TITLE: Does the fixed point corresponding to a Hopf bifurcation vary in the $(x, y)$ plane as the system parameter $\mu$ changes?
QUESTION [4 upvotes]: The question is for a 2D system, but for the sake of simplicity, let's consider a 1D system $\dot{x} = \mu + x^2$. Then for $\mu < 0$ the fixed point $x = \pm\sqrt{\mu}$ varies along the $x$-axis as $\mu$ varies. On the other hand for the system $\dot{x} = x (\mu + x^2)$, we have three fixed points for a given $\mu$, but one of them $x = 0$ does not vary with respect to $\mu$.
For a 2D system $(\dot{x}, \dot{y}) = f(x, y; \mu)$, a Hopf bifurcation is usually introduced as a bifurcation of a fixed point that changes from stable to unstable (as a stable spiral changes to an unstable spiral or the other way around).
My question is, can this fixed point vary in the $(x, y)$ plane, or does it always remain at a fixed point e.g. $(x_0, y_0) = (0, 0)$? So far, for the examples I have seen, the fixed point does indeed seem to stay fixed in the $(x, y)$ plane, but I don't how to show that this is always the case from the definition of a Hopf bifurcation, which is vague at the level of Strogatz's book.
Related question: As we vary $\mu$ just past the bifurcation point $\mu_c$, a limit cycle appears. Is this limit cycle always elliptical?

REPLY [2 votes]: Yes, the coordinates of the equilibrium can vary. Here is an example:
$$
\dot N=rN\left(1-\frac{N}{K}\right)-\frac{CNP}{A+N},\\
\dot P=P\left(-d+\frac{BN}{A+N}\right),
$$
where $r,K,C,A,B,d$ are some positive parameters.
I will leave all the computations to you, but probably the easiest way to see what happens in this system is to plot the null-clines and analyze their behavior depending on the parameters.


TITLE: Kinematics of 4 body decays for Higgs to 4 leptons
QUESTION [0 upvotes]: I would like to calculate the decay width of the Higgs boson $h$ to 4 leptons via $ZZ^*$ intermediate states. I just did a similar calculation for $h$ to $W^{\pm}$ and 2 fermions via $WW^*$, so I am trying to base this new calculation off of that.
I have constructed the matrix element squared, averaged over spins, $|\mathcal{M}|^2$. In the case of the 3-body decay, at this stage I simply used equation 49.22 from the kinematics document of the PDG guide:
\begin{equation}
d\Gamma=\frac{1}{(2\pi)^3}\frac{1}{32 M^3}|\mathcal{M}|^2dm_{12}^2dm_{23}^2
\end{equation}
where $M$ is the mass of the decaying particle and $m_{ij}^2=(p_i+p_j)^2$ where $i,j$ label the outgoing three particles. Then the integration bounds are given by 49.23. For fixed $m_{12}$:
\begin{equation}
    (m_{23}^2)_{max}=(E_2^{*}+E_3^{*})^2-\left(\sqrt{(E_2^{*})^2-m_2^2}-\sqrt{(E_3^{*})^2-m_3^2}\right)^2
\end{equation}
\begin{equation}
    (m_{23}^2)_{min}=(E_2^{*}+E_3^{*})^2-\left(\sqrt{(E_2^{*})^2-m_2^2}+\sqrt{(E_3^{*})^2-m_3^2}\right)^2
\end{equation}
where $E_i^{*}$ indicates the energy of the particle in the $m_{12}$ rest frame. Then $m_{12}^2$ gets integrated from $(m_1+m_2)^2$ to $(M-m_3)^2$. (For simplicity I am approximating all the fermion masses as zero in the eventual numerical calculation.)
So, I clearly need to generalize this now that I have four final states. The next section of the PDG simply says to treat multiple final states as "effective states" thus allowing you to use previous formulas. You are supposed to make the generalization:
\begin{equation}
m_{ijk...}=\sqrt{(p_i+p_j+p_k+...)^2}
\end{equation}
in place of e.g. $m_{23}$. Naively I would start with this:
$dm_{12}^2dm_{23}^2\to dm_{12}^2dm_{234}^2$
But then I do not know how to get the bounds for the integration. Can you assist me?

REPLY [0 votes]: I learned that for this particular decay we do not actually consider it as a 4-body problem in this way. Only one of the $Z$'s is off shell, $Z^*$, which decays right away to the 2 leptons. The other is on shell. So for the first step, we consider a 3-body decay using exactly the equations given in the question.
Then, you separately consider the decay of the on-shell $Z$ to two leptons. Combining the two steps gives the overall process $h$ to 4 leptons via $ZZ^*$.


TITLE: Prove: If points of a convergent sequence are reordered, the new sequence converges to the same limit.
QUESTION [0 upvotes]: I am having trouble wrapping my mind around this proof I found online.
Let $p_n$, $n ∈ \Bbb N$ be a sequence in a metric space $(E, d)$. Suppose $lim_{n→∞} p_n = p ∈ E$. Let $f : N → N$ be a bijection. Let $q_m = p_{f(m)}$
. Hence, the sequence $q_m$ is a
reordering of the points of $p_n$.
Let $ \epsilon > 0$. Since $p_n$ converges, there exists $N ∈ \Bbb N$ such that $d(p_n, p) < \epsilon $ for all $n > N$.
Let $M = max\{N ∪ \{f(n)\  |\ 0 ≤ n ≤ N\}\}$. Let $m > M$. Suppose $d(q_m, p) ≥ \epsilon$. Then for $n = f^{
−1}
(m)$, we have $d(p_n, p) ≥ \epsilon$. This implies that $0 ≤ n ≤ N$ which contradicts
the choice of $m > M$. Hence $d(q_m, p) < \epsilon$. Thus, $q_m$ also converges to $p = lim_{n→∞} p_n$.
First I am having trouble understanding the choice of $M$ and why this ensures that $p_m$ will be in the neighborhood of $p$. Next, I am confused how that is a contradiction. This probably stems from my confusion about $M$, but mainly I do not understand why that implies $0 \le n \le N$.

REPLY [1 votes]: First note that $q_m=p_n$. So $d(q_m,p)=d(p_n)\ge\epsilon$. If $n>N$,  then $d(p_n,p)<\epsilon$ by the choice of $N$. Therefore, $0\le n\le N$, but $m=f(n)>f(k)$ for all k such that $0\le k\le N$. This contradicts the fact that $M$ is the maximum of such numbers.
I'm thinking that $M$ should be $\max\{f(n):0\le n\le N\}$, instead of $\max\{N\cup\{f(n):0\le n\le N\}\}$.


TITLE: Variable Exchange $F(t) \to F(x)$? I want the value of the work done by drag
QUESTION [1 upvotes]: I got a nonlinear equation, which describes the magnitude of a force as a function of time, but I don't know how to calculate the work done by the force. Given:
$$F(t) = kv(t)$$
with
$$v(t)= \left(1-\frac{k}{m}\right)^t v_{0}$$
and I want to derive:
$$W = \int F(t) dx$$
and also want to convert $F(t)$ to $F(x)$

REPLY [1 votes]: $$ \begin{align}F(t)&=\space kv(t) \\
v&= \frac{dx}{dt}=(1-\large{\tfrac{k}{m}})^{\small{t}}\, v_0\\
{\rm d}x&=(1-{\tfrac{k}{m}})^{\small{t}} v_0 {\rm d}t\\
\int_0^x {\rm d}x&= \int_0^t (1-{\tfrac{k}{m}})^{\small{t}}\, v_0 {\rm d}t\\
x&=\frac{(1-{\frac{k}{m}})^{\small{t}}-1}{\ln(1-{\frac{k}{m}})} v_0\\
(1-{\tfrac{k}{m}})^{\small{t}} v_0&= v(t)= v_0 + x \space \ln(1-{\tfrac{k}{m}})\\ \end{align}$$
Substituting the values in the expression for $F(t)$ $$ \begin{aligned}
F(x)=k \left[v_0 + x \space \ln(1-{\tfrac{k}{m}})\right]
\end{aligned}$$

I hope you can calculate forward.Regards.


TITLE: Why exactly can we not know the position and momentum of a given particle?
QUESTION [0 upvotes]: I have heard that this is to do with the fact that a particle is also a wave. However, I do not understand why this would cause this. Surely, you could use the formula used for finding the velocity, and then use the formula used to find the position. Even better, why can you not just record this? It just seems to be a very abstract concept to grasp.
Also, yes I have seen the formula, $\sigma_x\sigma_p\ge\frac\hbar2$, but, I still don't understand why this is. Am I simply dense?
Disclaimer: I am not particularly experienced in physics, and I am just about to begin year 10, and I have began to start reading some books about physics in preparation. I am currently reading "The Feynman Lectures On Physics Vol. 1". This may be why I am more likely to not know about certain other parts of physics and perhaps math that may be necessary for this.

REPLY [1 votes]: The answer is that the wave changes shape depending upon how it interacts with  things. Sometimes the wave is very spread out and sometimes it is very localised. When it is spread out you can't say exactly where it is, obviously, and when it is vey localised it doesn't have a well-defined wavelength, and since the particle's velocity is determined by the wavelength of the associated wave, that means it doesn't have a well defined velocity when the wave is localised.
So, in a nutshell, when the wave is very localised it has a well-defined position but no definite wavelength (which equates to the velocity) and when the wave has a very well-defined wavelength (velocity) it doesn't have a well-defined position.
Incidentally, you shouldn't really think of the wave as being the particle. The wave is a mathematical function that encodes information about the position and momentum of the particle. The momentum is related to the wavelength of the function, and the position is given just as a probability as a function of the amplitude of the wave at any point in space.
Clearly you can record a particle's position and then measure its velocity, but the point is that when you measure its velocity it is no longer is in the position it was in when you recorded it. Likewise you can measure and record its velocity and then perform an experiment which nails down its position, but by performing the experiment you change the shape of the wave which means the particle no longer has the velocity you recorded.
What is especially lovely about quantum mechanics when you get into it is the way that the probabilities of getting specific experimental results arise from interrelationships between things known as eigenfunctions of operators. The idea is too complicated to explain slowly here, but it is very sweet when you get to understand it.


text:
\begin{document}

\begin{center}
{\Large Tempered Generalized Functions and Hermite Expansions\\}

\end{center}

\vspace{0.3cm}

\begin{center}
{\large Pedro Catuogno\footnote{Research partially supported by CNPQ 302704/2008-6.} and
Christian Olivera\footnote{Research partially supported by FAPESP 02/10246-2, 2007/54740-4.}}\\

\textit{Departamento de
 Matem\'{a}tica, Universidade Estadual de Campinas, \\ F. 54(19) 3521-5921  Fax 54(19)
3521-6094\\ 13.081-970 -
 Campinas - SP, Brazil. e-mail: pedrojc@ime.unicamp.br ; colivera@ime.unicamp.br}
\end{center}
\vspace{0.3cm}
\begin{center}
\begin{abstract}
\noindent In this work we introduce a new algebra of tempered
generalized functions. The tempered distributions are embedded in
this algebra via their Hermite expansions. The Fourier transform is
naturally extended to this algebra in such a way that the usual
relations involving multiplication, convolution and differentiation
are valid. We study the elementary properties of the association,
embedding, point values and Fourier transform. Furthermore, we give
a generalized It\^o formula in this context and some applications to
stochastic analysis.
\end{abstract}
\end{center}

\noindent {\bf Key words:} Hermite functions, Generalized function
algebras, Fourier transform, It\^o formula.

\vspace{0.3cm} \noindent {\bf MSC2000 subject classification:}
 46F30, 60H.

\section {Introduction}

\noindent The differential algebras of generalized functions of
Colombeau type were developed in connection with non linear
problems. These algebras provide a good framework to solve
differential equations with rough initial data or discontinuous
coefficients (see \cite{Biag}, \cite{gkos} and \cite{ober2}).
Recently there has been a great interest in developing a stochastic
calculus in algebras of generalized functions (see \cite{albe},
\cite{capa}, \cite{mar1}, \cite{mi-se}, \cite{ober4} and
\cite{Russo} and references).

\noindent  In order to develop a Fourier transform for generalized
functions, J. F. Colombeau in \cite{colb} introduced an algebra of
tempered generalized functions with the characteristic that most of
the properties involving the Fourier transform and convolution are
valid only in a weak sense. For a recent account of the theory and
applications we refer to  the reader to \cite{Del} and \cite{gkos}.

\noindent The following question naturally arises: is it possible to
construct an algebra of generalized functions in which the
operations of multiplication, convolution, differentiation and
Fourier transform are defined and have all the properties of the
ordinary operations?

\noindent In this work we introduced a new algebra of tempered
generalized functions that answers the above question. The
construction of this algebra is based on the Fourier-Hermite
expansion of tempered distributions. More precisely, the Hermite
representation theorem for $\mathcal{S}^{\prime}$ (see \cite{Miku},
\cite{Red1}, \cite{Schw}, and \cite{Red3}) establishes that every
$S\in\mathcal{S}^{\prime}$ can be represented by a Hermite series
\begin{equation}\label{repre}
S=\sum_{j=0}^{\infty} S(h_j)h_j
\end{equation}
where $\{h_j\}$ are the Hermite functions and the equality is in the
weak sense. The idea is to embed the tempered distributions into a
differential subalgebra $\mathcal{H}_{\mathbf{s}^{\prime}} \subset
\mathcal{S}^{\mathbb{N}_0}$ via the sequence of partial sums
\[
S_{n}=\sum_{j=0}^{n} S(h_j)h_j
\]
and define the algebra of tempered generalized functions as
\[
\mathcal{H}=\mathcal{H}_{\mathbf{s}^{\prime}}/\mathcal{H}_{\mathbf{s}}
\]
where $\mathcal{H}_{\mathbf{s}}$ is a differentiable ideal of
$\mathcal{H}_{\mathbf{s}^{\prime}}$ (see Section 3 for precise
definitions and details).

\noindent The operations in $\mathcal{H}$ are naturally introduced
in a pointwise way. By example the Fourier transform of $[f_n] \in
\mathcal{H}$ is defined to be $[\mathcal{F}(f_n)]$, where
$\mathcal{F}$ is the usual Fourier transform. Now, it is not
difficult to prove that the Fourier transform is an isomorphism
and that usual identities involving convolution, derivatives and
Fourier transform are satisfied.

\noindent Another important point of this paper is that we obtain an
It\^o formula under weak regularity conditions. We make use of
F\"{o}llmer's formulation of pathwise stochastic calculus \cite{Fol}
in order to establish an It\^o formula for tempered  generalized
functions. This formula extends the classical one via association
and the embedding of the tempered distributions into $\mathcal{H}$.

\noindent The paper is organized as follows: Section 2 contains
a brief summary without proofs of Hermite functions and the
Hermite representation theorems. In section 3, we introduce the
tempered algebra $\mathcal{H}$, this algebra contains the
tempered distributions and extends the product in $\mathcal{S}$.
Moreover, we study some elementary properties and we show that the
symmetric Hermite product of tempered distributions (see
\cite{calo} and \cite{Shen-Sun}) is associated with the product in
$\mathcal{H}$.

\noindent In section 4, in order to define the point value of
tempered generalized functions, we introduce the ring of tempered
numbers $\mathbf{h}$. Section 5 we deal with integration,
convolutions and the Fourier transform of tempered generalized
functions. We obtain the Fourier inversion theorem for
$\mathcal{H}$, the formula of interchange between the product and
the convolution and the rule of integration by parts. We remark that
the identities are in $\mathbf{h}$, this is in a strong sense.

\noindent In section 6 we introduce an elementary stochastic
calculus for generalized tempered functions. We derive an It\^o
formula for elements of $\mathcal{H}$. We observe that the
classical It\^o formula is associated to the generalized It\^o
formula, under suitable regularity conditions. Moreover, we obtain
a probabilistic representation of the solution of the heat
equation with data in $\mathcal{H}$. Finally we note that in
\cite{mar2} and \cite{mar1} is present an It\^o formula for
generalized functions containing some mistakes in definitions and
proofs; see \cite{capa}.

\section{N-Representation of tempered distributions.}
Let $\mathcal{S} \equiv \mathcal{S}(\mathbb{R})$ be the Schwartz
space of rapidly decreasing smooth real valued functions.

For each $m\in \mathbb{N}_0 \equiv \mathbb{N}\cup\{0\}$, we
consider $\|\cdot\|_{m}$ the norm of $\mathcal{S}$ given by
\[
\|\varphi\|_{m}=\Big(\int^{\infty}_{-\infty}|
(N+1)^m\varphi(x)\mid^2dx \Big)^{\frac{1}{2}},
\]
where $N+1=\frac{1}{2}(-\frac{d^2}{dx^2}+x^2+1)$.

We observe that $\mathcal{S}$ is provided with the natural topology
given by these norms is a sequentially complete locally convex
space and its dual space $\mathcal{S}^{\prime}$ is the space of
tempered distributions. The family of norms $\{ \|\cdot\|_{m}: m
\in \mathbb{N}_0\}$ is direct and equivalent to the family of
seminorms $\{ \|\cdot\|_{\alpha, \beta, \infty}: \alpha, \beta \in
\mathbb{N}_0\}$, given by
\[
\| \varphi \|_{\alpha, \beta, \infty}=\sup_x |(1+
|x|^{2})^{\alpha} D^{\beta} \varphi(x)|.
\]
We use often the following property of the multiplication on
$\mathcal{S}$. For all $m \in \mathbb{N}_0$ there exists $r, s \in
\mathbb{N}_0$ and a constant $C_m>0$ such that
\begin{equation}\label{eq1}
\|\varphi \psi \|_m \leq C_m \| \varphi \|_r \| \psi \|_s
\end{equation}
for all $\varphi, \psi \in \mathcal{S}$ (see for instance
\cite{radyno}, Theorem 2).

The {\it{Hermite polynomials}} $H_n(x)$ are defined by

\begin{equation}\label{forpolHe}
    H_n(x)=(-1)^n e^{\frac{x^2}{2}} \frac{d^n}{dx^n}  e^{-\frac{x^2}{2}}
\end{equation}
for $n\in \mathbb{N}_{0}$ or equivalently
\begin{equation}\label{forpolHp}
      H_n(x) =2^{-\frac{n}{2}}\sum_{k=0}^{[n/2]}\frac{(-1)^k
      n!( \sqrt{2}x)^{n-2k}}{k!(n-2k)!}.
\end{equation}
The {\it{Hermite functions}} $h_n(x)$ are defined by
\begin{equation}\label{relfunpolh}
    h_{n}(x)=  (\sqrt{2\pi}n!)^{-\frac{1}{2}}
    e^{{-\frac{1}{4}x^2}}H_n(x)
\end{equation}
for $n\in \mathbb{N}_{0}$. Some properties of the Hermite
functions that we will often use are:
\begin{itemize}
\item $h_{n}\in \mathcal{S}$ for all $n \in \mathbb{N}_{0}$,

\item $h_n$ is an even (odd) function if $n$ is even (odd),

\item $\sqrt{n+1}h_{n+1}(x)+2h_n^{\prime}(x)=\sqrt{n} h_{n-1}(x)$
for all $n \in \mathbb{N}_{0}$,

\item $\sqrt{n+1}h_{n+1}(x)=xh_n(x)-\sqrt{n} h_{n-1}(x)$ for all
$n \in \mathbb{N}_{0}$,

\item $\{h_{n}: n \in \mathbb{N}_{0} \}$ is an orthonormal basis
of $L^{2}(\mathbb{R})$,

\item $(N+1)h_{n}=(n+1)h_{n}$ for all $n \in \mathbb{N}_{0}$.
\end{itemize}

\noindent  From  the two last properties   we have

\[
\|\varphi\|^2_{m}=\sum_{n=0}^{\infty}(n+1)^{2m}<\varphi,h_n>^2,
\]
where $<\varphi,h_n>=\int \varphi(x)h_n(x)dx$ are the
Fourier-Hermite coefficients of the expansion of $\varphi$.

\noindent The Hermite representation theorem for $\mathcal{S}$
($\mathcal{S}^{\prime}$) states a topological isomorphism from
$\mathcal{S}$ ($\mathcal{S}^{\prime}$) onto the space of sequences
$\mathbf{s}$ ($\mathbf{s}^{\prime}$).

\noindent Let $\mathbf{s}$ be the space of rapidly decreasing
sequences
\[
\mathbf{s}=\{(a_n)\in \ell^2:\sum_{n=0}^{\infty}(n+1)^{2m}\mid
a_n\mid^2<\infty, \; \mbox{for all } \; m \in \mathbb{N}_0\}.
\]
The space $\mathbf{s}$ is a locally convex space with the sequence
of norms
\[
\| (a_n)\|_m = (\sum_{n=0}^{\infty}(n+1)^{2m}\mid
a_n\mid^2)^{\frac{1}{2}}
\]
or with the equivalent sequence of norms
\[
\mid(a_{n})\mid_{m,\infty} =  \sup_{n} (n+1)^{m}| a_n|.
\]
The topological dual space to $\mathbf{s}$, denoted by
$\mathbf{s}^{\prime}$, is given by
\[
\mathbf{s}^{\prime}=\{(b_n): \mbox{for some
}\;(C,m)\in\mathbb{R}\times \mathbb{N}_0, \; \mid b_n \mid \leq
C(n+1)^m \mbox{ for all } n \}.
\]
The natural pairing of elements from $\mathbf{s}$ and
$\mathbf{s}^{\prime}$, denoted by $\langle \cdot, \cdot \rangle$,
is given by
\[
\langle (b_n), (a_n) \rangle = \sum_{n=0}^{\infty}b_na_n
\]
for $(b_n) \in \mathbf{s}^{\prime}$ and $(a_n) \in \mathbf{s}$.

It is clear that $\mathbf{s}^{\prime}$ is an algebra with the
pointwise operations:
\begin{eqnarray*}
(b_n)+(b^{\prime}_n)& = & (b_n+b^{\prime}_n) \\
(b_n)\cdot(b^{\prime}_n)& = & (b_nb^{\prime}_n),
\end{eqnarray*}
and $\mathbf{s}$ is an ideal of $\mathbf{s}^{\prime}$.

The relation between $\mathbf{s}$ ($\mathbf{s}^{\prime}$) and
$\mathcal{S}$ ($\mathcal{S}^{\prime}$) is induced by the Hermite
functions, via Hermite coefficients (evaluation). The following
representation theorem is fundamental in our work, for the proof
see \cite{Red1} pp. 143.

\begin{theorem}[N-representation theorem for $\mathcal{S}$ and $\mathcal{S}^{\prime}$]\label{rpreS}
$\mathbf{a)}$ Let $\mathbf{h}:\mathcal{S}\rightarrow \mathbf{s}$
be the application
\[
\mathbf{h}(\varphi)=(<\varphi,h_n>).
\]
Then $\mathbf{h}$ is a topological isomorphism. Moreover,
\[
\|\mathbf{h}(\varphi)\|_m=\|\varphi\|_m
\]
for all $\varphi \in \mathcal{S}$.

\noindent $\mathbf{b)}$ Let
$\mathbf{H}:\mathcal{S}^{\prime}\rightarrow \mathbf{s}^{\prime}$
be the application $\mathbf{H}(T)=(T(h_n))$. Then $\mathbf{H}$ is
a topological isomorphism. Moreover, if $T \in
\mathcal{S}^{\prime}$ we have that
\[
T=\sum _{n=0}^{\infty} T(h_n)h_n
\]
in the weak sense and for all $\varphi \in \mathcal{S}$,
\[
T(\varphi)=\langle \mathbf{H}(T),\mathbf{h}(\varphi)\rangle.
\]
\end{theorem}
\noindent We say that the sequences $\mathbf{h}(\varphi)$ and
$\mathbf{H}(T)$ are the {\it Hermite coefficients} of the tempered
function $\varphi$ and the distribution $T$, respectively.

Next we provide the Hermite coefficients for some tempered
distributions.

\subsection{The delta distribution.} (see \cite{Miku} pp 191.)
\begin{equation}\label{delta} \delta (h_{n}) = h_{n}(0)  =
\left \{
\begin{array}{lll}

           \frac{(-1)^{\frac{n}{2}}}{\sqrt[4]{2\pi}}\sqrt{\frac{1}{2}\frac{3}{4}\cdots\frac{n-1}{n}}& & \mbox{for $n$ even},        \\
           0 & & \mbox{for $n$ odd}.
\end{array}
\right .
\end{equation}

\subsection{The constant distribution $\mathbf{1}$.} (see \cite{Miku} pp 190.)

\begin{equation}\label{uno}
1(h_{n})=\int_{-\infty}^{\infty} h_{n}(x) \; dx
  = \left \{
\begin{array}{lll}
          \sqrt[4]{8\pi}\sqrt{\frac{1}{2}\frac{3}{4}\cdots\frac{n-1}{n}} & & \mbox{for $n$
           even},\\
           0 & & \mbox{for $n$ odd.}

\end{array}
\right.
\end{equation}

\subsection{The $x_+^p$ distribution.} (see \cite{Sad1} pp 162.)

 We recall that $<x_+^p,\phi>= \int_{0}^{\infty}x^p \phi(x) \;
 dx$.
\begin{equation}\label{Heaviside}
x_+^p(h_{n}) = \left \{
\begin{array}{ll}
(\sqrt{2\pi}n!)^{- \frac{1}{2}} 2^p \Gamma(
\frac{p+1}{2})W_n(2p+1)
               & \mbox{for $n$ even},\\
   (\sqrt{2\pi}n!)^{-\frac{1}{2}}2^{p+1}\Gamma(\frac{p+2}{2})W_n(2p+1) & \mbox{for $n$ odd}
\end{array}
\right .
\end{equation}
where $W_n(x)$ are polynomials such that $W_0(x)=W_1(x)=1$ and
\[
W_{n+2}(x)=xW_n(x)+n(n-1)W_{n-2}(x).
\]
Note that if $p=0$, then $x_+^p$ is the Heaviside distribution
$H$.

\subsection{The $\delta^{\prime}$ distribution.}

\begin{equation}\label{deltaprima}
\delta^{\prime}(h_n) =-h_n^{\prime}(0)=\sqrt{n}h_{n-1}(0).
\end{equation}

\section{The Tempered  Algebra }

\noindent In order to introduce the tempered algebra we consider
$\mathcal{S}^{\mathbb{N}_0}$ the space of sequences of rapidly
decreasing smooth functions. It is clear that
$\mathcal{S}^{\mathbb{N}_0}$ has the structure of an associative,
commutative differential algebra with the natural operations:
\begin{eqnarray*}
(f_{n})+(g_{n})& = & (f_{n}+ g_{n}) \\
a(f_{n})& = & (af_{n}) \\
(f_{n})\cdot (g_{n})& = & (f_{n} g_{n}) \\
D(f_{n})& = & (Df_{n})
\end{eqnarray*}
where $(f_{n})$ and $(g_{n})$ are in $\mathcal{S}$ and $a \in
\mathbb{R}$.

\begin{definition}\label{defeal} Let
\begin{equation}
\mathcal{H}_{\mathbf{s}^{\prime}}=\{(f_{n})\in
\mathcal{S}^{\mathbb{N}_0} \ : \ \mbox{for each $m \in
\mathbb{N}_0$,} \ (\| f_{n} \|_{m})\in \mathbf{s}^{\prime} \ \}
\end{equation}
and
\begin{equation}
\mathcal{H}_{\mathbf{s}}=\{(f_{n})\in  \mathcal{S}^{\mathbb{N}_0}
\ : \ \mbox{for each $m \in \mathbb{N}_0$,} \  (\| f_{n}
\|_{m})\in \mathbf{s} \ \}.
\end{equation}

\end{definition}

\begin{lemma}\label{lema} $\mathcal{H}_{\mathbf{s}^{\prime}}$ is a subalgebra
of $\mathcal{S}^{\mathbb{N}_0}$ and $\mathcal{H}_{\mathbf{s}}$ is
a differential ideal of $\mathcal{H}_{\mathbf{s}^{\prime}}$.
\end{lemma}
\begin{proof} Let $(f_n),(g_n) \in \mathcal{H}_{\mathbf{s}^{\prime}}$ and $m \in
\mathbb{N}_0$. Applying the inequality (\ref{eq1}), there exists
$r, s \in \mathbb{N}_0$ and a constant $C_m>0$ such that
\[
\| f_n g_n\|_m  \leq  C_m \| f_{n} \|_r\| g_n \|_s.
\]
By definition, there exists constants $D, E>0$ and $p, q\in
\mathbb{N}_0$ such that
\begin{eqnarray*}
\| f_n \|_r & \leq & D(n+1)^p \\
\| g_n \|_s & \leq & E(n+1)^q.
\end{eqnarray*}
Combining these inequalities, we obtain
\[
\| f_n g_n\|_m  \leq C_mDE(n+1)^{p+q}.
\]
This proves that $(\| f_n g_n\|_m ) \in \mathbf{s}^{\prime}$, thus
$(f_n)\cdot(g_n) \in \mathcal{H}_{\mathbf{s}^{\prime}}$.

\noindent Now, we prove that $\mathcal{H}_{\mathbf{s}}$ is an
ideal of $\mathcal{H}_{\mathbf{s}^{\prime}}$. Let $(f_n) \in
\mathcal{H}_{\mathbf{s}^{\prime}}$, $(g_n) \in
\mathcal{H}_{\mathbf{s}}$ and $m \in \mathbb{N}_0$. From
(\ref{defeal}) we have that for each $r \in \mathbb{N}_0$ there
exists a constant $D>0$ and $p\in \mathbb{N}_0$ such that
\[
\| f_n \|_r  \leq  D(n+1)^p
\]
and for all $s,l \in \mathbb{N}_0$,
\[
\|(\| g_n\|_s)\|^2_l =\sum_{n=0}^{\infty}(n+1)^{2l}\| g_n\|^2_s <
\infty.
\]

\noindent Combining the inequality (\ref{eq1}) with the above
equations we obtain
\begin{eqnarray*}
\|(\| f_n g_n\|_m)\|^2_l & = & \sum_{n=0}^{\infty}(n+1)^{2l}\| f_n
g_n\|^2_m \\
& \leq &  C^2_m \sum_{n=0}^{\infty}(n+1)^{2l} \| f_{n} \|^2_r\| g_n \|^2_s \\
 & \leq & C^2_mD^2\sum_{n=0}^{\infty}(n+1)^{2(l+p)}\| g_n \|^2_s \\
 & < & \infty.
\end{eqnarray*}
Note that we have proved that $(\| f_n g_n\|_m) \in \mathbf{s}$,
for all $m \in \mathbb{N}_0$, therefore $(f_n)\cdot(g_n) \in
\mathcal{H}_{\mathbf{s}}$.

\noindent Finally, we prove that if $(f_n) \in
\mathcal{H}_{\mathbf{s}}$ then $(Df_n) \in
\mathcal{H}_{\mathbf{s}}$. In fact, let $m \in \mathbb{N}_0$.
Since $\{ \|\cdot\|_{m}: m \in \mathbb{N}_0\}$ is equivalent to
$\{ \|\cdot\|_{\alpha, \beta, \infty}: \alpha, \beta \in
\mathbb{N}_0\}$ we have that there exists $\alpha, \beta,
m_{\alpha, \beta} \in \mathbb{N}_0$ and a constants
$C_m,C_{\alpha,\beta+1}>0$ such that
\begin{eqnarray*}
\|Df_{n}\|_{m}& \leq & C_m\|Df_{n}\|_{\alpha, \beta, \infty} \\
& = & C_m\|f_{n}\|_{\alpha, \beta+1, \infty} \\
& \leq & C_mC_{\alpha,\beta+1}\|f_{n}\|_{m_{\alpha, \beta}}.
\end{eqnarray*}
As $(f_n) \in \mathcal{H}_{\mathbf{s}}$ we have $(\|Df_{n}\|_{m})
\in \mathbf{s}$, this implies that $(Df_n) \in
\mathcal{H}_{\mathbf{s}}$.
\end{proof}

\begin{proposition}\label{includ} Let $T\in \mathcal{S}^{\prime}$. Then
$(T_{n})\in  \mathcal{H}_{\mathbf{s}^{\prime}}$, where $T_n=\sum
_{j=0}^n T(h_j)h_j$.
\end{proposition}

\begin{proof} From Theorem \ref{rpreS}, there exists a constant $C>0$ and $p\in
\mathbb{N}_0$ such that
\[
| T(h_j) |  \leq  C(j+1)^p
\]
for all $j \in \mathbb{N}_0$. Then

\begin{eqnarray*}
\|T_{n}\|_{m}^{2}& = & \sum_{j=0}^{n}(j+1)^{2m} |T(h_{j})|^2 \\
& \leq & (n+1)^{2m} \sum_{j=0}^{n} |T(h_{j})|^2 \\
& \leq  & C(n+1)^{2(m+p+1)}.
\end{eqnarray*}
This completes the proof.
\end{proof}

\begin{definition}
The tempered algebra is defined as
\[
\mathcal{H}=\mathcal{H}_{\mathbf{s}^{\prime}}/\mathcal{H}_{\mathbf{s}}.
\]
The elements of $\mathcal{H}$ are called tempered generalized
functions.
\end{definition}
Let $(f_n) \in \mathcal{H}_{s^{\prime}}$ we will use $[f_n]$ to
denote the equivalent class $(f_n) + \mathcal{H}_{s}$.
\begin{proposition}\label{includ2}
Let $\iota:\mathcal{S}^{\prime}\rightarrow \mathcal{H}$ be the
application
\[
\iota(T)=[T_n].
\]
Then $\iota$ is a linear embedding. Moreover, we have that

$\mathbf{a)}$ For all $\varphi \in \mathcal{S}$,
\[
\iota(\varphi)=[\varphi].
\]

$\mathbf{b)}$ For all $T \in \mathcal{S}^{\prime}$
\[
\iota(DT)=D\iota(T).
\]
\end{proposition}

\begin{proof} It is clear from the previous Proposition, that $\iota$ is well defined and a linear
application. We claim that $\iota(T)=0$ implies $T=0$. Since
$(T_n) \in \mathcal{H}_{\mathbf{s}}$, we have
\[
\lim_{n \rightarrow \infty}\|T_{n}\|_{m}=0
\]
for all $m \in \mathbb{N}_0$, therefore the sequence $(T_{n})$
converges weakly to $0$, which proves the claim.

\noindent $\mathbf{a)}$ Let $\varphi\in \mathcal{S}$. We have that
$\iota (\varphi)=[\varphi_n]$ where $\varphi_{n}=\sum_{j=0}^{n}
<\varphi,h_j> h_{j}$. Then for all $m, s \in \mathbb{N}_0$ we have
that
\begin{eqnarray*}
\lim_{n \rightarrow
\infty}(n+1)^{2s}\|\varphi-\varphi_{n}\|_{m}^{2}& = & \lim_{n
\rightarrow\infty}(n+1)^{2s}\sum_{j=n+1}^{\infty}(j+1)^{2m} |<\varphi,h_j>|^{2} \\
& \leq & \lim_{n \rightarrow \infty}
\sum_{j=n+1}^{\infty}(j+1)^{2(m+s)} |<\varphi,h_j>|^{2}  \\ & = &
0
\end{eqnarray*}
where the last equality follows from $\|\varphi\|_{m+s} < \infty$.
Therefore, we conclude that $(\|\varphi-\varphi_{n}\|_m) \in
\mathbf{s}$. Since $(\varphi -\varphi_n) \in
\mathcal{H}_{\mathbf{s}}$, it follows that
$\iota(\varphi)=[\varphi]$.

\noindent $\mathbf{b)}$ Let $T \in \mathcal{S}^{\prime}$. Using the
above definitions and properties of Hermite functions it follows that:
\begin{eqnarray*}
DT_n & = & D(\sum_{j=0}^nT(h_j)h_j) \\
& = & \sum_{j=0}^nT(h_j)Dh_j \\
& = &
\sum_{j=0}^nT(h_j)\frac{1}{2}(\sqrt{j}h_{j-1}-\sqrt{j+1}h_{j+1})
\\
& = &
-\sum_{j=0}^n\frac{1}{2}(\sqrt{j}T(h_{j-1})-\sqrt{j+1}T(h_{j+1}))h_j
\\
& = &\sum_{j=0}^nDT(h_j)h_j \\
& = & (DT)_n.
\end{eqnarray*}
Therefore $D\iota(T)=[DT_n]=[(DT)_n]=\iota(DT)$.
\end{proof}

\begin{corollary} Let $\varphi,\psi\in \mathcal{S}$. Then
\[
\iota(\varphi\psi)=\iota(\varphi) \cdot \iota(\psi).
\]
\end{corollary}
\begin{proof}
We first observe that
\[
(\varphi\psi)-(\varphi_{n})\cdot(\psi_{n})=(\varphi)\cdot(\psi-\psi_{n})+(\varphi-\varphi_{n})\cdot(\psi).
\]
Applying Proposition \ref{includ2} and Lemma \ref{lema} we obtain
$(\varphi\psi)-(\varphi_{n})\cdot(\psi_{n}) \in
\mathcal{H}_{\mathbf{s}}$. Therefore
$\iota(\varphi\psi)=\iota(\varphi) \cdot \iota(\psi)$.
\end{proof}

\begin{remark}
Let $\mathcal{O}_{M}$ be the ring of multipliers of $\mathcal{S}$
(see \cite{Red1}). We have a natural multiplication from
$\mathcal{O}_{M}$ by $\mathcal{H}$ into $\mathcal{H}$, defined by
\[
g[f_n]:=[gf_n]
\]
where $g \in \mathcal{O}_{M}$ and $[f_n] \in \mathcal{H}$.

\noindent It is easy to check that the product is well defined and
that $\mathcal{H}$ is a $\mathcal{O}_{M}$-module.
\end{remark}

\noindent Now, we give two examples of tempered generalized
functions.

\begin{example} The distribution $\delta$. We have that $\iota(\delta)=[\delta_n]$, where
$\delta_n = \sum_{j=0}^{n}h_j(0)h_j$. Applying the formula
(\ref{delta}) and the following equality

\[
\sum_{j=0}^{n}h_j(x)h_j(y)=\frac{\sqrt{n+1}}{x-y}\Big(h_{n+1}(x)h_{n}(y)-h_{n+1}(y)h_{n}(x)\Big),
\]

we see that

\begin{equation}\label{delta1} \delta_n(x) =
\left \{
\begin{array}{lll}

           \sqrt{n+1}
\frac{(-1)^{\frac{n}{2}}}{\sqrt[4]{2\pi}}\sqrt{\frac{1}{2}\frac{3}{4}\cdots\frac{n-1}{n}}
 \  \ \frac{h_{n+1}(x)}{x}& & \mbox{for $n$ even},        \\
           \sqrt{n+1}
\frac{(-1)^{\frac{n+3}{2}}}{\sqrt[4]{2\pi}}\sqrt{\frac{1}{2}\frac{3}{4}\cdots\frac{n}{n+1}}
 \  \ \frac{h_{n}(x)}{x} & & \mbox{for $n$ odd}.
\end{array}
\right .
\end{equation}
\end{example}

\begin{example} The element $\delta^{2}$. We have that $\delta^{2}\equiv \iota(\delta)\cdot
\iota(\delta)=[\delta_n^2]$. From (\ref{delta1}) it follows that

\begin{equation}\label{deltaquadrado} \delta^2_n (x)=
\left \{
\begin{array}{lll}

           (n+1)
\frac{1}{\sqrt{2\pi}}(\frac{1}{2}\frac{3}{4}\cdots\frac{n-1}{n})
 \  \ \frac{h_{n+1}^{2}(x)}{x^{2}}& & \mbox{for $n$ even},        \\
           (n+1)
\frac{1}{\sqrt{2\pi}}(\frac{1}{2}\frac{3}{4}\cdots\frac{n}{n+1})
 \  \ \frac{h_{n}^{2}(x)}{x^{2}}& & \mbox{for $n$ odd}.
\end{array}
\right .
\end{equation}

\end{example}

\noindent We introduce the concept of association for tempered
generalized functions.

\begin{definition}\label{ass1} Let $[f_n]$ and $[g_n]$ be tempered generalized functions. We say
that $[f_n]$ and $[g_n]$ are associated, denoted by $ [f_n]
\approx [g_n]$, if for all $\varphi \in \mathcal{S}$
\[
\lim_{n\rightarrow\infty}<f_n-g_n,\varphi>=0.
\]
\end{definition}
\noindent We observe that the relation $\approx$ is well defined.
In fact, if $(l_n)\in \mathcal{H}_{\mathbf{s}}$ and $\varphi \in
\mathcal{S}$ we have that
$\lim_{n\rightarrow\infty}<l_n,\varphi>=0$. It follows immediately
that $\approx$ is an equivalence relation on $\mathcal{H}$.
\begin{proposition}\label{equias}
$\mathbf{a)}$ Let $[f_n], [g_n] \in \mathcal{H}$ such that $ [f_n]
\approx [g_n]$. Then $ D^{\alpha}[f_n] \approx D^{\alpha}[g_n]$
for all $\alpha \in \mathbb{N}$.

$\mathbf{b)}$ Let $[f_n], [g_n] \in \mathcal{H}$ such that $ [f_n]
\approx [g_n]$ and $l \in \mathcal{O}_M$. Then $ l[f_n] \approx
l[g_n]$.

$\mathbf{c)}$ Let $T, S \in \mathcal{S}^{\prime}$ such that
$\iota(T)\approx \iota(S)$. Then $T=S$.
\end{proposition}

\begin{proof}
\noindent $\mathbf{a)}$ By integration by parts and hypothesis,
\begin{eqnarray*}
\lim_{n\rightarrow\infty}<D^{\alpha}f_n -D^{\alpha}g_n, \varphi> &
= & \lim_{n\rightarrow\infty}<f_n -g_n, (-1)^{\alpha}D^{\alpha
}\varphi> \\ & = & 0
\end{eqnarray*}
for all $\varphi\in \mathcal{S}$, that is, $ D^{\alpha}
[f_n]\approx D^{\alpha}[g_n]$.

\noindent $\mathbf{b)}$ Let $\varphi\in \mathcal{S}$. As $l \in
\mathcal{O}_M$ we have $l\varphi \in \mathcal{S}$. By assumption,
\begin{eqnarray*}
\lim_{n\rightarrow\infty}<lf_n -lg_n, \varphi> & = &
\lim_{n\rightarrow\infty} <f_n -g_n, l\varphi> \\
& = & 0.
\end{eqnarray*}
We conclude that $ l[f_n] \approx l[g_n]$.

\noindent $\mathbf{c)}$ By definition, $\lim_{n\rightarrow\infty}
\int (T-S)_n(x) \ h_{k}(x) \ dx =(T-S)(h_k)$ for all $k\in
\mathbf{N}_0$. But $\lim_{n\rightarrow\infty} \int (T-S)_n(x) \
h_{k}(x) \ dx =0$ since $\iota(T)\approx \iota(S)$. Applying the
N-representation theorem we conclude that $T=S$.
\end{proof}

\begin{example}  $x \iota( \delta )\approx 0$. In fact,
\[
\lim_{n\rightarrow\infty}<x\delta_{n}, h_{k}> =(xh_{k})(0)=0.
\]
 \end{example}

\noindent Finally, we are interested in the relation between the
symmetric Hermite product of tempered distributions via Hermite
expansions and the product of tempered generalized functions. The
symmetric Hermite product of tempered distribution was introduced
by C. Shen and M. Sun in \cite{Shen-Sun} based on our previous
work \cite{calo}.
\begin{definition}\label{MulI}
Let $S$ and $T$ be tempered distributions. Suppose that for all
$k\in \mathbb{N}\cup\{0\}$ there exists
\[
c_k =\lim_{n\rightarrow\infty}<T_nS_n ,h_k>
\]
and that $(c_k) \in \mathbf{s}^{\prime}$. The symmetric Hermite
product of $S$ and $T$, denoted by $S \bullet T$, is defined to be
the tempered distribution
\begin{equation}\label{pro3}
\sum_{k=0}^{\infty}c_k h_k.
\end{equation}
\end{definition}
\begin{lemma} $\mathbf{a)}$ The symmetric Hermite
product is commutative and distributive.

\noindent $\mathbf{b)}$ The symmetric Hermite product verifies the
Leibnitz rule: Let $S$ and $T$ be in $\mathcal{S}^{\prime}$, then
\[
D(S \bullet T)=DS \bullet T +S \bullet DT.
\]
\noindent $\mathbf{c)}$ Let $S$ and $T$ be in
$\mathcal{S}^{\prime}$ such that there exists $S \bullet T$. Then
\[
\iota(S) \cdot \iota(T) \approx \iota(S \bullet T).
\]
\end{lemma}
\begin{proof}
$\mathbf{a)}$, $\mathbf{b)}$ The proofs are straightforward (see
\cite{calo}).

\noindent $\mathbf{c)}$ It is immediate from the definitions.

\end{proof}

\begin{remark} In order to work with ordinary differential equations
in the tempered generalized  functions setting, we introduce the
algebra $\mathcal{H}_T$ of time dependent tempered generalized
functions. We can proceed in a similar way to the construction of
the algebra $\mathcal{H}$, the details are left to the reader. Let
$\mathcal{S}_T$ be the set of functions $f:[0,T]\times\mathbb{R}
\rightarrow \mathbb{R}$ such that for each $t \in [0,T]$,
$f(t,\cdot) \in \mathcal{S}$ and for each $x \in \mathbb{R}$,
$f(\cdot, x) \in C^1([0,T])$. The set
$\mathcal{H}^T_{\mathbf{s}^{\prime}}$ is given by
\[
\{(f_{n})\in \mathcal{S}_T^{\mathbb{N}_0} \ : \ \mbox{for each $m
\in \mathbb{N}_0$,} \ (\sup_{t \in [0,T]}\| f_{n}(t, \cdot)
\|_{m}),\ (\sup_{t \in [0,T]}\| \frac{\partial f_{n}}{\partial
t}(t, \cdot) \|_{m})\in \mathbf{s}^{\prime} \ \}
\]
and the set $\mathcal{H}^T_{\mathbf{s}}$ is given by
\[
\{(f_{n})\in \mathcal{S}_T^{\mathbb{N}_0} \ : \ \mbox{for each $m
\in \mathbb{N}_0$,} \ (\sup_{t \in [0,T]}\| f_{n}(t, \cdot)
\|_{m}),\ (\sup_{t \in [0,T]}\| \frac{\partial f_{n}}{\partial
t}(t, \cdot) \|_{m})\in \mathbf{s} \  \}.
\]
It is clear that $\mathcal{H}^T_{\mathbf{s}}$ is a differentiable
ideal of the algebra $\mathcal{H}^T_{\mathbf{s}^{\prime}}$. We
define the algebra $\mathcal{H}_T$ as
$\mathcal{H}^T_{\mathbf{s}^{\prime}}/\mathcal{H}^T_{\mathbf{s}}$.
The elements of $\mathcal{H}_T$ are called time depended tempered
generalized functions. It follows immediately that for $[f_n] \in
\mathcal{H}_T$ we have that $\frac{\partial}{\partial t}[f_n(t,
\cdot )]$ is define by $[\frac{\partial}{\partial t}f_n (t, \cdot )]
\in \mathcal{H}$.

\end{remark}
\section{Tempered numbers and point values}

\begin{definition}\label{genseq} The ring of tempered numbers is defined as
\begin{equation}
\mathbf{h}=\mathbf{s}^{\prime}/ \mathbf{s}.
\end{equation}
The elements of $\mathbf{h}$ are called tempered numbers.
\end{definition}
Let $(b_n) \in \mathbf{s}^{\prime}$, we will use $[b_n]$ to denote
the equivalence class $(b_n) + \mathbf{s}$.

\begin{lemma} $\mathbf{a})$ Let $\iota:\mathbb{R}\rightarrow \mathbf{h}$ be the application
\[
\iota(a)=[a].
\]
Then $\iota$ is an embedding.

$\mathbf{b})$ $\mathcal{H}$ is a $\mathbf{h}$-module with the
natural operations.

$\mathbf{c})$ Let $[f_n] \in \mathcal{H}$ and $a \in \mathbb{R}$.
Then $[f_n(a)]\in \mathbf{h}$.
\end{lemma}
\begin{proof}
 $\mathbf{a})$ It is clear that $(a) \in \mathbf{s}^{\prime}$, then
$\iota(a)=[a]$ is well defined. Assuming that $\iota(a)=0$, we
have that $(a) \in \mathbf{s}$. In particular $\lim_{n \rightarrow
\infty}na=0$, it follows that $a=0$.

\noindent $\mathbf{b})$ We have divided the proof into two parts.
We first prove that for $(b_n) \in \mathbf{s}^{\prime}$ and $(f_n)
\in \mathcal{H}_{\mathbf{s}^{\prime}}$ we have $(b_nf_n) \in
\mathcal{H}_{\mathbf{s}^{\prime}}$. In fact, by definition there
exists constants $E, F>0$ and $p, q\in \mathbb{N}_0$ such that
\begin{eqnarray*}
\| f_n \|_m & \leq & E(n+1)^p \\
| b_n | & \leq & F(n+1)^q.
\end{eqnarray*}
Combining these inequalities, we obtain
\[
\| b_nf_n \|_m  \leq EF(n+1)^{p+q}.
\]
This proves that $(\| b_nf_n \|_m ) \in \mathbf{s}^{\prime}$, thus
$(b_nf_n) \in \mathcal{H}_{\mathbf{s}^{\prime}}$.

\noindent Finally, the proof is completed by showing that for
$(a_n) \in \mathbf{s}$ and $(f_n) \in
\mathcal{H}_{\mathbf{s}^{\prime}}$ or $(a_n) \in
\mathbf{s}^{\prime}$ and $(f_n) \in \mathcal{H}_{\mathbf{s}}$ we
have $(a_nf_n) \in \mathcal{H}_{\mathbf{s}}$.

\noindent $\mathbf{c})$ Since $\delta_a \in \mathcal{S}^{\prime}$,
there exists a constant $C>0$ and $m \in \mathbb{N}_0$ such that
\[
|f_n(a)|=|\delta_a(f_n)| \leq C\|f_n \|_{m},
\]
for all $n \in \mathbb{N}_0$.

\noindent Combining the above inequality with $(\|f_n \|_{m})\in
\mathbf{s}^{\prime}$ we conclude that $(f_n(a)) \in
\mathbf{s}^{\prime}$.
\end{proof}

\begin{remark}
We observe that $\mathbf{h}$ is not a field, since there exist
zero divisors in  $\mathbf{h}$. In fact, $[1+(-1)^n],
[1+(-1)^{n+1}] \in \mathbf{h}$ are non zero and its product is
zero.
\end{remark}
\begin{definition} The point value of $[f_n]\in \mathcal{H}$ in $a\in
\mathbb{R}$, denoted by $[f_n](a)$, is defined to be $[f_n(a)]$.
\end{definition}
\begin{example} The point value of $\delta$ in $a\in
\mathbb{R}$. From  (\ref{delta1}) we have that
$\iota(\delta)(a)=[a_n]$, where

\begin{equation}\nonumber a_n=
\left \{
\begin{array}{lll}

           \frac{\sqrt{n+1}}{a}h_{n+1}(a)h_{n}(0)& & \mbox{for $n$ even},        \\
           -\frac{\sqrt{n+1}}{a}h_{n}(a)h_{n+1}(0)& & \mbox{for $n$ odd}.
\end{array}
\right .
\end{equation}
\end{example}

\begin{example} The point value of $x_{+}$ in 0. It is easy to check that
$\iota(x_+)(0)=[a_n]$, where

\begin{equation}\nonumber a_n=
\left \{
\begin{array}{lll}

           \sqrt{n+1}h_{n}(0) \int_{0}^{\infty} h_{n+1}(x) \ dx& & \mbox{for $n$ even},        \\
           \sqrt{n}h_{n-1}(0) \int_{0}^{\infty} h_{n}(x) \ dx& & \mbox{for $n$ odd}.
\end{array}
\right .
\end{equation}
\end{example}

\noindent  We introduce the concept of association for tempered
numbers.
\begin{definition}\label{ass2}
Let $[a_n]$ and $[b_n]$ be tempered numbers. We say that $[a_n]$
and $[b_n]$ are associated, denoted by $ [a_n] \approx [b_n]$, if
\[
\lim_{n\rightarrow\infty}(a_n-b_n)=0.
\]
\end{definition}
\noindent We observe that the relation $\approx$ is well defined
and that $\approx$ is an equivalence relation on $\mathbf{h}$.
\begin{example} $\iota(x_+)(0) \approx 0$.
\end{example}

\section{Integration and Fourier transform}

In this section we present the integration theory of tempered
generalized functions and the Fourier transform.
\begin{definition}\label{defin} Let $[f_n]\in
\mathcal{H}$ and $A$ be a Lebesgue measurable set. The integral of
$[f_n]$ on $A$, denoted by $\int_A [f_n](x) \ dx$ is defined to be
\begin{equation}\label{integr}
[\int_Af_{n}(x) \ dx].
\end{equation}
\end{definition}

\noindent We observe that the integral is well defined as an
element of $\mathbf{h}$. In fact, as $1_A$ is a tempered
distribution there exists a constant $C>0$ and $m \in
\mathbb{N}_0$ such that
\[
|\int_{A} g(x)\ dx|\leq C \|g \|_{m},
\]
for all $g \in \mathcal{S}$. In particular, for $[f_n]\in
\mathcal{H}$ we have that
\[
|\int_{A} f_{n}(x)\ dx|\leq C \|f_{n} \|_{m}.
\]
As $(\|f_{n} \|_{m}) \in \mathbf{s}^{\prime}$, we conclude that
$\int_A [f_n](x) \ dx \in \mathbf{h}$.

In the next Lemma we collect some fundamental properties of the
integral of tempered generalized functions.

\begin{lemma} Let $[f_n]$ and $[g_n]$ be tempered
generalized functions, $a=[a_n] \in \mathbf{h}$ and $\alpha\in
\mathbb{N}$. Then

\noindent $\mathbf{a)}$ Let $A$ and $B$ disjoint Lebesgue
measurable sets. Then
\[
\int_{A\cup B} [f_n](x) \ dx= \int_{A} [f_n](x) \ dx + \ \int_{B}
[f_n](x) \ dx.
\]
$\mathbf{b)}$
\[
\int_{A} ([f_n]+a[g_n])(x) \ dx= \int_{A} [f_n](x) \ dx + \ a
\int_{A} [g_n](x) \ dx.
\]

\noindent $\mathbf{c)}$ Let $\varphi\in \mathcal{S}$. Then
\[
\iota(\int_{A} \varphi \ dx)= \int_{A} \iota(\varphi) \ dx.
\]

\noindent $\mathbf{d)}$ "Rule of integration by parts"
\[
\int_{\mathbb{R}}  [f_n] D^{\alpha} [g_n](x) \ dx=(-1)^{\alpha}
\int_{\mathbb{R}}( D^{\alpha}[f_n]) [g_n](x)\ dx.
\]

\noindent $\mathbf{e)}$ Let $\varphi\in \mathcal{S}$ and $T \in
\mathcal{S}^{\prime}$. Then
\[
\int_{\mathbb{R}} \iota(T) \cdot \iota (\varphi)(x) \ dx =
\iota(T(\varphi)).
\]
\end{lemma}
\begin{proof}
The proof of $\mathbf{a)}$, $\mathbf{b)}$ and $\mathbf{c)}$ are
immediate.

\noindent $\mathbf{d)}$ We have that
\begin{eqnarray*}
\int_{\mathbb{R}}  [f_n] D^{\alpha} [g_n](x) \ dx & = &
[\int_{\mathbb{R}}  f_n D^{\alpha} g_n(x) \ dx] \\
& = & [\int_{\mathbb{R}} (-1)^{\alpha} D^{\alpha}f_n (x) g_n(x) \
dx] \\ & = & (-1)^{\alpha} \int_{\mathbb{R}}( D^{\alpha}[f_n])
[g_n](x)\ dx.
\end{eqnarray*}
\noindent $\mathbf{e)}$ By definitions
\begin{eqnarray*}
\int_{\mathbb{R}} \iota(T) \cdot \iota (\varphi)(x) \ dx & = &
[\int_{\mathbb{R}} T_n(x) \varphi(x) \ dx]
\end{eqnarray*}
Let us prove that $(T(\varphi)-\int T_n(x) \varphi(x) \ dx)\in
\mathbf{s}$. Combining definitions, the  N-representation theorem
 and $\lim_{n \rightarrow \infty}<\varphi,h_j>=0$,
we obtain that there exists $n_0 \in \mathbb{N}$ such that if $n
\geq n_0$,
\begin{eqnarray*}
|T( \varphi)-\int_{\mathbb{R}} T_{n}(x) \varphi(x) \ dx| & = &
|T-T_{n}(
\varphi)| \\
& \leq & \sum_{j=n+1}^{\infty} |T(h_j)| |<\varphi,h_j>| \\
& \leq & \sum_{j=n+1}^{\infty} C(j+1)^p |<\varphi,h_j>| \\
& \leq & \sum_{j=n+1}^{\infty} C(j+1)^{2p} |<\varphi,h_j>|^2.
\end{eqnarray*}
Since $\varphi \in \mathcal{S}$, it follows that
\[
\lim_{n \rightarrow \infty} \sum_{j=n+1}^{\infty} (j+1)^{q}
|<\varphi,h_j>|^2=0
\]
for all $q \in \mathbb{N}_0$. Combining the above inequalities we
see that
\[
\lim_{n \rightarrow \infty}(n+1)^{r}|T( \varphi)-\int_{\mathbb{R}}
T_{n}(x) \varphi(x) \ dx|=0,
\]
for all $r \in \mathbb{N}_0$. This proves that $(T( \varphi)-\int
T_n(x) \varphi(x) \ dx)\in \mathbf{s}$, which completes the proof.
\end{proof}

\begin{example} Let $T\in \mathcal{S}^{\prime}$. Then
\begin{eqnarray*}
\int_{\mathbb{R}} \iota (T)(x) \ dx & = & [\int_{\mathbb{R}}
T_n(x) \ dx] \\
& = & [1( T_n)] \\
& = & [\sum_{ j \ even }^{n} T(h_j)h_j(0)(-1)^{\frac{j}{2}}],
\end{eqnarray*}
where the last equality follows from formula (\ref{uno}).
\end{example}

\begin{example} $\delta^{2}$. From formula (\ref{deltaquadrado}) we have that
\begin{eqnarray*}
\int_{\mathbb{R}} \delta^{2}(x) \ dx & = & [\int_{\mathbb{R}}
\delta_n^{2}(x) \ dx] \\
& = & [ \frac{(n+1)}{\sqrt{2\pi}}
\frac{1}{2}\frac{3}{4}\cdots\frac{n-1}{n}].
\end{eqnarray*}
\end{example}
\noindent The Fourier transform and the convolution are very
important tools of classical and modern analysis, our aim is
introduce these operations in the context of tempered generalized
functions. We recall that the Fourier transform
$\mathcal{F}:\mathcal{S}\rightarrow\mathcal{S}$ is defined by
\[
\mathcal{F}(\varphi)(t)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-itx}\varphi(x)
\; dx
\]
and the convolution $\ast: \mathcal{S} \times \mathcal{S}
\rightarrow\mathcal{S}$ is defined by
\[
\varphi \ast \psi (t) = \int_{\mathbb{R}}\varphi(t-x)\psi(x) \;
dx.
\]
For a fuller treatment of the issues discussed below,  we refer the reader to
\cite{Red2}.
\begin{definition} The Fourier transform of a generalized tempered function $[f_n]$, denoted by $\mathcal{F}([f_n])$,
is defined to be $[\mathcal{F}(f_n)]$.
\end{definition}
\noindent We observe that the above definition is independent of
the representatives, because for all $m \in \mathbb{N}_0$ and
$\varphi \in \mathcal{S}$ we have that $\|\mathcal{F}(\varphi)\|_m
= \| \varphi \|_m$.

\noindent Here are some elementary properties of the Fourier
transform and convolution.

\begin{theorem}
$\mathbf{a)}$ The Fourier transform $\mathcal{F}: \mathcal{H}
\rightarrow\mathcal{H}$ is a linear isomorphism and its inverse is
given by
\[
\mathcal{F}^{-1}([f_n])=[\mathcal{F}^{-1}(f_n)].
\]

\noindent $\mathbf{b)}$ Let $[f_n] \in \mathcal{H}$ and $\alpha
\in \mathbb{N}_0$. Then
\begin{eqnarray*}
\mathcal{F}(D^{\alpha}[f_n]) & = & (ix)^{\alpha}\mathcal{F}([f_n])
\\
\mathcal{F}(x^{\alpha}[f_n]) & = & i^{\alpha}D^{\alpha}
\mathcal{F}([f_n]).
\end{eqnarray*}

\noindent $\mathbf{c)}$ Let $T$ be a tempered distribution. Then
$\iota(\mathcal{F}(T))= \mathcal{F}(\iota (T))$.

\end{theorem}
\begin{proof}
\noindent $\mathbf{a)}$ Define $\mathcal{G}: \mathcal{H}
\rightarrow \mathcal{H}$ by
$\mathcal{G}([f_n])=[\mathcal{F}^{-1}(f_n)]$. We observe that
$\mathcal{G}$ is well defined, because for any $m \in
\mathbb{N}_0$ and $\varphi \in \mathcal{S}$ we have $\|\varphi\|_m
=\| \mathcal{F}^{-1} \varphi\|_m$.  It is clear that
$\mathcal{F}\circ \mathcal{G} = I_{\mathcal{H}}$ and
$\mathcal{G}\circ \mathcal{F} = I_{\mathcal{H}}$.

\noindent $\mathbf{b)}$ and  $\mathbf{c)}$. The proofs follows
from the definitions and properties of the Fourier transform in
$\mathcal{S}$.
\end{proof}
\begin{definition} Let $[f_n]$ and $[g_n]$ be  tempered generalized  functions. The convolution of $[f_n]$ and $[g_n]$,
denoted by $[f_n]\ast[g_n]$, is defined to be $\mathcal{F}^{-1}(
\sqrt{2 \pi} \mathcal{F}([f_n]) \cdot \mathcal{F}([g_n]))$.
\end{definition}
\begin{theorem}
\noindent $\mathbf{a)}$ Let $[f_n], [g_n] \in \mathcal{H}$. Then
\[
[f_n]\ast[g_n]=[f_n \ast g_n].
\]
\noindent $\mathbf{b)}$ Let $[f_n], [g_n], [h_n] \in \mathcal{H}$
and $\alpha \in \mathbb{N}_0$. Then
\begin{eqnarray*}
[f_n]\ast[g_n] & = & [g_n]\ast[f_n] \\
D^{\alpha}([f_n]\ast[g_n]) & = & (D^{\alpha}[f_n]) \ast[g_n] \\
([f_n]\ast[g_n]) \ast [h_n] & = & [f_n] \ast ([g_n] \ast [h_n])
\end{eqnarray*}

\noindent $\mathbf{c)}$  Let $[f_n], [g_n] \in \mathcal{H}$. Then
\begin{eqnarray*}
\mathcal{F}([f_n]\cdot[g_n]) & = & \frac{1}{\sqrt{2 \pi}}
\mathcal{F}([f_n]) \ast \mathcal{F}([g_n]) \\
\mathcal{F}([f_n]\ast[g_n]) & = & \sqrt{2 \pi} \mathcal{F}([f_n])
\cdot \mathcal{F}([g_n])
\end{eqnarray*}

\noindent $\mathbf{d)}$ Let $T\in \mathcal{S}^{\prime}$ and
$\varphi \in \mathcal{S}$. Then
\[
\iota(T) \ast\iota(\varphi)\approx\iota(T\ast \varphi).
\]
\end{theorem}
\begin{proof}
\noindent $\mathbf{a)}$ The proof is a consequence of the above
theorem and definitions.

\noindent $\mathbf{b)}$ and  $\mathbf{c)}$ The proofs follows from
the definitions and properties of the convolution in
$\mathcal{S}$.

\noindent $\mathbf{d)}$  We have that for all $\psi \in
\mathcal{S}$,
\begin{eqnarray*}
\lim_{n\rightarrow \infty}< (T_{n}\ast\varphi), \psi >
& = & T\ast\varphi(\psi) \\
& = & \lim_{n\rightarrow \infty}< (T \ast \varphi)_n , \psi>.
\end{eqnarray*}
This shows that $\iota(T) \ast\iota(\varphi)\approx\iota(T\ast
\varphi)$.
\end{proof}
\begin{example} The Fourier  transform of $\delta$. By formula (\ref{delta}) we
have that
\[
\mathcal{F}(\iota (\delta))=[\sum_{k=0}^{n} (-i)^{k} h_{k}(0)
h_{k}(x)].
\]
\end{example}

\section{Generalized Stochastic Calculus}

\noindent Let $(\Omega, \mathcal{F}, \{ \mathcal{F}_t: t \in [0,T]
\}, \mathbb{P})$ be a filtered probability space, which satisfies
the usual hypotheses. For a recent account of stochastic calculus
and notations we refer the reader to the book of  Ph. Protter
\cite{Protter}.

\begin{definition}
Let $[f_n] \in \mathcal{H}$, $X$ be a continuous jointly
measurable process and $V$ be a finite variation process. We
define the integral of $[f_n](X)$ in relation to $V$ from 0 to
$t$, denoted by $\int_0^t[f_n](X_s)dV_s$, to be
\[
[\int_0^tf_n(X_s)dV_s].
\]
\end{definition}
\noindent It is clear that for each $\omega \in \Omega$ and $t \in
[0,T]$ we have that $[\int_0^tf_n(X_s)dV_s(\omega)] \in
\mathbf{h}$, because
\[
|\int_0^tf_n(X_s)dV_s(\omega)| \leq \sup_x|f_n(x)||V|_t(\omega)
\]
where $|V|_t(\omega)$ is the total variation of $V$ in $[0,t]$.
\begin{definition}
$\mathbf{a})$ Let $[f_n] \in \mathcal{H}$ and $X$ be a random
variable. We define the expectation of $[f_n](X)$, denoted by
$\mathbb{E}([f_n](X))$, to be $[\mathbb{E}(f_n(X))]$.

$\mathbf{b})$ Let $[f_n] \in \mathcal{H}$, $X$ be a continuous
jointly measurable process and $V$ be a finite variation process
such that $|V|_t$ is integrable. We define the expectation of
$\int_0^t[f_n](X_s)dV_s$, denoted by
$\mathbb{E}(\int_0^t[f_n](X_s)dV_s)$, to be $[
\mathbb{E}(\int_0^tf_n(X_s)dV_s)]$.
\end{definition}
It is easy to check that the above notion is  well
defined. We also observe that the natural definition of expectation
doesn't work. In fact, let $Y_n: \Omega \rightarrow \mathbb{R}$ be
random variables such that $(Y_n(\omega)) \in \mathbf{s}^{\prime}$
for all $\omega \in \Omega$. We have that $[(\mathbb{E}(Y_n))]$ is
dependent of the representatives $(Y_n)$, because if $\{ A_n: n
\in \mathbb{N}\}$ is a partition measurable of $\Omega$ such that
$\mathbb{P}(A_n)=\frac{1}{2^n}$ for all $n \in \mathbb{N}$ and
$(b_n) \in \mathbf{s}^{\prime}$,
$[Y_n(\omega)]=[Y_n(\omega)+b_n2^n1_{A_n}(\omega)]$ and
$(\mathbb{E}(b_n2^n1_{A_n})) =(b_n) \notin \mathbf{s}$.

We can now prove the It\^o formula for tempered generalized
functions. Clearly, this formula is an extension of the classical
It\^o formula via infinite dimensional methods.
\begin{theorem}
Let $[f_n] \in \mathcal{H}$ and $X$ be a continuous
semimartingale. Then
\begin{equation}\label{ito}
[f_n](X_t)=[f_n](X_0)+\int_0^tD[f_n](X_s)dX_s +
\frac{1}{2}\int_0^tD^2[f_n](X_s)d<X>_s
\end{equation}
where $\int_0^tD[f_n](X_s)dX_s(\omega)$ defined by
$[\int_0^tDf_n(X_s)dX_s(\omega)]$ is the It\^o integral of
$[f_n](X)$ in relation to $X$ from 0 to $t$.
\end{theorem}
\begin{proof}
We first show that $[\int_0^tDf_n(X_s)dX_s(\omega)]$ is well
defined. In fact, let $(g_n) \in \mathcal{H}_{\mathbf{s}}$. Since
$(Dg_n) \in \mathcal{H}_{\mathbf{s}}$, we have
$(\int_0^tDg_n(X_s)d<X>_s(\omega)) \in \mathbf{s}$. We see that
$(g_n(X_t(\omega)))$ and $(g_n(X_0(\omega)))$ are in $\mathbf{s}$,
which is clear from the definition of point value. Combining these
facts with the It\^o formula we have
\[
(\int_0^tDg_n(X_s)dX_s(\omega))=(g_n(X_t(\omega)))-(g_n(X_0(\omega)))-
(\frac{1}{2}\int_0^tD^2g_n(X_s)d<X>_s(\omega))
\]
are in $\mathbf{s}$ . Finally we see that
$[\int_0^tDf_n(X_s)dX_s(\omega)] \in \mathbf{h}$ and the formula
(\ref{ito}) holds, this is clear from the It\^o formula applied to
$f_n$,
\[
f_n((X_t(\omega)))=
f_n((X_0(\omega)))+\int_0^tDf_n(X_s)dX_s(\omega)+\frac{1}{2}\int_0^tD^2f_n(X_s)d<X>_s(\omega).
\]
\end{proof}
\begin{remark} Let $f \in \mathcal{S}$. By Proposition \ref{includ2},
$[f]=[f_n]$. Then it is clear that
\[
[ f_n](X_t)  =  [f(X_t)]
\]
and
\[
\int_0^tD^2[f_n](X_s)d<X>_s =  [\int_0^tD^2f(X_s)d<X>_s].
\]
Consequently,
\[
\int_0^tD[f_n](X_s)dX_s = [\int_0^tDf(X_s)dX_s].
\]
In particular, the members of the It\^o formula for $f$ as
function are the same that the members of the It\^o formula for
$f$ as tempered generalized function.
\end{remark}
\begin{remark}
We observe that the members of the It\^o formula for $C^4$ functions
with appropriated decreasing at infinite are associated with the
corresponding members of the It\^o formula as tempered generalized
functions. In fact, we have that $(f_n)$ converges uniformly over
compacts whenever $f$ is twice continuously differentiable and
$O(e^{-cx^2})$ for some $c>1$ as $x \rightarrow \infty$ (see
\cite{stone} for more details). In particular,
\[
[f_n](x) \approx [f(x)]
\]
for all $x \in \mathbb{R}$. Thus
\[
[ f_n](X_t)  \approx  [f(X_t)]~~\mbox{ and }~~[ f_n](X_0)  \approx
[f(X_0)].
\]
If $D^2f \in \mathcal{C}^2$ and $D^2f$ is $O(e^{-cx^2})$ for some
$c>1$ as $x \rightarrow \infty$ we have that
\[
\int_0^tD^2[f_n](X_s)d<X>_s \approx [\int_0^tD^2f(X_s)d<X>_s].
\]
Combining the above identities with the classical It\^o formula
for $f$ we conclude that
\[
\int_0^tD[f_n](X_s)dX_s \approx [\int_0^tDf(X_s)dX_s].
\]
\end{remark}
\begin{remark}
We recall that S. Ustunel \cite{ustunel} and B. Rajeev \cite{rajeev}
obtained It\^o formula for tempered distributions, in the context of
stochastic integration in Hilbert spaces. Our formula (\ref{ito}) is
different from the It\^o formulas given in \cite{rajeev} and
\cite{ustunel}, in particular, the formulas in these references do
not extend the classical It\^o formula as ours does.

\end{remark}
\begin{corollary}
Let $X$ be a semimartingale. Then
\begin{equation}\label{itotanaka}
|X_t-a|=
|X_0-a|+\int_0^t\mathrm{sgn}(X_s-a)dX_s+\int_0^t\delta_a(X_s)d<X>_s.
\end{equation}

\end{corollary}
\begin{proof}
Applying the It\^o formula (\ref{ito}) to the tempered
distribution $|\cdot -a|$ we obtain (\ref{itotanaka}).

\end{proof}

\begin{corollary}
Let $[f_n] \in \mathcal{H}$ and $B$ be a Brownian motion such that
$B_0=0$. Then
\begin{equation}\label{dynkin}
\mathbb{E}([f_n](B_t+x))=[f_n](x)+
\frac{1}{2}\int_0^t\mathbb{E}(D^2[f_n](B_s+x))ds
\end{equation}
\end{corollary}
\begin{proof}
We have
\[
\mathbb{E}([\int_0^tD[f_n](B_s+x)dB_s])=[\mathbb{E}\int_0^tDf_n(B_s+x)dB_s]=0,
\]
because $\int Df_n(B_s+x)dB_s$ is a martingale.
\end{proof}
\begin{corollary}
Let $[f_n] \in \mathcal{H}$. Then
$g_t=[\mathbb{E}(f_n(B_t+\cdot))] \in \mathcal{H}_T$ solves the
Cauchy problem
\begin{eqnarray*}
D_tg & = & \frac{1}{2}D_x^2g \\
g_0 & = & [f_n].
\end{eqnarray*}
\end{corollary}
\begin{proof}
We observe that
\[
\mathbb{E}(f_n(B_t+x))=\int_{\mathbb{R}}f_n(y)p_t(x-y)~dy=f_n \ast
p_t(x)
\]
where $p_t(y)= \frac{1}{\sqrt{2 \pi t}}e^{-\frac{y^2}{2t}}$ is the
heat kernel. As $\ast$ is a continuous operation in $\mathcal{S}$
and $\lim_{n \rightarrow \infty}f_n \ast p_t=f_n$ in $\mathcal{S}$
we conclude that $[g_n] \in \mathcal{H}_T$ where the functions
$g_n:[0,T]\times\mathbb{R} \rightarrow \mathbb{R}$ are given by
$g_n(t,x)=\mathbb{E}(f_n(B_t+x))$. Applying the formula
(\ref{dynkin}) we completes the proof.
\end{proof}
\begin{theorem}
Let $[f_n] \in \mathcal{H}_T$ and $X$ be a continuous
semimartingale. Then
\begin{eqnarray*}
[f_n](X_t) & = &
[f_n](X_0)+\int_0^tD_t[f_n](s,X_s)ds+\int_0^tD_x[f_n](s,X_s)dX_s
\\& & + \frac{1}{2}\int_0^tD_x^2[f_n](s,X_s)d<X>_s.
\end{eqnarray*}
\end{theorem}
\begin{proof}
We observe that $\int_0^tD_t[f_n](s,X_s)ds$ is well defined and
proceed analogously to the proof of the extension of It\^o
formula.
\end{proof}

config: arxiv


TITLE: Manifolds with boundary and foliations
QUESTION [1 upvotes]: Is there a theory of foliations by manifolds with boundary? Particularly, Is there a generalization of the Frobenius theorem and the Stefan-Sussmann theorem in which the leaves are manifolds with boundary?
Thanks for your help!

REPLY [1 votes]: Once you assume appropriate boundary conditions, everything is the same as in the theory without boundary.
This preprint and this preprint give Frobenius and Stefan-Sussmann type theorems for manifolds with boundary assuming the distribution is normal: there is a vector in the distribution transverse to the boundary at each point. It is precisely this situation where the leaves have boundary.


TITLE: Taylor series of $\frac{x-2}{\sqrt{x^2-4x+8}}$ at $x = 2$. Where is the mistake?
QUESTION [3 upvotes]: $$\frac{x-2}{\sqrt{x^2-4x+8}} \quad \text{at}\ x_0=2$$
Let $t = x-2$. then as $x\to 2$, $t\to 0$ and we can find the Maclaurin Series in terms of $t$.
\begin{align*}
\frac{x-2}{\sqrt{x^2-4x+8}} &=\frac{x-2}{\sqrt{(x-2)^2+4}}\\
&=\frac{t}{\sqrt{t^2+4}}\\
&=\frac{1}{2}\frac{t}{\sqrt{1+\left(\frac{t}{2}\right)^2}}
\end{align*}
Since the Maclaurin Series
$$
\left(1+\left(\frac{t}{2}\right)^2\right)^{-1/2} = \sum_{k=0}^n{-1/2 \choose k}\left(\frac{t}{2}\right)^{2k} + o(t^{2n})$$
Then
$$\frac{t}{2}\left(1+\left(\frac{t}{2}\right)^2\right)^{-1/2} = \sum_{k=0}^n{-1/2 \choose k}\left(\frac{t}{2}\right)^{2k+1} + o(t^{2n})$$
Here we have
\begin{align*}
{-1/2 \choose k} &= \frac{-\frac{1}{2}\left(-\frac{1}{2}-1\right)\left(-\frac{1}{2}-2\right)...\left(-\frac{1}{2}-(k-1)\right)}{k!}\\
&=\frac{-\frac{1}{2}\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)...\left(-\frac{2k-1}{2}\right)}{k!}
\end{align*}
If we denote $1\cdot 3\cdot 5\cdot ... \cdot (2k-1) = (2k-1)!!$ we get
$${-1/2 \choose k} = \frac{(-1)^k(2k-1)!!}{2^k k!}$$
And finally
\begin{align*}\frac{t}{2}\left(1+\left(\frac{t}{2}\right)^2\right)^{-1/2} &= \sum_{k=0}^n{-1/2 \choose k}\left(\frac{t}{2}\right)^{2k+1} + o(t^{2n})\\
&= \frac{t}{2} + \sum_{k=1}^n\frac{(-1)^k(2k-1)!!}{2^kk!}\left(\frac{t}{2}\right)^{2k+1} + o(t^{2n})
\end{align*}
And in terms of $x$:
$$
\frac{x-2}{\sqrt{x^2-4x+8}}\Bigg\rvert_{x_0=2} = \frac{x-2}{2} + \sum_{k=1}^n\frac{(-1)^k(2k-1)!!}{2^kk!}\left(\frac{x-2}{2}\right)^{2k+1} + o((x-2)^{2n})$$
But the textbook's answer is
$$
\frac{x-2}{2} + \sum_{k=1}^{n-1}\frac{(-1)^k(2k-1)!!}{2^{2k+1}(2k)!!}(x-2)^{2k+1} + o((x-2)^{2n})$$
I don't get why the sum goes just to $n-1$ instead of $n$ and instead of $2^{3k+1}k!$ in the denominator, as I got, it is $2^{2k+1}(2k)!!$
What am I missing?

REPLY [2 votes]: There's no error in what you did, you only "wasted" two orders of magnitude in your remainder term.
The terms of the textbook's answer look different from yours, but are in fact the same, since
$$(2k)!! = 2^k\cdot k!$$
for one definition of the double factorial of even natural numbers (there's another definition that differs from the used one by a factor of $\sqrt{\frac{2}{\pi}}$), thus
$$2^{3k+1}\cdot k! = 2^{2k+1}(2k)!!$$
and consequently we see that the terms are identical.
You wasted one order of magnitude in the remainder term by writing
$$\biggl(1 + \Bigl(\frac{t}{2}\Bigr)^2\biggr)^{-1/2} = \sum_{k = 0}^n \binom{-\frac{1}{2}}{k} \biggl(\frac{t}{2}\biggr)^{2k} + o(t^{2n})\tag{$\ast$}$$
and interpreting the polynomial as the Taylor polynomial of order $2n$. It is also the Taylor polynomial of order $2n+1$, and therefore one can write the remainder term as $o(t^{2n+1})$. The next order of magnitude you wasted in the remainder term was when you multiplied $(\ast)$ with $\frac{t}{2}$, since $\frac{t}{2}o(t^m) = o(t^{m+1})$, and you didn't increase the exponent.
So with these orders taken into account, we find
$$\frac{x-2}{\sqrt{x^2-4x+8}} = \frac{x-2}{2} + \sum_{k = 1}^m \frac{(-1)^k (2k-1)!!}{2^{2k+1}(2k)!!}(x-2)^{2k+1} + o\bigl((x-2)^{2m+2}\bigr)$$
for any fixed $m\in \mathbb{N}$. So if we want to have a remainder term in $o(t^{2n})$, it suffices to take $m = n-1$.


TITLE: Deriving the autocorrelation function for the ARMA model
QUESTION [1 upvotes]: Definitions
The ARMA model
$$x_n=-\sum_{p=1}^P a_px_{n-p}+\sum_{q=0}^Qb_qw_{n-q} \tag{1}$$
where $w_n$ is zero mean stationary white noise with unit variance.
Question
To derive the autocorrelation function of the ARMA process, $R_{XX}[r]=E[x_nx_{n+r}]$ I think we can do this two ways:

Substitute eqn 1 into $x_{n+r}$

$$R_{XX}[r]=E\left[x_n\left\{-\sum_{p=1}^Pa_px_{n+r-p}+\sum_{q=0}^Qb_qw_{n+r-q}\right\}\right]$$
$$=-\sum_{p=1}^Pa_pE[x_nx_{n+r-p}]+\sum_{q=0}^Qb_qE[x_nw_{n+r-q}]$$
$$=-\sum_{p=1}^Pa_pR_{XX}[r-p]+\sum_{q=0}^Qb_qR_{XW}[r-q]$$

Substitute eqn 1 into $x_{n}$

$$R_{XX}[r]=E\left[\left\{-\sum_{p=1}^Pa_px_{n-p}+\sum_{q=0}^Qb_qw_{n-q}\right\}x_{n+r}\right]$$
$$=-\sum_{p=1}^Pa_pE[x_{n-p}x_{n+r}]+\sum_{q=0}^Qb_qE[w_{n-q}x_{n+r}]$$
$$=-\sum_{p=1}^Pa_pR_{XX}[r+p]+\sum_{q=0}^Qb_qR_{XW}[-r-q]$$
But $R_{XX}[r-p]\ne R_{XX}[r+p]$ and $R_{XW}[r-q]\ne R_{XX}[r+q]$, so what mistake have I made? Is only one method correct, and why?

REPLY [2 votes]: Both ways are correct, but only one way is useful.
Before, be careful with your definition of cross-correlation ($R_{XW}[t]\ne R_{WX}[t]$), it's a matter of convention, but you must be consistent. If we define $R_{XW}[t]=E[x_n w_{n-t}]$, then the  first derivation has an error, it should be:
$$R_{XX}[r]=-\sum_{p=1}^Pa_pR_{XX}[r-p]+\sum_{q=0}^Qb_qR_{XW}[q-r]\tag{1}$$
The second is ok:
$$R_{XX}[r]=-\sum_{p=1}^Pa_pR_{XX}[r+p]+\sum_{q=0}^Qb_qR_{XW}[r+q]\tag{2}$$
Both are correct (and the equivalence is non-trivial) but only $(1)$ is useful, in the Yule-Walker context (we want to write $R_{XX}[0]$, $R_{XX}[1]$ $\cdots$ and "solve" the system), because only the first one allows us to exploit causality, the fact that $x_n$ does not depend (is independent of) future values of $w$, that is, $E[x_n w_{n-t}]=R_{XW}[t]=0 $ for $t<0$.
Once more, I encourage you to get familiar with the AR case, which is conceptually similar, but simpler.


TITLE: Lachlan on topology for priority arguments
QUESTION [7 upvotes]: There is a set of notes by Lachlan from 1973 on casting priority arguments in topological language; references to these notes are few and far between, but one source refers to them as "Topology for Priority Arguments," which for now I'll assume is the title. Regardless, they seem to have never been published; my understanding is that their line of inquiry never caught on.
I did see them long ago, in a computability theory class; and I've recently become re-interested in them. The instructor, unfortunately, no longer has a copy. I was wondering if anyone knew how I could find these notes, or had a copy.

REPLY [5 votes]: After about my third priority argument in a graduate class, I too was interested in finding a general approach for handling them. You can find something similar from Lachlan in published form as The priority method for the construction of recursively enumerable sets in Lecture Notes in Mathematics 337.
The most developed general framework for priority arguments is A framework for priority arguments by Lerman. It's more order-theoretic than topological, but I doubt it would be too difficult to recast it in topological language.


TITLE: Proof regarding a divergent sequence
QUESTION [0 upvotes]: How to show that if $x_n$ diverges and $x_0$ is any number, then there exists $\epsilon$ and a subsequence ${x_n}_k$ such that $|{x_n}_k-x_0|>\epsilon$ for all k. Any hints or solutions welcome.

REPLY [1 votes]: Sketch:
Suppose $(x_n)$ diverges. Then by definition given any $x_0 \in \Bbb R $ there is an $\epsilon \gt 0$ such that for any $n \in \Bbb N$ there exists another natural number $m \ge n$ such that $ |x_m  - x_0 |\ge \epsilon$.
Now start iwth $1$. Let $r_1$ be the natural number ($m$) that satisfies this mentioned condition. Now go to $r_1 + 1$. Let $r_2$ be the natural number which satisfies this condition for $r_1 + 1$. Keep going and you will get a strictly increasing sequence of natural numbers $(r_n)$ such that $|x_{r_n} - x_0| \ge \epsilon$.


TITLE: Series, and Sequences $a_n = \frac{(2n-1)!}{(2n+1)!}$
QUESTION [2 upvotes]: Is the following sequence convergent?
  $$a_n = \frac{(2n-1)!}{(2n+1)!}$$

I have this sequence to calculate whether it converges or not, as well as where does it converge and I was wondering if the way I proceeded with the problem was optimal.
I have to admit I don't fully grasp the factorial precalculus methods, and would be open to suggestions.  I am getting the problem from the James Stewart Essential Calculus Early Transcendentals 2nd Edition.
This is the way I did the problem, and I am wondering
A. if its right, and
B. Are there resources I could have to better tackle factorials in my calculus problems.
\begin{align}
           a_n&= \frac{(2n-1)!}{(2n+1)!} \\
           \lim_\limits{n \to \infty}(a_n)&=\lim_\limits{n \to \infty}(\frac{\frac{(2n)!}{2n}}{(2n+1)(2n)!})\\
            &= \lim_\limits{n \to \infty} (\frac{1}{2n(2n+1)}) = 0
\end{align}
That is the answer in the back of the book, but I wonder if my process was right. Thank you.

REPLY [1 votes]: Your process is correct. Usually, you would simply notice that $(2n+1)!=(2n-1)! (2n)(2n+1)$ (in the same way that $8!=6!\times 7\times8$) so
$$
\frac{(2n-1)!}{(2n+1)!}=\frac{(2n-1)!}{(2n-1)! (2n)(2n+1)}=\frac{1}{(2n)(2n+1)}.
$$


TITLE: There exists a unique $\mathbb {C}[G] $-submodule of the regular module which is isomorphic to the trivial $\mathbb {C} [G] $-module.
QUESTION [0 upvotes]: Let $G $ a finite group. Show that there exists a unique $\mathbb {C}[G] $-submodule of the regular $\mathbb {C}[G] $-module which is isomorphic to the trivial $\mathbb {C} [G] $-module.
How to start?

REPLY [2 votes]: Hint: consider the module generated by $x=\sum_{g\in G}g$. Show after that if $y$ is fixed by $g$ then there exists $c\in\mathbb{C}$ such that $y=cx$ (remark that if $g\in G, y=u_1g_1+...+u_ng_n$, $gy$ permutes the coordinates $(c_1,...,c_n)$ of $y$.)


TITLE: Critic Loss in PPO
QUESTION [0 upvotes]: TL,DR: How precisely is the critic loss in PPO defined?
I am trying to understand the PPO algorithm so that I can implement it. Now I'm somewhat confused when it comes to the critic loss. According to the paper, in the objective that we want to maximize, there is a term
$$
-c_1 (V_\theta(s_t) - V_t^{targ})^2
$$
which is the loss for the critic ($"-"$ in the beginning, since the objective is maximized). I didn't really know what $V_t^{targ}$ was supposed to be at first, after some research online it looks like that's what we're getting from GAE for $\hat{A}_t$, so it's our advantage estimate. On the other hand, $V_\theta(s_t)$ seems to be the output of our critic network, which estimates the value of the state. This would mean that we are trying to get the value of the state close to the advantage estimate, which doesn't make any sense, as the advantage of an action is defined as the action's value minus the value of the state (hence we would end up with "(value of the action - value of the state) - value of the state" in the critic loss). This can also be seen in the definition of the (generalized) advantage estimate, where the value of the state is deducted already.
So I suppose I'm misunderstanding something about the critic loss?

REPLY [0 votes]: For anyone interested, I think I figured it out now. $V_\theta(s_t)$ is indeed the output of the critic network, i.e. our estimate of the state value according to our policy. $V_t^{targ}$, on the other hand, is an estimate for the action value based on the trajectory, i.e. the discounted sum of future rewards: $\sum_{i=t}^T \gamma^{i-t}R_\theta(s_i)$. (In particular, it's not $\hat{A}_t$ from GAE, but in case $\lambda=0$ we have $\hat{A}_t\approx V_t^{targ}-V(s_t)$.) But since we are choosing the action according to our policy, the state value (acc. to the policy) should equal the value of the action that's chosen according to that policy (in the trajectory). So it does make sense to use the MSE loss of $V_\theta(s_t)$ and $V_t^{targ}$.


TITLE: Understanding the last statement in the following paragraph. on pg.145 in Royden (4th edition)
QUESTION [0 upvotes]: The paragraph is given below:

But I do not understand the last statement, is it because by the definition of the $p-$norm if $\{f_{n}\} \rightarrow f$ then we have $$(\int_{E} |f_{n} - f|^p)^{1/p} \rightarrow 0 $$ then upon taking the $p^th$ power on both sides we get  $$(\int_{E} |f_{n} - f|^p) \rightarrow 0. $$
Is my justification correct?

REPLY [1 votes]: Of course $u_{n}^{p}\rightarrow u^{p}$ is equivalent to $u_{n}\rightarrow u$, here $u_{n},u\geq 0$. Such an equivalence follows by the continuities of the maps $a\rightarrow a^{p}$ and $b\rightarrow b^{1/p}$ for $a,b\geq 0$.


TITLE: Closed-form for $\sum _{k=4}^{\infty }{\frac { \left( -\ln \left( 2 \right) \right) ^ {k}\zeta \left( 4-k \right) }{k!}}$
QUESTION [4 upvotes]: Can anybody find a closed form for this infinite sum?
$$S = \sum _{k=4}^{\infty }{\frac { \left( -\ln  \left( 2 \right)  \right) ^
{k}\zeta  \left( 4-k \right) }{k!}},$$
where $\zeta$ is the Riemann zeta function.
An approximate value of $S$ is
$$S \approx -0.00469807827332540098459248437391306962194656968313196911104278149327118$$
I found nothing with Maple, Mathematica or ISC.

REPLY [5 votes]: Starting with the well-known expansion for the polylogarithm in terms of the zeta function,
$$ \mathrm{Li}_n(z) = \sum_{m\geq 0, m\neq n-1}\zeta(n-m)\frac{\log^m z}{m!} + \frac{\log^{n-1}z}{(n-1)!}(H_{n-1}-\log\log\tfrac1z), $$
which is valid for $|\log z|<2\pi$, we can substitute $n=4$, $z=\frac12$, and obtain
$$ S = \text{Li}_4(\tfrac{1}{2})+\zeta (3) \log2-\zeta (4)-\tfrac{1}{2} \zeta (2) \log^22-\tfrac{1}{6} \log^32 \log \log2+\tfrac{11}{36} \log^32. $$


TITLE: Verify that moments of gaussian variable are given by a formula
QUESTION [1 upvotes]: I would like to ask you to verify if the following statement is true.
Let $X$ be a normal-distributed R.V. with $0$ mean and $\sigma ^2$ variance. Then
$$
    \mathrm{E}\left[X^p\right] =
      \begin{cases}
        0 & \text{if }p\text{ is odd,} \\
        \sigma^p\,(p-1)!! & \text{if }p\text{ is even.}
      \end{cases}
  $$
Is that true?

REPLY [2 votes]: Yes.  First divide $X$ by $\sigma$ to get a standard normal, then use integration by parts noting that $\phi'(y)=-y\phi(y)$. Please note that !! has a special meaning,ie $3!!=3$, $5!!=3*5$, etcetera.


TITLE: Existence of continuous function satisfying a condition
QUESTION [2 upvotes]: After hearing about "locally constant". This question popped up in my mind.

Is there a continuous non-constant function $f: \mathbb{R} \to \mathbb{R}$ which is locally constant everywhere? That is given any $r \in \mathbb{R}$, there exists $\epsilon >0$ such that $f(B_{\epsilon}(r))=\{f(r)\}$. In words, given any $r \in \mathbb{R}$, there is a neighborhood of $r$ where such that the restriction of the function $f$ is constant.

I tried to solve this problem by assuming that $x \neq y$ and then showing that $f(x)=f(y)$, but it is possible that we cannot do that in finite number of steps, and even in countably infinite number of steps. then how can I solve this?

REPLY [1 votes]: Take $r$ in the image of $f$ and consider $A=\{x\in\mathbb{R}:f(x)\ne r\}$, which is open in $\mathbb{R}$. Let $x\in\mathbb{R}$ such that $f(x)=r$; then there is a neighborhood $U$ of $x$ such that $f(y)=r$, for every $y\in U$. Therefore also $B=\{x\in\mathbb{R}:f(x)=r\}$ is open. Since $\mathbb{R}$ is connected, the only possibility is that $A=\emptyset$, because $B\ne\emptyset$.
So the only case when the function is locally constant is when it is constant.
More generally, if $f\colon X\to Y$ is a locally constant continuous function between topological spaces (metric spaces, if you prefer) is constant in every connected component of the domain.


TITLE: Varieties with ample canonical bundle contain no rational curve
QUESTION [2 upvotes]: MMP tells that a general type variety with no rational curve has ample canonical bundle.
I wonder if the converse is true? i.e. a variety with ample canonical bundle is of general type (of course) and contains no rational curve.

REPLY [6 votes]: This is not true. Take a general hypersurface of degree at least five in $\mathbb{P}^3$ containing a fixed line.


TITLE: Harmonic oscillator differential equation solution
QUESTION [0 upvotes]: My math book explains how to solve second order equations like :
$$\ddot{x} + \omega^2x = 0$$
but I end up with the general solution :
$$A\cos(\omega t) + iB\sin(\omega t).$$
Now my physics book says the solution is
$$\rho\cos(\omega t + \phi)$$
How can I get there from the general solution?

REPLY [2 votes]: Using the sum rule for cosines, we find
$$ \rho \cos(\omega t + \phi) = \rho \cos(\phi) \cos(\omega t) - \rho \sin(\phi) sin(\omega t).$$
So we see that $\rho \cos(\omega t + \phi)$ is the same as $A\cos(\omega t) + iB\sin(\omega t)$ when
$$ A = \rho \cos(\phi)$$
and
$$B = i\rho \sin(\phi).$$


TITLE: How to determine if a calculated Raman mode in a crystal is active or not?
QUESTION [3 upvotes]: I have used an ab initio electronic structure code to calculate the harmonic Raman spectrum of a (molecular) crystal. However, no use of symmetry has been made, and it might well be that some of the vibrational modes that I obtained are actually Raman-inactive. As a matter of fact, when I consider a single cell, two of the vibrational modes I obtain have a high Raman intensity, but when I double the size of the cell, these two modes seem to fuse into a single mode of much lower intensity (not zero, but...).
My question(s) would thus be the following.

Looking at a specific calculated vibrational mode, how can I say whether it is Raman-active or not ?
I think I first have to determine the symmetry of my mode, and see if it has the same symmetry as one component of the polarizability tensor, but I'm not sure how to do this, as I am a bit confused with point/space groups. Maybe there is also another way.
Looking at a specific calculated vibrational mode in a single cell, can I determine what this mode would look like for the double-sized cell ?

Some more information:

My system contains 80 atoms (4 molecules) per unit cell, and its space group is P2$_1$/a.
The two modes I'm talking about are methyl group rotations.
The information about the vibrational modes is stored as an xyz file, containing, in order, the atom type, the position, and the eigenvector (basically, a vector "sits" on every atom). The file would look like this:
80
stable frequency at      155.045 1/cm Raman int. is 1.6100E+02 Ang^4/amu; red. mass is 1.807 a.m.u.; force const. is 0.026 mDyne/Ang.
  C             8.3725    5.4769    9.6155   -0.0071    0.0657   -0.0516
  H             8.2818    6.5453    9.3705    0.2118    0.0540   -0.1846
  H             7.6925    5.2982   10.4580   -0.1575    0.2755   -0.1301
  H             9.4066    5.2741    9.9201   -0.0804   -0.0702    0.1096
  ...
  C             6.3734   10.0506    7.9586    0.0071    0.0657    0.0516
  H             5.3393    9.8479    7.6541    0.0804   -0.0703   -0.1096
  H             7.0533    9.8719    7.1161    0.1575    0.2755    0.1301
  H             6.4642   11.1191    8.2036   -0.2118    0.0540    0.1846
  ...

Thank you in advance!

REPLY [2 votes]: In principle you should determine the irreducible representation of the mode, i.e. how its eigenvector transforms under the symmetry operation of the space group, and look up whether the irreducible representation is Raman active. The Raman tensors for all space groups can be found at the Bilbao crystallographic server. If your irreducible representation is listed your mode is Raman active. Most ab initio codes support a symmetry analysis of the Raman and IR modes, i.e. phonon modes at the $\Gamma$ point
For your space group P2$_1$/a with point group $C_{2h}$ there are two gerade, Raman active irreducible representations, $A_g$ and $B_g$ and two ungerade Raman inactive irreducible representations, $A_u$ and $B_u$.
Your second question is based on the misconception that the the super cell is a physically different system than the single cell, which is not true. The translational symmetry of crystals is baked into solid state theory and both descriptions, super and single cell must yield the same results. Now you might ask: "Why do I get twice as many modes with my super cell?".
Raman and IR modes are nothing more than phonon modes at the $\Gamma$ point, i.e. $k=0$. Solid state theory tells us that only phonon modes with almost (dipole) or exactly zero (Raman) contribute to dipole interaction and Raman scattering  Phonon modes are described through a phonon dispersion with respect to the k vector. The k vector basically tells you how the phase of a certain mode changes over several unit cells.
When you double the cell in real space you cut k space in half. All phonon branches that extend into the cut region are folded back as indicated by the figure below.

However the single cell description is still valid. Only the modes that originally were IR or Raman active (indicated by the red circle in the image) are IR or Raman active in the supercell calculation. The supercell has a new symmetry operation, namely a translation by a half lattice vector. The new modes at $\Gamma$ transform with -1 character under this transformation and are therefore light inactive.
You can also predict the shape of the new modes. It is simply
$$\begin{align}
l_{ij}' = l^X_{ij} \exp{\left(-i \mathbf{k}_X \mathbf{R}_i \right)}
\end{align}$$
with $l_{ij}$ the original eigenvector component, however at the X point, of nucleus $i$ and coordinate $j=x,y,z$, the nuclear positions $\mathbf{R}_i$ and k vector at the X point, e.g. $\mathbf{k}_X = \frac{\pi}{a} \mathbf{e}_x$