpha$, $x_i^2x_m^{\alpha-2}$, and $x_i^{\alpha-2}x_m^2$ are
 zero modulo $ I$. As the above list shows,  the monomials $x_i^{\alpha-1}x_m$ appear
 for $j=1$ and $j=2$. Hence, the sum of these monomials in $f \times \ell$  has a coefficient equal to 2, which by
 our assumption, this sum would become zero. As well as, whenever $j=\alpha-1$, each monomial $x_ix_m^{\alpha-1}$, where
 $1 \le i < j\le r$,  is equal to a monomial $x_kx_i^{\alpha-1}$ with $1\le k<i<m\le r$ and $j=\alpha-1$, and vice versa.
  Hence, their sum in $f \times \ell$ is
a multiple of 2. Finally,  for $j=2$ and $j=\alpha-1$ the monomials
$x_i^{\alpha-2}x_mx_t$  and $x_ix_m^{\alpha-2}x_t$ remain in $f
\times \ell$. We  count the number of occurrence of these monomials
in $f \times \ell$ in two different ways.

 {\it First method.} The number of distinct monomials in three distinct
variables of $\{x_1, \dots, x_r\}$ is equal to ${r \choose 3}$, and
since the exponent of only one of its variables is equal to
$\alpha-2$, hence the total number of these elements are equal to
\begin{align*}
3 {r \choose 3}=\frac{r(r-1)(r-2)}{2}.
\end{align*}
We label these monomials as $a_{1}, a_{2}, \dots, a_{3 {r \choose
3}}$ and set $M=a_{1}+ a_{2}+ \dots+ a_{3 {r \choose 3}}$.

{\it Second method.} Each of one of the monomials
$x_i^{\alpha-2}x_mx_t$ and $x_ix_m^{\alpha-2}x_t$ are obtained by
multiplying $x_t\in \{x_1, \dots, x_r\}$, where $ t\neq i, m$, to
terms  $x_i^{\alpha-2}x_m$ and $x_ix_m^{\alpha-2}$  of $f$. Hence,
the number of these monomials  is equal to $2(r-2){r \choose
2}=r(r-1)(r-2)$. We label these terms as $b_{1}, b_{2}, \dots,
b_{2(r-2){r \choose 2}}$.

By comparing the results of these two methods of counting, one can
observe that all monomials in  $f \times \ell$, have a coefficient
equal to two, since $b_{1}+ b_{2}+ \dots+ b_{2(r-2){r \choose
2}}=2(a_{1}+ a_{2}+ \dots+ a_{3 {r \choose 3}})=2M$. Therefore, $f
\times \ell=0$, and the weak Lefschetz property does not hold for
this algebra.
\end{proof}

\begin{rem}\label{special}
Let $J$ be as the above ideal and $A=\bigoplus_{i=0}^e A_i$ with
$h_i=\dim A_i$. In the above argument, we need to have $h_{\alpha-1}
\le h_{\alpha}$. This inequality does not always hold. In fact, from
the structures of $A_{\alpha-1}$ and $A_{\alpha}$, we can deduce
\begin{align*}
h_{\alpha-1}= {\alpha+r-2 \choose r-1} \le h_{\alpha}={\alpha + r-1
\choose r-1} -r -2{r \choose 2},
\end{align*}
where for calculating $h_{\alpha}$, the total number of
$x_i^{\alpha}s$, which is equal to $r$, and the total number of
monomials in the forms $x_i^{\alpha-2}x_j^{2}$ and $x_i^2
x_j^{\alpha -2}$, which  is equal to $2 {r \choose 2}$, should be
subtracted. The above inequality implies $r^2 \le
\frac{(\alpha+r-2)!}{(r-2)!\alpha!} $. But this inequality holds if
$r\ge 4$ and $\alpha\ge 5$.
\end{rem}

According to the method of proofs of the above theorems, we are able
to determine many examples of  monomial Artinian algebras without
the WLP. On the other hand, it is possible to specify some classes
of them with the WLP.
\begin{exm}
Consider the ideal $$I=(x_1^4, x_2^4, x_3^3, x_4^3, x_5^2,
x_1x_2x_3x_4x_5),$$ in $\mathbb{K}[x_1, \dots, x_5]$. The {\bf
h}-vector of $R/I$ is $(1,5,14,28,43,52,49,35,18,6,1)$. Then by
Theorem \ref{M}, $ A_{3} \longrightarrow A_{4 }$ is not injective.
Therefore, it doesn't have the WLP whenever the characteristic of
$\mathbb{K}$ is two.

\end{exm}
\begin{exm}
Let
$I=(x_1^5,x_2^5,x_3^5,x_4^5,x_1^3x_2^2,x_1^2x_2^2,\dots,x_3^3x_4^2,x_3^3x_4^2)$
be an ideal in $\mathbb{K}[x_1,x_2,x_3,x_4]$. Its $\textbf{h}$
vector is $(1, 4, 10, 20, 35, 40, 26, 8, 1)$. Then by Theorem
\ref{C}, $ A_{4} \longrightarrow A_5$  is not injective, so it
doesn't have the WLP whenever the characteristic of $\mathbb{K}$ is
two.
\end{exm}

config: arxiv


TITLE: If $B$ is an uncountable set and $A$ is a countable set, then prove that $B$ is similar to $B-A$.
QUESTION [0 upvotes]: If $B$ is an uncountable set and $A$ is a countable set, then  prove that $B$ is similar to $B-A$.
Attempt:
Two sets $A$ and $B$ are called similar $\iff$ thee exists a one to one function $F$ whose domain is the set $A$ and whose range is the set $B$.
$B$ is an uncountable set and $A$ is a countable set, then $B-A$ must also be uncountable and hence, an infinite set.
To prove this, let us suppose $B-A$ is countable. Since, $A$ is countable, hence, $(B-A) \cup A$ should also be countable. But $B \subseteq (B-A) \cup A$ should be countable as well which is a contradiction.
Hence, $B-A$ is an uncountable set as well. Now, we need to show that $(B-A) \sim B$.
How do I move ahead? I think we need to define a one-one onto function from $(B-A) \rightarrow B$ but I am not able to think of such a function.
Thank you for your help.

REPLY [1 votes]: You showed that $B-A$ is uncountable, so you can extract another countable set from it.
Let's say that $B=A\cup (B-A)=A\cup C \cup B' $, where $C$ is countable and $B'$ uncountable, with $C$ and $B'$ disjoint.
It's easy to construct a bijection between $C$ and $A\cup C$, since are both countable sets, so you have a bijection from $C \cup B'=B-A$ and $A\cup C \cup B'=B$
This works if $A\subseteq B$. If this isn't true, then let $A'$ be $A\cap B$, and repeat the same reasoning with $A'$, and it works since $B-A=B-A'$


TITLE: Counterexample to high dimensional Nielsen realization problem
QUESTION [5 upvotes]: Can someone give me an example of a closed smooth oriented manifold $M$ and an orientation preserving diffeomorphism $f:M\rightarrow M$ such that $f^k$ is isotopic to the identity for some $k\geq 1$, but $f$ is not isotopic to any finite order diffeomorphism?  Similarly for homeomorphisms, and also for homotopies.
Of course, Kerckhoff's famous theorem says that this is impossible in dimension $2$.

REPLY [5 votes]: This is discussed at great length (with copious references) in Jonathan Block's and Shmuel Weinberger's paper. The title of the paper is the suggestive "On the generalized Nielsen realization problem" (comm math helv, 2008)


TITLE: Identifying the group $G = \langle x,y \space | \space xy=yx, x^4=y^2 \rangle$ from the given presentation
QUESTION [4 upvotes]: I'm trying to solve a problem in my textbook which asks me to identify the groups $G_1 = \langle x,y \space | \space x^3y=y^2x^2=x^2y\rangle$ and $G_2 = \langle x,y \space | \space xy=yx, x^4=y^2 \rangle$ from the given presentations.
For $G_1$, I'm pretty sure I can say that $x^3y=x^2y \implies x^3y(x^2y)^{-1} = e \implies x = e$ (where $e$ is the identity), and then $x^3y=y^2x^2 \implies y = y^2$ as $x=e$ to get $G_1 = \langle x,y \space | \space x=y=e \rangle \cong \{e\}$ (though I'd appreciate it if you could tell me if I have this wrong).
What I'm struggling with is trying to do the same sort of thing for $G_2$ - I can't see any way of getting this into a form where I can see the represented group.
I'd appreciate any help you could offer.

REPLY [2 votes]: $$\begin{align}G_2 &= \langle x, y | xy = yx, x^4 = y^2 \rangle \\
&= \frac{ \langle x, y | xy = yx \rangle }{\langle \langle x^4y^{-2} \rangle \rangle} \\
&= \frac{\langle x \rangle_{\infty} \times \langle y \rangle_{\infty} }{\langle x^4y^{-2} \rangle} \\
&\cong \frac{\langle x \rangle_{\infty} \times \langle y \rangle_{\infty} }{\langle x^2y^{-1} \rangle} \times C_2 \\
&\cong C_{\infty} \times C_2.\end{align}$$


TITLE: $\dim V<\infty$. Show there exists $m$ so that $\ker T^m \cap T^m(V)=0$
QUESTION [2 upvotes]: Here's a question.

M7. Let $V$ be a finite dimensional vector space and $T$ be a linear transformation on $V$. Prove there exists an $m\ge1$ such that
  (a) $\ker T^n=\ker T^m$ for all $n\ge m$.
  (b) $\ker T^m\cap T^m(V)=0$.
  Note: $T^n$ denotes the composite map $T\circ T\circ T\circ\cdots\circ T$ ($n$ times); and $\ker$ denotes kernel.

Now I have done the first part. Here's what I have done

Now I'm stuck with the last part. Any help is appreciated.

REPLY [1 votes]: Your proof for (a) is correct.
(b): Let $x \in ker(T^m) \cap T^m(V)$. Then $T^mx=0$ and $x= T^mv$ for some $v \in V.$
Hence $0=T^mx=T^{2m}v$. This gives $v\in ker(T^{2m}).$ By (a) we have $ker(T^m)=ker(T^{2m})$, thus $v \in ker(T^m)$. We get $x=T^mv=0.$


TITLE: Given two algebraic conjugates $\alpha,\beta$ and their minimal polynomial, find a polynomial that vanishes at $\alpha\beta$ in a efficient way
QUESTION [10 upvotes]: Inspired by this question, I was wondering about the following problem.

$\alpha,\beta,\gamma,\ldots$ are the roots of an irreducible
  polynomial over $\mathbb{Q}$. How to compute the coefficients of $$
 (x-\alpha\beta)(x-\alpha\gamma)\cdot \ldots\cdot (x-\beta\gamma)\cdot\ldots, $$ i.e. the candidate minimal polynomial
  of $\alpha\beta$, in the most efficient way?

My approach is based on a simple lemma and a general fact. If $p_m$ is the power sum $\alpha^m+\beta^m+\gamma^m+\ldots$ and $e_i$ is the $i$-th elementary symmetric polynomial, the identity
$$ \exp\left(-\sum_{m\geq 1}\frac{p_m}{m}x^m\right) = \sum_{r\geq 0}(-1)^r e_r\,x^r$$
gives a way to compute $p_1,p_2,p_3,\ldots$ from $e_1,e_2,e_3,\ldots$ through the Taylor series of a logarithm ($\text{LOG}$) as well as $e_1,e_2,e_3,\ldots$ from $p_1,p_2,p_3,\ldots$ through the Taylor series of an exponential ($\text{EXP}$). Moreover, the characteristic polynomial of the sequence $\{p_k\}_{k\geq 0}$ is exactly the minimal polynomial of $\alpha$, hence we may compute $p_{n+1},p_{n+2},\ldots,p_{n+m}$ from $p_1,p_2,\ldots,p_n$ with a simple recursive approach (I will call this procedure $\text{CH}$, from Cayley-Hamilton). $\text{L}$ will be my previously mentioned lemma, i.e.
$$ p_k(\alpha\beta,\alpha\gamma,\ldots) = \frac{1}{2}\left(p_k^2-p_{2k}\right).$$
Now my algorithm goes as follows:
(1). $$ e_i(\alpha,\beta,\gamma,\ldots)_{i\leq n}\quad\xrightarrow{\text{LOG}}\quad p_i(\alpha,\beta,\gamma,\ldots)_{i\leq n}$$
 (2). $$ p_i(\alpha,\beta,\gamma,\ldots)_{i\leq n}\quad\xrightarrow{\text{CH}}\quad p_i(\alpha,\beta,\gamma,\ldots)_{i\leq n(n-1)}$$
 (3). $$ p_i(\alpha,\beta,\gamma,\ldots)_{i\leq n(n-1)}\quad\xrightarrow{\text{L}}\quad p_i(\alpha\beta,\alpha\gamma,\ldots)_{i\leq \binom{n}{2}}$$
 (4). $$ p_i(\alpha\beta,\alpha\gamma,\ldots)_{i\leq \binom{n}{2}}\quad\xrightarrow{\text{EXP}}\quad e_i(\alpha\beta,\alpha\gamma,\ldots)_{i\leq \binom{n}{2}}.$$

REPLY [2 votes]: I don't see that the assumption that $\alpha$ and $\beta$ are roots of the same polynomial is particularly helpful. Suppose they have minimal polynomials $f(\alpha), g(\beta)$. Write down the companion matrices of these polynomials, then the Kronecker product of these matrices, then compute the characteristic polynomial of the Kronecker product. This polynomial isn't guaranteed to be irreducible, but from here you can factor it.


text:
\begin{definition}[Definition:Fermat's Spiral]
'''Fermat's spiral''' is the [[Definition:Locus|locus]] of the [[Definition:Equation of Geometric Figure|equation]] expressed in [[Definition:polar Coordinates|Polar coordinates]] as:
:$r^2 = \dfrac a \theta$
:[[File:FermatSpiral-positive.png|600px]]
\end{definition}

config: wiki


text:
\begin{document}
\maketitle

\begin{abstract} We associate with an algebra ${\bf A}={\mathbb K}[[f_1,\ldots,f_s]]\subseteq {\mathbb K}[[x_1,\ldots,x_n]]$
over a field ${\mathbb K}$ a fan called the canonical fan of $\bf A$. This generalizes the notion of
the standard fan  of an ideal.
\medskip

\noindent {{Keywords}: {Canonical basis, ${\mathbb K}$-algebras, affine semigroups}}
\end{abstract}

\section*{Introduction}

\medskip

\noindent Let $\KK$ be a field and let $f_1,\ldots,f_s$ be nonzero elements of the ring $\CC={\mathbb K}[[x_1,\ldots,x_n]]$ of formal power series in $x_1,\ldots,x_n$ over ${\mathbb K}$. Let
 ${\bf A}=\mathbb{K}[[f_1,\ldots,f_s]]$ be the ${\mathbb K}$-algebra generated by $f_1,\ldots,f_s$. Set $U={{\mathbb R}_+^*}$  and let $a\in U^n$. If $a=(a_1,\ldots,a_n)$ then $a$ defines
a linear form on $\RR^n$ which maps
 $\alpha=(\alpha_1,\ldots,\alpha_n)\in {\RR}^n$ to the inner product

$$
<a,\alpha>=\sum_{i=1}^na_i\alpha_i
$$

\noindent  of $a$ with $\alpha$.  We denote abusively the linear form by $a$. Let $\underline{x}=(x_1,\ldots,x_n)$ and
let $f=\sum c_{\alpha}\underline{x}^{\alpha}$ be a nonzero element of ${\CC}$. We set
Supp$(f)=\lbrace \alpha| c_{\alpha}\not=0\rbrace$ and we call it the support of $f$. We set

$$
\nu(f,a)={\rm min}\lbrace <a,\alpha>|\alpha\in {\rm Supp}(f)\rbrace
$$

\noindent  and we call it the $a$-valuation of $f$). We set by convention $\nu(0,a)=+\infty$. Let

$$
{\mathrm in}(f,a)=\sum_{\alpha\in {\rm Supp}(f)|<a,\alpha>=\nu(f,a)}c_{\alpha}\underline{x}^{\alpha}.
$$

\noindent  We call ${\mathrm in}(f,a)$ the $a$-initial form of $f$. Note that ${\mathrm in}(f,a)$ is a polynomial. This notion can also be defined this way:  we associate with $f$ its Newton polyhedron defined to be  $\Gamma_+(f)=$ the convex hull in ${\mathbb R}^n_+$ of $\bigcup_{\alpha\in{\rm
supp}(f)}\alpha+{\mathbb R}^n_+$.  The set of compact faces of $\Gamma_+(f)$ is finite. Let $\lbrace \triangle_1,\ldots,\triangle_t\rbrace$ be this set. Given $i\in \lbrace 1,\ldots,t\rbrace$. We set  $f_{\triangle_i}=\sum_{\alpha\in{\mathrm Supp}(f)\cap \triangle_i}c_{\alpha}{\underline{x}}^{\alpha}$. Then $\lbrace {\mathrm in}(f,a)\mid a\in U^n\rbrace=\lbrace f_{\triangle_i}\mid 1\leq i\leq t\rbrace$.

\medskip

\noindent Let the notations be as above, and let $\prec$ be a well ordering on ${\mathbb N}^n$. We set ${\mathrm exp}(f,a)={\rm max}_{\prec}{\rm Supp}\\({\mathrm in}(f,a))$
 and ${\mathrm M}(f)=c_{{\rm exp}(f,a)}\underline{x}^{{\rm exp}(f,a)}$. We set ${\mathrm in}({\bf A},a)=\mathbb{K}[[{\mathrm in}(f,a)|f\in {\bf A}\setminus\lbrace 0\rbrace]]$. We also set ${\mathrm M}({\bf A})={\mathbb K}[[{\mathrm M}(f,a), f\in{\bf A}\setminus\lbrace 0\rbrace]]$. The set ${\mathrm exp}({\bf A},a)=\lbrace {\mathrm exp}(f,a)\mid f\in{\bf A}\setminus \lbrace 0\rbrace\rbrace$ is an affine subsemigroup of ${\mathbb N}^n$.
\medskip

\noindent If  $a\in{\mathbb R}^n$ then ${\mathrm in}(f,a)$ may not be a polynomial, hence ${\mathrm exp}(f)$ is not well defined. If $a=(a_1,\ldots,a_n)$ with $a_{i_1}=\ldots=a_{i_l}=0$ then we can avoid this difficulty in completing by the tangent cone order on $(x_{i_1},\ldots,x_{i_l})$. We shall however consider elements in $U^n$ in order to avoid technical definitions and results.

\medskip

\noindent The aim of this paper is to study the
stability of ${\mathrm in}({\bf A},a)$ and ${\mathrm M}({\bf A},a)$ when $a$ varies in $U^n$. Note that ${\mathrm exp}({\bf A},a)$ is not necessarily finitely generated (see example \ref{notfinite}). It becomes so If the length of $\dfrac{\CC}{\bf A}$ is finite. Under this condition, our main
results are the following:

\medskip

\noindent {\bf Theorem} The set $\lbrace {\mathrm in}({\bf A},a)| a\in U^n\rbrace$  is a finite set. The same holds for the set $\lbrace {\mathrm M}({\bf A},a)| a\in U^n\rbrace$.

 \medskip

\noindent {\bf Theorem} Let $S$ be a finitely generated affine semigroup. The set $E_S=\lbrace
a\in U^n\mid {\rm exp}({\bf A},a)=S\rbrace$  is a union of convex polyhedral cones and the set of $E_S, S$ defines a fan of $U^n$.

\medskip

\noindent These results generalize those of \cite{mr} for ideals in ${\mathbb K}[x_1,\ldots,x_n]$ and \cite{acg1}, \cite{acg2} for ideals in the ring of differential operators over ${\mathbb K}$.

\medskip

\noindent The paper is organized as follows. In Section 1. we recall the notion of canonical basis of ${\bf A}$ with respect to $a\in U^n$, and we give an algorithm that computes an $a$-canonical basis starting with a set of generators of ${\bf A}$ (see \cite{m} for the case $a=(-1,\ldots,-1)$). In Section 2. we prove the finiteness theorem, and in Section 3. we prove the existence of a fan associated with ${\bf A}$.

\medskip

\section{Preliminary results}

\medskip

\noindent Let ${\bf A}={\mathbb K}[[f_1,\ldots,f_s]]$ be a subalgebra of ${\CC}$ generated by $\lbrace f_1,\ldots,f_s\rbrace\subseteq {\CC}$ and let the notations be as above. In particular  $\prec$ is a total well ordering on $\NN^{n}$
compatible with sums. Let $a\in {U}^n$ and consider the total
ordering  on $\NN^{n}$
defined by:

$$
 \begin{array}{ccl}\alpha <_a \alpha' &
\Leftrightarrow & \left\{ \begin{array}{l} <a,\alpha> <
<a,\alpha'> \\ \mbox{or} \\ <a,\alpha>=
<a,\alpha'> \mbox{ and } \alpha \prec \alpha' \end{array} \right.
\end{array}
 $$

\noindent The total ordering $<_a$ is compatible with sums in
$\NN^{n}$. We shall use sometimes the notations $\alpha \succ \beta$ for $\beta\prec \alpha$ and $\alpha >_a \beta$ for $\beta <_a \alpha$. We have the following:

\begin{lema}\label{Dixon} There doesn't exist  infinite sequences $(\alpha_k)_{k\geq 0}$ such that

$$
\alpha_0 >_a \alpha_1 >_a \ldots >_a \alpha_k \ldots
$$

\noindent With $<a,\alpha_0>=<a,\alpha_k>$ for all $k\geq 1$.

\end{lema}

\begin{proof} This is a consequence of Dixon's Lemma, since such a sequence satisfies $\alpha_0 \succ \alpha_1 \succ \ldots \succ \alpha_k  \succ\ldots$
\end{proof}

\noindent Let  $a\in U^n$. Then $a$ defines a filtration on ${\CC}: \CC=\sum_{d\geq 0}\CC_d$ where ${\bf  F}_d$ is the ${\mathbb K}$-vector space generated by $\underline{x}^{\alpha}, <a,\alpha>=d$. Let $U_1={\mathbb Q^*}_+$. If $a\in U_1^n$ then,  given two indices $d_1<d_2$, the set of indices $d$ such that $d_1<d<d_2$ is clearly finite. In particular we get the following:

\begin{lema}\label{Dixon2} Let $\alpha,\beta \in U_1^n$ and assume that $\alpha >_a \beta$. There doesn't exist infinite sequences $(\alpha_k)_{k\geq 0}$ such that

$$
\alpha>_a\alpha_0 >_a \alpha_1 >_a \ldots >_a \beta
$$

\end{lema}

\begin{proof} The set of $<a,\gamma>, <a,\alpha> ><a,\gamma> > <a,\beta>$ is finite. Then the result is a consequence of Lemma \ref{Dixon}
\end{proof}

\begin{definicion} Let $a\in U^n$ and let $f= \sum
c_{\alpha}\underline{x}^\alpha$ be a nonzero element of ${\CC}$. We say that $f$ is $a$-homogeneous if $f\in \CC_d$ for some $d$. This is equivalent to $\nu(\alpha,a)=\nu(f,a)$ for all $\alpha\in{\mathrm Supp}(f)$. Note that if $a\notin U_1^n$ then $f$ is $a$-homogeneous if and only if $f$ is a monomial.
\end{definicion}

\begin{definicion} Let $a\in U^n$ and let ${\bf H}$ be a subalgebra of ${\CC}$. We say that $\bf H$ is $a$-homogeneous if it can be generated by $a$-homogeneous elements of ${\CC}$.
\end{definicion}

 \noindent Every nonzero element $f=\sum
c_{\alpha}\underline{x}^\alpha\in \CC$ decomposes as $f=\sum_{k\geq d}f_k$, with $f_d= {\mathrm in}(f,a)$ and  for all $k> d$, if $f_k\not=0$ then $f_k\in {\CC}_k$.

\medskip

\begin{definicion} Let $a\in U^n$ and let $\lbrace g_1,\ldots,g_r\rbrace\subseteq {\bf A}$. We say that  $\lbrace g_1,\ldots,g_r\rbrace$ is an $a$-canonical basis of ${\bf A}$ if $\M({\bf A},a)={\mathbb K}[[{\mathrm M}(g_1,a),\ldots,{\mathrm M}(g_r,a)]]$. Clearly $\lbrace g_1,\ldots,g_r\rbrace$ is an $a$-canonical basis of ${\bf A}$ if and only if ${\mathrm exp}({\bf A},a)$ is generated by $\lbrace {\mathrm exp}(g_1,a),\ldots,{\mathrm exp}(g_r,a)\rbrace$. In this case we write ${\mathrm exp}({\bf A},a)=\langle {\mathrm exp}(g_1,a),\ldots,{\mathrm exp}(g_r,a)\rangle$
\end{definicion}

\noindent  An  $a$-canonical basis $\lbrace g_1,\ldots,g_r\rbrace$
of $\bf A$ is said to be minimal if  $\lbrace \M(g_1,a),\ldots,\M(g_r,a)\rbrace$ is a minimal
set of generators of $\M({\bf A},a)$.  It is said to be reduced if the following conditions
are satisfied:

i) $\lbrace g_1,\ldots,g_r\rbrace$ is minimal.

ii) For all $1\leq i\leq r, c_{{\rm exp}(g_i,a)}=1$.

iii) For all $1\leq i\leq r$, if $g_i-\M(g_i,a)\not= 0$ then   $\underline{x}^{\underline{\alpha}}\notin {\mathbb K}[[\M({g_1,a}),\ldots, \M(g_r,a)]]$ for all
$\alpha\in$ Supp$(g_i-\M(g_i,a))$.

\medskip

\begin{lema} \label{reduced} If an $a$-reduced canonical basis exists, then it is unique.
 \end{lema}

\begin{proof} Let  $F=\lbrace g_1,\ldots,g_r\rbrace$ and $G=\lbrace g'_1,\ldots,g'_t\rbrace$ be two
$a$-reduced canonical bases of $\bf A$. Let $i=1$. Since $\M(g_1,a)\in {\mathbb K}[[\M({g'_1,a}),\ldots, \M(g'_t,a)]]$, then  $\M(g_1,a)=\M(g'_1,a)^{l_1}\cdots{\M(g'_t,a)}^{l_t}$ for some $l_1,\ldots,l_t\in {\mathbb N}$. Every $\M(g'_i,a), i\in\lbrace 1,\ldots,t\rbrace$ is in ${\mathbb K}[\M(g_1,a),\ldots,\M(g_r,a)]$. Then the equation above is possible only if $\M(g_1,a)=\M(g'_{k_1},a)$ for some $k_1\in\lbrace 1,\ldots,t\rbrace$. This gives an injective map from $\lbrace \M(g_1,a),\ldots,\M(g_r,a)\rbrace$ to $\lbrace \M(g'_1,a),\ldots,\M(g'_t,a)\rbrace$. We construct in the same way an injective map from $\lbrace \M(g'_1,a),\ldots,\M(g'_t,a)\rbrace$ to  $\lbrace \M(g_1,a),\ldots,\M(g_r,a)\rbrace$. Hence $r=t$ and both sets are equal. Suppose, without loss of generality that $\M(g_i,a)=\M(g'_i,a)$ for all $i\in\lbrace 1,\ldots,r\rbrace$. If $g_i\not= g'_i$ then $\M(g_i-g'_i)\in \M({\bf A},a)$ because $g_i-g'_i\in{\bf A}$. This contradicts iii).
\end{proof}

\medskip

 \noindent We now recall the division process in ${\bf A}$ (see \cite{m} for the tangent cone order $a=(-1,\ldots,-1)$ and \cite{agm} for $n=1$).

\medskip

\begin{teorema}\label{division} Let $a\in U_1^n$ and let $\lbrace F_1,\ldots,F_s\rbrace\subseteq {\mathbb K}[[\underline{x}]]$. Let $F$ be a nonzero element of
${\mathbb K}[[\underline{x}]]$. There exist  $H\in {\mathbb K}[[F_1,\ldots,F_s]]$ and $R\in {\CC}$ such that the following conditions hold:

\begin{enumerate}

 \item $F=H+R$

\item If $R=\sum_{\beta}b_{\beta}{\underline{x}}^{\beta}$, then for all $\alpha\in{\rm Supp}(R), \underline{x}^{\beta}\notin{\mathbb{K}}[[\M(F_1,a),\ldots,\M(F_s,a)]]$.

\item Set $H=\sum_{\alpha}c_{\alpha}F_1^{\alpha_1}.\ldots.F_s^{\alpha_s}$. If $H\not=0$ then
$\e(F,a)={\rm max}_{<_a}\lbrace \e(F_1^{\alpha_1}.\ldots.F_s^{\alpha_s},a),c_{\alpha}\not=0\rbrace$.

\end{enumerate}

\end{teorema}

\begin{proof} We define the sequences
$(F^k)_{k\geq 0},(h^k)_{k\geq 0}$, $(r^k)_{k\geq 0}$ in $\CC$ by $F^0=F, h^0=r^0=0$ and $\forall k\geq
0$~:

\medskip

{\it (i)} If $\M(F^k,a)\in {\mathbb K}[[\M(F_1,a),\ldots,\M(F_s,a)]]$, write
$\M(F^k,a)=c_{\alpha}\M(F_1,a)^{\alpha_1}\cdots\M(F_s,a)^{\alpha_s}$. We set

$$
F^{k+1}=F^k- c_{\alpha}F_1^{\alpha_1}\cdots F_s^{\alpha_s},\quad
h^{k+1}=h^k+c_{\alpha}F_1^{\alpha_1}\cdots F_s^{\alpha_s}, \quad r^{k+1}=r^k
$$

{\it (ii)} If $\M(F^k,a)\notin {\mathbb K}[[\M(F_1,a),\ldots,\M(F_s,a)]]$, we set

$$
F^{k+1}=F^k-{\mathrm M}(F^{k+1},a),\quad
h^{k+1}=h^k,\quad r^{k+1}=r^k+{\mathrm M}(F^k,a)
$$

\noindent in such a way that for all $k\geq 0$, ${\mathrm exp}(F^k,a)<_a {\mathrm exp}(F^{k+1},a)$ and $F=F^{k+1}+h^{k+1}+r^{k+1}$. If $F^l=0$ for some $l\geq 1$ then we set $H=h^l$ and $R=r^l$. We then easily verify that $H,R$ satisfy conditions (1) to (3). Suppose that $\lbrace F^{k}| k\geq 0\rbrace$ is an infinite set. Note that, by Lemma \ref{Dixon}, given $k\geq 1$, if $F^k\not=0$ then there exists $k_1 >k$ such that $\nu(F^k,a)<\nu(F^{k_1},a)$. Hence,  there exists a subsequence $(F^{j_l})_{l\geq 1}$ such that $\nu(F^{j_1},a)<\nu(F^{j_2},a)<\cdots$ In particular, if we set $G=\lim_{k\to+\infty}F^k, H=\lim_{k\to+\infty}h^k$, and  $R=\lim_{k\to+\infty}r^k$, then $G=0$, $F=H+R$, and $H,R$ satisfy conditions (1) to (3). This completes the proof.
\end{proof}

\begin{definicion} We call the polynomial $R$ of Theorem \ref{division} the $a$-remainder of the division of $F$ with respect to $\lbrace F_1,\ldots,F_s\rbrace$ and we
denote it by $R=R_a(F,\lbrace F_1,\ldots,F_s\rbrace)$.
\end{definicion}

\noindent Suppose that $\lbrace f_1,\ldots,f_s\rbrace$ is an $a$-canonical
basis of $\bf A$. If
M$(f_i,a)\in {\mathbb K}[[({\rm M}(f_j,a)|j\not= i)]]$ for some $1\leq i\leq s$, then obviously
$\lbrace f_j|j\not=i\rbrace$ is also an $a$-canonical basis of $\bf A$, consequently we can get this way a minimal $a$-canonical basis of ${\bf A}$. Assume  that
$\lbrace f_1,\ldots,f_s\rbrace$ is minimal and let $1\leq i\leq s$. If $a\in U_1^n$ then,
dividing  $f=f_i-\M(f_i,a)$ by $\lbrace f_1,\ldots,f_s\rbrace$, and
replacing $f_i$ by $\M(f_i,a)+R_a(f_i-\M(f_i,a),\lbrace f_1,\ldots,f_r\rbrace)$, we obtain  an $a$-reduced canonical basis of ${\bf A}$.

\medskip

\noindent The next proposition gives a criterion for a finite set of ${\bf A}$ to be an $a$-canonical basis of ${\bf A}$.

\begin{proposicion}\label{criterion} Let $a\in U_1^n$. The set $\lbrace f_1,\ldots,f_s\rbrace\subseteq {\bf A}$ is an $a$-canonical basis of $\bf A$ if and only if
$R_a(f,\lbrace f_1,\ldots,$\\$f_s\rbrace)=0$ for all $f\in {\bf A}$.
\end{proposicion}

\begin{proof} Suppose that  $\lbrace f_1,\ldots,f_s\rbrace$ is an $a$-canonical basis of $\bf A$ and let $f\in{\bf A}$. Let $R=R_a(f,\lbrace f_1,\ldots$\\$,f_s\rbrace)$. If $R\not=0$ then $\M(R,a)\notin \M({\bf A},a)$. This is a contradiction because $R\in{\bf A}$. Conversely, suppose that $R_a(f,\lbrace f_1,\ldots,f_s\rbrace)=0$ for all $f\in {\bf A}$ and let $f\in{\bf A}$. If $\M(f,a)\notin {\mathbb K}[[\M(f_1,a),\ldots,\M(f_s,a)]]$ then $\M(f,a)$ is  a monomial of $R_a(f,\lbrace f_1,\ldots,f_s\rbrace)$, which is $0$. This is a contradiction.
\end{proof}

\noindent The criterion given in Proposition \ref{criterion} is not effective since we have to divide infinitely many  elements of ${\CC}$. In the following we shall see that it is enough to divide a finite number of elements.

\medskip

\noindent Let $\phi:{\mathbb K}[X_1,\ldots,X_s]\longmapsto {\mathbb K}[\M(f_1,a),\ldots,\M(f_s,a)]$ be the morphism of rings defined by  $\phi(X_i)=\M(f_i,a)$ for all $1\leq i\leq s$. We have the following

\begin{lema} \label{binomial}The ideal Ker$(\phi)$ is a binomial ideal, i.e., it can be generated by binomials.

\end{lema}

\begin{proof} Suppose that $f_i$ is monic for all $i\in\lbrace 1,\ldots,s\rbrace$ and write $\M(f_i,a)=x_1^{\theta_1^i}\ldots x_n^{\theta_n^i}=\underline{x}^{\theta^i}$.
\medskip

\noindent Let $F$ be a polynomial of ${\mathrm Ker}(\phi)$ and write $F=M_1+\ldots+M_p$ where $M_i$ is a monomial for all $i\in \lbrace 1,\ldots,p\rbrace$. We shall prove by induction on $p$, that $F$ is a finite sum of binomials, each of them is in ${\mathrm Ker}(\phi)$. Write $M_i=b_iX_1^{\beta_1^i}\cdots X_s^{\beta_s^i}$. If $p=2$ then $F$ is a binomial. Suppose that $p\geq 3$. We have
$\phi(M_1)=b_1{\M(f_1,a)}^{\beta_1^1}\cdots {\M(f_s,a)}^{\beta_s^1}$, which is a monomial in $x_1,\ldots,x_n$. Write $\phi(M_1)=b_1x_1^{\theta_1}\ldots x_n^{\theta_n}$. Since $\phi(F)=0$ then $\phi(M_i)=b_ix_1^{\theta_1}\ldots x_n^{\theta_n}$ for some $i\in\lbrace 1,\ldots,p\rbrace$. whence $\phi (M_1-\dfrac{b_1}{b_i}M_i)=0$. Write

$$
F=M_1-\dfrac{b_1}{b_i}M_i+(b_i+\dfrac{b_1}{b_i})M_i+\sum_{j\not=1,i}M_j
$$

\noindent If $F_1=(b_i+\dfrac{b_1}{b_i})M_i+\sum_{j\not=1,i}M_j$ then the cardinality of monomials of $F_1$ is at most $p-1$. By induction, $F_1$ is a sum of binomials, each of them is in Ker$(\phi)$. Consequently the same holds for $F$.
\end{proof}

\noindent Let $\bar{S}_1,\ldots,\bar{S}_m$ be a system of generators of Ker$(\phi)$, and assume, by Lemma \ref{binomial}, that $\bar{S}_1,\ldots,\bar{S}_m$ are binomials in ${\mathbb K}[X_1,\ldots,X_s]$. Assume that $f_1,\ldots,f_s$ are monic with respect to $<_a$. For all $1\leq i\leq m$, we can write $S_i(X_1,\ldots,X_s)=X_1^{\alpha_1^i}\ldots X_s^{\alpha_s^i}- X_1^{\beta_1^i}\ldots X_s^{\beta_s^i}$. Let $S_i=\bar{S}_i(f_1,\ldots,f_s)$. We have
$S_i=f_1^{\alpha_1}\cdots f_s^{\alpha_s}-f_1^{\beta_1}\cdots f_s^{\beta_s}
$, and  ${\mathrm exp}(S_i)>_a {\mathrm exp}(f_1^{\alpha_1}\cdots f_s^{\alpha_s})={\mathrm exp}(f_1^{\beta_1}\cdots f_s^{\beta_s})$

\begin{proposicion}\label{S-polynomials} Let $a\in U_1^n$. With the notations above, the following conditions are equivalent:

\begin{enumerate}
    \item The set $\lbrace f_1,\ldots,f_s\rbrace$ is an $a$-canonical basis of ${\bf A}$.

    \item For all $i\in\lbrace 1,\ldots, m\rbrace, R_a(S_i,\lbrace f_1,\ldots,f_s\rbrace)=0$.
\end{enumerate}

\end{proposicion}

\begin{proof} (1) implies (2) by Proposition \ref{criterion}.

\medskip

(2) $\Longrightarrow$ (1): We shall prove that $R_a(f,\lbrace f_1,\ldots,f_s\rbrace)=0$ for all $f\in{\bf A}$. Let $f$ be a nonzero element of ${\bf A}$ and let $R=R_a(f,\lbrace f_1,\ldots,f_s\rbrace)$. Then $R\in{\bf A}$. If $R\not=0$ then $\M(R,a)\in \M({\bf A},a)$. Write

$$
R=\sum_{\theta}c_{\theta}f_1^{\theta_1}\cdots f_s^{\theta_s}
$$

\noindent and let $\alpha={\mathrm inf}_{\theta,c_{\theta}\not=0}({\mathrm exp}(f_1^{\theta_1}\cdots f_s^{\theta_s},a))$. Since ${\mathrm exp}(R,a)\notin {\mathrm exp}({\bf A})=\langle {\mathrm exp}(f_1,a),\ldots, {\mathrm exp}(f_s,a)\rangle$ then ${\mathrm exp}(R,a)>_a \alpha$. Let $\lbrace {\theta}^1,\ldots,{\theta}^l\rbrace$ be such that $\alpha= {\rm exp}(f_1^{\theta^i_1}\cdots  f_s^{\theta^i_s},a)$ for all
 $i\in\{1,\ldots, l\}$ (such a set is clearly finite). If $\sum_{i=1}^lc_{{\theta}^i}\M(f_1^{\theta^i_1}\cdots  f_s^{\theta^i_s},a)\not=0$, then ${\mathrm exp}(R,a)\in \langle {\mathrm exp}(f_1,a),\ldots,{\mathrm exp}(f_s,a) \rangle$, which is a contradiction. Hence,  $\sum_{i=1}^lc_{{\theta}^i}\M(f_1^{\theta^i_1}\cdots  f_s^{\theta^i_s},a)=0$, and consequently
 $\sum_{i=1}^lc_{\underline{\theta}^i}X_1^{\theta^i_1}\cdots  X_s^{\theta^i_s}$ is an element of  $\ker(\phi)$. In particular

 $$
 \sum_{i=1}^lc_{\underline{\theta}^i}X_1^{\theta^i_1}\cdots  X_s^{\theta^i_s}=\sum_{k=1}^m\lambda_k\bar{S}_k
 $$

\noindent  with $\lambda_k\in\KK[X_1,\ldots,X_s]$ for all $k\in\{1,\ldots,m\}$. Whence

 $$
 \sum_{i=1}^lc_{\underline{\theta}^i}f_1^{\theta^i_1}\cdots  f_s^{\theta^i_s}=\sum_{k=1}^m\lambda_k(f_1,\ldots,f_s)S_k.
 $$

\noindent From the hypothesis $\mathrm R_a(S_k,\lbrace f_1,\ldots,f_s\rbrace)=0$ for all $k\in\lbrace 1,\ldots,m\rbrace$. Hence there is an expression of $S_k$ of the form $S_k=\sum_{\underline{\beta}^k}c_{\underline{\beta}^k} f_1^{\beta^k_1}\cdots f_s^{\beta^k_s}$ with ${\mathrm exp}(f_1^{\beta^k_1}\cdots f_s^{\beta^k_s},a)>_a {\mathrm exp}(S_k,a)$. Replacing  $\sum_{i=1}^lc_{\underline{\theta}^i}f_1^{\theta^i_1}\cdots  f_s^{\theta^i_s}$ by $\sum_{k=1}^m\lambda_k(f_1,\ldots,f_s)\sum_{\underline{\beta}^k}c_{\underline{\beta}^k} f_1^{\beta^k_1}\cdots f_s^{\beta^k_s}$ in the expression of $R$, we can rewrite $R$ as

$$
R=\sum_{\underline{\theta}'}c_{\underline{\theta}'}f_1^{\theta'_1}\cdots  f_s^{\theta'_s}
$$

\noindent with $\alpha_1={\mathrm inf}_{\theta', c_{\theta'}\not=0}{\mathrm exp}(f_1^{\theta'_1}\cdots  f_s^{\theta'_s},a)) >_a\alpha$. Then we restart with this representation. We construct this way an infinite sequence ${\rm exp}(R,a)>_a\ldots >_a \alpha_1>_a \alpha $, which contradicts Lemma \ref{Dixon2}.
\end{proof}

\medskip

\noindent The characterization given in Proposition \ref{S-polynomials} suggests an algorithm that construct, starting with a set of generators of ${\bf A}$, an $a$-canonical basis of ${\bf A}$. However, such a canonical basis can be infinite as it is shown in the following example:

\begin{example} (see \cite{rs})\label{notfinite} Let ${\bf A}={\mathbb K}[[x+y,xy,xy^2]]$ and let $a=(2,1)$. Then $\M(x+y,a)=x, \M(xy,a)=xy$, and $\M(xy^2,a)=xy^2$. The kernel of the map:

$$
\phi:{\mathbb K}[X,Y,Z]\longmapsto {\mathbb K}[x,y], \phi(X)=x,\phi(Y)=xy,\phi(Z)=xy^2
$$

\noindent is generated by $\bar{S_1}=XZ-Y^2$. Hence $S=(x+y)xy^2-x^2y^2=-xy^3=R_a(-xy^3,\lbrace x+y,xy,xy^2\rbrace)$. Then $xy^3$ is a new element of the $a$-canonical basis of ${\bf A}$. If we restart with the representation ${\bf A}={\mathbb K}[[x+y,xy,xy^2,xy^3]]$, then a new element, $xy^4$, will be added to the $a$-canonical basis of ${\bf A}$. In fact, $xy^n$ is an element of the minimal reduced $a$-canonical basis of ${\bf A}$ for all $n\geq 1$. In particular the $a$-canonical basis of ${\bf A}$ is infinite.

\end{example}

\noindent In the following we shall assume that the length $l(\dfrac{\CC}{\bf A})$ is finite. This guarantees the finiteness of a canonical basis. Under this hypothesis, using the results above, we get the following algorithm:

\medskip

\noindent {\bf Algorithm}. Let ${\bf A}={\mathbb K}[[f_1,\ldots,f_s]]$ and let $a\in U_1^n$. Let $\lbrace \bar{S}_1,\ldots,\bar{S}_m\rbrace$
be a set of generators of the map $\phi$ of Proposition \ref{binomial} and let $S_i=\bar{S}_i(f_1,\ldots,f_s)$ for all $i\in\lbrace 1,\ldots,m\rbrace$.

\medskip

\begin{enumerate}
    \item If $R_a(S_i,\lbrace f_1,\ldots,f_s\rbrace)=0$ for all $i\in\lbrace 1,\ldots,m\rbrace$ then $\lbrace f_1,\ldots,f_s\rbrace$ is an $a$-canonical basis of ${\bf A}$.

    \item If $R=R_a(S_i,\lbrace f_1,\ldots,f_s\rbrace)\not=0$ for some $i\in\lbrace 1,\ldots,m\rbrace$ then we set $R=f_{s+1}$ and we restart with $\lbrace f_1,\ldots,f_r,f_{r+1}\rbrace$. Note that in this case, we have $\langle {\mathrm exp}(f_1,a),\ldots,{\mathrm exp}(f_s,a)\rangle \subset \langle {\mathrm exp}(f_1,a),\ldots,{\mathrm exp}(f_s,a),{\mathrm exp}(f_{s+1},a)\rangle \subseteq {\mathrm exp}({\bf A},a)$. By hypothesis, $l(\dfrac{\CC}{\bf A})<+\infty$, hence, after a finite number of operations, we get an $a$-canonical basis of ${\bf A}$.
\end{enumerate}

\section{A finiteness Theorem}

\noindent Let the notations be as in Section 1. In particular ${\bf A}={\mathbb K}[[f_1,\ldots,f_s]]$ with $\lbrace f_1,\ldots,f_s\rbrace \subseteq \CC$. We shall assume that $l(\dfrac{\bf F}{\bf A})<+\infty$. The aim of this section is to prove that the set of ${\mathrm in}(A,a), a\in U^n$ is finite. We first recall this result when ${\bf A}={\mathbb K}[f]$ then we
prove some preliminary results which will  also be used later in the paper.

\begin{lema}\label{onegenerator} Let $f$ be a nonzero element of ${\mathbb K}[[x_1,\ldots,x_n]]$. The set $\lbrace {\mathrm M}(f,a),a\in U^n\rbrace$
 (resp. $\lbrace {\mathrm in}(f,a),a\in U^n\rbrace$) is finite.
\end{lema}

\begin{proof} Write $f=\sum_{\alpha} c_{\alpha}{\underline{x}}^{\alpha}$ and let $E=\cup_{\alpha\in {\mathrm Supp}(f)}\alpha+{\mathbb N}^n$. Then $E+{\mathbb N}^n\subseteq E$, and consequently there exists a finite set of $E$, say $\lbrace \alpha_1,\ldots,\alpha_r\rbrace$, such that $E=\cup_{i=1}^r \alpha_i+{\mathbb N}^n$. By definition $\alpha_i\in {\mathrm Supp}(f)$ for all $i\in\lbrace 1,\ldots,r\rbrace$. If we choose the $\alpha_i$'s such that $\alpha_i\notin \cup_{j\not=i}\alpha_j+{\mathbb N}^n$, then $\lbrace {\mathrm exp}(A,a), a\in U^n\rbrace=\lbrace \alpha_1,\ldots,\alpha_r\rbrace$. This proves our assertion.

\end{proof}

\begin{lema} Let $a\not=b$ be two elements of $U^n$. Let $\lbrace g_1,\ldots,g_r\rbrace$ be an $a$-reduced canonical basis of ${\bf A}$. If $\M({\bf A}, a)=\M({\bf A},b)$ then
$\lbrace g_1,\ldots,g_r\rbrace$ is also a $b$-reduced canonical basis of ${\bf A}$.
\end{lema}

\begin{proof} Let $i\in\lbrace 1,\ldots,r\rbrace$ and write $g_i=\M(g_i,a)+\sum c_{\beta}\underline{x}^{\beta}$ where for all $\beta$, if $c_{\beta}\not=0$, then $\underline{x}^{\beta}\notin \M({\bf A},a)$. Since $\M({\bf A}, a)=\M({\bf A},b)$ then for all $\beta$, if $c_{\beta}\not=0$, then $\underline{x}^{\beta}\notin \M({\bf A},a)$. This implies that $\M(g_i,a)=\M(g_i,b)$. Hence $\lbrace g_1,\ldots,g_r\rbrace$ is a $b$-canonical basis of ${\bf A}$, and the same argument shows that this basis is also reduced.
\end{proof}

\begin{lema} \label{expequal}Let $a\not=b$ be two elements of $U^n$. If $\M({\bf A},a)\not=\M({\bf A},b)$, then  $\M({\bf A},a)\not\subseteq \M({\bf A},b)$.
\end{lema}

\begin{proof} Assume that $\M({\bf A},a)\subseteq \M({\bf A},b)$ and let
 $\lbrace g_1,\ldots,g_r\rbrace$ be
a $b$-reduced canonical basis of $\bf A$. By hypothesis, there is
$1\leq i\leq r$ such that  $\M(g_i,b)\in \M({\bf A},b)\setminus\M({\bf A},a)$.  Write $g_i=\M(g_i,b)+\sum c_{\beta}\underline{x}^{\beta}$.
For all $\beta$, if $c_{\beta}\not=0$, then $\underline{x}^{\beta}\notin \M({\bf A},b)$, hence $\underline{x}^{\beta}\notin \M({\bf A},a)$, which implies that $\M(g_i,a)\notin\M({\bf A},a)$ ($\M({\bf A},a)\subseteq \M({\bf A},b)$). This is a contradiction because $g_i\in \bf A$.
\end{proof}

\begin{corolario}\label{rational} If $a\in U^n\setminus U_1^n$ then there exists $b\in U_1^n$ such that $\M({\bf A},a)=\M({\bf A},b)$.
\end{corolario}

\begin{proof} Let
 $\lbrace g_1,\ldots,g_r\rbrace$ be
an $a$-reduced canonical basis of $\bf A$. By hypothesis, there exists $\epsilon>0$ such that for all $i\in \lbrace 1,\ldots,r\rbrace$ and for all $b\in B(a,\epsilon)$, $\M(g_i,a)=\M(g_i,b)$ (where $B(a,\epsilon)$ denotes the ball of ray $\epsilon$ centered at $a$). Take $b\in U^n\cap B(a,\epsilon)$. We have $\M({\bf A},a)\subseteq \M({\bf A},b)$. By Lemma \ref{expequal}, $\M({\bf A},a)=\M({\bf A},b)$
\end{proof}

\begin{corolario}\label{equalin} Let $a\not=b$ be two elements of $U^n$. If in$({\bf A},a)\not={\mathrm in}({\bf A},b)$, then  ${\mathrm in}({\bf A},a)\not\subseteq {\mathrm in}({\bf A},b)$

\end{corolario}

\begin{proof} We shall prove that if ${\mathrm in}({\bf A},a)\subseteq {\mathrm in}({\bf A},b)$ then ${\mathrm in}({\bf A},a)={\mathrm in}({\bf A},b)$. Let to this end $\lbrace g_1,\ldots,g_r\rbrace$ be an $a$-reduced canonical basis of ${\bf A}$ and let $i\in\lbrace 1,\ldots,r\rbrace$. Write ${\mathrm in}(g_i,a)=M_1+\ldots+M_t$ where $M_j$ is $b$-homogeneous for all $i\in\lbrace 1,\ldots, t\rbrace$. By hypothesis ${\rm in}(g_i,a)\in{\mathrm in}({\bf A},b)$,  hence $M_j\in {\mathrm in}({\bf A},b)$ for all $j\in\lbrace 1,\ldots,t\rbrace$. Suppose that $\M(g_i,a)$ is a monomial of $M_1$. We have $M_1\in{\mathrm in}({\bf A},b)$. But $M_1$ is also $a$-homogeneous. It follows that $\M(M_1,b)=\M(M_1,a)=\M(g_i,a)$, in particular $\M(g_i,a)\in \M({\bf A},b)$. This proves that $\M({\bf A},a)\subseteq \M({\bf A},b)$. By Lemma \ref{expequal} we have $\M({\bf A},a)=\M({\bf A},b)$ and $\lbrace g_1,\ldots,g_r\rbrace$ is also a $b$-reduced canonical basis of ${\bf A}$. Finally in$({\bf A},a)$ is generated by $\lbrace {\mathrm in}(g_1,a),\ldots,{\mathrm in}(g_s,a)\rbrace$ (resp. in$({\bf A},b)$ is generated by $\lbrace {\mathrm in}(g_1,b),\ldots,{\mathrm in}(g_r,b)\rbrace$). Now the argument above shows that for all $i\in\lbrace 1,\ldots,r\rbrace$, ${\rm in}(g_i,a)={\rm in}(g_i,b)$. This proves the equality.

\end{proof}

\begin{nota} \label{rationalin} 1. \label{in=exp}Let $a\not=b$ be two elements of $U^n$. The proof of Corollary \ref{equalin} implies that if ${\mathrm in}({\bf A},a)={\mathrm in}({\bf A},b)$ and if $\lbrace g_1,\ldots,g_r\rbrace$ is an $a$-reduced canonical basis of ${\bf A}$ then $\lbrace g_1,\ldots,g_r\rbrace$ is also a $b$-reduced canonical basis of ${\bf A}$.
\medskip

\noindent 2. Corollary \ref{equalin} implies the following: if $a\in U^n\setminus U_1^n$ then there exists $b\in U_1^n$ such that ${\mathrm in}{\bf A},a)=\M({\bf A},a)=\M({\bf A},b)={\mathrm in}({\bf A},b)$..

\end{nota}

\noindent We can now state and prove the following finiteness theorem:

\begin{teorema}\label{finiteness} Let $A={\mathbb K}[[f_1,\ldots,f_s]]$ and let the notations be as above.  The set
$\M(A)=\lbrace \M(A;a)| a\in U^n\rbrace$ is finite. In particular
the set $I(A)=\lbrace {\mathrm in}({\bf A},a)| a\in U^n\rbrace$ is finite.
 \end{teorema}

\begin{proof} We need only to prove that $\M(A)$ is a finite set. Assume that $M$ is infinite. By Lemma \ref{onegenerator} there is an infinite set $U_1=\lbrace a_1,a_2,\ldots\rbrace$ in
$U^n$ such that
for all $1\leq k\leq s$ and for all $a\in U_1$, M$(f_k,a)=m_k$ where $m_k$ is a nonzero monomial of $f_k$. Let $J_1={\mathbb K}[[m_1,\ldots,m_s]]$: $J_1\subseteq \M({\bf A},a)$
for all $a\in U_1$. Obviously $J_1\not= \M({\bf A},a_1)$ (otherwise,
$\M({\bf A},a_1)\subset \M({\bf A},a_2)$,
for example. This contradicts Lemme \ref{expequal}).  We claim that there is $f_{s+1}\in {\bf A}$ such that for all $\beta\in{\rm Supp}(f_{s+1}), \underline{x}^{\beta}\notin J_1$. Let to this end $m\in \M({\bf A},a_1)\setminus J_1$ and let $f\in {\bf A}$ such that $\M(f,a_1)=m$. Obviously
$m\notin J_1$. We set $f_{s+1}=R_{a_1}(f,\lbrace f_1,\ldots,f_s\rbrace)$.  By lemma \ref{onegenerator}, there is monomial  $m_{s+1}$ of $f_{s+1}$ and an infinite subset $U_2\subseteq U_1$ such that for
all $a\in U_2, \M(f_{s+1},a)=m_{s+1}$.

\vskip0.2cm

\noindent Let $J_2={\mathbb K}[[m_1,\ldots,m_s,m_{s+1}]]$:
$J_1\subset J_2$. The same process applied
to $\lbrace f_1,\ldots,f_{s+1}\rbrace, J_2$ and
$U_2$ will construct $m_{s+2}\notin J_2$,
$f_{s+2}\in \bf A$, and an infinite subset $U_3\subseteq U_2$
such that for
all $a\in U_3, \M(f_{s+2},a)=m_{s+2}$. We get this way an
infinite increasing sequence
$J_1\subset J_2\subset J_3\subset \ldots$
and for all $i$, there is $a_i\in U^n$ such that $J_i\subseteq \M({\bf A},a_i)$. This is a contradiction because $l(\dfrac{\bf F}{\bf A})$ is finite.
\end{proof}

\begin{definicion} The set $\lbrace g_1,\ldots,g_r\rbrace$ of ${\bf A}$ which is an $a$- canonical basis of ${\bf A}$ for all $a\in U^n$, is called the universal canonical basis of ${\bf A}$.

\end{definicion}

\section{The Newton fan}

\medskip

\noindent Let ${\bf A}={\mathbb K}[[f_1,\ldots,f_s]]$ and let the notations be as in Section 2. In this section we aim to study the stabitily of ${\mathrm exp}({\bf A},a)$ and ${\mathrm in}({\bf A},a)$ when $a$ very in $U^n$. Let $S$ be a finitely generated affine semigroup of ${\mathbb N}^n$. Let

$$
E_S=\lbrace a\in U^n\mid {\mathrm exp}({\bf A},a)=S\rbrace
$$

\noindent We have the following:

\begin{teorema}\label{fan} There exists a partition $\mathcal{P}$ of $U^n$ into convex rational polyherdal cones such that for all $\sigma\in\mathcal{P}$, exp$({\bf A},a)$ and in$({\bf A},a)$ do not depend on $a\in \sigma$.

\end{teorema}

\noindent In order to prove Theorem \ref{fan} we start by fixing some notations. Let $S$ be a finitely generated affine semigroup of ${\mathbb N}^n$ and let $a\in E_S$. Let $\lbrace g_1,\ldots,g_r\rbrace$ be the $a$-reduced canonical basis of ${\bf A}$. By Lemma \ref{expequal}, Lemma \ref{equalin}, and Remark \ref{in=exp}, $\lbrace g_1,\ldots,g_r\rbrace$ is also the $b$-reduced canonical basis of ${\bf A}$ for all $b\in E_S$. Denote by $\sim$ the equivalence relation on $U^n$ defined from $\lbrace g_1,\ldots,g_r\rbrace$ by

$$
a\sim b \Longleftrightarrow  {\mathrm in}(g_i,a)={\mathrm in}(g_i,b) \quad \text {for  all}\quad i\in\lbrace 1,\ldots,r\rbrace,
$$

\begin{proposicion} $\sim$ defines on $U^n$ a finite partition into convex rational polyhedral cones
and $E_S$ is a union of a part of these cones.

\end{proposicion}

\begin{proof} Let $c,d\in U^n$ such that $c\sim d$ and let $e$ in the segment $[c, d]$, also let
$\theta \in [0, 1]$ such that $e= \theta c + (1-\theta))d$. Then $a\in U^n$ and  in$(g_i,e)={\mathrm in}(g_i,c)={\mathrm in}(g_i,d)$
by an immediate verification. Moreover, $c\sim t\cdot c$ for all $c\in U^n$ and $t>0$. Therefore the equivalence classes are convex rational polyhedral cones (the rationality results from Corollary \ref{rational} and Remark \ref{rationalin}). On the other hand, if $c\sim d$ and $c\in E_S$, then $d\in E_S$. This proves that $E_S$ is a union of classes for $\sim$, the number of classes
being finite by Theorem \ref{finiteness}
\end{proof}

\noindent {\bf Proof of Theorem \ref{fan}} We define $\mathcal{P}$ in the following way: for each $S$ we consider the restriction ${\mathcal {P}}_S$ on $E_S$ of the above partition, and then ${\mathcal {P}}$ is the finite union of the $\mathcal{P}_S$'s. On each cone of the partition, ${\mathrm in}({\bf A},a)$ and ${\mathrm exp}({\bf A},a)$ are fixed. Conversely assume that ${\mathrm in}({\bf A},a)$ is fixed and let $b$ such that ${\mathrm in}({\bf A},a)={\mathrm in}({\bf A},b)$. By Corollary \ref{equalin} and Remark \ref{in=exp}, an $a$-reduced canonical basis $\lbrace g_1,\ldots,g_r\rbrace$ of ${\bf A}$ is also a $b$-reduced canonical basis of ${\bf A}$. Moreover,  ${\mathrm in}(g_i,a)={\mathrm in}(g_i,b)$ and ${\mathrm exp}({\bf A},a)= {\mathrm exp}({\bf A},b)$. This ends the proof of the theorem except for the convexity of $E_S$ proved below.

\begin{lema} $E_S$ is a convex set: if $a\not=b\in E_S$ then $[a,b]\subseteq E_S$
\end{lema}
\begin{proof} Let $a,b\in E_S$ and let $\lambda\in]0,1[$. Let $\lbrace g_1,\ldots,g_r\rbrace$ be an $a$ (and then $b$) reduced canonical basis of ${\bf A}$. Let $i\in\lbrace 1,\ldots,r\rbrace$ and set $M={\rm in}(g_i,a)$. Write $M=M_1+\ldots+M_t$ where $M_k$ is $b$-homogeneous for all $k\in\lbrace 1,\ldots,t\rbrace$ and $\nu(M_1,b)>\nu(M_k,b)$ for all $k\in\lbrace 2,\ldots,t\rbrace$. We have $\nu(g_i,a)=\nu(M_1,a)=\nu(M_k,a)$ and  $\nu(M_1,b)>\nu(M_k,b)$ for all $k\in\lbrace 2,\ldots,t\rbrace$. This implies that $\nu(g_i,\theta a+(1-\theta)b)=\nu(M_1,\theta a+(1-\theta)b)>\nu(M_k,\theta a+(1-\theta)b))$ for all $k\in\lbrace 2,\ldots,t\rbrace$, hence ${\mathrm exp}(g_i,\theta a+(1-\theta)b)={\mathrm exp}(g_i,a)={\rm exp}(g_i,b)$. In particular ${\mathrm exp}({\bf A},a)\subseteq {\mathrm exp}({\bf A},  \theta a+(1-\theta)b)$. By Lemma \ref{expequal} we get the equality. This proves that ${\mathrm exp}({\bf A},  \theta a+(1-\theta)b)=S$.
\end{proof}

\noindent In the following we shall give some precisions about the partition above. Obviously if $S_1$ and $S_2$ are two distinct finitely generated affine semigroups and if $a\in E_{S_1}$ and $b\in E_{S_2}$ then ${\mathrm in}({\bf A},a)\not= {\mathrm in}({\bf A},b)$ and ${\mathrm exp}({\bf A},a)\not={\mathrm exp}({\bf A},b)$, hence, by Lemma \ref{expequal} and Corollary \ref{equalin}, neither ${\mathrm in}({\bf A},a)\subseteq {\mathrm in}({\bf A},b)$ nor ${\mathrm in}({\bf A},b)\subseteq{\mathrm in}({\bf A},a)$, and the same conclusion holds for ${\mathrm exp}({\bf A},a)$ and ${\mathrm exp}({\bf A},b)$.

\medskip

\noindent  Next we shall characterize open cones of the partition with maximal dimension. Let $f$ be a nonzero element of ${\mathbb K}[[x_1,\ldots,x_n]]$ and let $a\in U^n$. We say that ${\mathrm in}(f,a)$ is multihomogeneous if ${\mathrm in}(f,a)$ is $b$-homogeneous for all $b\in U^n$. This is equivalent to saying that ${\mathrm in}(f,a)$ is a monomial.

\begin{definicion} \label{multi}Let $a\in U^n$. We say that ${\mathrm in}({\bf A},a)$ is a multihomogeneous algebra if it is generated by multihomogeneous elements. Note that in this case, If $g\in {\mathrm in}({\bf A},a)$ then every monomial of $g$ is also in ${\mathrm in}({\bf A},a)$.
\end{definicion}

\begin{lema} Let $a\in U^n$ and let $\lbrace g_1,\ldots,g_r\rbrace$ be an $a$-reduced canonical basis of ${\bf A}$. Then ${\mathrm in}({\bf A},a)$ is a multihomogeneous algebra if and only if  ${\mathrm in}(g_i,a)$ is a monomial for all $i\in\lbrace 1,\ldots,r\rbrace$.
\end{lema}

\begin{proof} We only need to prove the if part. Let $i\in\lbrace 1,\ldots,r\rbrace$ and write ${\mathrm in}(g_i,a)=M_1+\ldots+M_t$ with ${\mathrm exp}(g_i,a)={\mathrm exp}(M_1,a)$. Assume that $t>1$. Since ${\mathrm in}({\bf A},a)$ is multihomogeneous then $M_i\in {\mathrm in}({\bf A},a)$ for all $i\in\lbrace 2,\ldots,t\rbrace$. But $\lbrace g_1,\ldots,g_r\rbrace$ is reduced. This is a contradiction. Hence $t=1$ and ${\mathrm in}(g_i,a)$ is a monomial.
\end{proof}

\begin{proposicion} The set of $a\in U^n$ for which ${\mathrm in}({\bf A},a)$ is multihomogeneous defines the open cones of dimension $n$ of ${\mathcal{P}}$.

\end{proposicion}

\begin{proof} Let $a\in U^n$ and let $\lbrace g_1,\ldots,g_r\rbrace$ be an $a$-reduced canonical basis of ${\bf A}$. For all $i\in\lbrace 1,\ldots,r\rbrace$, ${\mathrm in}(g_i,a)$ is a monomial, hence there exists $\epsilon >0$ such that ${\mathrm in}(g_i,b)={\mathrm in}(g_i,a)$ for all $b\in B(a,\epsilon)$ (where $B(a,\epsilon)$ is the ball centered at $a$ of ray $\epsilon$). This proves, by Lemma \ref{expequal} and Corollary \ref{equalin},  that $\lbrace g_1,\ldots,g_r\rbrace$ is also a $b$-reduced canonical basis of ${\bf A}$ for all $b\in B(a,\epsilon)$. Conversely, if $a$ is in an open cone of $E_S$ for some $S$, then for all $b$ in a neighbourhood, an $a$-reduced canonical basis $\lbrace g_1,\ldots,g_r\rbrace$ of ${\bf A}$ is also a $b$-reduced canonical basis of ${\bf A}$. This implies that ${\mathrm in}(g_i,a)$ is a monomial. This proves our assertion.
\end{proof}

\begin{definicion} ${\mathcal P}$ is called the standard fan of ${\bf A}$.
\end{definicion}

\begin{nota} (1) Although we do not have a proof for the existence of a fan for subalgebras ${\bf A}$ with $l(\dfrac{\CC}{\bf A})$ not necessarily finite, we think that this fan does exist. This is true of course if ${\mathrm exp}({\bf A},a)$ is finitely generated for all $a\in U^n$, but it is not easy to verify this condition a priori.

\medskip

\noindent (2) (see \cite {agm}) Suppose that $n=1$, i.e. ${\bf A}={\mathbb K}[[f_(t),\ldots,f_s(t)]]\subseteq {\mathbb K}[[t]]$. Then $X_1=f_1(t),\ldots,X_s=X_s(t)$ represents the expansion of a curve in ${\mathbb K}^s$ near the origin. In tis case, ${\mathrm in}({\bf A},a)$ does not depend on $a\in U^n$ and if the parametrization is primitive then $l(\dfrac{{\mathbb K}[[t]]}{\bf A})<+\infty$. If $s=2$ then ${\mathrm in}({\bf A},a)$ is a free numerical semigroup and the arithmetic of this semigroup contains a lot of information about the singularity of the curve at the origin.

\medskip

\noindent (3)
(see also \cite{j}) If ${\bf A}={\mathbb K}[f_1,\ldots,f_s]$ is a subalgebra of ${\bf P}={\mathbb  K}[x_1,\ldots,x_n]$ then Theorem \ref{finiteness} and Theorem \ref{fan} remain valid when we vary $a\in {\mathbb R}^n_+$, under the assumption that $l(\dfrac{\bf P}{\bf A})<+\infty$ (note that in this case, ${\mathrm in}(f,a)$ is a polynomial for all $a\in{\mathbb R}^n_+$ and for all $f\in \DD$).
 \end{nota}

\begin{example} Let $f(X_1,\ldots,X_n,Y)\in{\mathbb K}[[X_1,\ldots,X_n]][Y]$ and suppose that $f$ has a parametrization of the form $X_1=t_1^{e_1},\ldots, X_n=t_n^{e_n}, Y=Y(t_1,\ldots,t_n)\in{\mathbb K}[[t_1,\ldots,t_n]]$ (for instance, this is true if $f$ is a quasi-ordinary polynomial, i.e. the discriminant of $f$ is of the form $X_1^{N_1}\cdots X_s^{N_s}(c+\phi(X_1,\ldots,X_s))$ with $c\in{\mathbb K}^*$ and $\phi(0,\ldots,0)=0$). Then $\dfrac{{\mathbb K}[[X_1,\ldots,X_n]][Y]}{f}\simeq {\bf A}={\mathbb K}[[t_1^{e_1},\ldots,t_n^{e_n},$\\$Y(t_1,\ldots,t_n)]]$. In this case, $(e_1,0,\ldots,0),\ldots,(0,\ldots,0,e_n)$ belong to ${\mathrm exp}({\bf A},a)$ for all $a\in U^n$. Moreover, ${\rm exp}({\bf A}, a)$ is is a free finitely generated affine semigroup in the sense of \cite{a}. In this case, $l(\dfrac{\CC}{\bf A})$ need not to be finite, but results of Theorem \ref{finiteness} and Theorem \ref{fan} are valid.
\end{example}


TITLE: Calculating the following integral $\int^{\infty}_{2}\frac{x-1}{x^3+4x+7}$
QUESTION [1 upvotes]: I'm trying to find the solution of the following integral, to check for convergence or divergence. Though, I'm finding the integral itself difficult to resolve. I welcome any feedback towards my approach.
$$\int^{\infty}_{2}\frac{x-1}{x^3+4x+7}$$
My approach:
$$\int^{\infty}_{2}\frac{x-1}{x^3+4x+7} \implies \frac{1}{2}\int^{\infty}_{2}\frac{2x-4}{x^3+4x+7}+\frac{1}{2}\int^{\infty}_{2}\frac{2}{x^3+4x+7}$$
Then to solve for the left-side first, I'm substiting u for $x^2-4x+7$, so the denominator looks like this: $xu$
Then to find x, such that: $x = \frac{u-7}{x-4}$
To get:
$$\frac{1}{2}\int^{\infty}_{2}\frac{du}{\frac{u-7}{x-4}\cdot u}$$
Though, I'm not sure if my apporach is valid, hence I would really enjoy some support towards my approach.

REPLY [3 votes]: Write $x^3+4x+7=(x-a)((x-b)^2+c^2)$ with $a\approx-1.25538,\,b\approx0.627692,\,c\approx2.2764$. Then$$\begin{align}\frac{x-1}{x^3+4x+7}&\equiv\frac{A}{x-a}+\frac{B(x-b)+C}{(x-b)^2+c^2}\\\iff x-b+b-1&\equiv A((x-b)^2+c^2)+(x-b+b-a)(B(x-b)+C)\\&=(A+B)(x-b)^2+((b-a)B+C)(x-b)+Ac^2+(b-a)C\\\iff A=-B&=\frac{a-1}{(b-a)^2},\\C&=\frac{b-1}{b-a}.\end{align}$$One antiderivative is $A\ln\tfrac{|x-a|}{\sqrt{(x-b)^2+c^2}}+\tfrac{C}{c}\arctan\tfrac{x-b}{c}$, so the integral is $A\ln\tfrac{\sqrt{(2-b)^2+c^2}}{|2-a|}+\tfrac{C}{c}\arctan\tfrac{c}{2-b}$.


TITLE: Order of eigenvalues for converting quadratic equations
QUESTION [3 upvotes]: In my textbook there is a way of converting a quadratic equation to its canonical form using eigenvalues. I'm going to give you a rough description of the method:
The quadratic equation is of the form:
$a_{11}x^2+a_{22}y^2+a_{33}z^2+2a_12xy+2a_{23}yz+2a_{31}xz+2a_{14}x+a_{24}y+a_{34}z+a_{44}=0$
Which can be express as following
$\begin{pmatrix}x & y & z\end{pmatrix}$$\begin{pmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{pmatrix}$$\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}$$+2(a_{14}+a_{24}+a_{34})$$\begin{pmatrix}
x\\
y\\
z
\end{pmatrix}$$+a_{44}=0$
Let $A=\begin{pmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{pmatrix}$
because A is symmetric there exists an orthogonal matrix $P$ such that
$P^{-1}AP=$ $\begin{pmatrix}
t_{1} & 0 & 0\\
0 & t_{2} & 0\\
0 & 0 & t_{3}
\end{pmatrix}$, $t_{1},t_{2},t_{3}$ the eigenvalues of $A$
Now let $B=PB'$, $B'=$$\begin{pmatrix}
x'\\
y'\\
z'
\end{pmatrix}$
then $\begin{pmatrix}x' & y' & z'\end{pmatrix}$$\begin{pmatrix}
t_{1} & 0 & 0\\
0 & t_{2} & 0\\
0 & 0 & t_{3}
\end{pmatrix}$$\begin{pmatrix}
x'\\
y'\\
z'
\end{pmatrix}$$+2(a_{14}+a_{24}+a_{34})P$$\begin{pmatrix}
x'\\
y'\\
z'
\end{pmatrix}$$+a_{44}=0$$\Leftrightarrow$
$t_{1}x'^2+t_2y'^2+t_3z'^2$$+2(a_{14}+a_{24}+a_{34})P$$\begin{pmatrix}
x'\\
y'\\
z'
\end{pmatrix}$$+a_{44}=0$
My question now is the following, is there a specific order which determines which eigenvalue is number 1,2 etc? Because if for example an eigenvalue is zero this means that one of the $x^2,y^2,z^2$ is zeroed as well. How do you determine what eigenvalue to asign to each of the $x^2,y^2,z^2$?

REPLY [1 votes]: The order of the eigenvalues is determined by how you want the figure to be oriented in space. To give you an example, consider a simple case: ellipse in 2D ($a_{33} = a_{23} = a_{31} = a_{34} = 0$) centered at the origin ($a_{14} = a_{24} = 0$ and $a_{44} = -1$) with major axis along the $x$ axis ($a_{11} = 1, a_{22} = 1 / b^2$, $a_{12} = 0$, $b<1$), that is
$$
x^2 + \frac{y^2}{b^2} = 1 \tag{1}
$$
In this case
$$
A = \begin{pmatrix} 1 & 0 \\ 0 & 1/b^2 \end{pmatrix} \tag{2}
$$
This is a trivial example in the sense that the matrix is already diagonal,
$$
P = \begin{pmatrix}1 & 0 \\ 0 & 1 \end{pmatrix}, ~~ T = \begin{pmatrix}1 & 0 \\ 0 & 1/b^2 \end{pmatrix}  \tag{3}
$$
But here's the idea, image that you decide to organize your eigenvalues as
$$
P_{\rm new} = \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}, ~~ T_{\rm new} = \begin{pmatrix}1/b^2 & 0 \\ 0 & 1 \end{pmatrix} \tag{4}
$$
In this case your coordinate system is defined as
$$
\begin{pmatrix}x' \\ y' \end{pmatrix} = P_{\rm new} \begin{pmatrix}x \\ y \end{pmatrix} = \begin{pmatrix}y \\ x \end{pmatrix} \tag{5}
$$
So that the ellipse is written as
$$
y'^2 + \frac{x'^2}{b^2} = 1 \tag{6}
$$
Which is now an ellipse with major axis along the $y'$ axis! so changing the order of the eigenvalues we change the orientation of the figure in space


TITLE: cohomological representations of GL(N)
QUESTION [4 upvotes]: I am trying to understand cohomology of $G := GL(N)$. For this I need to understand representations of $G(\mathbb{R})$ with nontrivial $(\mathfrak{g},K_\infty)$-cohomology, where $\mathfrak{g}$ is the Lie algebra of $G(\mathbb{R})$ and $K_\infty$ is $O(N,\mathbb{R})$. Is there a classification of such representations?
Thank you.

REPLY [5 votes]: Yes. (That's assuming your definition of “representation” gives at least a $(\mathfrak g, K_\infty)$-module.)


TITLE: Singular points , Poles
QUESTION [0 upvotes]: I am having trouble with these 2 parts. In the back of the book the answer for the parts is
For part a I expanded out sin^2 z using the taylor series expansion and found there is 0 term for the first third fifth term etc so I don't see how they are getting a pole of order 2. My view the it is a essential singularity .
For part b I changed it into exponential form but that didnt help.

REPLY [0 votes]: For (a) and (b):
If $f $ has a zero of order $k $ at $z_0$, then $1/f $ has a pole of order $k $ at $z_0$


TITLE: Use of infinitesimals in physics
QUESTION [0 upvotes]: I want to ask about infinitesimals and non-standard analysis. In physics we always use $\mathrm dx,~\mathrm dv,~\mathrm dt$ etc. as infinitesimal quantities. When we deduce equations in physics, when we set up integrals and in many other instances we use infinitesimals. For example in the first law of thermodynamics we have: $đQ=\mathrm dU+đW$.
Now when we ask teachers about what is $đQ$ here, they say it is an infinitesimal amount of thermal energy. So my question is, is it necessary to treat the whole non-standard analysis as an axiom for physics or do we have standard calculus explanation for physicist's use of $\mathrm dy/\mathrm dx$ as a quotient in all cases?

REPLY [1 votes]: Physicists do you the rules of calculus but can be sloppy in their notation.
The symbols $\delta, \Delta$ and $d$ often seem to be used interchangeably to mean a (small or infinitesimal) change in something or better still a final value minus an initial value.
So in your equation $dU$ is the change in internal energy of a system or final internal energy minus initial internal energy $U_f-U_i$.
However with this notation the use of $dW$ and $dQ$ is often frowned on as there is no initial amount of heat and final amount of heat it is just the amount of heat leaving or entering the system and the same is true of the work done.
To indicate this instead of using $d$ a new symbol $d$ with a line/bar through it is used (I do not known how to write this but it like $\hbar$).  That being said if you look through many textbooks $\delta$ and $d$ are used but often with an explanation that they represent inexact differentials.


TITLE: Proving $2\left(b^2+c^2\right)-a^2\leqslant 12$ with some condition.
QUESTION [2 upvotes]: Problem. Let $a,b,c\in\mathbb{R}$ such that $a+b+c=6,$ $a^2+b^2+c^2\in\left[12,\frac{68}3\right]$ and $a\geq b\geq c.$ Prove
$$2\left(b^2+c^2\right)-a^2\leqslant 12.$$
When do we have equality?
I can only prove it for $a,b,c \geqslant 0.$
From $a+b+c=6$ we have $a=6-b-c \geqslant 0.$ We need to prove the inequality when $$\Big[6-b-c\geqslant 0, \left( 6-b-c \right) ^{2}+{b}^{2}+{c}^{2}-12\geqslant 0,\\{\frac {68}{3}}-
 \left( 6-b-c \right) ^{2}-{b}^{2}-{c}^{2}\geqslant 0,6-c-2\,b\geqslant 0,b-c\geqslant 0\Big]$$
By computer we have$:$ $$12+ \left( 6-b-c \right) ^{2}-2\,{b}^{2}-2\,{c}^{2}$$
$$= \left[  \left( 6-b-c \right) ^{2}+{b}^{2}+{c}^{2}-12 \right]  \Big[\left(
b-c \right)  \left( {\dfrac {5}{32}}\,{c}^{2}+\dfrac{1}{16}\,bc \right)  +\left(
6-c-2\,b \right)  \left( {\frac {3}{64}}\,{b}^{2}+\dfrac{1}{4}b+{\frac {5}{32
}}\,bc \right) \Big]
$$
$$+{\frac {3}{32}} \left[ {\frac {68}{3}}- \left( 6-b-c \right) ^{2}-{b
}^{2}-{c}^{2} \right] \left( b-c \right) {c}^{2}+\dfrac{1}{2} \left( 6-b-c
 \right)  \left( 6-c-2\,b \right) c$$
$$+\dfrac{1}{6}\, \left( 6-c-2\,b \right) ^{2}
{b}^{2}+ \left(  \left( 6-b-c \right) ^{2}+{b}^{2}+{c}^{2}-12 \right) ^{2}
 \left( \dfrac{1}{12}+{\frac {3}{256}}\,{c}^{2}+\dfrac{1}{32}c+{\frac {3}{128}}\,bc
 \right) +\dfrac{1}{6} \left( 6-c-2\,b \right)  \left( b-c \right) bc+\dfrac{1}{6}\,
 \left( b-c \right) ^{2}{c}^{2}$$
$$+ \left[  \left( 6-b-c \right) ^{2}+{b}^{2}+{c}^{2}-12 \right]  \left[
{\frac {68}{3}}- \left( 6-b-c \right) ^{2}-{b}^{2}-{c}^{2} \right]
 \left( {\frac {3}{256}}\,{c}^{2}+{\frac {3}{128}}\,bc \right) $$
$$+ \left[ {\frac {68}{3}}- \left( 6-b-c \right) ^{2}-{b}^{2}-{c}^{2}
 \right]  \left( 6-c-2\,b \right)  \left( {\frac {3}{64}}\,{b}^{2}+{
\frac {3}{32}}\,bc \right)\geqslant 0 $$
For the text of the above decomposition, please see in my file: Click here.

REPLY [2 votes]: Let $s = b - c \ge 0$. From $a = 6 - b - c \ge b$, we have $c \le 2 - \frac{2s}{3}$.
We have $a^2 + b^2 + c^2 = (6-b-c)^2 + b^2 + c^2 = 6(c - 2 + s/2)^2 + s^2/2 + 12$.
From $a^2 + b^2 + c^2 \in [12, 68/3]$, we have
$$6(c - 2 + s/2)^2 + s^2/2 + 12 \le \frac{68}{3}$$
and thus
$$0 \le s \le \frac{8}{\sqrt{3}}, \quad 2 - s/2 - \frac{1}{6}\sqrt{64-3s^2}
\le c \le 2 - s/2 + \frac{1}{6}\sqrt{64-3s^2}.$$
Clearly, $2 - \frac{2s}{3} \le 2 - s/2 + \frac{1}{6}\sqrt{64-3s^2}$
for $0 \le s \le \frac{8}{\sqrt{3}}$.
From $2 - s/2 - \frac{1}{6}\sqrt{64-3s^2} \le 2 - \frac{2s}{3}$,
we have $0 \le s \le 4$.
Then, we have $2(b^2+c^2) - a^2 = s^2+24c+12s-36
\le s^2 + 24(2 - \frac{2s}{3}) + 12s - 36 = 12 - s(4-s) \le 12$.
We are done.


TITLE: Integration with respect to the Lebesgue-Stieltjes measure associated with floor function
QUESTION [8 upvotes]: For two different measures, I am trying to identify $L^1(\mu)$(the set of integrable functions, i.e. $\{f:\int \lvert f\rvert du<\infty\}$, and trying to compute $\int fd\mu$ for $f\in L^1(\mu)$.

on $\mathbb R, \mu=\mu_F$, the Lebesgue-Stieltjes measure with $F(x)=\lfloor x\rfloor$.

Attempts

I know that the unique measure derived from the floor function would measure the number of integers between the end points, or in the set. But I'm not sure how to build on that to use to calculate the integral for a general $f$, or what would make a function integrable.

Update: $$\int 1_{E}(x)d\mu(x)=\mu(E)=\sum_{n\in\mathbb Z}1_E(n)$$
should be true for any characteristic function. And for this integral to be finite, in any Borel set $E$, there should only be finite number of integers in the set. But I'm not sure how to bootsrap up from this point to simple functions, and then functions in general.
Any help would be greatly appreciated!

REPLY [2 votes]: In your update you have correctly identified that $\mu=\sum_{k\in\mathbb{Z}}\delta_k$.
Take a simple function $f(x)=\sum_{i=1}^n a_i \chi_{A_i}(x)$ such that $\mu(A_i)<\infty$ for all $i$. Its integral is defined as
$$
\int f\,d\mu = \sum_{i=1}^n a_i \mu(A_i)
= \sum_{i=1}^n a_i \sum_{k\in\mathbb{Z}} \delta_k(A_i)
= \sum_{k\in\mathbb{Z}} \sum_{i=1}^n a_i \delta_k(A_i)
= \sum_{k\in\mathbb{Z}} \sum_{i=1}^n a_i \chi(A_i)(k)
= \sum_{k\in\mathbb{Z}} f(k) .
$$
If now $f$ is a non-negative Borel function and $(f_n)$ are simple functions converging monotonically to $f$, you have
$$
\int f\,d\mu = \lim_{n\to\infty} \sum_{k\in\mathbb{Z}} f_n(k)
= \sum_{k\in\mathbb{Z}} f(k)
$$
by monotone convergence.
Such a non-negative function is integrable iff $\sum_{k\in\mathbb{Z}} f(k) < \infty$.
Finally, if $f=f^+-f^-$ is a generic Borel function, it belongs to $L^1(\mu)$ if and only if $\int f^\pm\,d\mu < \infty$ and in such case you have
$$
\int f\,d\mu = \sum_{k\in\mathbb{Z}} f^+(k) - \sum_{k\in\mathbb{Z}} f^-(k)
= \sum_{k\in\mathbb{Z}} f(k) .
$$
So $f\in L^1(\mu)$ iff
$$
\int |f|\,d\mu = \sum_{k\in\mathbb{Z}} |f(k)| < \infty .
$$


TITLE: How many different strings can be made using the letters of ABBCCCDDDDEEEFF such that all the letters D must appear before all of letters F?
QUESTION [2 upvotes]: There are all together $15$ letters. $1$ A, $2$ B's, $3$ C's, $4$ D's, $3$ E's and $2$ F's.
I only know that the total different strings that can be made from those $15$ letters is $$\frac{15!}{1!2!3!4!3!2!} = 378,378,000$$ and I'm stuck there.
These are the methods that I'm using following my lecture's eg but I actually don't understand the reasoning behind them.
Method1 :  15c1 ( for the A) x 14c2 (for the Bs) x 12c3 (for the Cs) x 9c3 (for the Es) x 1 (no. of ways to place Ds and Fs) and I would get 25,225,200.
Method2:  I could use 15p9/ (2!3!3!1!) to get 25,225,200
As mentioned, I'm just blindly following the method but I dun understand them. Please explain them if you could. Thanks!

REPLY [2 votes]: There are six letters that are either a $D$ or an $F$.
A string is acceptable if the last two of the six are $F$s.
So only one in every $6\choose2$ of the 378378000 are acceptable.


TITLE: Role of $K$ be a compact space in showing that $L$ is positive.
QUESTION [0 upvotes]: I found this question. I am going to just copy it here again.
Ley $K$ be a compact Hausdorff space and let $X = C(K)$ be the space of continuous functions defined on $K$ with values in $\mathbb{K}$ alongside the supremum norm $\|\cdot\|_{\infty}$. Show that $L \in X^*$ and $\|L\| = L(\textbf{1})$, if and only if $L$ is a positive operator, that is, $L(f) \geq 0$ when $f \geq 0$.
By $\textbf{1} \in X$ we are refering to the constant function $1$.
The proof provided by a user is the following:
Let $a$ be a positive element such that $I(a)<0$. For a small enough $\beta$, $\|1-\beta a\|< 1$, but we have $I(1-\beta a)=I(1)-\beta I(a)>I(1)=\|I\|\geq \|I\| \|1-\beta a\|$, hence we get a contradiction, then $I$ is positive.
I understand that proof, but there is one piece missing. The fact that $K$ was a compact Hausdorff space was never used. My question is then, how does that fact play in?
I think it plays in, in that it gives boundedness of $I$, namely, since $L\in X^*$, then $L<\infty$. But I am not sure of that. Any hint would be appreciated.

REPLY [1 votes]: You're using it explicitly to get that $\|1-\beta a\|$ is strictly smaller than $1$. Say $K=\mathbb{R}$ (and we're considering the set of bounded continuous functions on $K$, so that the norm makes sense) and $a(x)=\exp(-|x|)$, then $\|1-\beta a\|\geq 1-\lim_{x\to \infty}a(x)=1$, so that part of the argument collapses, even though $a$ is strictly positive.


TITLE: Optimizing blockchain token purchase. Finding maximum value of a function.
QUESTION [2 upvotes]: I am working on smart contract on ethereum blockchain. The idea is that using some specified amount of money I have to buy maximum number of tokens from two different token smart contracts. Those two smart contracts calculate number of tokens to be sold using two different functions.

Contract 1 - C-ORG by Fairmint.

$$F1(x) = \sqrt{\frac{2x}{b} + a^2} - a$$
where b and a are constant values.

Contract 2 - Uniswap Token Traiding Contract.

$$F2(x) = \frac{x*d*997}{g*1000 + x * 997}$$
where d and g are constant.
What I need to do is split my aount of money I have between those two contracts so in total I can get maximum number of tokens possible from those two contracts combined.
So my understanding is that I have to find a maximum value of function
$$F3(x) = F1(z) + F(x-z)$$
where z is amount of money spend on contract 1 and
$z <= x$ and $z>0$
So I can say that
$$F3(x) = \sqrt{\frac{2z}{b} + a^2} - a  + \frac{(x-z)*d*997}{g*1000 + (x-z) * 997} $$
What I need to do is I have to find value of z, knowing value of x, for which this function will give maximum result.
I am not good in calculus so I used Integral-calculator and it came back with equation like this:
$$\dfrac{1000dg\ln\left(\left|997\left(x-z\right)-1000g\right|\right)}{997}+\dfrac{\left(2x+a^2b\right)^\frac{3}{2}}{3\sqrt{b}}+dx-ax$$
And now I understand that I have to find value of z where this function will be equal zero. Right ?
$$\dfrac{1000dg\ln\left(\left|997\left(x-z\right)-1000g\right|\right)}{997}+\dfrac{\left(2x+a^2b\right)^\frac{3}{2}}{3\sqrt{b}}+dx-ax = 0$$

I have following questions:

Is my thinking correct. Should I do it this way ?
Is result of calculus correct ?
How I can be sure I found maximum and not minimum ?
How I can simplify last function so it is easier to calculate. I am working in coding language ( solidity ) which only support basic mathematical functions (add,sub,mul,div and  binary operations). I don't have anything for $\sqrt a$ or $\ln$

Any help will be more than appreciated. Thank you.

REPLY [0 votes]: To determine the maximums and/or minimums of a function, you should take the derivative of it, not the integral. As you basically stated, your function is
$$F3(z; x) = \sqrt{\frac{2z}{b} + a^2} - a  + \frac{(x-z)*d*997}{g*1000 + (x-z) * 997} \tag{1}\label{eq1}$$
In this case, $x$ is some given amount and $z$ is your variable. Thus, you want to take the derivative of \eqref{eq1} with respect to $z$. Doing this gives
$$\begin{equation}\begin{aligned}
\frac{d\left(F3(z; x)\right)}{dz} & = \frac{1}{b\sqrt{\frac{2z}{b} + a^2}} - \frac{997d}{1000g + 997(x-z)} + \frac{997^2d(x - z)}{\left(1000g + 997(x-z)\right)^2} \\
& = \frac{1}{b\sqrt{\frac{2z}{b} + a^2}} + \frac{-997d\left(1000g + 997(x-z)\right) + 997^2d(x - z)}{\left(1000g + 997(x-z)\right)^2} \\
& = \frac{1}{b\sqrt{\frac{2z}{b} + a^2}} - \frac{997d(1000g)}{\left(1000g + 997(x-z)\right)^2}
\end{aligned}\end{equation}\tag{2}\label{eq2}$$
The maximum and minimum values will occur at the end-points, i.e., where $z = 0$ or $z = x$, or where the expression in \eqref{eq2} is $0$. As for determining the latter, you can equate both terms to each other, cross-multiply, square and collect the terms in $z$ together to get a quartic polynomial. In particular, first let $c = 997d(1000g)$ to make the algebra a bit simpler. You then get
$$\begin{equation}\begin{aligned}
\frac{1}{b\sqrt{\frac{2z}{b} + a^2}} & = \frac{c}{\left(1000g + 997(x-z)\right)^2} \\
\left(1000g + 997(x-z)\right)^2 & = cb\sqrt{\frac{2z}{b} + a^2} \\
\left(1000g + 997(x-z)\right)^4 & = \left(cb\right)^2\left(\frac{2z}{b} + a^2\right)
\end{aligned}\end{equation}\tag{3}\label{eq3}$$
I'll leave it to you to do the multiplications and collection of terms if you wish. As for finding its roots, there's actually an analytic solution, such as described in Wikipedia's Quartic function article, but it's somewhat long & messy. As an easier alternative, there are many root finding algorithms, such as described in Wikipedia's Root-finding algorithm article. I don't believe you'll need anything fancy, so even on those basic numerical methods will likely work well enough for you.
Regarding whether you have a maximum or minimum at some point, say $z_1$, you would normally check the second derivative at $z_1$, with it being negative meaning $f$ is a maximum at $z_1$, and it being positive meaning $f$ is a minimum at $z_1$. In your case, as there won't likely be an issue of these points being too close together, you may wish to just check the $f$ values at $z_1$, just a bit less than $z_1$, and just a bit more than $z_1$, to compare them & determine if it's a maximum or minimum.
As for handling the expression on a computer, there's apparently just the complication of determining the square root. Fortunately, there are relatively simple, but quite efficient, square root determination algorithms. For example, Wikipedia's Methods of computing square roots has some excellent suggestions, with likely even any fairly basic one they mention being sufficient for your needs.


TITLE: Showing that supermartingale is a u.i. martingale
QUESTION [1 upvotes]: Let $M_t$ be a non-negative supermartingale, $EM_0 < \infty$ and $M_t \to M_\infty$ a.s. I want to show that if $E M_\infty = EM_0$, then $M$ is a uniformly integrable martingale. I thought about doing so by showing that $M_t = E ( M_\infty | \mathcal{F}_t )$ but couldn't figure out the necessary steps. Any hints? Just somewhere to start would be great.
Edit: An additional question: Can I follow from the above that $E (sup_{t\geq0} M_t) < \infty$? That would also give uniform integrability, right?

REPLY [1 votes]: By the optional stopping theorem, we know that for any stopping time $T$, $$ M_T \ge E[M_\infty \mid \mathcal F_T]. $$ Moreover, since $M$ is a supermartingale, again via the optional stopping theorem, $$ E[M_0] \ge E[M_T]. $$ Thus, $$E[M_0] \ge E[M_T] \ge E[M_\infty], $$ which means that $$ E[M_0] = E[M_T] $$ for all stopping times $T$. The fact that $M$ is a UI martingale now follows from the following result.

Theorem. Suppose $\{M_t\}_{t\in[0,\infty]}$ is a right-continuous process adapted to the right-continuous filtration $\{\mathcal F_t\}_{t\in[0,\infty]}$. Then, $M$ is a uniformly integrable martingale if and only if for every stopping time $T$ we know that $E[\vert M_T\vert]<\infty$ and $E[M_T]=E[M_0]$.

Clearly, you only need the "if" part of this result. You don't explicitly assume right-continuity, but this is a standard assumption, so I hope it works for you.
Reference. Cohen, S. N., & Elliott, R. J. (2015). Stochastic calculus and applications (Vol. 2). New York: Birkhäuser.


TITLE: How to solve this integration?
QUESTION [0 upvotes]: I want to solve this
$$\int_0^w (b/x)^{a+1} e^{(cx-(b/x)^a)} dx$$
where $a$, $b$, and $c$ are arbitrary positive real numbers.
Do i have to solve it numerically? I have no clue to solve this analitically.
If yes, which numerical integration i should try?

REPLY [0 votes]: First, one can simplify the integral by
$$
\int_0^w e^{cx - (b/x)^a}\left(\frac{b}{x}\right)^{a+1} dx =
\begin{vmatrix}
u = \left(\frac{b}{x}\right)^a\\
x = \frac{b}{u^{1/a}}\\
du = -\frac{a}{b}\left(\frac{b}{x}\right)^{a+1}dx
\end{vmatrix} =\\=
\frac{b}{a}\int_{(b/w)^a}^\infty \exp\left(\frac{cb}{u^{1/a}} - u\right) du.
$$
So the problem reduces to the computation of the following integral
($\gamma = (b/w)^a > 0$, $\beta = cb > 0$, $\alpha = 1/a > 0$)
$$
\int_\gamma^\infty \exp\left(\beta u^{-\alpha} - u\right) du =
\begin{vmatrix}
z = u - \gamma\\
u = z + \gamma\\
dz = du
\end{vmatrix} = \\ =
\int_0^\infty \exp\left(
\beta (z + \gamma)^{-\alpha} - z - \gamma
\right)
dz = e^{-\gamma}
\int_0^\infty \exp\left(
\beta (z + \gamma)^{-\alpha}
\right)e^{-z} dz
$$
To compute the last integral numerically you may try Gauss-Laguerre quadrature.