TITLE: identity of $(I-z^nT^n)^{-1} =\frac{1}{n}[(I-zT)^{-1}+(I-wzT)^{-1}+...+(I-w^{n-1}zT)^{-1}]$
QUESTION [3 upvotes]: I am trying to understand the identity
$$(I-z^nT^n)^{-1} =\frac{1}{n}[(I-zT)^{-1}+(I-wzT)^{-1}+...+(I-w^{n-1}zT)^{-1}] \quad (*),$$
where $T \in \mathbb{C}^{n\times n},z\in \mathbb{C}$ and the spectral radius of $T$, $\rho(T)=\max\{|\lambda|: \exists v, Tv=\lambda v\}\leq 1$ and $|z|<1$ and $w$ is a primitive $n$th root of 1,i.e. $w =e^{i2\pi/n}$.
I have tried using the identity   $$[I-A]^{-1} = I+A+A^2+A^3+... $$ which holds when $A^n$ becomes zero matrix. This  works as  $\rho(T)=\max\{|\lambda|: \exists v, Tv=\lambda v\}\leq 1$ and $|z|<1$ implies the convergence of $A^n$. So I expand both sides of $(*)$ and then compared the coefficient of the two power series.
But this approach seems to be a bit tedious, I wonder whether there is a more clean and simpler proof.

REPLY [3 votes]: First we establish the identity:
$$
\frac{n}{1-z^n}=\sum_{i}\frac{1}{1-\omega^iz},\qquad z\in \mathbb{C}.
$$
(Hypothetically) expanding the RHS, we observe that the numerator is a polynomial of degree $\leq n-1$ and the denominator agrees with the LHS. Thus we are reduced to verifying a polynomial identity of degree $n-1$, so evaluating at $n$ points will suffice to establish equality.
For any $0\leq i\leq n-1$, evaluate both sides at $z=\omega^i$. The RHS has a pole with residue $-w^i$, coming from the term
$$
\frac{1}{1-\omega^{-i}z}=\frac{w^i}{w^i-z}.
$$
Meanwhile, the residue on the LHS is
$$
\left.\frac{n}{(1-z^n)'}\right|_{z=w^{i}}=\left.\frac{n}{-nz^{n-1}}\right|_{z=w^{i}}=-w^i.
$$
Thus the two expressions agree at $n$ points, whereupon the identity is established.
Applying the holomorphic functional calculus, we may formally replace $1$ with $I$ and $z$ with $zT$ (since $\rho(T)\leq 1$ and $|z|< 1$ everything converges), yielding your result.


TITLE: Does a measurable function to the reals have to map every set to a Borel set?
QUESTION [1 upvotes]: If we have a measure space $(X,S,\mu)$ and an $S$-measurable function $f:X\to \mathbb R$, then by the definition of measurability $f^{-1}(B)\in S$ for every Borel set $B$. Does this work the other way around? Is $f(E)$ a Borel set for every $E\in S$?

REPLY [4 votes]: Let $X=\mathbb R,$ let $\mathcal S$ be the Lebesgue measurable sets, and let $\mu$ be Lebesgue measure on $(\mathbb R,\mathcal S).$ The function $f(x)=x$ is Borel measurable (since every Borel set is Lebesgue measurable). However there are Lebesgue measurable sets $E$ that are not Borel measurable. For such an $E,$ $f(E)=E$ is not Borel measurable


TITLE: How to find this formula in this dihedral group of transformations of the plane?
QUESTION [1 upvotes]: In the group of all the bijections of the Euclidean plane onto itself, let $f(x,y) \colon = (-x,y)$ and $g(x,y) \colon = (-y,x)$ for all points $(x,y)$ in the plane. Let $$G:= \{f^i g^j | i=0,1; \  g=0,1,2,3 \}.$$ Then $G$ is the dihedral group of order $8$. Now given $i$, $j$, $s$, $t$, how to find $a$ and $b$ in terms of $i$, $j$, $s$, $t$ such that $$(f^i g^j)(f^s g^t) = f^a g^b?$$

REPLY [0 votes]: Noticing that

$fg=-gf$,
$- \mathrm{Id}$ commute with $f$ and $g$,
$g^2=- \operatorname{Id}$,

you should be able to show that $(f^ig^j)(f^sg^t)=f^{i+s}g^{(2s+1)j+t}$.
For an example, $$ \begin{array}{ll} f^3gf^2g^4 & =f^3(gf)fg^4=-f^3(fg)fg^4, \\ & =-f^3f(gf)g^4 = f^3f(fg)g^4, \\ & = f^5g^{5}. \end{array}$$


TITLE: A slicker proof that an object must be initial
QUESTION [8 upvotes]: If $\mathcal{C}$ is a category and $\lambda:\Delta_D \to id_{\mathcal{C}}$ is a cone for the identity functor, and $F:J \to \mathcal{C}$ is a functor such that $F\lambda:\Delta_D \to F$ is a limiting cone, then it follows that $D$ is an initial object of $\mathcal{C}$. This is Lemma 1 on page 230 in Maclane's Categories for the Working Mathematician. The proof given however is pretty down to earth and does not generalize immediately to higher category theory. Is there a more high-browed way of proving this that works in a more general context than 1-categories?

REPLY [1 votes]: The definition of object being initial is obviously equivalent to the statement "the inclusion of one-point category is final". Here by final functor I mean $p: I\to C$, such that for any $F: C\to D$ we have $$\mathrm{Lim}\ Fp \simeq \mathrm{Lim}\ F$$
The criterion of being final is proved in MacLane 9.3 (there such functors are called "initial", and "final" is used for colimits). Then $p: \ast \to C$ is the initial object iff $$\mathrm{Lim}\ 1_C \simeq \mathrm{Lim}\ 1_C\circ p = \mathrm{Lim} p = p$$
This is the theorem you mention. In higher category theory the concepts of limits and final functors work just the same, so we have a similar theorem. Note that the criterion of finalness differs: all corresponding overcategories must be contractible.


TITLE: Convolution of indicator functions with values in a finite field
QUESTION [1 upvotes]: This is something I haven't seen online yet, indicator functions with values in a finite field. Probably for a good reason, but I would like to know why, and if there are still things that can be said. For instance what can we say of the relationship (if any) between the support of the convolution of two indicator functions with values in a finite field vs the addition of the sets they indicate.
More precisely: Let $F$ be a finite field and let $A, B$ be sets with additive cyclic structure, say $A, B \subseteq \mathbb{Z}_N$ with $N$ coprime to the characteristic of $F$. Define the "characteristic functions" $1_A, 1_B : \mathbb{Z}_N \to F$ by $1_A(x) = 1$ if $x \in A$, and $1_A(x) = 0$ otherwise. Similarly for $1_B$. Is there any relationship/proposition we can infer between the set $A + B$ and the support of the (cyclic) convolution
$$
1_A \ast 1_B(k) = \sum_{j \in \mathbb{Z}_N} 1_A(j) 1_B(k-j)?
$$
For instance, if $1_A \ast 1_B(k) = 0$ for all $k \in \mathbb{Z}_N$, would this say anything at all about $A + B$?
Note in the case of the usual characteristic functions with values in $\mathbb{R}$ we have the nice property that $A + B$ is precisely the support of $1_A \ast 1_B$. So I wondered whether we can at least get something (although probably not quite this I imagine) when the values of the characteristic functions are either the additive or multiplicative identity in a finite field. Thanks!
Remark: Recall that the convolution here is taken modulo $p$, where $p$ is the characteristic of the field. Moreover note that the support of $1_A \ast 1_B$ is contained in $A + B$, since, if the convolution is non-zero modulo $p$, it is non-zero in $\mathbb{R}$ and so the support lies in $A + B$ by the fact mentioned above. This however does not say anything on whether or not the support is empty modulo $p$...

REPLY [3 votes]: Note that $1_A(j)1_B(k-j)$ is nonzero if and only if $j\in A$ and $k-j\in B$, which implies that $k$ is in $A+B$. Even if $1_A(j)1_B(k-j)$ is nonzero, their sum could be zero if one takes the finite field aspect into account. Thus the support of the convolution is no bigger than $A+B$. Here is an example that shows that things break down.
Let $\mathbb{F}$ be the field of three elements and $n=5$. Let $A = \{0,1,2\}$ and $B = \{0,1,2,3\}$. Then $A+B = \{0,1,2,3,4\}$ and the above analysis shows that $1_A(j)1_B(k-j)$ is nonzero if and only if $k\in \{0,\ldots,4\}$. However in that case
\begin{align}
(1_A\ast 1_B)(k) &= \sum_{j=0}^{4} 1_A(j)1_B(k-j) \\
&= 1_A(0)1_B(k) + 1_A(1)1_B(k-1) + 1_A(2)1_B(k-2).
\end{align}
If $k = 4$, then the first term is zero but the second two are nonzero so the convolution is nonzero (since $\mathbb{F}$ is characteristic $3$). Taking instead $k=3$, all three terms are nonzero and so they sum to one modulo $3$. Same goes for $k=2$. You can play this game whenever $n > \operatorname{char}(\mathbb{F})$. You just have to modify it a bit but the idea is that you want to have $A$ and $B$ to have at an overlap of $\operatorname{char}(\mathbb{F})$ when shifting. (Meaning let $A$ and $B$ have at least $\operatorname{char}(\mathbb{F})$ elements.) If both $A$ and $B$ have support less than $\operatorname{char}(\mathbb{F})$, then the support of their convolution should be $A+B$.

In the case that $n < \operatorname{char}(\mathbb{F})$, you get the analogous result from $\mathbb{R}$. This is easy to see by breaking $A$ and $B$ into a union of singletons. This will make it clear how things shake out. In general, $1_A = \sum_{a\in A} 1_{\{a\}}$. Thus your convolution is nothing more than
\begin{align}
(1_A\ast 1_B)(k) &= \sum_{j\in\mathbb{Z}_n} \left(\sum_{a\in A}1_{\{a\}}\right)(j)\left(\sum_{b\in B}1_{\{b\}}\right)(k-j) \\
&= \sum_{j\in\mathbb{Z}_n}\sum_{a\in A}\sum_{b\in B} 1_{\{a\}}(j)1_{\{b\}}(k-j)
\end{align}
If $k\in A+B$, then $k=a_0+b_0$ for some $a_0\in A$ and $b_0\in B$. Thus
$$ (1_A\ast 1_B)(k) = \sum_{j\in\mathbb{Z}_n}\sum_{a\in A}\sum_{b\in B} 1_{\{a\}}(j) 1_{\{b\}}(a_0+b_0-j).$$
Letting $j = a_0 = a$ and $b = b_0$, we see that the convolution is nonzero since $n<\operatorname{char}(\mathbb{F})$. Thus the support of $1_A\ast 1_B$ contains $A+B$. The question then is whether or not the support is only $A+B$. The convolution is nonzero at $k$ if $j\in A$ and $k-j\in B$ for some $j$. Thus $k\in A+B$. Again you need that $n<\operatorname{char}(\mathbb{F})$ to conclude that it can't sum to $0$.


TITLE: Omega Notation and Average Running Time Problem
QUESTION [0 upvotes]: if we have an algorithm that average running time of randomized algorithm A for input of size n is equal to $\theta(n^2)$.
why there would be an input data such that A solve it in $\Omega(n^{3n})$?

REPLY [0 votes]: Because $\Theta\left(n^2\right)$ is characterizing average running time across all possible inputs, - it just means that your algorithm runs extremely well on vast majority of inputs but has a small margin of pathological cases.
EDIT Note that in such a case, the worst-case complexity would be $\Omega \left(n^{3n}\right)$, but that does not contradict that average-case complexity is quadratic - you could be dividing by a large denominator when taking the average...


TITLE: Properties of the matrix rank
QUESTION [0 upvotes]: Let $I$ be the identity matrix $n\times n$, B a matrix $n{\times} n$ and $c$ a constant. Are these properties correct?$\DeclareMathOperator{rank}{rank}$
$$\rank(cB)=c\rank(B),$$
$$\rank(I-B)=n-\rank(B).$$

REPLY [1 votes]: Both are wrong:
The first would imply, if the rank of $B$ is positive, that the rank of $cB$ tends to $\infty$ as $c$ gets larger and larger.
Actually, $\operatorname{rank}cB=\operatorname{rank}B\;$ if $\:c\ne 0$.
The second  would imply that if $B$ is invertible, i.e. has full rank, that $I-B$ has rank $0$, i.e. $B=I$.


TITLE: How to prove $A^{n\times n}=I_n\Rightarrow A^n=A^{f(n\times n)}$?
QUESTION [1 upvotes]: Let $A\in M_2(\mathbb{Z})$ s.t. there is a positive integer $n$ satisfying $A^n=I_2$.
Show that $A^{12}=I_2$.
I have no idea where to start. Suggestions?

REPLY [0 votes]: Let $\alpha,\beta$ be the eigenvalues of $A$. Then $\alpha ^n=\beta ^n=1$; moreover $A$ is diagonalizable over $\mathbb{C}$.
Case 1. $\alpha,\beta\in\{\pm 1\}$. Then $A^2=I$.
Case 2. $\alpha,\beta\notin \mathbb{R}$. Then $\alpha=e^{2ik\pi/n},\beta=e^{-2ik\pi/n}$ where $k$ is a positive integer $<n$. Note that $\alpha+\beta=2\cos(2k\pi/n)\in\mathbb{Z}$ and $\cos(2k\pi/n)\in\{0,\pm 1/2\}$. Finally $2k\pi/n\in\{\pi/2,3\pi/2,\pi/3,2\pi/3,4\pi/3,5\pi/3\}$, that (resp.) gives: $A^4=I,A^4=I,A^6=I,A^3=I,A^3=I,A^6=I$. In each case, $A^{12}=I$.


TITLE: Expected number of vertices a distance $k$ away in a random graph?
QUESTION [4 upvotes]: Given a random (undirected and unweighted) graph $G$ on $n$ vertices where each of the edges has equal and independent probability $p$ of existing (see Erdős–Rényi model). Fix some vertex $u\in G$. I want to know what is the expected number of vertices a distance $k$ away (where distance means the shortest distance). I can only figure out that we expect there to be $(n-1)p$ vertices a distance 1 away since each of the possible $(n-1)$ edges outgoing from $u$ have probability $p$ of existing, but I am having trouble generalizing from this local property. Maybe someone can handle the task or perhaps point me in the right direction via a paper?
If it makes it easier, assume that $G$ is connected (as it almost surely is as $n\rightarrow \infty$, which is the case I am interested about). As a subproblem of almost equal importance, I would in particular like to know what is the expected longest shortest distance from vertex $u$ and how many vertices are away from vertex $u$ at that distance (subproblem with the maximum possible $k$).
Note: (1) The graph does not have a bound on the degree of the vertex (there could be as many as $n-1$ adjacent vertices to $u$).

REPLY [5 votes]: Let $\lambda=np$. Then the average distance $k$ between two randomly picked vertices from  the giant component is
$$
k\sim \frac{\log n}{\log \lambda},
$$
given that the giant component exists ($\lambda>1$). A proof can be found in, e.g., Durrett's Random graph dynamics.
This result implies that the actual number for your first question is
$$
\lambda^k.
$$
However, this only works if $pn\to\lambda<\infty$ when $n\to\infty$. What about the case $p=const$? Here is an exercise: Show that the diameter of the E-R random graph in this case is 2 a.a.s.
For the diameter (the largest among the shortest paths) the question is more subtle. A nice review is given in Complex Graphs and Networks bu Lu and Chung, and the answer depends on the parameters.


TITLE: find minimal polynomial in ${GF(3^2)}$
QUESTION [0 upvotes]: is this solution right ... when ${\alpha =g }$    and ${ \alpha = g^2 }$

REPLY [1 votes]: Here is a computer check, using sage:
sage: R.<x> = PolynomialRing( GF(3) )
sage: R
Univariate Polynomial Ring in x over Finite Field of size 3
sage: F.<g> = GF( 3^2, modulus=x^2+2*x+2 )
sage: g.minpoly()
x^2 + 2*x + 2
sage: F
Finite Field in g of size 3^2
sage: for k in [1..8]:
....:     print( "g^%s = %s has minimal polynomial %s"
....:            % ( k, g^k, (g^k).minpoly() ) )
....:
g^1 = g has minimal polynomial x^2 + 2*x + 2
g^2 = g + 1 has minimal polynomial x^2 + 1
g^3 = 2*g + 1 has minimal polynomial x^2 + 2*x + 2
g^4 = 2 has minimal polynomial x + 1
g^5 = 2*g has minimal polynomial x^2 + x + 2
g^6 = 2*g + 2 has minimal polynomial x^2 + 1
g^7 = g + 2 has minimal polynomial x^2 + x + 2
g^8 = 1 has minimal polynomial x + 2
sage:

I could not understand some lines from the picture.
(For instance, $g^4=2$, the minimal polynomial is $(x-2)=x+1$...)

Later edit:
Let us get all the minimal polynomials using explicit computations. I will denote by $\Phi$ the Frobenius morphism $\Phi(h)=h^3$, $h\in F:=\Bbb F_9=\Bbb F_3[g]$.

$g$ has minimal polynomial $f(x)=(x^2+2x+2)$, this is give. The two roots are $g$ and $\Phi g=g^3=g(g^2)=g(g+1)=g^2+g=(g+1)+g=2g+1$. Vieta check: the sum of $g,2g+1$ is $1$, and the coefficient of $x$ in $(x^2+2x+2)$ is $2=-1$, the product of  $g,2g+1$ is $g(2g+1)=2g^2+g=-(g+1)+g=-1=2$.
$g^2=g+1$, so the polynomial $f(x-1)$ annihilates it. (Here $f$ is the minimal polynomial of $g$ above.) We compute $f(x-1)= (x-1)^2+2(x-1)+2=(x^2-2x+1)+(2x-2)+2 =x^2+1$. Alternatively, we can compute explicitly $(x-g^2)(x-\Phi g^2)=(x-g^2)(x-g^6))=x^2 -(g^2+g^6)x+g^8$. Now $g$ is a generator of the unit group, so it has multiplicative order $8$, so $g^8=1$, and $g^4=-1$. This gives $g^2+g^6=g^2(1+g^4)=g^2(1+(-1))=0$.
$g^3$ is conjugated to $g$, same minimal polynomial as computed for $g$.
$g^5=2g$, so $f(x/2)=f(x/(-1))=f(-x)=x^2-2x+2=x^2+x+2$ is the minimal polynomial.
$g^6$ is conjugated to $g^2$, same minimal polynomial as computed for $g^2$.
$g^7$ is either $1/g$, so we take the reciprocal of $f$, which is $1+2x+2x^2$, adjust it to be monic, get $x^2+x+2$ as minimal polynomial for it. Or we use $g^7=g+2=g-1$, so $f(x+1)$ is this minimal polynomial, so we compute $(x+1)^2 + 2(x+1)+2$. Or we use the fact that it is conjugated to $g^5$, because $\Phi g^5=g^{15}=g^7\cdot g^8=g^7$, so the minimal polynomial coincides with the one of $g^5$.


TITLE: On energy levels and emission of photons
QUESTION [1 upvotes]: This is a very basic question but I cannot seem to find the answer anywhere.
Say we have an atom in ground state. Its first energy level is 2 eV. An incoming photon of energy 2.5 eV hits an electron in the atom (with the lowest energy level) which is excited and moves up one enery level. However it does not have enough energy to reach the next energy level. When it loses energy it emits a photon of energy 2 eV. What has happened to the remaining 0.5 eV? Is it some other type of energy in the atom such as kinetic energy?

REPLY [1 votes]: The probability of a photon of energy $E$ — or corresponding wavelength $\lambda = h c / E$ — being absorbed by an atom and bringing an electron from level $i$ to level $j$ is given by the cross section $\phi_{ij}(\lambda)$, which is a sharply peaked function of wavelength (a so-called Lorentzian function). For instance, to bring a hydrogen atom from its ground state to its first excited state, a photon of $\lambda$ = 1215.67 Å, or $E$ = 10.2 eV  is needed. If the energy is just a little smaller, or a little larger, say 10.21 eV, corresponding the a wavelength half an Ångström shorter, the probability of being absorbed is many orders of magnitude smaller. The photon is said no longer to be in resonance with the energy level.
Thermal broadening
In reality, when a photon enters an ensemble of atoms, the temperature of the gas causes the atoms to have a (Gaussian) distribution of velocities. This means that although the incoming photon has a "wrong" energy of 10.21 eV, there are plenty of atoms that happen to be moving away from the photon at exactly the right speed, such that in the reference frame of the atom, the photon is redshifted into resonance. That is, the probability of being absorbed is large.
The effect is that the absorption cross section of the energy level is broadened, becoming a convolution of the "natural" (Lorentzian) and the thermal (Gaussian) profiles, called a Voigt profile.


TITLE: Prove that $\gamma$ is a measure
QUESTION [0 upvotes]: Let $Q = \{A \subset \mathbb{R}: A \text{ countable or } A^c \text{ countable} \}$ and $\gamma: Q \to \{0,1\}$, where $\gamma(A) = 0$ if $A$ is countable, $\gamma(A) = 1$ otherwise.
I struggle with showing one of the properties that this is a measure,
$$
\gamma(\cup_{i=1}^{n} A_i) = \sum_{i=1}^{n} \gamma (A_i)
$$
This I want I came up with:
Let $A_i \in Q$ be pairwise disjoint, $i\in \mathbb{N}$ and order them s.t for $i = 1,\dots,k \ k\leq n$ are not countable and for $i=k,\dots,n$ countable. Then
$$
\gamma (\cup_{i=1}^{n} A_i) = \gamma (\cup_{i=1}^{k} A_k) \cup \gamma ( \cup_{i=k}^{n} A_i)
$$
I know that a countable union of countable sets is countable, hence $\gamma ( \cup_{i=k}^{n} A_i) = 0$. But this is all I got.

REPLY [3 votes]: If the $A_k$ are pairwise disjoint, then at most one of them can be the complement of a countable set.
To see this, suppose $A_1$ is the complement of a countable set, then we must have $A_k \subset A_1^c$ for all $k>1$, and since $A_1^c$ is countable, so is $A_k$.
In particular, if $A_k$ are pairwise disjoint, then there are only two cases to consider: (i) All are countable. (ii) Exactly one is the complement of a countable set and the rest are countable.
The desired result follows from these two cases.


text:
\begin{document}

\begin{frontmatter}
\title{On a generalization of the global attractivity for a periodically
forced Pielou's equation}
\author[waseda]{Keigo Ishihara}
\ead{keigo.i.123@gmail.com}
\author[waseda]{Yukihiko Nakata\corauthref{cor1}}
\ead{yunayuna.na@gmail.com}
\corauth[cor1]{Corresponding author}
\address[waseda]{Department of Pure and Applied Mathematics, Waseda University, 3-4-1
Ohkubo, Shinjuku-ku, Tokyo 169-8555, Japan}
\begin{abstract}
In this paper, we study the global attractivity for a class of periodic
difference equation with delay which has a generalized form of Pielou's
difference equation. The global dynamics of the equation is characterized
by using a relation between the upper and lower limit of the solution.
There are two possible global dynamics: zero solution is globally
attractive or there exists a periodic solution which is globally attractive.
Recent results in {[}E. Camouzis, G. Ladas, Periodically forced Pielou's
equation, J. Math. Anal. Appl. 333 (1) (2007) 117-127{]} is generalized.
Two examples are given to illustrate our results.\end{abstract}
\begin{keyword}
Periodically forced difference equation; global attractivity; Pielou's
equation
\end{keyword}
\end{frontmatter}

\section{Introduction }

Several authors have studied difference equations for mathematical
models in population biology (see \cite{MR0252051,MR1162273,MR1247956,MR2100727,MR2151684,MR2314577,MR2397325,MR2323480,MR2413754,MR2447183,Liz200626}
and the references therein). Pielou's equation was proposed by Pielou
in \cite{MR0252051} as a discrete analogue of the logistic equation
with delay.

Camouzis and Ladas \cite{MR2323480} studied the following Pielou's
equation with a periodic coefficient, \begin{equation}
x_{n+1}=\frac{\beta_{n}x_{n}}{1+x_{n-1}},n=0,1,2,\dots,\label{periodicpielou}\end{equation}
where $\beta_{n},n=0,1,2,\dots$ is a periodic sequence with an arbitrary
positive integer $k$. Initial conditions are given by $x_{0}>0,x_{-1}\geq0$.
They proved that every solution converges to $0$, if $\Pi_{n=1}^{k}\beta_{n}\leq1$,
while there exists a $k$-periodic solution which is globally attractive,
if $\Pi_{n=1}^{k}\beta_{n}>1$ by using an interesting relation between
the upper and lower limit of the solution. Recently, Nyerges \cite{MR2413754}
studied the global dynamics of a general autonomous difference equation
by extending their idea. Let us introduce the main result in Camouzis
and Ladas \cite{MR2323480}.

\begin{theorema} (see Camouzis and Ladas \cite[Theorems 3.2, 3.3 and 3.4]{MR2323480})
If \[
\Pi_{n=1}^{k}\beta_{n}>1,\]
then there exists a $k$-periodic solution $x_{n}^{*}$ such that
$x_{n}^{*}=x_{n+k}^{*}$ which is globally attractive, that is, for
any solution of (\ref{periodicpielou}), it holds that \[
\lim_{n\to+\infty}(x_{n}-x_{n}^{*})=0.\]
\end{theorema}

In this paper, we further generalize Theorem A. The present paper
is focused on nonautonomous difference equations, different from Nyerges
\cite{MR2413754}. We shall study the following difference equation,
\begin{equation}
x_{n+1}=x_{n}f_{n}(x_{n-1}),n=0,1,2,\dots,\label{eq:Mother_generalPielou}\end{equation}
where $f_{n}(x),n=0,1,2,\dots,$ is continuous, bounded and positive
function on $x\in[0,+\infty)$ and $k$-periodic on $n$, that is,\[
f_{n}(x)=f_{n+k}(x)\text{ where }k\text{ is an arbitrary positive integer}.\]
It is assumed that the initial conditions are given by $x_{0}=x^{0}>0,x_{-1}=x^{-1}\geq0$.
For the function $f_{n}(x),n=1,2,\dots,k$, we impose the following
monotonicity property\begin{equation}
f_{n}(x)\text{ is strictly decreasing on }x\in[0,+\infty)\text{ for }n=1,2,\dots,k,\label{eq:ass00}\end{equation}
and\begin{equation}
xf_{n}(x)\text{ is strictly increasing on }x\in[0,+\infty)\text{ for }n=1,2,\dots,k.\label{eq:assumption}\end{equation}
Under the assumption (\ref{eq:ass00}), $\Pi_{n=1}^{k}f_{n}(x)$ is
also strictly decreasing on $x\in[0,+\infty)$ and hence, there exists
a some constant $c\geq0$ such that \[
c=\lim_{x\to+\infty}\Pi_{n=1}^{k}f_{n}(x).\]
The following theorem is a generalized version of Theorem A.
\begin{thm}
\label{thm:mainthm}Assume that (\ref{eq:ass00}) and (\ref{eq:assumption}).
If \begin{equation}
\Pi_{n=1}^{k}f_{n}(0)>1,\text{ and }\lim_{x\to+\infty}\Pi_{n=1}^{k}f_{n}(x)<1,\label{eq:theorem11}\end{equation}
 then there exists a $k$-periodic solution $x_{n}^{*}$ such that
$x_{n}^{*}=x_{n+k}^{*}$ which is globally attractive, that is, for
any solution of (\ref{eq:Mother_generalPielou}), it holds that \[
\lim_{n\to+\infty}(x_{n}-x_{n}^{*})=0.\]

\end{thm}
One can see that the assumptions (\ref{eq:ass00}) and (\ref{eq:assumption})
are nice properties to obtain the global character of the periodic
difference equation which has a form (\ref{eq:Mother_generalPielou}).
Motivated by Camouzis and Ladas \cite{MR2323480}, we also find a
useful relation between the upper and lower limit of the solution.
Then we can obtain the existence of $k$-periodic solution which is
globally attractive if (\ref{eq:theorem11}) holds.

The paper is organized as follows. At first, we consider the case
where every solution approaches to zero solution in Section 2. In
Section 3, we show that every solution is bounded above and below
by a positive constant, respectively. This makes possible to consider
a set of the upper and lower limit of the solutions which are positive
constants (see (\ref{eq:defsh})). In Sections 4 and 5, we consider
the existence of a $k$-periodic solution which is globally attractive.
We divide the discussion in two cases, $k$ is an even integer in
Section 4 and $k$ is an odd integer in Section 5. It is important
to establish the relation (see Lemmas \ref{lem:Lemma_even} and \ref{lem:Lemma_odd})
between the upper and lower limit of the solution in these sections.
In Section 6, we apply Theorems \ref{thm:mainthm} to two nonautonomous
difference equations. The global attractivity for a delayed Beverton-Holt
equation with a periodic coefficient is established.

\section{Global attractivity of zero solution}

First of all, we consider the case where every solution approaches
to zero. Let us introduce the following result which generalizes Theorem
3.1 in Camouzis and Ladas \cite{MR2323480}.
\begin{thm}
\label{thm:main0}Assume that (\ref{eq:ass00}). If\begin{equation}
\Pi_{n=1}^{k}f_{n}(0)\leq1,\label{eq:theorem12}\end{equation}
then, for any solution of (\ref{eq:Mother_generalPielou}), it holds
that\[
\lim_{n\to+\infty}x_{n}=0.\]
\end{thm}
\begin{pf}
We have \[
\Pi_{n=1}^{k}f_{n}(x_{n})\begin{cases}
<\Pi_{n=1}^{k}f_{n}(0) & \text{ if there exists \ensuremath{n\in\left\{ 1,2,\dots,k\right\} } such that \ensuremath{x_{n}>0,}}\\
=\Pi_{n=1}^{k}f_{n}(0) & \text{ for }x_{n}=0,n=1,2,\dots,k,\end{cases}\]
from (\ref{eq:ass00}). Since $x_{n}>0$ for $n=1,2,\dots,$ for given
the initial conditions $x_{0}=x^{0}>0,x_{-1}=x^{-1}\geq0$, it holds
\[
x_{n}=x_{n-k}\Pi_{j=1}^{k}f_{n-j}(x_{n-j-1})<x_{n-k},\text{ for }n=k+1,k+2,\dots.\]
Thus, we obtain the conclusion and the proof is complete.\qed
\end{pf}

\section{Permanence}

In this section, we show that every solution is bounded above and
below by a positive constant, respectively, if (\ref{eq:theorem11})
holds. Based on the following result, we investigate the existence
of the periodic solution which is globally attractive for any solution
in Sections 4 and 5.
\begin{thm}
\label{thm:Permanence}Assume that (\ref{eq:ass00}). If (\ref{eq:theorem11}),
then, for any solution of (\ref{eq:Mother_generalPielou}), it holds
that\[
0<\tilde{x}\Pi_{n=1}^{k}f_{n}(\overline{x})\leq\liminf_{n\to+\infty}x_{n}\leq\limsup_{n\to+\infty}x_{n}\leq\tilde{x}\Pi_{n=1}^{k}f_{n}(0)<+\infty,\]
where $\overline{x}=\tilde{x}\Pi_{n=1}^{k}f_{n}(0)$ and $\tilde{x}$
is a unique positive solution of $\Pi_{n=1}^{k}f_{n}(x)=1$.\end{thm}
\begin{pf}
We see that there exists $\tilde{x}<+\infty$ such that $\Pi_{n=1}^{k}f_{n}(\tilde{x})=1$
by (\ref{eq:theorem11}).

Suppose that $\limsup_{n\to+\infty}x_{n}=+\infty$. Then there exists
a subsequence $\{\overline{n}_{m}\}_{m=1}^{+\infty}$ such that\begin{equation}
x_{\overline{n}_{m}}=\max_{0\leq n\leq\overline{n}_{m}}x_{n}\text{ and }\lim_{m\to+\infty}x_{\overline{n}_{m}}=\limsup_{n\to+\infty}x_{n}=+\infty.\label{eq:suppose1}\end{equation}
From (\ref{eq:Mother_generalPielou}), it holds that\begin{align}
x_{\overline{n}_{m}} & =x_{\overline{n}_{m}-1}f_{\overline{n}_{m}-1}(x_{\overline{n}_{m}-2})\nonumber \\
 & =x_{\overline{n}_{m}-2}f_{\overline{n}_{m}-2}(x_{\overline{n}_{m}-3})f_{\overline{n}_{m}-1}(x_{\overline{n}_{m}-2})\nonumber \\
 & =\dots\nonumber \\
 & =x_{\overline{n}_{m}-k}\Pi_{j=1}^{k}f_{\overline{n}_{m}-j}(x_{\overline{n}_{m}-j-1}).\label{eq:ite1}\end{align}
From (\ref{eq:suppose1}) and (\ref{eq:ite1}), we see\[
\Pi_{j=1}^{k}f_{\overline{n}_{m}-j}(x_{\overline{n}_{m}-j-1})\geq1,\]
and, by (\ref{eq:ass00}), it follows \[
\Pi_{j=1}^{k}f_{\overline{n}_{m}-j}(x_{\overline{n}_{m}-\underline{j}-1})\geq\Pi_{j=1}^{k}f_{\overline{n}_{m}-j}(x_{\overline{n}_{m}-j-1})\geq1,\]
where \[
x_{\overline{n}_{m}-\underline{j}-1}=\min_{1\leq j\le k}x_{\overline{n}_{m}-j-1}.\]
Then, we see $x_{\overline{n}_{m}-\underline{j}-1}\leq\tilde{x}$
and hence, from (\ref{eq:ite1}), we obtain \begin{align*}
x_{\overline{n}_{m}} & =x_{\overline{n}_{m}-\underline{j}-1}\Pi_{j=1}^{\underline{j}+1}f_{\overline{n}_{m}-j}(x_{\overline{n}_{m}-j-1})\\
 & \leq\tilde{x}\Pi_{n=1}^{k}f_{n}(0)<+\infty,\end{align*}
which leads a contradiction to our assumption. Thus \[
\limsup_{n\to+\infty}x_{n}<+\infty.\]

Moreover, similar to the above discussion, we obtain that\[
\limsup_{n\to+\infty}x_{n}\leq\tilde{x}\Pi_{n=1}^{k}f_{n}(0)<+\infty.\]

Next, suppose that $\liminf_{n\to+\infty}x_{n}=0$. Then there exists
a subsequence $\{\underline{n}_{m}\}_{m=1}^{+\infty}$ such that \begin{equation}
x_{\underline{n}_{m}}=\min_{0\leq n\leq\underline{n}_{m}}x_{n}\text{ and }\lim_{m\to+\infty}x_{\underline{n}_{m}}=\liminf_{n\to+\infty}x_{n}=0.\label{eq:suppose2}\end{equation}
From (\ref{eq:Mother_generalPielou}), it holds that\begin{align}
x_{\underline{n}_{m}} & =x_{\underline{n}_{m}-1}f_{\underline{n}_{m}-1}(x_{\underline{n}_{m}-2})\nonumber \\
 & =x_{\underline{n}_{m}-2}f_{\underline{n}_{m}-2}(x_{\underline{n}_{m}-3})f_{\underline{n}_{m}-1}(x_{\underline{n}_{m}-2})\nonumber \\
 & =\dots\nonumber \\
 & =x_{\underline{n}_{m}-k}\Pi_{j=1}^{k}f_{\underline{n}_{m}-j}(x_{\underline{n}_{m}-j-1}).\label{eq:ite2}\end{align}
 From (\ref{eq:suppose2}) and (\ref{eq:ite2}), we see \[
\Pi_{j=1}^{k}f_{\underline{n}_{m}-j}(x_{\underline{n}_{m}-j-1})\leq1,\]
and, by (\ref{eq:ass00}), it follows \[
\Pi_{j=1}^{k}f_{\underline{n}_{m}-j}(x_{\underline{n}_{m}-\overline{j}-1})\leq\Pi_{j=1}^{k}f_{\underline{n}_{m}-j}(x_{\underline{n}_{m}-j-1})\leq1,\]
where \[
x_{\underline{n}_{m}-\overline{j}-1}=\max_{1\leq j\le k}x_{\underline{n}_{m}-j-1}.\]
Then, we see $x_{\underline{n}_{m}-\overline{j}-1}\geq\tilde{x}$.
Hence, from (\ref{eq:ite2}), we obtain \begin{align*}
x_{\underline{n}_{m}} & \geq\tilde{x}\Pi_{n=1}^{k}f_{n}(\overline{x})>0,\end{align*}
where $\overline{x}=\tilde{x}\Pi_{n=1}^{k}f_{n}(0)$. This leads a
contradiction to our assumption. Thus, we obtain \[
\liminf_{n\to+\infty}x_{n}>0.\]

Moreover, similar to the above discussion, we obtain \[
\liminf_{n\to+\infty}x_{n}\geq\tilde{x}\Pi_{n=1}^{k}f_{n}(\overline{x})>0.\]
Hence, the proof is complete.\qed \end{pf}
\begin{rem}
The assumption (\ref{eq:assumption}) is not needed for the permanence
of the solution.
\end{rem}
Hereafter, we assume (\ref{eq:ass00}) and (\ref{eq:assumption})
hold. We are interested in the existence of $k$-periodic solution
which is globally attractive. Let us introduce some notations which
are used throughout the paper. At first, we set \begin{equation}
\begin{cases}
S_{h} & =\limsup_{n\to+\infty}x_{kn+h},\\
I_{h} & =\liminf_{n\to+\infty}x_{kn+h},\end{cases}\text{ for }h=\dots,-1,0,1,\dots,\label{eq:defsh}\end{equation}
and will show \[
S_{h}=I_{h},\text{ for }h=1,2,\dots,k,\]
holds in order to establish the existence of the periodic solution
which is globally attractive. Obviously, by Theorem \ref{thm:Permanence},
we see $(S_{h},I_{h})\in(0,\infty)\times(0,\infty)$ if (\ref{eq:theorem11})
holds. Further, from (\ref{eq:defsh}), it follows \begin{equation}
\begin{cases}
S_{h}=\limsup_{n\to+\infty}x_{kn+h} & =\limsup_{n\to+\infty}x_{k\left(n+j\right)+h}=S_{kj+h},\\
I_{h}=\liminf_{n\to+\infty}x_{kn+h} & =\liminf_{n\to+\infty}x_{k\left(n+j\right)+h}=I_{kj+h},\end{cases}\label{periodicityofsi}\end{equation}
where $j$ is a some integer and the relation (\ref{periodicityofsi})
will be used if necessary. We, further, introduce two sets of subsequences
$\{\overline{n}_{m}^{h}\}_{m=0}^{+\infty}$ and $\{\underline{n}_{m}^{h}\}_{m=0}^{+\infty}$
for $h=1,2,\dots,k$ such that\[
\begin{cases}
\lim_{m\to+\infty}x_{k\overline{n}_{m}^{h}+h} & =\limsup_{n\to+\infty}x_{kn+h}=S_{h},\\
\lim_{m\to+\infty}x_{k\underline{n}_{m}^{h}+h} & =\liminf_{n\to+\infty}x_{kn+h}=I_{h}.\end{cases}\text{ for }h=1,2,\dots,k.\]
Finally, for simplicity of the proof, we also define $f_{h}(x)$ as
\[
f_{h}(x)=f_{k+h}(x)\text{ for }h=-1,-2,\dots,-k.\]

\section{Global attractivity for the case where $k$ is an even integer}

In this section, we show that Theorem \ref{thm:mainthm} holds when
$k$ is an even integer. For the reader, we first consider the case
$k=2$ in Section 4.1. (\ref{eq:concl_Lem_2}) in Lemma \ref{lem:Lemma_2}
has an important role in this subsection. Then, we give Theorem \ref{thm:Theorem_2}
which shows that there exists a $2$-periodic solution which is globally
attractive. In Section 4.2, we generalize these results to the case
where $k$ is an arbitrary even integer.

\subsection{Case: $k=2$}

We introduce the following lemma which plays a crucial role in the
proof of Theorem \ref{thm:Theorem_2}.
\begin{lem}
\label{lem:Lemma_2} Let $k=2$. Assume that (\ref{eq:ass00}) and
(\ref{eq:assumption}). If (\ref{eq:theorem11}), then it holds that
\begin{equation}
f_{h}(I_{h-1})f_{h-1}(S_{h})=1,\label{eq:concl_Lem_2}\end{equation}
for $h=1,2$. \end{lem}
\begin{pf}
At first, we see that \begin{align}
x_{k\overline{n}_{m}^{h}+h} & =x_{k\overline{n}_{m}^{h}+h-1}f_{k\overline{n}_{m}^{h}+h-1}(x_{k\overline{n}_{m}^{h}+h-2})\nonumber \\
 & =x_{k\overline{n}_{m}^{h}+h-2}f_{k\overline{n}_{m}^{h}+h-2}(x_{k\overline{n}_{m}^{h}+h-3})f_{k\overline{n}_{m}^{h}+h-1}(x_{k\overline{n}_{m}^{h}+h-2})\nonumber \\
 & =x_{k\overline{n}_{m}^{h}+h-2}f_{h-2}(x_{k\overline{n}_{m}^{h}+h-3})f_{h-1}(x_{k\overline{n}_{m}^{h}+h-2}),\label{eq:lemma2_ite}\end{align}
 for $h=1,2$, from (\ref{eq:Mother_generalPielou}) and by considering
the limiting equation of (\ref{eq:lemma2_ite}) (letting $m\to+\infty$)
and using (\ref{eq:ass00}) and (\ref{eq:assumption}), it follows
\begin{align}
S_{h} & \leq S_{h-2}f_{h-2}(I_{h-3})f_{h-1}(S_{h-2})=S_{h}f_{h}(I_{h-1})f_{h-1}(S_{h}).\label{eq:lemma2Sh-1}\end{align}
Similarly, (by considering the limiting equation of (\ref{eq:lemma2_ite})
for the subsequences $\{\underline{n}_{m}^{h}\}_{m=0}^{+\infty},h=1,2$
and using (\ref{eq:ass00}) and (\ref{eq:assumption})), we can obtain
\begin{equation}
I_{h}\geq I_{h-2}f_{h-2}(S_{h-3})f_{h-1}(I_{h-2})=I_{h}f_{h}(S_{h-1})f_{h-1}(I_{h}).\label{eq:lemma2Ih}\end{equation}

Consequently, by (\ref{eq:lemma2Sh-1}) and (\ref{eq:lemma2Ih}),
the following holds \[
\begin{cases}
1 & \leq f_{h}(I_{h-1})f_{h-1}(S_{h}),\\
1 & \geq f_{h}(S_{h-1})f_{h-1}(I_{h}),\end{cases}\text{ for }h=1,2,\]
from which, we obtain (\ref{eq:concl_Lem_2}) and the proof is complete.
\qed
\end{pf}
Then, we obtain the following result.
\begin{thm}
\label{thm:Theorem_2}Let $k=2$. Assume that (\ref{eq:ass00}) and
(\ref{eq:assumption}). If (\ref{eq:theorem11}), then there exists
a $2$-periodic solution $x_{n}^{*}$ such that $x_{n}^{*}=x_{n+2}^{*}$
which is globally attractive, that is, for any solution of (\ref{eq:Mother_generalPielou}),
it holds that\[
\lim_{n\to+\infty}(x_{n}-x_{n}^{*})=0.\]
\end{thm}
\begin{pf}
In order to obtain the conclusion, we will show \begin{equation}
S_{h}=I_{h},h=1,2.\label{eq:concl2}\end{equation}
 Let\[
U_{h,h-j}=\lim_{m\to+\infty}x_{k\overline{n}_{m}^{h}+h-j}\text{ for }j=1,2,\dots,5,\]
and we claim

\begin{equation}
U_{h,h-j}=\begin{cases}
S_{h-j} & \text{ for }j=2,4,\\
I_{h-j} & \text{ for }j=1,3,5,\end{cases}\label{eq:claim01}\end{equation}
for $h=1,2$.

At first, we see that it holds \begin{align}
x_{k\overline{n}_{m}^{h}+h-j} & =x_{k\overline{n}_{m}^{h}+h-j-2}f_{k\overline{n}_{m}^{h}+h-j-2}(x_{k\overline{n}_{m}^{h}+h-j-3})f_{k\overline{n}_{m}^{h}+h-j-1}(x_{k\overline{n}_{m}^{h}+h-j-2})\nonumber \\
 & =x_{k\overline{n}_{m}^{h}+h-j-2}f_{h-j-2}(x_{k\overline{n}_{m}^{h}+h-j-3})f_{h-j-1}(x_{k\overline{n}_{m}^{h}+h-j-2}),\label{eq:ite2h-j}\end{align}
for $h=1,2$ and $j=0,1,2,\dots$. from (\ref{eq:Mother_generalPielou}).

Firstly, we show \begin{equation}
U_{h,h-j}=\begin{cases}
S_{h-j}=S_{h} & \text{ for }j=2,\\
I_{h-j}=I_{h-1} & \text{ for }j=3,\end{cases}\label{eq:U1}\end{equation}
for $h=1,2$. Suppose that there exists some $\overline{h}\in\left\{ 1,2\right\} $
such that $U_{\overline{h},\overline{h}-2}<S_{\overline{h}}$ or $U_{\overline{h},\overline{h}-3}>I_{\overline{h}-1}$.
By considering the limiting equation of (\ref{eq:ite2h-j}) with $j=0$,
it follows\[
S_{h}=U_{h,h-2}f_{h}(U_{h,h-3})f_{h-1}(U_{h,h-2}).\]
Then, by (\ref{eq:ass00}) and (\ref{eq:assumption}), it follows
$S_{\overline{h}}<S_{\overline{h}}f_{\overline{h}}(I_{\overline{h}-1})f_{\overline{h}-1}(S_{\overline{h}})$,
which implies \[
1<f_{h}(I_{h-1})f_{h-1}(S_{h}),\text{ for }h=\overline{h}.\]
This leads a contradiction to (\ref{eq:concl_Lem_2}) in Lemma \ref{lem:Lemma_2}.
Thus, (\ref{eq:U1}) holds.

Next, we show \begin{equation}
U_{h,h-j}=\begin{cases}
S_{h-j}=S_{h} & \text{ for }j=4,\\
I_{h-j}=I_{h-1} & \text{ for }j=5,\end{cases}\label{eq:U2}\end{equation}
for $h=1,2$. Suppose that there exists some $\overline{h}\in\left\{ 1,2\right\} $
such that $U_{\overline{h},\overline{h}-4}<S_{\overline{h}}$ or $U_{\overline{h},\overline{h}-5}>I_{\overline{h}-1}$.
By considering the limiting equation of (\ref{eq:ite2h-j}) with $j=2$
and substituting (\ref{eq:U1}), it follows\[
U_{h,h-2}=S_{h}=U_{h,h-4}f_{h}(U_{h,h-5})f_{h-1}(U_{h,h-4}),\]
for $h=1,2$. Then, by (\ref{eq:ass00}) and (\ref{eq:assumption}),
it follows $S_{\overline{h}}<S_{\overline{h}}f_{\overline{h}}(I_{\overline{h}-1})f_{\overline{h}-1}(S_{\overline{h}})$
which implies\[
1<f_{h}(I_{h-1})f_{h-1}(S_{h}),\text{ for }h=\overline{h}.\]
This gives a contradiction to (\ref{eq:concl_Lem_2}) in Lemma \ref{lem:Lemma_2}.
Thus, (\ref{eq:U2}) holds.

By considering the limiting equation of (\ref{eq:ite2h-j}) with $j=1$
and using (\ref{eq:U1})-(\ref{eq:U2}), we see\[
U_{h,h-1}=U_{h,h-3}f_{h-1}(U_{h,h-4})f_{h}(U_{h,h-3})=I_{h-1}f_{h-1}(S_{h})f_{h}(I_{h-1}).\]
Hence, it follows \begin{equation}
U_{h,h-1}=I_{h-1},\label{eq:U4}\end{equation}
for $h=1,2$, by (\ref{eq:concl_Lem_2}) in Lemma \ref{lem:Lemma_2}.
Consequently, (\ref{eq:claim01}) holds from (\ref{eq:U1}), (\ref{eq:U2})
and (\ref{eq:U4}).

From (\ref{eq:Mother_generalPielou}), it holds \[
x_{k\overline{n}_{m}^{h}+h}=x_{k\overline{n}_{m}^{h}+h-1}f_{k\overline{n}_{m}^{h}+h-1}(x_{k\overline{n}_{m}^{h}+h-2})=x_{k\overline{n}_{m}^{h}+h-1}f_{h-1}(x_{k\overline{n}_{m}^{h}+h-2}),\]
and by considering the limiting equation and using (\ref{eq:claim01}),
we obtain\[
S_{h}=U_{h,h-1}f_{h-1}(U_{h,h-2})=I_{h-1}f_{h-1}(S_{h}).\]
By (\ref{eq:concl_Lem_2}) in Lemma \ref{lem:Lemma_2}, it holds \begin{equation}
S_{h}f_{h}(I_{h-1})=I_{h-1}f_{h-1}(S_{h})f_{h}(I_{h-1})=I_{h-1},\label{eq2s}\end{equation}
for $h=1,2$, On the other hand, similar to the above discussion,
it also holds \begin{equation}
I_{h}f_{h}(S_{h-1})=S_{h-1}f_{h-1}(I_{h})f_{h}(S_{h-1})=S_{h-1},\label{eq2i}\end{equation}
for $h=1,2$. Consequently, it holds \[
I_{h-1}=S_{h}f_{h}(I_{h-1})\leq I_{h}f_{h}(S_{h-1})=S_{h-1},\]
by (\ref{eq2s}) and (\ref{eq2i}), and hence, (\ref{eq:concl2})
holds. Then, from (\ref{eq:defsh}), we see \[
\liminf_{n\to+\infty}x_{kn+h}=\limsup_{n\to+\infty}x_{kn+h}=\lim_{n\to+\infty}x_{kn+h},\]
for $h=1,2$ and there exist two positive constants $x_{1}^{*}=S_{1}=I_{1}$
and $x_{2}^{*}=S_{2}=I_{2}$ such that \[
x_{2}^{*}=\lim_{n\to+\infty}x_{2n},\]
and \[
x_{1}^{*}=\lim_{n\to+\infty}x_{2n+1}.\]
The proof is complete.\qed
\end{pf}

\subsection{Case: $k$ is an even integer}

In this subsection, we generalize results in Section 4.1 to the case
where $k$ is an arbitrary even integer.
\begin{lem}
\label{lem:Lemma_even}Let $k$ be an even integer. Assume that (\ref{eq:ass00})
and (\ref{eq:assumption}). If (\ref{eq:theorem11}), then it holds
that\begin{equation}
\left(f_{h-k}(I_{h-k-1})f_{h+1-k}(S_{h-k})\right)\dots\left(f_{h-4}(I_{h-5})f_{h-3}(S_{h-4})\right)\left(f_{h-2}(I_{h-3})f_{h-1}(S_{h-2})\right)=1,\label{eq:concl_lem_even}\end{equation}
 for $h=1,2,\dots,k.$ \end{lem}
\begin{pf}
At first, from (\ref{eq:Mother_generalPielou}), we see that it holds\begin{align}
x_{k\overline{n}_{m}^{h}+h} & =x_{k\overline{n}_{m}^{h}+h-2}f_{k\overline{n}_{m}^{h}+h-2}(x_{k\overline{n}_{m}^{h}+h-3})f_{k\overline{n}_{m}^{h}+h-1}(x_{k\overline{n}_{m}^{h}+h-2})\nonumber \\
 & =x_{k\overline{n}_{m}^{h}+h-2}f_{h-2}(x_{k\overline{n}_{m}^{h}+h-3})f_{h-1}(x_{k\overline{n}_{m}^{h}+h-2}),\label{eq:lemma3_ite}\end{align}
 for $h=1,2,\dots,k$, and by considering the limiting equation of
(\ref{eq:lemma3_ite}) and using (\ref{eq:ass00}) and (\ref{eq:assumption}),
it follows\begin{equation}
S_{h}\leq S_{h-2}f_{h-2}(I_{h-3})f_{h-1}(S_{h-2}),\label{eq:h-j}\end{equation}
for $h=1,2,\dots,k$. From (\ref{eq:h-j}) we obtain\begin{align*}
S_{h} & \leq S_{h-2}\left(f_{h-2}(I_{h-3})f_{h-1}(S_{h-2})\right)\\
 & \leq S_{h-4}\left(f_{h-4}(I_{h-5})f_{h-3}(S_{h-4})\right)\left(f_{h-2}(I_{h-3})f_{h-1}(S_{h-2})\right)\\
 & \leq\dots\\
 & \leq S_{h-k}\overline{G}_{h},\end{align*}
where \begin{align}
\overline{G}_{h} & =\left(f_{h-k}(I_{h-k-1})f_{h+1-k}(S_{h-k})\right)\dots\left(f_{h-4}(I_{h-5})f_{h-3}(S_{h-4})\right)\left(f_{h-2}(I_{h-3})f_{h-1}(S_{h-2})\right).\label{eq:overg}\end{align}
Then \[
1\leq\overline{G}_{h},\]
holds for $h=1,2,\dots,k$, because we have $S_{h}=S_{h-k}$ from
(\ref{periodicityofsi}).

Similarly (by considering the limiting equation of (\ref{eq:lemma3_ite})
for the subsequences $\{\underline{n}_{m}^{h}\}_{m=0}^{+\infty},h=1,2,\dots,k$
and using (\ref{eq:ass00}) and (\ref{eq:assumption})), it follows\begin{equation}
I_{h}\geq I_{h-2}f_{h-2}(S_{h-3})f_{h-1}(I_{h-2}),\label{eq:h-j2}\end{equation}
for $h=1,2,\dots,k$. From (\ref{eq:h-j2}) we obtain\begin{align*}
I_{h} & \geq I_{h-2}\left(f_{h-2}(S_{h-3})f_{h-1}(I_{h-2})\right)\\
 & \geq I_{h-4}\left(f_{h-4}(S_{h-5})f_{h-3}(I_{h-4})\right)\left(f_{h-2}(S_{h-3})f_{h-1}(I_{h-2})\right)\\
 & \geq\dots\\
 & \geq I_{h-k}\underline{G}_{h},\end{align*}
 where \begin{align}
\underline{G}_{h} & =\left(f_{h-k}(S_{h-k-1})f_{h+1-k}(I_{h-k})\right)\dots\left(f_{h-4}(S_{h-5})f_{h-3}(I_{h-4})\right)\left(f_{h-2}(S_{h-3})f_{h-1}(I_{h-2})\right).\label{eq:underg}\end{align}
Then \[
1\geq\underline{G}_{h},\]
holds for $h=1,2,\dots,k$, because we have $I_{h}=I_{h-k}$ from
(\ref{periodicityofsi}). Consequently, it holds that\begin{equation}
\underline{G}_{h}\leq1\leq\overline{G}_{h},\text{ for }h=1,2,\dots,k.\label{eq:Grel1}\end{equation}
By (\ref{eq:overg}) and (\ref{eq:underg}), we see\begin{equation}
\overline{G}_{1}=\underline{G}_{k}\text{ and }\overline{G}_{h}=\underline{G}_{h-1}\text{ for }h\in\left\{ 2,3,\dots,k\right\} .\label{eq:Grel2}\end{equation}
Therefore, we obtain (\ref{eq:concl_lem_even}) from (\ref{eq:Grel1})
and (\ref{eq:Grel2}). The proof is complete.\qed
\end{pf}
We show that there exists a $k$-periodic solution which is globally
attractive for any solution for the case where $k$ is an even integer.
\begin{thm}
\label{thm:Theore,_even}Let $k$ be an even integer. Assume that
(\ref{eq:ass00}) and (\ref{eq:assumption}). If (\ref{eq:theorem11}),
then there exists a $k$-periodic solution $x_{n}^{*}$ such that
$x_{n}^{*}=x_{n+k}^{*}$ which is globally attractive, that is, for
any solution of (\ref{eq:Mother_generalPielou}), it holds that\[
\lim_{n\to+\infty}\left(x_{n}-x_{n}^{*}\right)=0.\]
\end{thm}
\begin{pf}
To obtain the conclusion, we will show \begin{equation}
S_{h}=I_{h}\text{ for }h=1,2,\dots,k.\label{eq:concl_even}\end{equation}
Let\[
U_{h,h-j}=\lim_{m\to+\infty}x_{k\overline{n}_{m}^{h}+h-j}\text{ for }j=2,3\dots,k+1,\]
and we claim that it holds

\begin{equation}
U_{h,h-j}=\begin{cases}
S_{h-j} & \text{ for }j=2,4,\dots,k,\\
I_{h-j} & \text{ for }j=3,5,\dots,k+1,\end{cases}\label{inductive}\end{equation}
for $h=1,2,\dots,k$.

From (\ref{eq:Mother_generalPielou}), it holds\begin{align}
x_{k\overline{n}_{m}^{h}+h-j} & =x_{k\overline{n}_{m}^{h}+h-j-2}f_{k\overline{n}_{m}^{h}+h-j-2}(x_{k\overline{n}_{m}^{h}+h-j-3})f_{k\overline{n}_{m}^{h}+h-j-1}(x_{k\overline{n}_{m}^{h}+h-j-2})\nonumber \\
 & =x_{k\overline{n}_{m}^{h}+h-j-2}f_{h-j-2}(x_{k\overline{n}_{m}^{h}+h-j-3})f_{h-j-1}(x_{k\overline{n}_{m}^{h}+h-j-2}).\label{eq:gp2_even_gen}\end{align}

Firstly, we show \begin{equation}
U_{h,h-j}=\begin{cases}
S_{h-j} & \text{ for }j=2,\\
I_{h-j} & \text{ for }j=3,\end{cases}\label{eq:U1_gen1}\end{equation}
for $h=1,2,\dots,k$. Suppose that there exists $\overline{h}$ such
that $U_{\overline{h},\overline{h}-2}<S_{\overline{h}-2},$ or $U_{\overline{h},\overline{h}-3}>I_{\overline{h}-3}$.
By letting $m\to+\infty$ and considering the limiting equation of
(\ref{eq:gp2_even_gen}) with $j=0$, we obtain\begin{equation}
S_{h}=U_{h,h-2}f_{h-2}(U_{h,h-3})f_{h-1}(U_{h,h-2}),\label{eq:Sh1_gen}\end{equation}
for $h=1,2,\dots,k$. Then, we see that one of the following holds\[
\Pi_{i=1}^{p}S_{2i}<\Pi_{i=1}^{p}\left[S_{2i-2}f_{2i-2}(I_{2i-3})f_{2i-1}(S_{2i-2})\right],\]
or\[
\Pi_{i=0}^{p-1}S_{2i+1}<\Pi_{i=0}^{p-1}\left[S_{2i-1}f_{2i-1}(I_{2i-2})f_{2i}(S_{2i-1})\right],\]
where $p=\frac{k}{2}$, since \[
\begin{cases}
\Pi_{i=1}^{p}S_{2i} & =\Pi_{i=1}^{p}\left[U_{2i,2i-2}f_{2i-2}(U_{2i,2i-3})f_{2i-1}(U_{2i,2i-2})\right],\\
\Pi_{i=0}^{p-1}S_{2i+1} & =\Pi_{i=0}^{p-1}\left[U_{2i+1,2i-1}f_{2i-1}(U_{2i+1,2i-2})f_{2i}(U_{2i+1,2i-1})\right],\end{cases}\]
follows from (\ref{eq:Sh1_gen}). Then, we obtain \[
1<\Pi_{i=1}^{p}\left[f_{2i-2}(I_{2i-3})f_{2i-1}(S_{2i-2})\right],\]
or \[
1<\Pi_{i=0}^{p-1}\left[f_{2i-1}(I_{2i-2})f_{2i}(S_{2i-1})\right],\]
and this gives a contradiction to (\ref{eq:concl_lem_even}) with
$h=k$ and $h=k-1$, respectively, in Lemma \ref{lem:Lemma_even}.
Thus (\ref{eq:U1_gen1}) holds.

Next, we assume that \begin{equation}
U_{h,h-j}=\begin{cases}
S_{h-j} & \text{ for }j=j_{1},\\
I_{h-j} & \text{ for }j=j_{1}+1,\end{cases}\label{eq:U1_gen2}\end{equation}
for $h=1,2,\dots,k$ where $j_{1}$ is a positive even number. Under
the assumption (\ref{eq:U1_gen2}), we show \begin{equation}
U_{h,h-j}=\begin{cases}
S_{h-j} & \text{ for }j=j_{1}+2,\\
I_{h-j} & \text{ for }j=j_{1}+3,\end{cases}\label{eq:U1_gen}\end{equation}
for $h=1,2,\dots,k$. Suppose that there exists $\overline{h}$ such
that $U_{\overline{h},\overline{h}-j_{1}-2}<S_{\overline{h}-j_{1}-2},$
or $U_{\overline{h},\overline{h}-j_{1}-3}>I_{\overline{h}-j_{1}-3}$.
By considering the limiting equation of (\ref{eq:gp2_even_gen}) with
$j=j_{1}$ and substituting (\ref{eq:U1_gen2}),\[
U_{h,h-j_{1}}=S_{h-j_{1}}=U_{h,h-j_{1}-2}f_{h-j_{1}-2}(U_{h,h-j_{1}-3})f_{h-j_{1}-1}(U_{h,h-j_{1}-2}),\]
for $h=1,2,\dots,k$. Then, similar to the above discussion, we can
show that (\ref{eq:U1_gen}) holds. Thus, (\ref{inductive}) holds
by mathematical induction.

From (\ref{eq:Mother_generalPielou}), it holds \begin{align*}
x_{k\overline{n}_{m}^{h}+h} & =x_{k\overline{n}_{m}^{h}+h-(k-1)}f_{k\overline{n}_{m}^{h}+h-(k-1)}(x_{k\overline{n}_{m}^{h}+h-k})\dots f_{k\overline{n}_{m}^{h}+h-2}(x_{k\overline{n}_{m}^{h}+h-3})f_{k\overline{n}_{m}^{h}+h-1}(x_{k\overline{n}_{m}^{h}+h-2})\\
 & =x_{k\overline{n}_{m}^{h}+h-(k-1)}f_{h-(k-1)}(x_{k\overline{n}_{m}^{h}+h-k})\dots f_{h-2}(x_{k\overline{n}_{m}^{h}+h-3})f_{h-1}(x_{k\overline{n}_{m}^{h}+h-2}),\end{align*}
and by considering the limiting equation and using (\ref{inductive}),
we obtain \begin{align*}
S_{h} & =U_{h,h-(k-1)}f_{h-(k-1)}(U_{h,h-k})\dots f_{h-2}(U_{h,h-3})f_{h-1}(U_{h,h-2})\\
 & =I_{h-(k-1)}f_{h-(k-1)}(S_{h-k})\dots f_{h-2}(I_{h-3})f_{h-1}(S_{h-2}).\end{align*}
By (\ref{eq:concl_lem_even}) in Lemma \ref{lem:Lemma_even}, we see
\[
f_{h-(k-1)}(S_{h-k})\dots f_{h-2}(I_{h-3})f_{h-1}(S_{h-2})=\frac{1}{f_{h-k}(I_{h-k-1})},\]
hence, it holds that \begin{equation}
S_{h}f_{h-k}(I_{h-k-1})=I_{h+1},\label{Sendproof2p}\end{equation}
for $h=1,2,\dots,k$. Similar to the above discussion, it also holds
\begin{equation}
I_{h}f_{h-k}(S_{h-k-1})=S_{h+1},\label{Iendproof2p}\end{equation}
for $h=1,2,\dots,k$. Consequently, it holds \[
I_{h+1}=S_{h}f_{h-k}(I_{h-k-1})\leq I_{h}f_{h-k}(S_{h-k-1})=S_{h+1},\]
by (\ref{Sendproof2p}) and (\ref{Iendproof2p}), and hence, (\ref{eq:concl_even})
holds. Then, from (\ref{eq:defsh}), we see \[
\liminf_{n\to+\infty}x_{kn+h}=\limsup_{n\to+\infty}x_{kn+h}=\lim_{n\to+\infty}x_{kn+h},\]
for $h=1,2,\dots,k$, and there exist $k$ positive constants $x_{h}^{*}=S_{h}=I_{h},h=1,2,\dots,k$
such that \[
x_{h}^{*}=\lim_{n\to+\infty}x_{kn+h}.\]
The proof is complete.\qed
\end{pf}

\section{Global attractivity for the case where $k$ is an arbitrary odd integer}

In this section, we show that Theorem \ref{thm:mainthm} holds when
$k$ is an odd integer. For the reader, we first consider the case
$k=3$ in Section 5.1. (\ref{eq:concl_lem_3}) in Lemma \ref{lem:Lemma_3}
has an important role in this subsection. Then, we give Theorem \ref{thm:Theorem_3}
which states that there exists a $3$-periodic solution which is globally
attractive. In Section 5.2, we generalize these results to the case
where $k$ is an arbitrary odd integer.

\subsection{Case: $k=3$}

First, we introduce the following lemma which plays a crucial role
in the proof of Theorem \ref{thm:Theorem_3}.
\begin{lem}
\label{lem:Lemma_3}Let $k=3$. Assume that (\ref{eq:ass00}) and
(\ref{eq:assumption}). If (\ref{eq:theorem11}), then it holds that\begin{equation}
\begin{cases}
S_{h} & =S_{h-2}f_{h-2}(I_{h-3})f_{h-1}(S_{h-2}),\\
I_{h} & =I_{h-2}f_{h-2}(S_{h-3})f_{h-1}(I_{h-2}),\end{cases}\label{eq:concl_lem_3}\end{equation}
for $h=1,2,3$.\end{lem}
\begin{pf}
From (\ref{eq:Mother_generalPielou}), it holds\begin{align*}
x_{kn+h} & =x_{kn+h-2}f_{kn+h-2}(x_{kn+h-3})f_{kn+h-1}(x_{kn+h-2})\\
 & =x_{kn+h-2}f_{h-2}(x_{kn+h-3})f_{h-1}(x_{kn+h-2}),\end{align*}
for $h=1,2,3$. Then it follows that\[
\begin{cases}
x_{k\overline{n}_{m}^{h}+h} & =x_{k\overline{n}_{m}^{h}+h-2}f_{h-2}(x_{k\overline{n}_{m}^{h}+h-3})f_{h-1}(x_{k\overline{n}_{m}^{h}+h-2}),\\
x_{k\underline{n}_{m}^{h}+h} & =x_{k\underline{n}_{m}^{h}+h-2}f_{h-2}(x_{k\underline{n}_{m}^{h}+h-3})f_{h-1}(x_{k\underline{n}_{m}^{h}+h-2}),\end{cases}\]
for $h=1,2,3$. By considering the limiting equation and using (\ref{eq:ass00})
and (\ref{eq:assumption}), we get\begin{equation}
\begin{cases}
S_{h} & \leq S_{h-2}f_{h-2}(I_{h-3})f_{h-1}(S_{h-2}),\\
I_{h} & \geq I_{h-2}f_{h-2}(S_{h-3})f_{h-1}(I_{h-2}),\end{cases}\label{3lemma}\end{equation}
for $h=1,2,3$. In order to obtain the conclusion, we show that (\ref{3lemma})
holds with equality.

From (\ref{3lemma}), we see that \begin{equation}
\begin{cases}
\frac{S_{h}f_{h-1}(I_{h-2})}{I_{h}f_{h-1}(S_{h-2})} & \leq\frac{S_{h-2}}{I_{h}}f_{h-2}(I_{h-3})f_{h-1}(I_{h-2}),\\
\frac{S_{h-2}f_{h-2}(I_{h-3})}{I_{h-2}f_{h-2}(S_{h-3})} & \geq\frac{S_{h-2}}{I_{h}}f_{h-2}(I_{h-3})f_{h-1}(I_{h-2}),\end{cases}\label{3lemma4}\end{equation}
for $h=1,2,3$. It then follows \begin{equation}
\frac{S_{h}f_{h-1}(I_{h-2})}{I_{h}f_{h-1}(S_{h-2})}\leq\frac{S_{h-2}f_{h-2}(I_{h-3})}{I_{h-2}f_{h-2}(S_{h-3})},\label{3lemma5}\end{equation}
 for $h=1,2,3$. By multiplying (\ref{3lemma5}), we obtain \[
\left(\frac{S_{h}f_{h-1}(I_{h-2})}{I_{h}f_{h-1}(S_{h-2})}\right)\left(\frac{S_{h+1}f_{h}(I_{h-1})}{I_{h+1}f_{h}(S_{h-1})}\right)\leq\left(\frac{S_{h-2}f_{h-2}(I_{h-3})}{I_{h-2}f_{h-2}(S_{h-3})}\right)\left(\frac{S_{h-1}f_{h-1}(I_{h-2})}{I_{h-1}f_{h-1}(S_{h-2})}\right),\]
and hence, it holds that\[
\frac{S_{h}S_{h+1}f_{h}(I_{h-1})}{I_{h}I_{h+1}f_{h}(S_{h-1})}\leq\frac{S_{h-2}S_{h-1}f_{h-2}(I_{h-3})}{I_{h-2}I_{h-1}f_{h-2}(S_{h-3})},\]
for $h=1,2,3$. Therefore, the following inequalities hold\begin{align*}
\begin{cases}
\frac{S_{1}S_{2}f_{1}(I_{3})}{I_{1}I_{2}f_{1}(S_{3})} & \leq\frac{S_{2}S_{3}f_{2}(I_{1})}{I_{2}I_{3}f_{2}(S_{1})},\\
\frac{S_{2}S_{3}f_{2}(I_{1})}{I_{2}I_{3}f_{2}(S_{1})} & \leq\frac{S_{3}S_{1}f_{3}(I_{2})}{I_{3}I_{1}f_{3}(S_{2})},\\
\frac{S_{3}S_{1}f_{3}(I_{2})}{I_{3}I_{1}f_{3}(S_{2})} & \leq\frac{S_{1}S_{2}f_{1}(I_{3})}{I_{1}I_{2}f_{1}(S_{3})},\end{cases}\end{align*}
and it follows\[
\frac{S_{1}S_{2}f_{1}(I_{3})}{I_{1}I_{2}f_{1}(S_{3})}\leq\frac{S_{2}S_{3}f_{2}(I_{1})}{I_{2}I_{3}f_{2}(S_{1})}\leq\frac{S_{3}S_{1}f_{3}(I_{2})}{I_{3}I_{1}f_{3}(S_{2})}\leq\frac{S_{1}S_{2}f_{1}(I_{3})}{I_{1}I_{2}f_{1}(S_{3})}.\]
Hence, we see that \begin{equation}
\frac{S_{1}S_{2}f_{1}(I_{3})}{I_{1}I_{2}f_{1}(S_{3})}=\frac{S_{2}S_{3}f_{2}(I_{1})}{I_{2}I_{3}f_{2}(S_{1})}=\frac{S_{3}S_{1}f_{3}(I_{2})}{I_{3}I_{1}f_{3}(S_{2})}=\frac{S_{1}S_{2}f_{1}(I_{3})}{I_{1}I_{2}f_{1}(S_{3})}.\label{3lemma7}\end{equation}
(\ref{3lemma7}) implies that (\ref{3lemma4}) holds with equality
and thus, (\ref{3lemma}) also holds with equality for $h=1,2,3$.
Therefore, we obtain the conclusion. The proof is complete.\qed
\end{pf}
Then, we obtain the following result.

\begin{thm}
\label{thm:Theorem_3}Let $k=3$. Assume that (\ref{eq:ass00}) and
(\ref{eq:assumption}). If (\ref{eq:theorem11}), then there exists
a $3$-periodic solution $x_{n}^{*}$ such that $x_{n}^{*}=x_{n+3}^{*}$
which is globally attractive, that is, for any solution of (\ref{eq:Mother_generalPielou}),
it holds that\[
\lim_{n\to+\infty}\left(x_{n}-x_{n}^{*}\right)=0.\]
\end{thm}
\begin{pf}
In order to obtain the conclusion, we will show \begin{equation}
S_{h}=I_{h},h=1,2,3.\label{eq:eq:concl_3}\end{equation}
Let\[
U_{k,k-j}=\lim_{m\to+\infty}x_{k\overline{n}_{m}^{k}+k-j}\text{ for }j=1,2,\dots,5,\]
and we claim\begin{equation}
U_{k,k-j}=\begin{cases}
S_{k-j} & \text{ for }j=2,4,\\
I_{k-j} & \text{ for }j=1,3,5.\end{cases}\label{eq:claim3_0}\end{equation}

At first, we see that it holds\begin{align}
x_{k\overline{n}_{m}^{k}+k-j} & =x_{k\overline{n}_{m}^{k}+k-j-2}f_{k\overline{n}_{m}^{k}+k-j-2}(x_{k\overline{n}_{m}^{k}+k-j-3})f_{k\overline{n}_{m}^{k}+k-j-1}(x_{k\overline{n}_{m}^{k}+k-j-2})\nonumber \\
 & =x_{k\overline{n}_{m}^{k}+k-j-2}f_{k-j-2}(x_{k\overline{n}_{m}^{k}+k-j-3})f_{k-j-1}(x_{k\overline{n}_{m}^{k}+k-j-2}),\label{eq:ite3}\end{align}
for $j=0,1,2,\dots,$ from (\ref{eq:Mother_generalPielou}).

Firstly, we show\begin{equation}
U_{k,k-j}=\begin{cases}
S_{k-j} & \text{ for }j=2,\\
I_{k-j} & \text{ for }j=3.\end{cases}\label{eq:claim3_1}\end{equation}
Suppose that $U_{k,k-2}<S_{k-2}$ or $U_{k,k-3}>I_{k-3}$. By considering
the limiting equation of (\ref{eq:ite3}) with $j=0$, it follows\[
S_{k}=U_{k,k-2}f_{k-2}(U_{k,k-3})f_{k-1}(U_{k,k-2}).\]
Then, by (\ref{eq:ass00}) and (\ref{eq:assumption}), it follows
\[
S_{k}<S_{k-2}f_{k-2}(I_{k-3})f_{k-1}(S_{k-2}).\]
This gives a contradiction to (\ref{eq:concl_lem_3}) with $h=k$
in Lemma \ref{lem:Lemma_3}. Thus, (\ref{eq:claim3_1}) holds.

Next, we show \begin{equation}
U_{k,k-j}=\begin{cases}
S_{k-j} & \text{ for }j=4,\\
I_{k-j} & \text{ for }j=5.\end{cases}\label{eq:claim3_2}\end{equation}
Suppose that $U_{k,k-4}<S_{k-4}$ or $U_{k,k-5}>I_{k-5}$. By considering
the limiting equation of (\ref{eq:ite3}) with $j=2$ and substituting
(\ref{eq:claim3_1}), it follows\[
U_{k,k-2}=S_{k-2}=U_{k,k-4}f_{k-4}(U_{k,k-5})f_{k-3}(U_{k,k-4}).\]
Then, by (\ref{eq:ass00}) and (\ref{eq:assumption}), it follows
\[
S_{k-2}<S_{k-4}f_{k-4}(I_{k-5})f_{k-3}(S_{k-4}).\]
This gives a contradiction to (\ref{eq:concl_lem_3}) with $h=k-2$
in Lemma \ref{lem:Lemma_3}. Thus, (\ref{eq:claim3_2}) holds.

By considering the limiting equation of (\ref{eq:ite3}) with $j=1$
and using (\ref{eq:claim3_1})-(\ref{eq:claim3_2}), it follows \begin{align*}
U_{k,k-1} & =U_{k,k-3}f_{k-3}(U_{k,k-4})f_{k-2}(U_{k,k-3})\\
 & =I_{k-3}f_{k-3}(S_{k-4})f_{k-2}(I_{k-3}).\end{align*}
Hence, it holds \begin{equation}
U_{k,k-1}=I_{k-1},\label{eq:claim3_3}\end{equation}
by (\ref{eq:concl_lem_3}) in Lemma \ref{lem:Lemma_3}. Consequently,
(\ref{eq:claim3_0}) holds from (\ref{eq:claim3_1}), (\ref{eq:claim3_2})
and (\ref{eq:claim3_3}).

From (\ref{eq:Mother_generalPielou}), it holds\[
x_{k\overline{n}_{m}^{k}+k}=x_{k\overline{n}_{m}^{k}+k-1}f_{k\overline{n}_{m}^{k}+k-1}(x_{k\overline{n}_{m}^{k}+k-2})=x_{k\overline{n}_{m}^{k}+k-1}f_{k-1}(x_{k\overline{n}_{m}^{k}+k-2}),\]
 and by considering the limiting equation and using (\ref{eq:claim3_0}),
we obtain \begin{equation}
S_{k}=I_{k-1}f_{k-1}(S_{k-2}).\label{s1}\end{equation}
Similar to the above discussion, it also holds\begin{equation}
I_{k}=S_{k-1}f_{k-1}(I_{k-2}).\label{i1}\end{equation}
Consequently, it holds \[
I_{k}=S_{k-1}f_{k-1}(I_{k-2})\leq I_{k-1}f_{k-1}(S_{k-2})=S_{k},\]
by (\ref{s1}) and (\ref{i1}), and hence (\ref{eq:eq:concl_3}) holds.
Then, \[
\liminf_{n\to+\infty}x_{kn+h}=\limsup_{n\to+\infty}x_{kn+h}=\lim_{n\to+\infty}x_{kn+h},\]
for $h=1,2,3$ and there exist $3$-positive constants $x_{h}^{*}=S_{h}=I_{h},h=1,2,3$
such that \[
x_{h}^{*}=\lim_{n\to+\infty}x_{kn+h}.\]
The proof is complete.\qed
\end{pf}

\subsection{Case: $k$ is an odd integer}
\begin{lem}
\label{lem:Lemma_odd} Let $k$ be an odd integer. Assume that (\ref{eq:ass00})
and (\ref{eq:assumption}). If (\ref{eq:theorem11}), then it holds
that\begin{equation}
\begin{cases}
S_{h} & =S_{h-2}f_{h-2}(I_{h-3})f_{h-1}(S_{h-2}),\\
I_{h} & =I_{h-2}f_{h-2}(S_{h-3})f_{h-1}(I_{h-2}),\end{cases}\label{eq:concl_lem_odd}\end{equation}
for $h=1,2,\dots,k$.\end{lem}
\begin{pf}
From (\ref{eq:Mother_generalPielou}), it holds \begin{align*}
x_{kn+h} & =x_{kn+h-2}f_{kn+h-2}(x_{kn+h-3})f_{kn+h-1}(x_{kn+h-2})\\
 & =x_{kn+h-2}f_{h-2}(x_{kn+h-3})f_{h-1}(x_{kn+h-2}),\end{align*}
for $h=1.2.\dots,k$. Then, it follows that\[
\begin{cases}
x_{k\overline{n}_{m}^{h}+h} & =x_{k\overline{n}_{m}^{h}+h-2}f_{h-2}(x_{k\overline{n}_{m}^{h}+h-3})f_{h-1}(x_{k\overline{n}_{m}^{h}+h-2}),\\
x_{k\underline{n}_{m}^{h}+h} & =x_{k\underline{n}_{m}^{h}+h-2}f_{h-2}(x_{k\underline{n}_{m}^{h}+h-3})f_{h-1}(x_{k\underline{n}_{m}^{h}+h-2}),\end{cases}\]
for $h=1,2,\dots,k$. By considering the limiting equation and using
(\ref{eq:ass00}) and (\ref{eq:assumption}), we get\begin{equation}
\begin{cases}
S_{h} & \leq S_{h-2}f_{h-2}(I_{h-3})f_{h-1}(S_{h-2}),\\
I_{h} & \geq I_{h-2}f_{h-2}(S_{h-3})f_{h-1}(I_{h-2}),\end{cases}\label{3@lemma}\end{equation}
for $h=1,2,\dots,k$. In order to obtain the conclusion, we show that
(\ref{3@lemma}) holds with equality.

From (\ref{3@lemma}), we see that \begin{equation}
\begin{cases}
\frac{S_{h}f_{h-1}(I_{h-2})}{I_{h}f_{h-1}(S_{h-2})} & \leq\frac{S_{h-2}}{I_{h}}f_{h-2}(I_{h-3})f_{h-1}(I_{h-2}),\\
\frac{S_{h-2}f_{h-2}(I_{h-3})}{I_{h-2}f_{h-2}(S_{h-3})} & \geq\frac{S_{h-2}}{I_{h}}f_{h-2}(I_{h-3})f_{h-1}(I_{h-2}),\end{cases}\label{3@lemma4}\end{equation}
for $h=1,2,\dots,k$. Thus, it follows that \begin{equation}
\frac{S_{h}f_{h-1}(I_{h-2})}{I_{h}f_{h-1}(S_{h-2})}\leq\frac{S_{h-2}f_{h-2}(I_{h-3})}{I_{h-2}f_{h-2}(S_{h-3})},\label{3@lemma5}\end{equation}
for $h=1,2,\dots,k$. By multiplying (\ref{3@lemma5}), we obtain
\[
\left(\frac{S_{h}f_{h-1}(I_{h-2})}{I_{h}f_{h-1}(S_{h-2})}\right)\left(\frac{S_{h+1}f_{h}(I_{h-1})}{I_{h+1}f_{h}(S_{h-1})}\right)\leq\left(\frac{S_{h-2}f_{h-2}(I_{h-3})}{I_{h-2}f_{h-2}(S_{h-3})}\right)\left(\frac{S_{h-1}f_{h-1}(I_{h-2})}{I_{h-1}f_{h-1}(S_{h-2})}\right),\]
and hence, it holds that\[
\frac{S_{h}S_{h+1}f_{h}(I_{h-1})}{I_{h}I_{h+1}f_{h}(S_{h-1})}\leq\frac{S_{h-2}S_{h-1}f_{h-2}(I_{h-3})}{I_{h-2}I_{h-1}f_{h-2}(S_{h-3})},\]
for $h=1,2,\dots,k$. Therefore, the following inequalities hold\[
\begin{cases}
\frac{S_{1}S_{2}f_{1}(I_{k})}{I_{1}I_{2}f_{1}(S_{k})} & \leq\frac{S_{k-1}S_{k}f_{k-1}(I_{k-2})}{I_{k-1}I_{k}f_{k-1}(S_{k-2})},\\
\frac{S_{k-1}S_{k}f_{k-1}(I_{k-2})}{I_{k-1}I_{k}f_{k-1}(S_{k-2})} & \leq\frac{S_{k-3}S_{k-2}f_{k-3}(I_{k-4})}{I_{k-3}I_{k-2}f_{k-3}(S_{k-4})},\\
\dots & \dots\\
\frac{S_{k}S_{k+1}f_{k}(I_{k-1})}{I_{k}I_{k+1}f_{k}(S_{k-1})} & \leq\frac{S_{k-2}S_{k-1}f_{k-2}(I_{k-3})}{I_{k-2}I_{k-1}f_{k-2}(S_{k-3})},\\
\dots & \dots\\
\frac{S_{3}S_{4}f_{3}(I_{2})}{I_{3}I_{4}f_{3}(S_{2})} & \leq\frac{S_{1}S_{2}f_{1}(I_{k})}{I_{1}I_{2}f_{1}(S_{k})}.\end{cases}\]
Thus, we obtain \begin{equation}
\frac{S_{1}S_{2}f_{1}(I_{k})}{I_{1}I_{2}f_{1}(S_{k})}=\dots=\frac{S_{k-1}S_{k}f_{k-1}(I_{k-2})}{I_{k-1}I_{k}f_{k-1}(S_{k-2})}=\frac{S_{k}S_{k+1}f_{k}(I_{k-1})}{I_{k}I_{k+1}f_{k}(S_{k-1})}.\label{3@lemma7}\end{equation}
(\ref{3@lemma7}) implies that (\ref{3@lemma4}) holds with equality
and thus, (\ref{3@lemma}) also holds with equality for $h=1,2,\dots,k$.
Then, we obtain the conclusion. Hence, the proof is complete.\qed\end{pf}
\begin{thm}
\label{thm:Theorem_odd}Let $k$ be an odd integer. Assume that (\ref{eq:ass00})
and (\ref{eq:assumption}). If (\ref{eq:theorem11}), then there exists
a $k$-periodic solution $x_{n}^{*}$ such that $x_{n}^{*}=x_{n+k}^{*}$
which is globally attractive, that is, for any solution of (\ref{eq:Mother_generalPielou}),
it holds that\[
\lim_{n\to+\infty}\left(x_{n}-x_{n}^{*}\right)=0.\]
\end{thm}
\begin{pf}
In order to obtain the conclusion, we will show \begin{equation}
S_{h}=I_{h},h=1,2,\dots,k.\label{eq:concl_odd}\end{equation}
 Let\[
U_{k,k-j}=\lim_{m\to+\infty}x_{k\overline{n}_{m}^{k}+k-j}\text{ for }j=1,2,\dots.,k,\]
and we claim \begin{equation}
U_{k,k-j}=\begin{cases}
S_{k-j}, & \text{ for }j=2,4,\dots,k-1,\\
I_{k-j}, & \text{ for }j=1,3,5,\dots,k.\end{cases}\label{inductive2p+1}\end{equation}

At first, we see that it holds\begin{align}
x_{k\overline{n}_{m}^{k}+k-j} & =x_{k\overline{n}_{m}^{k}+k-j-2}f_{k\overline{n}_{m}^{k}+k-j-2}(x_{k\overline{n}_{m}^{k}+k-j-3})f_{k\overline{n}_{m}^{k}+k-j-1}(x_{k\overline{n}_{m}^{k}+k-j-2})\nonumber \\
 & =x_{k\overline{n}_{m}^{k}+k-j-2}f_{k-j-2}(x_{k\overline{n}_{m}^{k}+k-j-3})f_{k-j-1}(x_{k\overline{n}_{m}^{k}+k-j-2}),\label{eq:ite_odd}\end{align}
for $j=0,1,2,\dots$.

Firstly, we show \begin{equation}
U_{k,k-j}=\begin{cases}
S_{k-j} & \text{ for }j=2,\\
I_{k-j} & \text{ for }j=3.\end{cases}\label{inductive1}\end{equation}
Suppose that $U_{k,k-2}<S_{k-2}$ or $U_{k,k-3}>I_{k-3}$. By considering
the limiting equation of (\ref{eq:ite_odd}) with $j=0$, it follows\[
S_{k}=U_{k,k-2}f_{k-2}(U_{k,k-3})f_{k-1}(U_{k,k-2}).\]
Then, by (\ref{eq:ass00}) and (\ref{eq:assumption}), it follows
\[
S_{k}<S_{k-2}f_{k-2}(I_{k-3})f_{k-1}(S_{k-2}).\]
This gives a contradiction to (\ref{eq:concl_lem_odd}) with $h=k$
in Lemma \ref{lem:Lemma_odd}. Thus, (\ref{inductive1}) holds.

Next we assume that\begin{equation}
U_{k,k-j}=\begin{cases}
S_{k-j} & \text{ for }j=j_{1},\\
I_{k-j} & \text{ for }j=j_{1}+1,\end{cases}\label{inductive2}\end{equation}
where $j_{1}$ is a positive even integer. Under the assumption (\ref{inductive2}),
we show\begin{equation}
U_{k,k-j}=\begin{cases}
S_{k-j} & \text{ for }j=j_{1}+2,\\
I_{k-j} & \text{ for }j=j_{1}+3.\end{cases}\label{inductive3}\end{equation}
Suppose that $U_{k,k-j_{1}-2}<S_{k-j_{1}-2}$ or $U_{k,k-j_{1}-3}>I_{k-j_{1}-3}$.
By considering the limiting equation of (\ref{eq:ite_odd}) with $j=j_{1}$,
it follows\[
U_{k,k-j_{1}}=S_{k-j_{1}}=U_{k,k-j_{1}-2}f_{k-j_{1}-2}(U_{k,k-j_{1}-3})f_{k-j_{1}-1}(U_{k,k-j_{1}-2}).\]
Then, by (\ref{eq:ass00}) and (\ref{eq:assumption}), it follows
\[
S_{k-j_{1}}<S_{k-j_{1}-2}f_{k-j_{1}-2}(I_{k-j_{1}-3})f_{k-1}(S_{k-j_{1}-2}).\]
This gives a contradiction to (\ref{eq:concl_lem_odd}) with $h=k-j_{1}$
in Lemma \ref{lem:Lemma_odd}. Thus, (\ref{inductive3}) holds.

By considering the limiting equation of (\ref{eq:ite_odd}) with $j=1$
and substituting (\ref{inductive1}) and (\ref{inductive3}) with
$j_{1}=2$, it follows \begin{align*}
U{}_{k,k-1} & =U_{k,k-3}f_{k-3}(U_{k,k-4})f_{k-2}(U_{k,k-3})\\
 & =I_{k-3}f_{k-3}(S_{k-4})f_{k-2}(I_{k-3}).\end{align*}
Hence, \begin{equation}
U_{k,k-1}=I_{k-1},\label{eq:inductive4}\end{equation}
by (\ref{eq:concl_lem_odd}) with $h=k-1$ in Lemma \ref{lem:Lemma_odd}.
Hence, (\ref{inductive2p+1}) holds by (\ref{eq:inductive4}) and
mathematical induction.

From (\ref{eq:Mother_generalPielou}), it holds\[
x_{k\overline{n}_{m}^{k}+k-j}=x_{k\overline{n}_{m}^{k}+k-j-1}f_{k\overline{n}_{m}^{k}+k-j-1}(x_{k\overline{n}_{m}^{k}+k-j-2})=x_{k\overline{n}_{m}^{k}+k-j-1}f_{k-j-1}(x_{k\overline{n}_{m}^{k}+k-j-2}),\]
for $j=0,1,2,\dots,k-1$, and by considering the limiting equation
and using (\ref{inductive2p+1}), we obtain \begin{equation}
\begin{cases}
I_{k-j} & =S_{k-j-1}f_{k-j-1}(I_{k-j-2}),j=1,3,\dots,k,\\
S_{k-j} & =I_{k-j-1}f_{k-j-1}(S_{k-j-2}),j=2,4,\dots,k-1.\end{cases}\label{s}\end{equation}

Similar to the above discussion, it also holds\begin{equation}
\begin{cases}
S_{k-j} & =I_{k-j-1}f_{k-j-1}(S_{k-j-2}),j=1,3,\dots,k,\\
I_{k-j} & =S_{k-j-1}f_{k-j-1}(I_{k-j-2}),j=2,4,\dots,k-1.\end{cases}\label{i}\end{equation}
Consequently, it holds \[
I_{k-j}=S_{k-j-1}f_{k-j-1}(I_{k-j-2})\leq I_{k-j-1}f_{k-j-1}(S_{k-j-2})=S_{k-j},\]
for $j=1,2,\dots,k$, by (\ref{s}) and (\ref{i}), and hence (\ref{eq:concl_odd})
holds. Then, \[
\liminf_{n\to+\infty}x_{kn+h}=\limsup_{n\to+\infty}x_{kn+h}=\lim_{n\to+\infty}x_{kn+h},\]
for $h=1,2,\dots,k$ and there exist $k$ positive constants $x_{h}^{*}=S_{h}=I_{h},h=1,2,\dots,k$
such that \[
x_{h}^{*}=\lim_{n\to+\infty}x_{kn+h}.\]
The proof is complete.\qed

Finally, we establish Theorem \ref{thm:mainthm} from Theorems \ref{thm:Theorem_2},
\ref{thm:Theore,_even}, \ref{thm:Theorem_3} and \ref{thm:Theorem_odd}.
\end{pf}

\section{Applications}

In this section, we give two examples to demonstrate our result. At
first, we introduce the following example.

\begin{equation}
f_{n}(x)=\frac{\lambda}{1+(\lambda-1)\frac{x}{K_{n}}},n=0,1,2\dots,x\in[0,+\infty),\label{eq:beverton}\end{equation}
where $\lambda>1$ and $K_{n}=K_{n+k}>0$ for $n=0,1,2,\dots,$ ($k$
is an arbitrary positive integer). (\ref{eq:Mother_generalPielou})
with (\ref{eq:beverton}) is called a delayed Beverton-Holt equation
(see also \cite{MR2397325,MR2151684}). Since $f_{n}(x)$ is periodic
on $n$ and satisfies (\ref{eq:ass00}) and (\ref{eq:assumption}),
the following result is derived from Theorem \ref{thm:mainthm}.
\begin{cor}
\label{cor:bev}There exists a $k$-periodic solution $x_{n}^{*}$
such that $x_{n}^{*}=x_{n+k}^{*}$ which is globally attractive, that
is, for any solution of (\ref{eq:Mother_generalPielou}) with (\ref{eq:beverton}),
it holds that \[
\lim_{n\to+\infty}(x_{n}-x_{n}^{*})=0.\]

\end{cor}
One can see (\ref{eq:Mother_generalPielou}) with (\ref{eq:beverton})
is equivalent to the Pielou's equation by a simple transformation.
Therefore, Corollary \ref{cor:bev} is also derived from Theorem A.

Secondly, we introduce the following example.\begin{equation}
f_{n}(x)=\frac{\beta_{n}}{1+\alpha_{n}^{1}\frac{x}{1+\alpha_{n}^{2}x}},n=0,1,2,\dots,x\in[0,+\infty),\label{eq:Ex2}\end{equation}
where $\beta_{n},\alpha_{n}^{1},\alpha_{n}^{2}$ are positive periodic
sequences with a period $k$. It is not necessary that $\beta_{n},\alpha_{n}^{1},\alpha_{n}^{2}$
share the same period and, in this case, we can also easily find such
a $k$. We obtain the following results by applying Theorems \ref{thm:main0}
and \ref{thm:mainthm}, respectively.
\begin{cor}
If $\Pi_{n=1}^{k}\beta_{n}\leq1$, then, for any solution of (\ref{eq:Mother_generalPielou})
with (\ref{eq:Ex2}), it holds that\[
\lim_{n\to+\infty}x_{n}=0.\]

\end{cor}
On the other hand, for the case $\Pi_{n=1}^{k}\beta_{n}>1$, we establish
the following result.
\begin{cor}
If \[
\Pi_{n=1}^{k}\beta_{n}>1,\text{ and }\Pi_{n=1}^{k}\frac{\beta_{n}}{1+\frac{\alpha_{n}^{1}}{\alpha_{n}^{2}}}<1,\]
then, there exists a $k$-periodic solution $x_{n}^{*}$ such that
$x_{n}^{*}=x_{n+k}^{*}$ which is globally attractive, that is, for
any solution of (\ref{eq:Mother_generalPielou}) with (\ref{eq:Ex2}),
it holds that \[
\lim_{n\to+\infty}(x_{n}-x_{n}^{*})=0.\]
\end{cor}
\begin{ack}
The authors are very grateful to the anonymous referee for carefully
reading and valuable comments which led to a significant improvement
of the original manuscript. This work was partially done during the
second author's stay as a member of International Research Training
Group (IGK 1529) at the Technical University Darmstadt. The authors
would like to express their gratitude to Professor Yoshiaki Muroya
for his valuable comments to this paper.
\end{ack}

config: arxiv


TITLE: Do we draw a distincton between a number as an element of the reals, and an element of the naturals?
QUESTION [1 upvotes]: I see in some explanations of attempts to formalize numbers such as Von Neumann's ordinals like in this rather philosophical question that we can draw a distinction between a real number '1' and a natural number '1', obviously in mathematical contexts I have encountered, we describe these as one and the same object, is this distinction one of a simply philosophical nature, And does it affect the rigor of mathematical foundations to treat them as one and the same, as we tend to do?
Do we ever see the number '2' as a natural number differently than '2' as a real number? Are they considered different objects? Is there any definitive answer?
To what extent is it fine to treat there as being one 'number' that exists?
As you may see in the answers below (and in this question) some seem to believe they are the same object with different set based representations, two different objects.
try to avoid going too much into advanced set theory formalism as they are far above my level

REPLY [1 votes]: As you mentioned $\sqrt2$ in the question I want mention one subtlety that comes up with number systems you won't meet early in your studies, namely the $p$-adics.
The reason I bring this up is because when we extend our number system from $\Bbb{Q}$, things really branch out. We can extend $\Bbb{Q}$ into infinitely many different directions. The most popular one is surely $\Bbb{R}$ because that's where physic, engineering and a number of other things take place. When construct the real number called $\sqrt2$ we do it (for example) as a limit $S$ (or the equivalence class of the Cauchy sequence if you want to be pedantic) of the sequence $1$, $14/10$, $141/100$, $\ldots$. By the carefully designed arithmetic, we can then show that the equation $S^2=2$ holds.
In this limiting process we measure how far two rational numbers $q_1$ and $q_2$ are from each other by calculating the absolute value of their difference $|q_1-q_2|$. As we advance in the above sequence the differences eventually go below any power of $1/10$, for example $|141/100-1414/1000|=4/1000$ is below $1/100$. This leads to a working concept of a limit. Hopefully you have seen that.
The other ways of measuring how far $q_1$ and $q_2$ are from each other is number theoretic. We start out by picking a prime number $p$. Then we, again, look at the difference $q_1-q_2$. This time we write the difference, using factorization into primes, in the form $q_1-q_2=p^a\cdot m/n$, where neither of the integers $m,n$ is divisible by $p$. The integer $a$ can be positive or negative as the case may be. We then define the $p$-adic distance $d_p(q_1,q_2)=p^{-a}$. In other words, the numbers are close to each other if a high power of $p$ divides their difference.
The surprising fact is that this concept of a distance still leads to a similar extension of rational numbers by means of equivalence classes of Cauchy sequences, the field of $p$-adic numbers $\Bbb{Q}_p$.
Since I have been motivated by $\sqrt2$ let us select $p=7$. We see that $3^2=9$ differs from $2$ by seven, now a small difference, so according to the $7$-adic metric $3$ is close to being a square root of two. We then observe that
$3+1\cdot7=10$ is an even better approximation of $\sqrt2$ as $10^2-2=98$ is divisible by $7^2$. $3+1\cdot7+2\cdot7^2=108$ is better still, as $108^2-2=2\cdot7^3\cdot17$ is divisible by $7^3$.
We can keep going on forever (requires a proof, but it isn't difficult). Observe that my "approximate square roots" also differed from the preceding one by something that is divisible by a higher and higher power of seven. Therefore they form a $7$-adic Cauchy sequence. We can (the proof is similar to the real case) again show that the limit
$$S=3+1\cdot7+2\cdot7^2+6\cdot7^3+\cdots$$
satisfies $S^2=2$.
At this point we can may be agree that $1.41421356\ldots$ and
$S=3+1\cdot7+2\cdot7^2+6\cdot7^3+\cdots$ don't really have much in common. Yet they can both claim to be $\sqrt2$.
Closing with a few remarks

In the real numbers we can single out $1.414\cdots$ as the square root of two for it is positive.  However, in the $7$-adic domain there is no similar division to positive and negative numbers, and $S$ and $-S$ have equal claims to being called $\sqrt2$.
You know that in the reals only positive numbers have square roots. As I said that there is no positive/negative dichotomy in the $p$-adic realm, the existence of square roots in $\Bbb{Q}_7$ is different. For example, it turns out that there is no $7$-adic $\sqrt3$. The above process leading to $\sqrt2$ fails at the first hurdle because $n^2-3$ is not divisible by seven for any integer $n$. For the same reason there is no $5$-adic $\sqrt2$.
One reason $p$-adic numbers are not used (much) in natural sciences is that their size and addition behave very differently. In calculus we arrive at interesting stuff like integrals by the process of adding together more and more very tiny numbers. This won't work in the $p$-adics. No matter how many numbers divisible by $7^5$ you add together, the end result is still divisible by $7^5$. It never "grows" to be divisible only by a lower power of seven.

Summary: When we extend from $\Bbb{Q}$ to directions other than $\Bbb{R}$, the number systems really branch out, and things like roots of the equation $x^2=2$ (should they exist) have nothing else in common (an exaggeration, but I'm not gonna discuss how we might bring these different extension back together again).