text:
\begin{document}

\title{Bare Advanced Demo of IEEEtran.cls\\ for Computer Society Journals}

\author{Michael~Shell,~\IEEEmembership{Member,~IEEE,}
        John~Doe,~\IEEEmembership{Fellow,~OSA,}
        and~Jane~Doe,~\IEEEmembership{Life~Fellow,~IEEE}
\IEEEcompsocitemizethanks{\IEEEcompsocthanksitem M. Shell is with the Department
of Electrical and Computer Engineering, Georgia Institute of Technology, Atlanta,
GA, 30332.\protect\\

E-mail: see http://www.michaelshell.org/contact.html
\IEEEcompsocthanksitem J. Doe and J. Doe are with Anonymous University.}
\thanks{Manuscript received April 19, 2005; revised January 11, 2007.}}

\markboth{Journal of \LaTeX\ Class Files,~Vol.~6, No.~1, January~2007}
{Shell \MakeLowercase{\textit{et al.}}: Bare Advanced Demo of IEEEtran.cls for Journals}

\IEEEcompsoctitleabstractindextext{
\begin{abstract}

The abstract goes here.
\end{abstract}

\begin{IEEEkeywords}
Computer Society, IEEEtran, journal, \LaTeX, paper, template.
\end{IEEEkeywords}}

\maketitle

\IEEEdisplaynotcompsoctitleabstractindextext

\IEEEpeerreviewmaketitle

\section{Introduction}

\IEEEPARstart{T}{his} demo file is intended to serve as a ``starter file''
for IEEE Computer Society journal papers produced under \LaTeX\ using
IEEEtran.cls version 1.7 and later.

I wish you the best of success.

\hfill mds

\hfill January 11, 2007

\subsection{Subsection Heading Here}
Subsection text here.

\subsubsection{Subsubsection Heading Here}
Subsubsection text here.

\section{Conclusion}
The conclusion goes here.

\appendices
\section{Proof of the First Zonklar Equation}
Appendix one text goes here.

\section{}
Appendix two text goes here.

\ifCLASSOPTIONcompsoc

  \section*{Acknowledgments}
\else

  \section*{Acknowledgment}
\fi

The authors would like to thank...

\ifCLASSOPTIONcaptionsoff
  \newpage
\fi


TITLE: Let X and Y be topological spaces. A function f: X → Y is continuous if and only if $f^{-1}$ (C) is closed in X for every closed set C ⊂ Y.
QUESTION [1 upvotes]: I need help proving this theorem
This is the first part of the biconditional, I think if I can prove this. Proving the converse shouldn't be nearly as difficult.
Assume f is continuous ⇒ $f^{-1}$ (C) is closed in X for every closed set C ⊂ Y.
Referencing Theorem 4.7,
Let f: X → Y.  f is continuous at a point x ∈ X iff for each neighborhood M of f(a), $f^{-1}$ (M)  is a neighborhood of a
Referencing Theorem 6.3,
Let f: X → Y.  Then f is continuous iff for each open set O of Y, the subset $f^{-1}$ (O) is an open subset of X.
Referencing Theorem 6.5,
A subset F of a topological space is said to be closed iff its compliment, C(F), is open.
Referencing the definition of a continuous function,
Let X and Y be topological spaces. A function f : X → Y is continuous if, for every x ∈ X and every open set U containing f(x), there exists a neighborhood V of x such that f(V) ⊂ U.
Proof:
Let C be a closed subset of Y, s.t, C ⊂ Y. Clearly, if C is closed, the set Y-C  is open since the compliment of a closed set is an open set(Theorem 6.5). Thus, since the inverse image of an open set is open, $f^{-1}$ (C) is open. Notice, that the topological space X can be written as  $f^{-1}$ (Y–C)  ⋃ $f^{-1}$ (C).

REPLY [1 votes]: Let $O \subseteq Y$ be open, we need to show that $f^{-1}[O]$ is open, which means exactly that we have to show that $X \setminus f^{-1}[O]$ is closed.
Now, $Y \setminus O$ is closed, so the assumption on $f$ gives us that $f^{-1}[Y \setminus O]$ is closed.
Now set theory in general tells us that $f^{-1}[Y \setminus O] = X \setminus f^{-1}[O]$ ($x$ is in the left hand side iff $f(x) \in Y \setminus O$ and this happens iff $f(x) \notin O$, which is saying exactly that $x \notin f^{-1}[O]$, or $x \in X \setminus f^{-1}[O]$).
Combining these three statements gives us that indeed $f^{-1}[O]$ is open, as required.


TITLE: Tensor product for real numbers ?
QUESTION [2 upvotes]: What is the tensor product of two real numbers ?
$\otimes: \mathbb R \times \mathbb R\rightarrow  \mathbb R \otimes \mathbb R, \, \otimes(a,b) \mapsto a \otimes b$
I think $\mathbb R \otimes \mathbb R$ is just $\mathbb R$, no ? But I still don't understand the operation $\otimes$. What is $3 \otimes 4$ for example ? The classical multiplication ?

REPLY [3 votes]: We have, for any commutative ring $R$ and $R$-module $M$ a canonical isomorphism:
\begin{align}
R\otimes_RM&\longrightarrow M, \\
a\otimes m&\longmapsto am
\end{align}
and by linearity, $\;\sum_{i=1}^n a_i\otimes m_i\longmapsto \sum_{i=1}^n a_i m_i$.


TITLE: In the Schrödinger equation, can I have a Hamiltonian without a kinetic term?
QUESTION [11 upvotes]: To find out the stationary states of Hamiltonian, we will be finding the eigenvalues and eigenstates. Is there any condition that form of the Hamiltonian should be like, $$\hat{H}=\hat{T}(\hat{p})+\hat{V}(\hat{x}).$$ I mean the sum of Kinetic operator and potential operator term. Can I have Hamiltonian without a kinetic term, $$\hat{H}=\hat{x}^{2}$$ or in other words? Can I have a Hamiltonian with just potential term alone? $$\hat{H}=\hat{V}(\hat{x}).$$
Does quantum mechanics allow that?

REPLY [3 votes]: In studying the quantum Hall effect, it is common to assume that all particles have the same kinetic energy (sometimes people describe this by saying the kinetic energy is "quenched"). The assumption is reasonable when all of the particles occupy a single Landau level and the energy spacing between Landau levels (the cyclotron energy) is large relative to other energy scales in the problem. In this case, the kinetic energy just gives a constant shift in the energy spectrum, so we can ignore it and consider a hamiltonian with only an interaction term (projected to a Landau level), $H = PVP$.


TITLE: Roots of polynomials with bounded integer coefficients
QUESTION [21 upvotes]: For $a\in\mathbb N$, let $R_a\subseteq\mathbb{R}$ be the set of real roots of polynomials whose coefficients are integers with absolute value at most $a$.
$$R_a=\left\{r\in\mathbb{R}\middle|\sum_{i=0}^na_ir^i=0,n\in\mathbb N,a_i\in\mathbb Z,|a_i|\leq a, a_n\neq 0\right\}$$
(Note that the degrees of the allowed polynomials aren't bounded.)
I want to know which numbers can be arbitrarily well approximated by such roots. In particular, is $R_a$ dense on some interval for any $a$?

Discussion:
If $r\in R_a$ and $r\neq 0$ then $1/r\in R_a$, since by reversing the order of the coefficients of the polynomial of which $r$ was a root one obtains a polynomial of which $1/r$ is a root.
In the comments, mathworker21 noted that by considering the polynomial $1 - aX -\dots-aX^n$ one can show that every nonzero $r\in R_a$ satisfies $|r|>\frac{1}{a+1}$. Therefore we also have $|r|<a+1$, so $R_a$ is bounded.
We could also ask about the set $R_a^\mathbb C$ of the roots of such polynomials in $\mathbb C$. John Baez has a page about such sets here, which contains some beautiful pictures.

He's also written a paper, which gives a reference to a paper by Thierry Bousch in which it is proved that the roots of the Littlewood polynomials (those with coefficients $\pm1$ only) are dense on the annulus $2^{-1/4}<|z|<2^{1/4}$.
Since $\{-1,1\}\subseteq\{-1,0,1\}$, this answers the complex analogue of my question: the set $R_1^\mathbb C$ is dense on some ball. Unfortunately the paper is written in French, so I can't tell if the proof generalises to the reals. (The real result isn't an immediate corollary of the complex version, because there might be a sequence in $R_a^\mathbb C$ tending to $x\in\mathbb R$ without there being such a sequence in $R_a^\mathbb C\cap\mathbb R$.)

REPLY [5 votes]: It seems that the following paper (linked by Baez on the webpage you link above)
https://www.e-periodica.ch/digbib/view?pid=ens-001:1993:39::181#568
of Odlyzko and Poonen implies that there is an interval in which $R_1$ is dense. Namely, there is some $\delta>0$ so that the interval $[-\phi^{-1}-\delta,-\phi^{-1}]$, where
$$\phi=\frac{1+\sqrt{5}}{2}$$ is the golden ratio, is contained in the closure of $R_1$. See the bottom of page 318 to page 319 (see also page 330 for specific values of $\delta$ that work, and page 325 for remarks on results specific to $R_1$). In fact, the density is achieved already by numbers which are roots of polynomials with coefficients all $0$ or $1$ (no $-1$'s necessary).


TITLE: Calculate polynomial equation for curve based on known gradient?
QUESTION [0 upvotes]: I am trying to figure out how to calculate the polynomial equation representing a curve. I only know that the gradient is represented by a linear equation.
To be more specific: I know that the gradient of the curve increases by $1$ (degree, if that is important?) every $10$ units. I.e. at $x=0$ it is $0$. At $x=10$ it is $1$, at $x=20$ the inclination of the gradient is $2$ (again, degree if that matters). If I go by $y=ax$, I get gradient $a$ as being $0.1$, and therefor my equation representing the gradient is $y=0.1*x$.
Now I know that the gradient is the derivative of the curve. So the equation of the curve should be the integral of the gradient equation, which is $Y=(a/2)*x^2$, and with $a=0.1$ it should be $Y=0.05*x^2$. But the resulting curve doesn't fit at all. It is much too steep. The graphical solution (created by using 10 segments of each 10 units) yields ca. $f(x=100)=~10$.
What am I overseeing?

REPLY [0 votes]: As it turned out, I can't just consider an angle as an arbitrary number. I actually had to use trigonometric functions to calculate hypothenuse, opposite side, and adjacent side of the triangle to make it work.


TITLE: Prove that PDE $u_t+x^2u_x=0$ has infinite solutions
QUESTION [2 upvotes]: the text of the exercise I am takling is this one.

Given
  \begin{cases}
u_t+x^2u_x=0 \text{ in } (0, +\infty)\times \mathbf{R}\\
u(0,x)=0 \text{ in } \mathbf{R}
\end{cases}
  prove that for every $\delta>0$ there are infinite solutions in $[0,\delta]\times \mathbf{R}$.

I tried using the method of characteristic. The system i solve is
\begin{cases}
\dot{v}=v^2 \\
\dot{w}=0 \\
v(0)=r \\
w(0)=0
\end{cases}
Even before solving the system I see that there is a problem: I will obatin $w(t)\equiv0 $, and any solution of the type $u(t,x)=w(t,r(t,x))$ will be the one constantly zero. Drawing a picture I find out that the vector field $(1,x^2)$ in the plane $(t,x)$ is directed as the $x$-axe, which is the line where I have my initial value for the problem.
Anyway solving the system for $v$ I obtain $v(t)=\frac{1}{c-t}$ with $c=\frac{1}{r}$. In this way I can find a family of characteristic curves $x_r(t)$. I see that for every fixed $\delta$ I can choose $r$ small enough to make the curve exist until time $\delta$.
My problem is how to define anon-zero solution on this curve.
Beyond this single problem, what is the reason for the difficuty of this exercise? Is there a general method to solve problems like this?

REPLY [3 votes]: The problem is that when you consider the characteristics backward, not all of them intersect the line $t = 0$.
For $x > 0$, even though the characteristics curves blow up in finite time, this is not actually a problem. What we care is that we can extend the characteristic backwards in time. In order to not get confuse, lets use $s$ as the parameter for the curve
$$ x(s) = \frac{r}{1 - rs}$$
and lets take a fixed point $(t,x) \in [0,\delta]\times \mathbb R_+$. Then the curve that intersects $(t,x)$ satisfies
$$ x = x(t) =
\frac{r}{1 - rt}$$
so
$$ r = \frac{x}{xt + 1}$$
This value is always well defined and more importantly, $x(s)$ is well defined in $[0,t]$ due to the fact that
$$ 1 - rs = 1 - \frac{xs}{xt + 1} = \frac{x(t - s) + 1}{xt + 1} > 0 $$
This forces then $u(t,x) = u_0(r) = 0$
For $x < 0$ on the other hand, we may encounter some problems. The first one is the possibility that $xt + 1 = 0$ in which case there characteristic curve is $x(s) = -1/s$ that never touches the $t = 0$ line. Even worse is the case $xt + 1 < -1$ where all the computation of the first part remain valid, except that the characteristic curve has a singularity at $s = \frac{xt + 1}{x} = t + \frac1x \in (0,t)$ (this value was always bigger than $t$ in the previous case)
The problem then is that in this case the characteristic comes from $x = -\infty$ rather than the line $t = 0$.
Finally if $xt + 1 > 0$ then all the analysis done in the first case is still valid so $$u(t,x) = 0$$ as well.
With all this analysis we can construct the non trivial solutions:
Lets take $\phi \in C^1(\mathbb R)$ such that
$$\lim_{y \to -\infty} y \phi(y) = 0$$
Then a solution is given by
$$ u(t,x) = \begin{cases}
0 & tx + 1\geq 0\\
\phi \left( \frac{x}{xt + 1}\right )& tx + 1 < 0
\end{cases}
$$
To prove that it is a solution, in the case $tx +1 \not=0$ is just direct computations.
In the case $tx + 1 = 0$ you can prove that $\partial_t u = \partial_x u = 0$ using the limit condition on $\phi$.
As we mentioned earlier, this solutions correspond to a wave coming from $-\infty$ so we would need a condition in $-\infty$ to have uniqueness.


TITLE: For which $a>0$ does $\int_a^\infty \frac{\mathrm{d}x}{(x^2-a)^{4a}}$ converge?
QUESTION [2 upvotes]: As the title suggests, I need  help finding  $a>0$ for which the following improper integral converges:
$$\int_a^\infty \frac{\mathrm{d}x}{(x^2-a)^{4a}}$$
So, at first I thought I would just do this:
$$ f(x)=\frac{1}{(x^2-a)^{4a}}$$
Then I wanted to find a function $$g(x)$$ such that $f(x)\leq g(x)$ for any $x\in [a, \infty)$.
I thought I could use something like $$g(x)=\frac{1}{(a^2-a)^{4a}}$$
but I am not sure if that works?
So  if I could prove that $g(x)$ converges for some $a$, I could state that $f(x)$ also converges.
Correct me if I am wrong.
Any help would be appreciated.

REPLY [5 votes]: Your integral converges at "infinity" if $8a>1$, i.e., $a>\frac{1}{8}.$
To see what happens at finite points, first write
$$\frac{1}{x^2-a}=\frac{1}{x-\sqrt a} \cdot \frac{1}{x+\sqrt a}.$$
The second fraction does not have any impact on convergence/divergence since it defines a bounded continuous function on $[0,\infty)$.
If $\sqrt a<a$, then the improper integral does not have any singularities at finite points from where it follows that the improper integral definitively converges whenever $a>1.$ Suppose now $0<a\leq 1$. Then $\sqrt a\geq a$ and in this case, the condition that the improper integral converges at $x=\sqrt a$ is $4a<1.$ From this we conlcude that the improper integral converges for
$a\in (\frac{1}{8},\frac{1}{4})\cup (1,\infty)$.


TITLE: Getting an intuition of a "Vector Field" in General relativity
QUESTION [2 upvotes]: In GR the way we define a vector field is :
$$ v^a=\sum_\mu v^\mu (\partial_\mu)^a $$
The a in the superscript is the abstract index notation.
I understand this attachment of a vector field if there is function at every point to act upon or along a curve.
Many times in GR there is a vector field attached to the space time without the specification of any functions or curve. What does it mean by this intuitively?
An example:
For a time orientable spacetime (M,$g_{ab})$ we always have a timelike vector field.
 - Okay I get the part that we have a vector field which is timelike. But what are these vectors at each point, rather what is the physical dimension of these vectors ?
 - I am not sure if this is a right thinking or not but I am having trouble understanding the idea of a vector field when we talk about vectors as directional derivatives attached to every point.
How can this vector field be understood intuitively.
For example when we talk about Vector fields in Classical field theory we attach a vector to every point in the space like an electric field vector (Having a physical dimension [Newtons/Coulomb] ) for a Electric vector field.
Edit: What is the r vector here in the example? I got everything else!

REPHRASED QUESTION:
We define vectors by directional derivatives in Diff.Geo. Then at every point on the manifold we have a unique tangent space where these vectors live. When we have a curve going through these points we can
take the directional derivative at every point and have a vector field in coordinate basis at every point along the curve using the parameter of the curve.
But many times in GR we define a vector field without talking about any curve, i.e.
we associate a vector to every point on the manifold without talking about any curve. So how do we make the choice of a specific vector on every point on the manifold and called it a vector field if we have no curve and a curve parameter.

REPLY [1 votes]: Quick Answers:
One need not specify a function to talk about an object which eats functions and spits out real numbers.  We can talk about how such an object would act on any function which we choose to feed it, but we need not specify one function in particular.
It turns out that given any choice of vector, there exists a curve such that the action of the vector on a function can be regarded as the directional derivative of the function along that curve.  I expand on this more later in my answer.
The dimensions of a vector are usually Hz (assuming that we parameterize our curves by something with dimensions of time).  However, the components of a vector have dimensions of length over time (assuming that the coordinates of the manifold have dimensions of length).  This changes if you use different coordinate charts.
Here's a more complete explanation.
Motivation
In terms of raw intuition, the picture that you have in your head is not actually a bad one.  Given some manifold $M$ (a sphere, say), at each point $p\in M$ you attach a tangent plane $T_p M$.  Each tangent plane is simply a copy of $\mathbb{R}^d$, where $d$ is the topological dimension of $M$ (for a sphere, $d=2$), and vectors can be thought of as the familiar "arrows" which live in $\mathbb{R}^d$.
Picture that in your mind.  What you are imagining is called an embedding of the sphere into a 3D space.  The notion of gluing a plane to a point on a sphere makes real sense only if you're picturing the sphere as a subset of a larger space.
We would like to avoid this.  In particular, if $M$ is supposed to represent the entire universe, it's a bit odd to try to model dynamics on it by appealing to a "top down" view from outside $M$.  Instead, we would much prefer to take an intrinsic approach - to define vectors in a space without leaving the space.
The Rough Idea
Imagine that you're walking through your house, and you're measuring the temperature at each point.  Let $p$ be a point in your house (notice that I haven't defined the point by some coordinate system or anything - $p$ genuinely represents a point, not a pair of numbers).  The function
$$T : \text{House} \rightarrow \mathbb{R}$$
$$ p \mapsto \text{Temperature at } p$$
maps points in your house to the temperature at that point.
As you walk along some path through $p$, the temperature you measure will be constantly changing.  You can therefore define the rate of change of the temperature as you walk through $p$.  If you follow the same path at twice the speed, then the rate of change will double; if you follow a different path through $p$, then the rate of change will be different as well.
This is the central idea.  If $f$ is a function on the manifold (i.e. your house), then the rate of change of $f$ at a particular point $p$ contains information about the direction at which you move through $p$, as well as the rate at which you walk - which amount to "direction" and "magnitude".
Our job now is to make this notion mathematically precise.

Manifold Charts
Let $M$ be a smooth manifold.  A chart $(U,x)$ is an open subset $U\subset M$ and a continuous, invertible map
$$ x : M \rightarrow \mathbb{R}^d$$
$$ p \mapsto x(p)$$
which takes a point $p\in U$ to $x(p)=(x^1,x^2,\ldots,x^d) \in \mathbb{R}^d$.  We call $x(p)$ the coordinates of the point $p$ in the chart.
We should pause here to appreciate what this means.  Up until now, we've typically defined a point by its coordinates, which means we have been implicitly choosing particular charts.  Now, we recognize that a point exists as a point rather than as a list of numbers, and that the coordinates of a point completely depend on what labels we decide to apply to it.
Curves on Manifolds
We define a curve $\gamma:\mathbb{R} \rightarrow M$ to be a map which takes a real number and maps it to a point in the manifold $M$.  If you'd like, the real number might represent time, but we aren't talking about physics yet, so for now it's just a parameter.
Derivatives along a Curve
Let $f$ be a smooth function $f : M \rightarrow \mathbb{R}$ (note: the set of smooth functions on $M$ is denoted $C^\infty(M)$), and let $\gamma$ be a curve which passes through the point $p\in M$ such that $\gamma(0) = p$ (this is just for convenience).
Now consider the function
$$ f \circ \gamma : \mathbb{R} \rightarrow \mathbb{R}$$
$$ t \mapsto f\big( \gamma(t) \big) $$
This function takes a real number $t$ and maps it to the value of $f$ at the point in the manifold given by $\gamma(t)$.  In other words, it is the value of the function along the curve $\gamma$.
As a function from $\mathbb{R}$ to $\mathbb{R}$, it makes perfect sense to differentiate this object - we define $(f \circ \gamma)'(0)$ to be the directional derivative of $f$ along $\gamma$ at the point $p$.
Such a thing deserves a new notation - we will call it $V_{\gamma,p}$.
$$ V_{\gamma,p} : C^\infty(M) \rightarrow \mathbb{R} $$
$$ f \mapsto (f\circ \gamma)'(0)$$
It also deserves a name - we say that $V_{\gamma,p}$ is the tangent vector to the curve $\gamma$ at the point $p$.  The space of all such objects defined at a point $p$ is denoted $T_pM$, and called the tangent space to $M$ at $p$.
Tangent Spaces
Having defined one tangent vector, we might now embark on a quest for others.  Given some curve $\gamma$, is it possible to find another curve $\sigma$ such that
$$ V_{\sigma,p}f = c V_{\gamma,p}f $$
for any choice of smooth function $f$ and any real number $c$?  The answer is yes - therefore, we can multiply these tangent vectors by real numbers to get new tangent vectors.
Can we add them?  That is, given two curves $\gamma$ and $\sigma$, does there always exist a third curve $\delta$ such that
$$ V_{\gamma,p}f+V_{\sigma,p}f = V_{\delta,p}f$$
for any smooth function $f$?  The answer again is yes.  I won't prove these two facts (the first is very simple, the latter requires a bit of though) - you can think about them yourself, or simply look them up.
In any case, this implies that $T_pM$ is indeed a vector space - we can add vectors and multiply them by real numbers.
Coordinate Representation of Tangent Vectors
This is the meat and potatoes of your question.  Up to this point, hopefully things have been reasonable - but they have been very abstract.  It's time to drop to the level of a chart and to see what these things look like when we define a coordinate system.
Recall that a coordinate chart $x$ is continuous and invertible, so $x^{-1}:\mathbb{R}^d \rightarrow U \subset M$ is perfectly well-defined.  Therefore, let's choose a chart $(U,x)$.  Then we can write
$$ V_{\gamma,p}f = (f \circ \gamma)'(0) = (f \circ x^{-1} \circ x \circ \gamma)'(0)$$
We can regard this as the composition of two maps:

$(x \circ \gamma):\mathbb{R} \rightarrow \mathbb{R}^d$ takes a real number $t$ and maps it to the coordinates of the corresponding point $\gamma(t)$ in the chart.
$(f \circ x^{-1}) : \mathbb{R}^d \rightarrow \mathbb{R}$ takes the coordinates of the point $\gamma(t)$ in the chart and maps them to the value of $f$ at that point.

When we differentiate, we do so using the chain rule:
$$(f \circ x^{-1} \circ x \circ \gamma)'(0) = \partial_a \big(f \circ x^{-1}\big)(x(p)) \cdot \big(x^a \circ \gamma\big)'(0)$$
Note: The symbol $\partial_a$ applies to functions $g:\mathbb{R}^d\rightarrow \mathbb{R}$, and means "the partial derivative of $g$ with respect to the $a^{th}$ slot."
From here, we define a new symbol:
$$ \left(\frac{\partial}{\partial x^a}\right)_p $$
Here's the idea:  The symbol $\partial_a$ is used to differentiate functions on $\mathbb{R}^d$, which take $d$ real numbers as inputs.  However, $f$ is a function on the manifold.  Its only input is a point $p$, so how do we differentiate it?
The answer is as follows:  We take a point $p$, we map it to its coordinates by choosing a chart $x$.  We then differentiate the function $f \circ x^{-1}$ (which we know how to do) and evaluate at $x(p)$.
In other words,
$$\left(\frac{\partial}{\partial x^a}\right)_p f := \partial_a\big(f \circ x^{-1}\big) (x(p))$$
and so we find that in the chart $(U,x)$,
$$V_{\gamma,p}f = V^a \left(\frac{\partial }{\partial x^a}\right)_p f$$
where we have defined the real numbers
$$V^a \equiv (x^a \circ \gamma)'(0)$$
to be the components of $V$ in the chart.

That's most of the major technical machinery.  In the old days, our definition of a vector was the list of numbers $(V^1,V^2,\ldots,V^d)$.  Through the lens of differential geometry, these are just a particular choice of label for a more abstract object which exists independent of any particular choice of coordinate system.
You can also calculate what must happen to the components of a vector under a change of chart - this is a useful exercise, so not one that I will reproduce here.


text:
\begin{document}
\maketitle
\begin{abstract}
In order to extend the study of uniqueness property of multi-dimensional systems of stochastic differential equations, in this paper, we look at the following three-dimensional system of equations, of which the two-dimensional case was well-studied before: $dX_t=Y_tdt\quad, dY_t=Z_tdt,\quad dZ_t=|X_t|^{\alpha}dB_t$. We proved that if $(X_0,Y_0,Z_0)\neq(0,0,0)$, and $\frac{3}{4}<\alpha<1$, then the system of equations has a unique solution in the strong sense.
\end{abstract}.
\section{Introduction and main results}
 The uniqueness of ordinary differential equations (ODE) has been extensively studied, see for example \cite{ODE}. In
 particular, if $F(u)$ is Lipschitz continuous, then $$u'(t)=F\left(u(t)\right),\quad u(0)=u_0$$ has a unique
solution for all $t\geq 0$. In the case above, $F$, $u(t)$, and $u_0$ take value in $\mathbf{R}^d$, $d\geq1$. The stochastic differential equation (SDE) realm, on the contrary,
has different criteria for uniqueness of solutions, see for example \cite{SDE}. One of the most well-known results
regarding strong uniqueness of SDE is due to Yamada and
Watanabe's paper \cite{ST}. The result states that if
$f(x)$ is locally H\"older continuous with index $\alpha\in[\frac{1}{2},1]$ and with linear growth, then $$dX=f(X)dW,\quad   X_0=x_0$$ has
a unique strong solution for all time $t\geq 0.$ Yamada and Watanabe's theory essentially focuses on
one-dimensional SDE.

One motivation for studying higher-dimensional SDE comes from the wave equation:
\begin{equation}\nonumber
\begin{array}{rcl}
    \partial_t^2u & =&\Delta u \\
  u(0,x)&=&u_0(x)\\
  \partial_{t}u(0,x)&=&u_1(x).
\end{array}
\end{equation}
 In this equation, we have $$ \partial_t^2u  =\partial_x^2u=\Delta u.$$ If we let $$ v=\partial_{t} u,$$
 then we can rewrite the wave equation as the following system of equations:
\begin{equation}\nonumber
\begin{array}{rcl}
  \partial_{t} u &= & v \\
    \partial_{t} v&=&\Delta u.
\end{array}
\end{equation}

The original wave equation includes no noise. However, many physical systems are affected by noise. Hence, a modification
of the wave equation which includes white noise is also studied:
\begin{equation}
\label{wave_eq}
    \begin{array}{rcl}
         \partial_{t}^2u & =&\Delta u +f(u)\dot{W}\\
  u(0,x)&=&u_0(x)\\
  \partial_{t}u(0,x)&=&u_1(x).
\end{array}
\end{equation}
    Note: $x\in\mathbf{R} $ and $\dot{W}=\dot{W}(t,x)$ is white noise.

One well-known point is that Lipschitz continuity is sufficient for the uniqueness of SDE. Thus, many
mathematicians have studied whether H\"older continuity can still ensure the uniqueness property of SDE.
In Gomez, Lee, Mueller, Neuman, and Salins\cite{GLMNS}, the uniqueness property of the following two-dimensional model of SDE was studied:
\begin{equation}
\label{2-dimensions}
\begin{array}{rcl}
dX&=&Ydt\\
dY&=&|X|^{\alpha}dB\\
(X_0,Y_0)&=&(x_0,y_0).
\end{array}
\end{equation}
The results focused on $f(x)=|x|^{\alpha}$ since it is a prototype of an equation with H\"older continuous coeffients.
Moreover, \eqref{2-dimensions} is a version of \eqref{wave_eq} when we drop the dependence on $x$, which allows us to study the modified wave equation
with more simplicity. We can notice that if we take the differential $dY$ of the first derivative of $X$, which is $Y$
in the system of equations, it resembles
the second derivative in time in the stochastic wave equation.

It has been proven in \cite{GLMNS} that if $\alpha>\frac{1}{2}$ and $(x_0,y_0)\neq(0,0)$, then \eqref{2-dimensions} has a
unique solution in the strong sense up to the time $\tau$ at which $(X_t,Y_t)$ first hits the
origin $(0,0).$

Since much is still unknown about higher-dimensional SDE, we wish to continue the study of uniqueness property
of \eqref{2-dimensions} in the three-dimensional case, which is:

\begin{equation}
\label{3-dimensions}
    \begin{array}{rcl}
    dX&=&Ydt \\
   dY&=&Zdt\\
   dZ&=&|X|^ {\alpha}dB\\
   (X_0,Y_0,Z_0)&=&(x_0,y_0,z_0).
\end{array}
\end{equation}

\begin{thm}
If \(\frac{3}{4}<\alpha<1\) and $(X_0,Y_0,Z_0)\neq0$, then \eqref{3-dimensions} has a unique strong solution, up to the time \(\tau\) at
which the solution $(X_t,Y_t,Z_t)$ first hits the value (0,0,0) or blows up.
\end{thm}

Moreover, we say the solution $(X_t,Y_t,Z_t)$ blows up in finite
time, with positive probability, if there is a random time
$\tau<\infty$ such that $$P\left(\lim_{t\uparrow\tau}|(X_t,Y_t,Z_t)|_{l^\infty}=\infty\right)>0.$$
\section{proof of theorem 1}
Let $(X_t^i,Y_t^i,Z_t^i),i=1,2$ be two solutions to \eqref{3-dimensions} with $(x_0,y_0,z_0)\neq(0,0,0),$ in other words, $(X_t^i,Y_t^i,Z_t^i), i=1,2$ have the same initial condition $(X_0,Y_0,Z_0)\neq (0,0,0)$. Note that since $(X_t^i,Y_t^i,Z_t^i),i=1,2$ have the same initial conditions, from now, we use $X_0,$ $Y_0,$ $Z_0$ instead of $X_0^{i,n},$ $Y_0^{i,n},$ $Z_0^{i,n}.$

Let $\tau$ be the first time $t$ that either $(X_t^1,Y_t^1,Z_t^1)$ or $(X_t^2,Y_t^2,Z_t^2)$
hits the origin $(0,0,0)$ or blows up. We let $\tau$ be infinity if there is no such time.

First of all, if  $X_0\neq0,$ then \eqref{3-dimensions} will have Lipschitz continuity up to the time that $X_t=0$, thus enable uniqueness
to hold. Suppose after a certain amount of time, $X_t$ hits zero, where Lipschitz continuity no longer holds, then due to strong Markov property, we begin the process again with $X_0=0.$ So our goal is to
prove pathwise uniqueness between excursion of X up to the time $\tau$ starting from $X_0=0.$

For any fixed $n$, let $\tau_n$ be the first time that either
$$|(X_t^1,Y_t^1,Z_t^1)|_{l^{\infty}}\wedge |(X_t^2,Y_t^2,Z_t^2)|_{l^{\infty}}\leq 2^{-n}$$
or $$|(X_t^1,Y_t^1,Z_t^1)|_{l^{\infty}}\vee |(X_t^2,Y_t^2,Z_t^2)|_{l^{\infty}}\geq 2^{n}.$$
\vspace{0.3cm}
In which:

The infinity norm (also known as the $L_\infty$-norm, $l_\infty$-norm, max norm, or uniform norm) of a vector $\vec{v}$ is denoted $|\vec{v}|_{l_\infty}$ and is defined as the maximum of the absolute values of its components:
$$|\vec{v}|_{l_\infty}=\max{ |v_i| : i=1,2,..n},$$

and
\begin{equation}\nonumber
\begin{array}{rcl}
   a\wedge b  & =& \min(a,b)\\
   a\vee b  &=& \max(a,b).
\end{array}
\end{equation}
If there is no such time, we let $\tau_n$ be infinity. Note that: $\lim_{n \uparrow\infty}(\tau_{n})=\tau$.

Now, for each fixed $n$, we will show uniqueness up to time $\tau_n$ in the following system of equations:
\begin{equation}
\label{3-dim-i,n}
 \begin{array}{rcl}
 dX_t^{i,n}&=&Y_t^{i,n}dt \\
   dY_t^{i,n}&=&Z_t^{i,n}dt\\
   dZ_t^{i,n}&=&|X_t^{i,n}|^ \alpha\mathbf{1}_{[0,\tau_n]}(t)dB_t\\
   (X_0,Y_0,Z_0)&=&(x_0,y_0,z_0),
   \end{array}
\end{equation}
 In other words, after the time $\tau_n,$ $dZ_t^{i,n}=0$, which makes $Z_t^{i,n}$ become constant. Specifically, given
 $m,n \in N,$ we need: $$(X_t^n, Y_t^n, Z_t^n)= (X_t^m, Y_t^m, Z_t^m)$$ for all $t\leq\tau_n\wedge\tau_m.$

Now, before continuing the proof of uniqueness, by contradiction method, we show that the times that $X_t$ hits zero do not accumulate before the time $\tau_n$, almost surely.

For each $n$, let $A_n$ be the event on which the times that $X_t^{i,n}=0$, $i=1$ or $i=2$, accumulate before $\tau_n$, and assume
$P(A_n)>0.$ Then, on $A_n$, suppose $\sigma_n$ is an accumulation point of the
times $t$ at which $X_t^{i,n}=0,$ i.e, there
exits a sequence of time $\rho_{1,n}<\rho_{2,n}<\ldots$ that converges to
$\sigma_n,$ and $X_{\rho_{k,n}}^{i,n}=0.$ Hence, on $A_n,$ $\lim_{k\to\infty}\rho_{k,n}=\sigma_n.$

We have $X_t^{i,n}$ is almost surely continuous, and that $X_{\rho_{k,n}}^{i,n}=0$ on $A_n$, so: $$\lim_{\rho_{k,n}\to\sigma_n}X_{\rho_{k,n}}^{i,n}=X_{\sigma_n}^{i,n}=0$$ on $A_n.$

Note that $dX_t^{i,n}=Y_t^{i,n}dt,$ and $Y_t^{i,n}$ is almost surely continuous. So if $Y_{\sigma_n}^{i,n}\neq0$ on $A_n,$ then there exists a
random interval $[\sigma_n(\omega)-\epsilon(\omega),\sigma_n(\omega)]$ of positive length for which $X_t^{i,n}\neq0$ on $[\sigma_n(\omega)-\epsilon(\omega),\sigma_n(\omega)].$ This contradicts the hypothesis of $\rho_{k,n}$ converging to $\sigma_n.$

If $Y_{\sigma_n}^{i,n}=0$, there are two cases, which are $$\sigma_n\geq\tau_n,$$ or $$\sigma_n<\tau_n.$$
If $\sigma_n\geq\tau_n,$ then it means that the times at which $X_t^{i,n}$ hit zero do not accumulate before $\tau_n$.

If $\sigma_n<\tau_n,$ then with $X_{\sigma_n}^{i,n}=0$ and $Y_{\sigma_n}^{i,n}=0$, we have $|Z_{\sigma_n}^{i,n}|>2^{-n}$.
Now suppose $Z_{\sigma_n^{i,n}}>2^{-n},$ as the case $Z_{\sigma}^{i,n}<2^{-n}$ is approached similarly due to symmetry.

So, if $Z_{\sigma_n}^{i,n}>2^{-n},$ since $Z_t$ is almost surely continuous, there exists a time interval $[\sigma_{n}(\omega)-\epsilon'(\omega),\sigma_{n}(\omega)]$ on which
$Z_t^{i,n}>\frac{2^{-n}}{2}.$ Hence, for all $t\in[\sigma_n(\omega)-\epsilon'(\omega),\sigma_n(\omega)],$ we almost surely have:

\begin{equation}\nonumber
 \begin{array}{rcl}
   Y_t^{i,n}&=&Y_{\sigma_n}^{i,n} - \int_t^{\sigma_n}Z_s^{i,n}ds  \\[10pt]
     &< & \frac{2^{-n}}{2}(\sigma_n-t).\\[10pt]
 \end{array}
\end{equation}

Hence, integrating over $X_t^{i,n}$ for $t\in[\sigma_n-\epsilon',\sigma_n],$ we have:
\begin{equation}\nonumber
    \begin{array}{rcl}
       X_t^{i,n}& = & X_{\sigma_n}^{i,n}-\int_t^{\sigma_n}Y_s^{i,n}ds \\[10pt]
        &> & -\int_t^{\sigma_n}Y_s^{i,n}ds\\[10pt]
        &>&\int_t^{\sigma_n}\frac{2^{-n}}{2}(\sigma_n-s)ds\\[10pt]
        &=&\frac{2^{-n}}{2}(\sigma_ns-\frac{s^2}{2})\left. \right\vert_t^{\sigma_n}\\[10pt]
        &=&\frac{2^{-n}}{4}(\sigma_n-t)^2\\[10pt]
        &>&0,
        \end{array}
\end{equation}
as $t<\sigma_n.$ Hence, this contradicts the hypothesis of $\rho_{k,n}$ converging to $\sigma_n$ on $A_n.$
So in conclusion, $P(A_n)=0,$ which means the times at which $X_t$ hits zero does not accumulate before the time $\tau_n.$

One more point we need to address before continuing with the proof of uniqueness is the existence of solutions. This problem is resolved in Theorem 21.7, Theorem 21.8, and Lemma 21.17 of \cite{kallbook},
which prove that for all time $t\geq0$, solutions of multi-dimensional SDE exist with probability one provided the coefficients are continuous and bounded.

Specifically, in our problem, since $\alpha\in(0,1),$ we have the coefficients of the system of equation \eqref{3-dimensions} are continuous and bounded by $(2^n)^{\alpha}\vee2^n=2^n$ up
to the $\tau_n$ for each $n$. which satisfies the condition stated in the existence theory in \cite{kallbook}. Hence, existence of solution
holds for all $t\leq\tau_n$ for all n. Therefore, up to the
time $\tau=\sup\tau_n$,  existence of solutions is ensured.

From now, $X_t^{i,n}$ means $X_t^{1,n}$ and $X_t^{2,n}$. Similar interpretation for $Y_t^{i,n}$ and $Z_t^{i,n}$. Back to the proof of uniqueness, we have, with all $t\in[0,\tau_n]:$ $$|Z_t^{i,n}|\vee|Y_t^{i,n}|\leq2^n.$$
So:
\begin{equation}\nonumber
    \begin{array}{rcl}
     |Y_t^{i,n}|& =& |Y_0 +\int_0^t Z_s^{i,n}ds| \\[10pt]
     &\leq&|Y_0|+\int_0^t|Z_s^{i,n}|ds\\[10pt]
      & \leq& 2^n+2^nt.
    \end{array}
\end{equation}
Therefore:
\begin{equation}\nonumber
    \begin{array}{rcl}
     |X_t^{i,n}|&\leq& |X_0+\int_0^tY_s^{i,n}ds|\\[10pt]
         & \leq&\int_0^t|Y_s^{i,n}|ds\\[10pt]
         &\leq&\int_0^t(2^n+2^ns)ds\\[10pt]
         &=&2^n(t+\frac{t^2}{2}).
    \end{array}
\end{equation}
Now let  \begin{equation}
\label{eq:t0}
       t_{0,n}=\frac{2^{-2n}}{2}.
   \end{equation}
Then:
\begin{equation}\nonumber
  \begin{array}{rcl}
      t_{0,n}+\frac{t_{0,n}^2}{2}&= &\frac{2^{-2n}}{2} +\frac{2^{-2n}}{8} \\[10pt]
       &\leq& \frac{2^{-2n}}{2}+\frac{2^{-2n}}{2}\\[10pt]
       &=&2^{-2n}.
  \end{array}
\end{equation}
So
\begin{equation}\nonumber
  \begin{array}{rcl}
      2^n(t_{0,n}+\frac{t_{0,n}^2}{2})&\leq &2^n.2^{-2n}  \\[10pt]
      &= & 2^{-n}.
  \end{array}
\end{equation}
Since the quadratic function $2^n(t+\frac{t^2}{2})$ is increasing when $t
\geq0,$ we have:
\begin{equation}\nonumber
   \begin{array}{rcl}
     |X_t^{i,n}|&\leq&2^n(t+\frac{t^2}{2})    \\[10pt]
        &\leq& 2^{-n}
   \end{array}
\end{equation}
for all $t\in[0,t_{0,n}].$

Since $|X_t^{1,n}|$ and $|X _t^{2,n}|$ belong in $ [0,2^{-n}]$ for $t\in [0,t_{0,n}]$, based on the definition of $\tau_n$ above, either
\begin{equation}
\label{case_6}
 |Y_0|\geq2^{-n}
\end{equation}
or
\begin {equation}
\label{case_7}
|Z_0|\geq 2^{-n}
\end{equation}
for each fixed $n$. This is due to the fact
that the solutions have the same initial condition $(X_0,Y_0,Z_0)$ and for each time $t\in[0,t_{n,0}]$, either :
\begin{equation}\nonumber
    |Y_t^{i,n}|\geq2^{-n}
\end{equation}
or
\begin{equation}\nonumber
   |Z_t^{i,n}|\geq2^{-n}.
\end{equation}
First, we deal with $Y_0 > 0$ since due to symmetry, we could deal with the case $Y_0<0$ with similar methods.

Now, with $Y_0 > 0$, we look at other subcases based on $Z_0^{i,n}.$

\textbf{Case I:} $Y_0>0,$ $|Z_0|\leq 2^{-n}$ .

If $|Z_0|\leq 2^{-n}$, then \eqref{case_6} takes place. We are looking at the case $Y_0>0,$ thus $Y_0>2^{-n}.$  Also, note that $dZ_t^{i,n}=0\quad\forall t>\tau_n$, and $|Z_t^{i,n}|\leq 2^{n}$ for $\forall t\in[0,\tau_n].$  Hence, $|Z_t^{i,n}|\leq2^n\quad\forall t$, which means $Z_t^{i,n}\geq-2^n \quad \forall t.$
Next, we have:
\begin{equation}\nonumber
    \begin{array}{rcl}
    Y_t^{i,n}&=&Y_0 +\int_0^t Z_sds\\[10pt]
    &\geq&2^{-n} - \int_0^t 2^nds\\[10pt]
    &=&2^{-n}-2^nt.\\
    \end{array}
\end{equation}

If  $$0<t<t_{0,n}=\frac{2^{-2n}}{2},$$ where $t_{0,n}$ is defined as in \eqref{eq:t0}, then:

    $$ 2^nt< \frac{2^{-n}}{2} $$
     thus:
      $$2^{-n}-2^nt> \frac{2^{-n}}{2}$$
$ \forall t\in[0,t_{0,n}].$
In other words, $Y_t^{i,n}>\frac{2^{-n}}{2}$ for $t\in[0,t_{0,n}].$
So, for all $t\in[0,t_{0,n}],$ we have:$$ Y_t^{i,n}\geq\frac{2^{-n}}{2}.$$ Hence:
\begin{equation}
\label{8}
    \begin{array}{rcl}
    X_t^{i,n}& \geq & X_0^{i,n} + \int_{0}^{t}\frac{2^{-n}}{2}ds \\[10pt]
    &\geq& \frac{2^{-n}}{2}t.
    \end{array}
\end{equation}

Furthermore, based on \eqref{8} and $|X_t^{i,n}|\leq 2^{-n}$ for all $t\in[0,t_{0,n}]$, it leads to  $t_{0,n}\leq 2,$ otherwise $ X_t^{i,n}>2^{-n},$ which means that $t>t_{0,n},$ hence a contradiction.

Note that
\begin{equation}\nonumber
\begin{array}{rcl}
 X_t^{i,n}& =&X_0+\int_{0}^{t}Y_s^{i,n}ds \\[10pt]
 Y_s^{i,n}&=    &Y_0+\int_{0}^{s}Z_k^{i,n}dk\\[10pt]
 Z_k^{i,n}&=&Z_0+\int_{0}^{k}|X_r^{i,n}|^{\alpha}\mathbf{1}_{[0,\tau_n]}(t)dB_r.
\end{array}
\end{equation}
Hence:
\begin{equation}\nonumber
    \begin{array}{rcl}
     X_t^{i,n}&  =   &X_0+Y_{0}t+\displaystyle\int_{0}^{t}\int_{0}^{s}\left(Z_0+\int _{0}^{k}|X_r^{i,n}|^{\alpha}\mathbf{1}_{[0,\tau_n]}(t)dB_r\right)dkds   \\[10pt]

         &=& X_0+Y_0t+\displaystyle\int_{0}^{t}\int_{0}^{s} Z_0dkds+\int_{0}^{t}\int_{0}^{s}\int_{0}^{k}|X_r^{i,n}|^{\alpha}\mathbf{1}_{[0,\tau_n]}(t)dB_{r}dkds\\[10pt]

         &=&X_0+Y_0t +Z_0\frac{t^2}{2} +\displaystyle\int_{0}^{t}\int_{0}^{s}\int_{0}^{k}|X_r^{i,n}|^{\alpha}\mathbf{1}_{[0,\tau_n]}(t)dB_rdkds.\\
    \end{array}
\end{equation}

Hence:
$$(X_t^{1,n}-X_t^{2,n})^2=\left(\int_{0}^{t}\int_{0}^{s}\int_{0}^{k}\left(|X_{r}^{1,n}|^{\alpha}-|X_{r}^{2,n}|^{\alpha}\right)\mathbf{1}_{[0,\tau_n]}(r)dB_rdkds\right)^2.$$
Apply the Cauchy-Schwarz inequality twice:
$$\leq t\int_{0}^{t}
\left(\int_{0}^{s}\int_{0}^{k}\left(|X_{r}^{1,n}|^{\alpha}-|X_{r}^{2,n}|^{\alpha}\right)\mathbf{1}_{[0,\tau_n]}(r)dB_rdk\right)^2ds $$
$$\leq t\int_{0}^{t}s\int_{0}^{s}\left(\int_{0}^{k}\left(|X_{r}^{1,n}|^{\alpha}-|X_{r}^{2,n}|^{\alpha}\right)\mathbf{1}_{[0,\tau_n]}(r)dB_r\right)^2dkds$$
Thus:
$$ E\left[(X_t^{1,n}-X_t^{2,n})^2\right]\leq tE\int_{0}^{s}s\int_{0}^{s}\left(\int_{0}^{k}\left(|X_{r}^{1,n}|^{\alpha}-|X_{r}^{2,n}|^{\alpha}\right)\mathbf{1}_{[0,\tau_n]}(r)dB_r\right)^2dkds.$$
By Ito's isometry:
\begin{equation}
\begin{array}{rcl}\nonumber
 E\left[(X_t^{1,n}-X_t^{2,n})^2\right] &\leq& t E\displaystyle\int_{0}^{t}s\int_{0}^{s}\int_{0}^{k}\left((|X_{r}^{1,n}|^{\alpha}-|X_{r}^{2,n}|^{\alpha})\mathbf{1}_{[0,\tau_n]}(r)\right)^2drdkds \\[10pt]
 &\leq&  tE \displaystyle\int_{0}^{t}t\int_{0}^{s}\int_{0}^{k}\left(|X_{r}^{1,n}|^{\alpha}-|X_{r}^{2,n}|^{\alpha}\right)^2drdkds \\[10pt]
 &\leq& t^2E \displaystyle\int_{0}^{t}\int_{0}^{s}\int_{0}^{k}\left(|X_{r}^{1,n}|^{\alpha}-|X_{r}^{2,n}|^{\alpha}\right)^2drdkds \\[10pt]
 &\leq& t^2E \displaystyle\int_{0}^{t}\int_{0}^{t}\int_{0}^{t}\left(|X_{r}^{1,n}|^{\alpha}-|X_{r}^{2,n}|^{\alpha}\right)^2drdkds \\[10pt]
 &=&t^4E\displaystyle\int_{0}^{t}\left(|X_{r}^{1,n}|^{\alpha}-|X_{r}^{2,n}|^{\alpha}\right)^2dr.

\end{array}
\end{equation}
Now we apply the Mean Value Theorem for the function $f(x)=x^\alpha, 0< \alpha<1$, and $a<b$ :
$$b^\alpha -a^\alpha =\alpha c^{\alpha -1}(b-a)\leq \alpha a^{\alpha -1}(b-a)$$
for $c\in(a,b).$
Then for $r\in[0,t_{0,n}],$ where $t_0$ is determined in \eqref{eq:t0}, we apply \eqref{8}:
\begin{equation}
\label{method}
 \Big||X_{r}^{1,n}|^{\alpha}-|X_{r}^{2,n}|^{\alpha}\Big|\leq \alpha\left(\frac{2^{-n}}{2}r\right)^{\alpha-1}\Big||X_{r}^{1,n}|-|X_{r}^{2,n}|\Big|.
 \end{equation}

Now let $$D_t=E\Big[\left(|X_{t}^{1,n}|-|X_{t}^{2,n}|\right)^2\Big].$$
Since $t_{0,n}\leq2$, we have for all $t\in[0,t_{0,n}],$
$$D_t\leq E\left[(X_{t}^{1,n}-X_{t}^{2,n})^2\right]\leq C_n\int_{0}^{t}r^{2\alpha -2}D_rdr.$$

for some $C_n$ depending on $n$. Since $\alpha > \frac{3}{4},  r^{4\alpha-4}$ is integrable on $ [0,t_{0,n}].$ At this stage we apply Gronwall's lemma:
\begin{lem*}
Let $I$ denote an interval of the real line of the form $[a,\infty)$ or $[a,b]$ or $[a,b)$ with $a<b.$ Let $\beta$ and
$u$ be real-valued continuous functions defined on $I$. If $u$ is differentiable on the interior $I^{o}$ of $I$ (the
interval $I$ without the endpoint $a$ and possibly $b$) and satisfies the differential inequality
$$ u^{'}(t)\leq\beta(t)u(t),\quad t\in I^{o},$$
then $u$ is bounded by the solution of the corresponding differential equation $v^{'}(t)=\beta(t)v(t):$
$$u(t)\leq u(a)\exp\left(\int_{a}^{t}\beta(s)ds\right)$$
for all $t\in I.$
\end{lem*}
\vspace{0.5cm}
 Hence, with $D_0=0$, we have $D_t=0$ for all $t\in[0,t_{0,n}].$ Therefore, \eqref{3-dimensions} has unique strong solution in $[0,t_{0,n}].$

Since $\alpha\geq\frac{3}{4}$, we have $2\alpha -2 \geq -1$. Hence, $r^{2\alpha -2}$ is integrable on $[0,t_{0,n}]$
Note: in this case since $X_t\geq 2^{-n},$ $\eta\leq 1.$ Applying Gronwall's lemma, with $D_0=0,$ we have $D_t=0$ for
all $t\in[0,t_{0,n}]$. Therefore, \eqref{3-dimensions} has a unique solution in strong sense up to time $t_{0,n}$.

Note that for all $t\in[0,t_{0,n}],$ we have $Y_t^{i,n}>\frac{2^{-n}}{2}>0.$ Hence $X_t^{i,n}$ is strictly increasing, which, leads to $X_{t_{0,n}}$ strictly positive. Therefore, by strong Markov property, we have uniqueness till the time $X$ next hits zero.
Now we proceed to the next case.

\textbf{Case II:} $Y_0>0$, $Z_0< -2^{-n}.$

Since $Z_0$ starts negative, $Y_t^{i,n}$ decreases for an amount of time. Since $Y_0$ is positive, let say $Y_0=\beta >0.$
Note that $\forall t,$ we have $|Z_t^{i,n}|<2^{n},$ which means $Z_t^{i,n}>-2^n.$ First, we have;
\begin{equation}\nonumber
\begin{array}{rcl}
    Y_t^{i,n}&= & \beta +\int_0^t Z_sds \\[10pt]
    &\geq& \beta-\int_0^t 2^nds\\[10pt]
    &=&\beta-2^nt.\\[10pt]
\end{array}
\end{equation}

Let
\begin{equation}
\label{t'}
t_{0,n}^{'}=\frac{\beta}{2^{n+1}}.
\end{equation}
If $$0<t<t_{0,n}^{'}=\frac{\beta}{2^{n+1}},$$ then:
$$2^nt< \frac{\beta}{2}$$
thus
   $$\beta-2^nt>\frac{\beta}{2} $$
for $\forall t\in[0,t_{0,n}{'}].$ In other words, $Y_t^{i,n}>\frac{\beta}{2}$ for $\forall t\in[0,t_{0,n}^{'}].$

So , $\forall t\in[0,t_{0,n}\wedge t_{0,n}^{'}],$ where $t_{0,n}$ and $t_{0,n}^{'}$ are determined in \eqref{eq:t0} and \eqref{t'} respectively, we have:

  $$ Y_t^{i,n}\geq\frac{\beta}{2}$$
  Hence:
\begin{equation}\nonumber
\begin{array}{rcl}
      X_t^{i,n}&\geq & X_0^{i,n} + \int_{0}^{t}\frac{\beta}{2}ds \\[10pt]
     &\geq& \frac{\beta}{2}t.\\
     \end{array}
 \end{equation}

Applying the same method \eqref{method} above, we use Mean Value Theorem  for the new lower bound of $X_t^{i,n}$:
$$\Big||X_{r}^{1,n}|^{\alpha}-|X_{r}^{2,n}|^{\alpha}\Big|\leq \alpha\left(\frac{\beta}{2}r\right)^{\alpha-1}\Big||X_{r}^{1,n}|-|X_{r}^{2,n}|\Big|.$$
Hence:
$$ D_t\leq E\left[(X_{t}^{1,n}-X_{t}^{2,n})^2\right]\leq C_n\int_{0}^{t}r^{2\alpha -2}D_rdr.$$
Again, applying Gronwall's lemma, with $D_0=0$, we have $D_t=0$ for all $t\in[0,t_{0,n}\wedge t_{0,n}^{'}].$ Therefore, \eqref{3-dimensions} has unique strong solution in $[0,t_{0,n}\wedge t_{0,n}^{'}].$
As in the previous cases, we have $Y_t^{i,n}>\frac{\beta}{2}>0$ for all $t\in[t_{0,n}\wedge t_{0,n}^{'}],$ which makes $X_t^{i,n}$ strictly increasing. So $X_{t_{0,n}\wedge t_{0,n}^{'}}$ is strictly positive. Thus, by strong Markov property, we have uniqueness till the next time $X$ hits zero.

\textbf{Case III: }
$Y_0=0$, $Z_0 > 2^{-n}$.

Now we let $T_n$ be the first time that either $Z_t^{1,n}$ or $Z_t^{2,n}$ hits the value $\frac{2^{-n}}{2}.$ Since both $Z_t^{1,n}$ and $Z_t^{2,n}$ are continuous, we have $T_n>0$ with probability one. So we now prove uniqueness up
to the time $t_{0,n}\wedge T_n,$ where $t_{0,n}$ is defined in \eqref{eq:t0}.

 Then for all $t$ in $[0,t_{0,n}\wedge T_{n}],$ we have: $$Z_t^{i,n}\geq\frac{2^{-n}}{2}, $$ therefore:
\begin{equation}
\label{case_11}
\begin{array}{rcl}
Y_t^{i,n}&\geq& Y_0 + \int_{0}^{t}\frac{2^{-n}}{2}ds\\[10pt]
  &\geq& \frac{2^{-n}}{2}t.
    \end{array}
\end{equation}
(Because $Y_0^{i,n}\geq 0$).

Based on \eqref{case_11} for $t\in[0,t_{0,n}\wedge T_n]:$
\begin{equation}
\label{12}
    \begin{array}{rcl}
      X_t^{i,n}&\geq   & X_0+\int_{0}^{t}\frac{2^{-n}}{2}s ds \\[10pt]
     & \geq&\frac{2^{-n}}{4}t^{2}.
    \end{array}
\end{equation}
Now we denote:
\begin{equation}\nonumber
   \begin{array}{rcl}
  X_t^{i,n}&=&\tilde{X}_t^{i,n}\\[10pt]
  Y_t^{i,n}&=&\tilde{Y}_t^{i,n}\\[10pt]
  Z_t^{i,n}&=&\tilde{Z}_t^{i,n}\\
   \end{array}
\end{equation}
for $i=1,2$ and for $t\leq\tau_n\wedge\ T_{n}\wedge t_{0,n},$ as $t_{0,n}$ defined as in \eqref{eq:t0} above.

Hence, the following system of equations holds up to the stopping time $\tau_n\wedge T_{n}\wedge t_{0,n}.$
\begin{equation}
\label{tilde-system}
    \begin{array}{rcl}
      d\tilde{X}_t^{i,n}&=   &\tilde{Y}_t^{i,n}dt  \\[10pt]
        d\tilde{Y}_t^{i,n} & =&\tilde{Z}_t^{i,n}dt\\[10pt]
        d\tilde{Z}_t^{i,n}&=&|\tilde{X}_t^{i,n}|^{\alpha}\mathbf{1}_{[0,\tau_n\wedge T_{n}\wedge t_{0,n}]}(t)dB_t,
    \end{array}
\end{equation}
with $(\tilde{X}_0^{i,n},\tilde{Y}_0^{i,n},\tilde{Z}_0^{i,n})=(X_0,Y_0,Z_0),$ for $i=1,2.$ Furthermore, using \eqref{tilde-system}, $\tilde{X}_t^{i,n}, \tilde{Y}_t^{i,n},$ and
$\tilde{Z}_t^{i,n}$ can be defined for all time.

Use Ito's isometry as above with $\tilde{X}_t^{i,n}$, $\tilde{Y}_t^{i,n}$, and $\tilde{Z}_t^{i,n}$:
\begin{equation}
\begin{array}{rcl}\nonumber
 E\left[(\tilde{X}_t^{1,n}-\tilde{X}_t^{2,n})^2\right] &\leq& t E\displaystyle\int_{0}^{t}s\int_{0}^{s}\int_{0}^{k}\left((|\tilde{X}_r^{1,n}|^{\alpha}-|\tilde{X}_r^{2,n}|^{\alpha})\mathbf{1}_{[0,\tau_n\wedge T_{n}\wedge t_{0,n}]}(r)\right)^2drdkds \\[10pt]
 &\leq&  tE \displaystyle\int_{0}^{t}t\int_{0}^{s}\int_{0}^{k}\left(|\tilde{X}_{r}^{1,n}|^{\alpha}-|\tilde{X}_{r}^{2,n}|^{\alpha}\right)^2\mathbf{1}_{[0,\tau_n\wedge T_{n}\wedge t_{0,n}]}(r)drdkds \\[10pt]
 &\leq& t^2E \displaystyle\int_{0}^{t}\int_{0}^{s}\int_{0}^{k}\left(|\tilde{X}_{r}^{1,n}|^{\alpha}-|\tilde{X}_{r}^{2,n}|^{\alpha}\right)^2\mathbf{1}_{[0,\tau_n\wedge T_{n}\wedge t_{0,n}]}(r)drdkds \\[10pt]
 &\leq& t^2E \displaystyle\int_{0}^{t}\int_{0}^{t}\int_{0}^{t}\left(|\tilde{X}_{r}^{1,n}|^{\alpha}-|\tilde{X}_{r}^{2,n}|^{\alpha}\right)^2\mathbf{1}_{[0,\tau_n\wedge T_{n}\wedge t_{0,n}]}drdkds \\[10pt]
 &=&t^4E\displaystyle\int_{0}^{t}\left(|\tilde{X}_{r}^{1,n}|^{\alpha}-|\tilde{X}_{r}^{2,n}|^{\alpha}\right)^2\mathbf{1}_{[0,\tau_n\wedge T_{n}\wedge t_{0,n}]}dr.
\end{array}
\end{equation}

Using \eqref{12} and Mean Value Theorem, for $r\in[0,\tau_n\wedge T_{n}\wedge t_{0,n}],$ we have:
\begin{equation}\nonumber
\Big||\tilde{X}_{r}^{1,n}|^{\alpha}-|\tilde{X}_{r}^{2,n}|^{\alpha}\Big|\leq \alpha\left(\frac{2^{-n}}{4}r^{2}\right)^{\alpha-1}\Big||\tilde{X}_{r}^{1,n}|-|\tilde{X}_{r}^{2,n}|\Big|,
\end{equation}

Hence:
$$ E\left[(\tilde{X}_t^{1,n}-\tilde{X}_t^{2,n})^2\right]\leq t^4\alpha^2\left(\frac{2^{-n}}{4}\right)^{2(\alpha-1)}E\displaystyle\int_{0}^{t}r^{4(\alpha-1)}(|\tilde{X}_{r}^{1,n}|-|\tilde{X}_{r}^{2,n}|)^2dr.$$

So if  we let $$D_t=E\Big[\left(|\tilde{X}_{t}^{1,n}|-|\tilde{X}_{t}^{2,n}|\right)^2\Big],$$ then:

$$D_t\leq E\left[(\tilde{X}_{t}^{1,n}-\tilde{X}_{t}^{2,n})^2\right]\leq C_n\int_{0}^{t}r^{4\alpha -4}D_rdr.$$
Again, applying Gronwall's lemma, with $D_0=0$, we have $D_t=0.$
 Note that at the time $t_{0,n}\wedge T_{n},$ since we have $Z_t^{i,n}>0$ for all $t\in[0,t_{0,n}\wedge T_{n}]$, and also initial $Y_0>0$, it leads to $Y_t^{i,n}>0$ for all
 $t\in[0,t_{0,n}\wedge T_{n}].$ Thus $X_t^{i,n}$ is strictly increasing, which makes $X_{t_{0,n}\wedge
 T_{n}}$ must be strictly greater than zero. Therefore, by strong Markov property, we obtain uniqueness of the process till $X$ next hits zero.
Now we continue with the proof of the theorem with the next case.

\vspace{0.5cm}

For $Y_0 < 0,$ as stated above, is solved similarly due to symmetry.
\vspace{0.5cm}

Lastly, we complete the proof of Theorem 1 with:

\textbf{Case IV:} $Y_0=0.$

If $Y_0=0$, then based on the definition of $\tau_n,$
$|Z_0|>2^{-n}.$ We will first deal with the case $Z_0> 2^{-n},$ and the case $Z_0< 2^{-n}$ is
approached the same way due to symmetry. As in \textbf{ Case III}, let $T_n$ be the first
time that either $Z_t^{1,n}$ or $Z_t^{2,n}$ hits the value $\frac{2^{-n}}{2}.$ Due to the continuity of $Z_t^{1,n}$ and $Z_t^{2,n}$, $T_{n}>0$ with probability one. So with $\forall t\in[0,t_0\wedge T_{n}],$ where $t_0$ is determined in \eqref{eq:t0}, we have:
\begin{equation*}\nonumber
    \begin{array}{rcl}
    Y_t^{i,n}&\geq& Y_0 + \int_{0}^{t}\frac{2^{-n}}{2}ds\\[10pt]
  &=&\frac{2^{-n}}{2}t.\\[10pt]
  \end{array}
\end{equation*}
Then
\begin{equation*}
    \begin{array}{rcl}
        X_t^{i,n}&\geq   & X_0+\int_{0}^{t}\frac{2^{-n}}{2}s ds \\[10pt]
     & \geq&\frac{2^{-n}}{4}t^{2}.

    \end{array}
\end{equation*}
We now apply the same method as in \textbf{Case III} by looking at $\tilde{X}_t^{i,n},\tilde{Y}_t^{i,n},$ and $\tilde{Z}_t^{i,n},$ which are defined as:
\begin{equation}\nonumber
   \begin{array}{rcl}
  X_t^{i,n}&=&\tilde{X}_t^{i,n}\\[10pt]
  Y_t^{i,n}&=&\tilde{Y}_t^{i,n}\\[10pt]
  Z_t^{i,n}&=&\tilde{Z}_t^{i,n}\\
   \end{array}
\end{equation}
for $i=1,2$ and for $t\leq\tau_n\wedge\ T_{n}\wedge t_{0,n},$ as $t_{0,n}$ defined as in \ref{eq:t0} above.

Hence, the following system of equations holds up to the stopping time $\tau_n\wedge T_{n}\wedge t_{0,n}.$
\begin{equation}\nonumber
    \begin{array}{rcl}
      d\tilde{X}_t^{i,n}&=   &\tilde{Y}_t^{i,n}dt  \\[10pt]
        d\tilde{Y}_t^{i,n} & =&\tilde{Z}_t^{i,n}dt\\[10pt]
        d\tilde{Z}_t^{i,n}&=&|\tilde{X}_t^{i,n}|^{\alpha}\mathbf{1}_{[0,\tau_n\wedge T_{n}\wedge t_{0,n}]}(t)dB_t,
    \end{array}
\end{equation}
with $(\tilde{X}_0^{i,n},\tilde{Y}_0^{i,n},\tilde{Z}_0^{i,n})=(X_0,Y_0,Z_0),$ for $i=1,2.$ Furthermore, using \eqref{tilde-system}, $\tilde{X}_t^{i,n}, \tilde{Y}_t^{i,n}, \tilde{Z}_t^{i,n}$ can be defined for all time.

Again, using the same strategy in \textbf{Case III} and Mean Value Theorem, we have:
$$ \Big||\tilde{X}_{r}^{1,n}|^{\alpha}-\tilde{X}_{r}^{2,n}|^{\alpha}\Big|\leq \alpha\left(\frac{2^{-n}}{4}r^{2}\right)^{\alpha-1}\Big||\tilde{X}_{r}^{1,n}|-|\tilde{X}_{r}^{2,n}|\Big|.$$
Hence, if we let $$D_t=E\Big[\left(|\tilde{X}_{t}^{1,n}|-|\tilde{X}_{t}^{2,n}|\right)^2\Big],$$ then :
$$D_t\leq E\left[(\tilde{X}_{t}^{1,n}-\tilde{X}_{t}^{2,n})^2\right]\leq C_n\int_{0}^{t}r
^{4\alpha -4}D_rdr.$$
Hence, using Gronwall's lemma with $D_0=0$ yields $D_t=0$, which completes the proof of Theorem 1.

In this casee, we also have $Y_t^{i,n}>\frac{2^{-n}}{2}t>0$ for all $t\in[0,t_{0,n}\wedge T_{n}]$. Hence $X_t^{i,n}$ is strictly increasing, which yields $X_{t_{0,n}\wedge T_{n}}$ strictly positive. So, by strong Markov property, we have uniqueness up to the time  $X$ next hits zero.

Now with uniqueness proved, we actually can even strengthen the proof by showing that $\tau_n^1=\tau_n^2$ for all $n$, where $\tau_n^1$ and $\tau_n^2$ respectively stand for the stopping time at the critical value for $X_t^1$ and $X_t^2.$ Without loss of generality, suppose $\tau_n^2>\tau_n^1:$
So at the time $\tau_n^1$, $X_t^2$ has not yet reached the critical values, which are $2^{-n}$ or $2^n,$ as stated above. But since we have uniqueness up to $\tau_n^1\wedge\tau_n^2,$ this implies $X_t^1$ has also not reached the critical value at the time $\tau_n^1,$ which is a contradiction to the definition of $\tau_n^1.$ Hence, $\tau_n^1$ and $\tau_n^2$ must be equal.
\vskip.5in

\bibliography{mybibliography}
\bibliographystyle{plain}

config: arxiv


TITLE: How to prove that $g$ or $g+p$ is a primitive root modulo $p^a$ for a primitive root $g$ modulo $p$?
QUESTION [3 upvotes]: I wish to prove the following:

If $p$ is an odd prime and $g$ is a primitive root modulo $p$, then either $g$ or $g+p$ is a primitive root modulo every power of $p$.

The only reference I can find to this is on page 26 of "A Course in Computational Algebraic Number Theory" by Henri Cohen (https://books.google.no/books?id=hXGr-9l1DXcC&pg=PP1&dq=A+Course+in+Computational+Algebraic+Number+Theory&ei=98TxSsRYjeSTBP7p6egL&redir_esc=y#v=snippet&q=primitive&f=false)
I'm struggling to understand the dense proof presented there. I understand all the steps and statements, with the exception of the last part of the following:

For any $m$ we have $m^p\equiv m\pmod{p}$, hence it follows that for every prime $l$ dividing $p-1$, $g^{p^{a-1}(p-1)/l}\equiv g^{(p-1)/l}\not\equiv 1\pmod{p}$. So for $g$ to be a primitive root, we need only that $g^{p^{a-2}(p-1)}\not\equiv 1\pmod{p^a}$.

Here is my understanding: $m^p\equiv m\pmod{p}$ is just Fermat's little theorem, and by induction it also follows that $m^{p^k}\equiv m\pmod{p}$. By inserting $a-1$ for $k$ and $g^{(p-1)/l}$ for $m$, we obtain $g^{p^{a-1}(p-1)/l}\equiv g^{(p-1)/l}\pmod{p}$. Further, since $g$ is a primitive root modulo $p$, $g^k\not\equiv 1\pmod{p}$ holds for all $k<p-1$, in particular for $k=(p-1)/l$. Going from there to say that it is sufficient that $g^{p^{a-2}(p-1)}\not\equiv 1\pmod{p^a}$ for $g$ to be a primitive root modulo $p^a$, is beyond me. Could anybody please help me understand this step?
The rest of the proof after that point is pretty straightforward, although it omits a lot of details.

REPLY [0 votes]: The order of $g$ mod $p^a$ has to divide $p^{a-1}(p-1)$ by Fermat-Euler, and you have shown you can't take out factors from $p-1$.  So the order must be $p^j(p-1)$ for some $0\leq j\leq a-1$.  Hence it suffices to impose the constraint $g^{p^{a-2}(p-1)}\not\equiv 1\pmod{p^a}$.


TITLE: Smallest positive integer to make integers monotonic
QUESTION [0 upvotes]: I saw the following programming problem in https://www.ohjelmointiputka.net/postit/tehtava.php?tunnus=lukut:
A positive integer $n$ will be called monotonic, if its decimal digits are in ascending or descending order. Suppose we are given a list of numbers:
363 726 1089 1313 1452 1717 1798 1815 1919 2121 2156 2178 2189 2541 2626 2805
2904 2997 3131 3267 3297 3434 3630 3838 3993 4037 4092 4107 4191 4242 4257 4312
4334 4343 4356 4378 4407 4532 4646 4719 4747 4807 4949 5011 5055 5071 5082 5151
5214 5353 5423 5445 5454 5495 5610 5665 5731 5808 5819 5858 5951 5989 5994 6171
6248 6281 6429 6446 6468 6523 6534 6565 6567 6594 6721 6767 6868 6897 6919 7051
7077 7128 7139 7171 7227 7260 7381 7424 7474 7513 7623 7678 7831 7858 7878 7881
7909 7986 8041 8063 8074 8088 8107 8129 8162 8173 8184 8195 8214 8283 8316 8349
8382 8415 8453 8484 8514 8624 8649 8712 8756 8778 8814 8932 8987 8989 8990 8991
9053 9064 9075 9099 9101 9119 9141 9156 9191 9213 9251 9292 9309 9328 9361 9393
9438 9493 9515 9546 9595 9597 9603 9614 9667 9678 9757 9797 9801 9802 9834 9890
9898 9909.

I heard that one can find an algorithm to find the least possible positive multipliers for every case that makes the numbers monotonic in less than second. What kind of algorithm would be suitable? I solved the problem by Python but the solution takes too much time:
def generate_all_numbers(length):
    l = list()
    for one in range(0,length):
        for two in range(0,length-one):
            for three in range(0,length-one-two):
                for four in range(0,length-one-two-three):
                    for five in range(0,length-one-two-three-four):
                        for six in range(0,length-one-two-three-four-five):
                            for seven in range(0,length-one-two-three-four-five-six):
                                for eight in range(0,length-one-two-three-four-five-six-seven):
                                    for nine in range(0,length-one-two-three-four-five-six-seven-eight):
                                        for ten in range(0,length-one-two-three-four-five-six-seven-eight-nine):
                                            if max(one,two,three,four,five,six,seven,eight,nine) > 0:
                                                num1 = "1"*one+"2"*two+"3"*three+"4"*four+"5"*five+"6"*six+"7"*seven+"8"*eight+"9"*nine
                                                num2 = "9"*one+"8"*two+"7"*three+"6"*four+"5"*five+"4"*six+"3"*seven+"2"*eight+"1"*nine+"0"*ten
                                                l.append(int(num1))
                                                l.append(int(num2))
    return(list(set(l)))

def find_smallest_increasing(number, length):
    ehd = -1
    num = "0"
    length += 1
    for one in range(0,length):
        for two in range(0,length-one):
            for three in range(0,length-one-two):
                for four in range(0,length-one-two-three):
                    for five in range(0,length-one-two-three-four):
                        for six in range(0,length-one-two-three-four-five):
                            for seven in range(0,length-one-two-three-four-five-six):
                                for eight in range(0,length-one-two-three-four-five-six-seven):
                                    for nine in range(0,length-one-two-three-four-five-six-seven-eight):
                                        if max(one,two,three,four,five,six,seven,eight,nine) > 0:
                                            num = "1"*one+"2"*two+"3"*three+"4"*four+"5"*five+"6"*six+"7"*seven+"8"*eight+"9"*nine
                                            if int(num) % number == 0:
                                                if ehd == -1:
                                                    ehd = int(num)
                                                if int(num) < ehd:
                                                    ehd = int(num)
    return(ehd)

def find_smallest_decreasing(number, length):
    ehd = -1
    num = "0"
    length += 1
    for one in range(0,length):
        for two in range(0,length-one):
            for three in range(0,length-one-two):
                for four in range(0,length-one-two-three):
                    for five in range(0,length-one-two-three-four):
                        for six in range(0,length-one-two-three-four-five):
                            for seven in range(0,length-one-two-three-four-five-six):
                                for eight in range(0,length-one-two-three-four-five-six-seven):
                                    for nine in range(0,length-one-two-three-four-five-six-seven-eight):
                                        for zero in range(0,length-one-two-three-four-five-six-seven-eight-nine):
                                            if max(one,two,three,four,five,six,seven,eight,nine) > 0:
                                                num = "9"*one+"8"*two+"7"*three+"6"*four+"5"*five+"4"*six+"3"*seven+"2"*eight+"1"*nine+"0"*zero
                                                if int(num) % number == 0:
                                                    if ehd == -1:
                                                        ehd = int(num)
                                                    if int(num) < ehd:
                                                        ehd = int(num)
    return(ehd)

numbers = [363,726, 1313, 1452, 1717, 1798, 1815, 1919, 2121, 2156, 2189, 2541, 2626, 2805,
2904, 2997, 3131, 3297, 3434, 3630, 3838, 3993, 4037, 4092, 4107, 4191, 4242, 4257, 4312,
4334, 4343, 4378, 4407, 4532, 4646, 4719, 4747, 4807, 4949, 5011, 5055, 5071, 5082, 5151,
5214, 5353, 5423, 5454, 5495, 5610, 5665, 5731, 5808, 5819, 5858, 5951, 5989, 5994, 6171,
6248, 6281, 6429, 6446, 6468, 6523, 6565, 6567, 6594, 6721, 6767, 6868, 6897, 6919, 7051,
7077, 7128, 7139, 7171, 7227, 7260, 7381, 7424, 7474, 7513, 7678, 7831, 7858, 7878, 7881,
7909, 7986, 8041, 8063, 8074, 8088, 8107, 8129, 8162, 8173, 8184, 8195, 8214, 8283, 8316, 8349,
8382, 8415, 8453, 8484, 8514, 8624, 8649, 8756, 8778, 8814, 8932, 8987, 8989, 8990, 8991,
9053, 9064, 9075, 9099, 9101, 9119, 9141, 9156, 9191, 9213, 9251, 9292, 9309, 9328, 9361, 9393,
9438, 9493, 9515, 9546, 9595, 9597, 9603, 9614, 9667, 9678, 9757, 9797, 9802, 9834, 9890,
9898, 9909]

hardnumbers = [1089, 2178, 3267, 4356, 6534, 7623, 8712, 9801, 5445]

l = generate_all_numbers(20)
A = list()
for i in range(len(l)):
    for j in range(len(numbers)):
        if l[i] % numbers[j] == 0:
             A.append(l[i])
B = list()
for j in range(len(numbers)):
 best = int("9" * 2000)
 for i in range(len(A)):
    if A[i] % numbers[j] == 0:
        if A[i] < best:
            best = A[i]
 print(str(numbers[j])+" "+str(best/numbers[j])+ " " + str(best))

for k in range(0,len(hardnumbers)):
    number = hardnumbers[k]
    a = -1
    b = -1
    i= 1
    j= 1
    while a == -1:
        if a % 5 != 0:
            a = find_smallest_increasing(number,i)
        i = i + 1
    b = -1
    j = 1
    while b == -1:
        b = find_smallest_decreasing(number,max(i,j))
        j = j + 1
    print(str(number)+" "+str(min(a,b)/number)+" " + str(min(a,b)))

The existence of a such number is proved in Does every positive number $n$ have a multiple, which has a monotonic digits in decimal basis? .

REPLY [0 votes]: Suppose, a number $N>1$ is given. Write $N$ in the form $$\large N=2^a\cdot 5^b\cdot n$$ with $\gcd(n,10)=1$. Then, the number $$\large (10^{\phi(n)}-1)\cdot 10^{max(a,b)}$$ is monotonic (it only has nines followed by zeros or only nines) and divisible by $N$. $\phi(n)$ denotes the totient-function.