TITLE: Convergent sequences must have bounded range?
QUESTION [0 upvotes]: I am currently reading Baby Rudin and I am having trouble understanding why convergent sequences must have a bounded range. Specifically, I am thinking of the following counterexample: $f(n)=1/(n-1)$
At $n=1$, this sequence diverges, but at infinity, it "converges" to zero. Any insight into the above theorem and my "counterexample" would be appreciated. Also, I apologize for the lack of $\LaTeX$. I am doing this from my phone.

REPLY [0 votes]: The terms have to be (well) defined, that's not the case with your 'first term', $f(1)$.
You can always split the terms of a sequence into a finite first part (say the first $k$ terms) and the "tail" (all the others).
If a sequence converges, it has a limit $L \in \mathbb{R}$. Because it converges, we know there is a number $N$ such that all the terms $a_i$ are closer to $L$ than any $\varepsilon$ we want, as long as $i>N$ (definition!).
In this way, you can split all the terms into the first $N$ (this a finite number of terms and thus bounded) and all the other terms, which are bounded below by $L-\varepsilon$ and bounded above by $L+\varepsilon$. Hence all the terms are finite and the range of the whole sequence is bounded.


text:
\begin{document}

\date{\vspace{-5ex}}

\maketitle

\begin{abstract}
We use the lace expansion to give a simple
proof that the critical two-point function  for weakly self-avoiding
walk on $\mathbb{Z}^d$ has decay $|x|^{-(d-2)}$ in dimensions $d>4$.
The proof uses elementary Fourier analysis and the Riemann--Lebesgue Lemma.
\end{abstract}

\section{Introduction and main result}

The lace expansion has been used to prove $|x|^{-(d-2)}$ decay for
the long-distance behaviour of critical two-point
functions in a variety of statistical mechanical models on $\Z^d$
above their upper critical
dimensions, including self-avoiding walk
for $d>4$ \cite{HHS03,Hara08,BHH19,BHK18}, percolation for $d>6$
\cite{HHS03,Hara08}, lattice trees and lattice animals for $d>8$ \cite{HHS03,Hara08},
the Ising model for $d>4$ \cite{Saka07}, and the $\varphi^4$ model
for $d>4$ \cite{Saka15,BHH19}.  For weakly self-avoiding walk and oriented
percolation in dimensions $d>4$, local central limit theorems have
also been proved \cite{BR01,ABR16,HS02}.
Related results for long-range models are proved in \cite{CS15}.

Typically, $|k|^{-2}$ behaviour
for the Fourier transform of the critical two-point function (near $k=0$) had been proved first (as in, e.g, \cite{BS85,Slad06}).  However,
this does not directly imply $|x|^{-(d-2)}$ behaviour for the inverse
Fourier transform;
see \cite[Example~1.6.2]{MS93} for a counterexample and \cite[Appendix~A]{Soka82} for
further discussion of this point.

Our purpose here is to use the lace expansion to give
a simple proof that the critical two-point function
for weakly self-avoiding walk in dimensions $d>4$
has Gaussian decay $|x|^{-(d-2)}$.
Apart from the derivation of the lace expansion, which is well documented in the literature
and not repeated here, our proof uses little more than
elementary Fourier analysis, the Riemann--Lebesgue Lemma, and
the product rule for differentiation.
Although the realm of application of our convergence proof for the lace expansion
appears to be less general than other methods,
its application to weakly self-avoiding walk
is strikingly simple and provides a new tool for problems of this genre.

To make the presentation as simple as possible,
we restrict attention to the
two-point function of the nearest-neighbour weakly self-avoiding walk.
For background we refer to \cite{MS93,Slad06}.
We follow the approach in \cite{HHS03} apart from one key ingredient which is significantly
simplified.

Let $D:\Z^d \to \R$ be given by
$D(x)=\frac{1}{2d}$ if $\|x\|_1=1$ and otherwise $D(x)=0$.  Let $D^{*n}$ denote the $n$-fold
convolution of $D$ with itself.
For $n \in \N$, let $\Wcal_n(x)$ denote the set of $n$-step walks from $0$ to $x$, i.e., the
set of $\omega = (\omega(0),\omega(1),\ldots,\omega(n))$ with each $\omega(i)\in \Z^d$,
$\omega(0)=0$, $\omega(n)=x$,
and $\|\omega(i)-\omega(i-1)\|_1=1$ for $1 \le i \le n$.
The set $\Wcal_0(x)$ consists of the zero-step walk $\omega(0)=0$ when $x=0$,
and otherwise it is the empty set.  We write $\degree = 2d$
for the degree of the nearest-neighbour graph.
The simple random walk two-point function is
defined, for $z \in [0,1/\degree]$, by
\begin{equation}
    C_z(x) = \sum_{n=0}^\infty \sum_{\omega\in \Wcal_n(x)} z^n
    =
    \sum_{n=0}^\infty (z\degree)^n D^{*n}(x).
\end{equation}
The Green function is $C_{1/\degree}(x)$.

For $\omega \in \Wcal_n(x)$ and $0 \le s < t \le n$, we define
\begin{equation}
    U_{st}(\omega) =
    \begin{cases}
        -1 & (\omega(s)=\omega(t))
        \\
        0 & (\text{otherwise}).
    \end{cases}
\end{equation}
Given $\beta \in (0,1)$, $z \ge 0$, and $x \in \Z^d$,
the \emph{weakly self-avoiding walk two-point function} is then
defined by
\begin{equation}
    G_z(x) = \sum_{n=0}^\infty \sum_{\omega\in \Wcal_n(x)} z^n \prod_{0\le s<t\le n}(1+\beta U_{st}(\omega)).
\end{equation}
The \emph{susceptibility} is defined by $\chi(z)=\sum_{x\in \Z^d} G_z(x)$.
A standard subadditivity argument implies the existence of $z_c=z_c(\beta) \ge
z_c(0)=1/\degree$ such that
$\chi(z)$ is finite if and only if $z \in [0,z_c)$; also $\chi(z) \ge z_c/(z_c-z)$
so $\chi(z_c)=\infty$ (see, e.g., \cite[Theorem~2.3]{Slad06}).
In particular, $G_z(x)$ is finite if $z\in [0,z_c)$; in fact it decays exponentially in $x$.
We will prove the following theorem.  The constant $a_d$ in the theorem is
$a_d=\frac{d\Gamma(\frac{d-2}{2})}{2 \pi^{d/2}}$.

\begin{theorem}
\label{thm:wsaw}
Let $d>4$, and let $\beta>0$ be sufficiently small.  There is a constant
$c_d = a_d (1 +O(\beta))$ such that
\begin{equation}
    G_{z_c}(x) = c_d \frac{1}{|x|^{d-2}} + o\left( \frac{1}{|x|^{d-2}} \right).
\end{equation}
For $d > 5$, the error term is improved to $o(|x|^{-(d-1)})$.
\end{theorem}

In \cite{Slad20_wsaw}, the method of proof of Theorem~\ref{thm:wsaw} is extended to
analyse the near-critical two-point function.  Namely, it is proved in \cite{Slad20_wsaw}
that for $d>4$ and for $\beta$ sufficiently small, there are constants $\kappa_0>0$
and $\kappa_1 \in (0,1)$ such that for all $z \in (0,z_c)$ and all $x \in \Z^d$,
\begin{equation}
\label{e:Gmr}
    G_z(x) \le \kappa_0 \frac{1}{1\vee |x|^{d-2}} e^{-\kappa_1 (z_c-z)^{1/2}|x|}.
\end{equation}
The estimate \eqref{e:Gmr} is applied in \cite{Slad20_wsaw} to prove existence of
a ``plateau'' for the weakly self-avoiding walk two-point function on a large discrete torus
in dimensions $d>4$.

An alternate proof of Theorem~\ref{thm:wsaw} is given in \cite{BHK18}, based on
Banach algebras and a fixed-point theorem.  That proof avoids explicit use of the Fourier transform, though
it does rely on an expansion of the Green function
$C_{1/\degree}(x)$ which is proved using the Fourier transform.
Theorem~\ref{thm:wsaw} is proved for the strictly self-avoiding walk (the case $\beta =1$)
in \cite{Hara08}, and for spread-out strictly self-avoiding walk in \cite{HHS03}.
Thus Theorem~\ref{thm:wsaw} is not new or best possible; our goal here is to present
a new and simple method of proof rather than to obtain a new result.
A sample consequence of Theorem~\ref{thm:wsaw} is that the bubble condition holds for $d>4$,
and this implies the matching upper bound $\chi(z) \le O((z_c-z)^{-1})$ (see \cite[Theorem~2.3]{Slad06}).

We use the Fourier transform.
Let $\mathbb{T}^d = (\R / 2 \pi \Z )^d$ denote the continuum torus of period $2\pi$.
For a summable function $f:\Z^d \to \C$ we define its Fourier transform by
\begin{equation}
    \hat f(k) = \sum_{x\in\Z^d}f(x)e^{ik\cdot x}
    \qquad
    (k \in \T^d).
\end{equation}
The inverse Fourier transform is
\begin{equation}
    f(x) = \int_{\T^d}\hat{f}(k) e^{-i k \cdot x } \frac{dk}{(2\pi)^{d}}
    \qquad
    (x \in \Z^d).
\end{equation}

\section{Lace expansion}

The lace expansion was introduced by Brydges and Spencer \cite{BS85} to prove
that the weakly self-avoiding walk is diffusive in dimensions $d>4$.
In the decades since 1985, the lace expansion has been adapted and extended to a broad range of
models and results.

For the weakly self-avoiding walk, the lace expansion
\cite{BS85,MS93,Slad06}
produces an explicit formula for the $\Z^d$-symmetric function $\Pi_z: \Z^d \to \R$
which satisfies, for $z\in [0,z_c)$,
\begin{equation}
    G_z(x) = \delta_{0,x} + z\degree (D*G_z)(x) + (\Pi_z* G_z)(x) \qquad (x \in \Z^d),
\end{equation}
or equivalently,
\begin{equation}
    \hat G_z(k) = \frac{1}{1-z\degree \hat D(k) - \hat \Pi_z(k)}
    \qquad (k \in \T^d).
\end{equation}
Let $\delta:\Z^d\to\R$ denote the Kronecker delta $\delta(x)=\delta_{0,x}$.
Then $\hat\delta(k)=1$.
We define
\begin{equation}
    F_z  = \delta - z\degree D  - \Pi_z , \qquad \hat F_z = 1 -z\degree\hat D -\hat\Pi_z.
\end{equation}
Then
\begin{equation}
     (G_z * F_z)(x)  = \delta_{0,x} , \qquad \hat G_z(k) = \frac{1}{\hat F_z(k)}.
\end{equation}

\section{Proof of main result}

\subsection{Diagrammatic estimate}

As in many applications of the lace expansion, we use a bootstrap argument.
We define the \emph{bootstrap function}
\begin{equation}
    b(z) = \sup_{x\in \Z^d} \frac{G_z(x)}{C_{1/\degree}(x)} \qquad (z \in [0,z_c]).
\end{equation}
The bootstrap function can be seen to be finite and
continuous in $z\in [0,z_c)$, using the fact that $G_z(x)$ is continuous and decays exponentially for large $|x|$.  We do not know {\it a priori} that $G_{z_c}(x)$ is finite.
By definition, $b(z) \le 1$ for $z \in [0,\frac{1}{\degree}]$.
The next proposition gives consequences of the assumption that $b(z) \le 3$.
We will not need to know more about the function $\Pi_z$ than
Proposition~\ref{prop:diagram}, so we do not give
its explicit formula here.  The formula can be found in \cite{BS85,MS93,Slad06}.

\begin{prop}
\label{prop:diagram}
Let $d>4$ and let $\beta$ be sufficiently small.  Fix $z \in [\frac{1}{\degree},z_c]$.
If $b(z) \le 3$ then there is a constant $K$ depending only on $d$ (and on ``3'') such that
\begin{equation}
\label{e:Pibd}
    |\Pi_z(x)| \le K\beta \frac{1}{1+|x|^{3(d-2)}}
    \qquad
    (x \in \Z^d)
\end{equation}
and hence $\hat\Pi_z \in C^{s}(\T^d)$
for any nonnegative integer $s< 2d-6$, in particular $\hat\Pi_z \in C^{d-2}(\T^d)$.
In addition, the \emph{infrared bound} holds, i.e., there exists $c>0$ (independent
of $\beta,z,k$) such that
\begin{equation}
\label{e:irbd}
    \hat F_z(k) \ge  c  |k|^{2}
    \qquad (k \in \T^d).
\end{equation}
\end{prop}

\begin{proof}
The bound \eqref{e:Pibd} is a diagrammatic estimate proved via well-developed
technology (e.g., \cite[(12)]{BHK18})
and we omit its proof.  It follows immediately from \eqref{e:Pibd}
that $|x|^s|\Pi_z(x)|$ is summable for all $s< 2d-6$ and hence
that $\hat\Pi_z \in C^{s}(\T^d)$
for any nonnegative integer $s< 2d-6$.
For \eqref{e:irbd}, we use
\begin{align}
    \hat F_z(k) & = \hat F_z(0) +[\hat F_z(k) -\hat F_z(0)]
    = \hat F_z(0) + z\degree [1-\hat D(k)] + [\hat\Pi_z(0)-\hat\Pi_z(k)].
\end{align}
The first term on the right-hand side is  $\hat F_z(0) = \chi(z)^{-1} \ge 0$.
By definition, the second term obeys $1-\hat D(k) = d^{-1}\sum_{j=1}^d(1-\cos k_j) \ge \frac{4|k|^2}{\pi^2\degree}$.
By \eqref{e:Pibd}, the second derivative of $\hat\Pi_z(k)$ with respect to $k$ is $O(\beta)$,
and \eqref{e:irbd} then follows by a Taylor estimate on $\hat\Pi_z(0)-\hat\Pi_z(k)$
(by symmetry there is no linear term in $k$).
\end{proof}

\subsection{Isolation of leading term}

Following \cite{HHS03}, we isolate the leading term by writing $G_z$ as a $z$-dependent
multiple of the random walk two-point function $C_\mu$ at a $z$-dependent value of $\mu$.
Let  $A_z = \delta -z\degree D$, $\lambda  > 0$ and $\mu\in[0,\frac{1}{\degree}]$.
Since $C_\mu * A_\mu=\delta$ and $G_z *F_z = \delta$,
we have
\begin{align}
    G_z & = \lambda C_\mu + \delta*G_z - \lambda C_\mu * \delta
    \nnb & =
    \lambda C_\mu + C_\mu * A_\mu*G_z - \lambda C_\mu  *F_z * G_z
    \nnb & =
    \lambda C_\mu + C_\mu * E_{z,\lambda,\mu}*G_z
    ,
\end{align}
with
\begin{align}
     E_{z,\lambda,\mu} &  = A_\mu - \lambda F_z.
\end{align}

Given $z \in [\frac{1}{\degree},z_c)$,
we choose $\lambda=\lambda_z$ and $\mu=\mu_z$ in order to achieve
\begin{equation}
\label{e:Emoments}
    \sum_{x\in\Z^d} E_{z,\lambda_z,\mu_z}(x) = \sum_{x\in\Z^d} |x|^2 E_{z,\lambda_z,\mu_z}(x) =0,
\end{equation}
namely (since $\sum_x |x|^2 D(x)=1$)
\begin{align}
    \lambda_z & = \frac{1}{\hat F_z(0) -
    \sum_x |x|^2 F_z(x)}
    =
    \frac{1}{1-\hat \Pi_z(0) +
    \sum_x |x|^2 \Pi_z(x)} ,
    \\
\label{e:muzdef}
    \mu_z \degree & = 1-\lambda_z \hat F_z(0)
        = \frac{z\degree + \sum_x |x|^2 \Pi_z(x)}
    {\hat F_z(0) + z\degree + \sum_x |x|^2 \Pi_z(x)}
    .
\end{align}
By Proposition~\ref{prop:diagram}, if we assume $b(z) \le 3$
then the second moment of $\Pi_z$ is $O(\beta)$, and hence
the above formulas are well defined, $\lambda_{z} = 1+O(\beta)$, and $\mu_z\degree \in [0,1)$.
In particular, if $b(z_c)\le 3$
(as we will eventually show to be the case), then, since $\hat F_{z_c}(0)=\chi(z_c)^{-1}=0$,
we see from \eqref{e:muzdef} that $\mu_{z_c} =1/\degree$ is the critical
value for $C_\mu$.

With these choices of $\lambda_z,\mu_z$, we have
\begin{align}
\label{e:GSEG}
    G_z & =
    \lambda_z C_{\mu_z} + f_z, \qquad f_z = C_{\mu_z} * E_{z}*G_z
    ,
\end{align}
with
\begin{equation}
\label{e:Ezdef}
    E_z = E_{z,\lambda_z,\mu_z} =
    (1-\lambda_z )(\delta-D) - \lambda_z \hat \Pi_z(0) D + \lambda_z \Pi_z.
\end{equation}
By definition,
\begin{equation}
\label{e:fhdef}
    \hat f_z(k) =  \hat C_{\mu_z}(k) \hat E_z(k) \hat G_z(k).
\end{equation}
Roughly, since we have arranged
via \eqref{e:Emoments}
that the Taylor expansion of $\hat E_z(k)$ has no constant
term or term of order $|k|^2$, we expect it to be of order $\beta |k|^4$.
On the other hand, according to the
infrared bound, the Fourier transform of $\hat G_z(k)$
will be of order $|k|^{-2}$ for small $|k|$.  The same is true for $\hat C_{\mu_z}(k)$,
so $\hat f_z(0)=O(\beta)$.
We will show that
this less singular behaviour of $\hat f_z(k)$
translates into better decay than $|x|^{-(d-2)}$ for $f_z(x)$.
This will permit
the bootstrap argument to be completed by proving $b(z) \le 2$, and the proof will
essentially be complete.  The details in this rough sketch are given below.

\subsection{The bootstrap}

The bootstrap argument is encapsulated in the following proposition.

\begin{prop}
\label{prop:boot}
Fix $z \in [\degree^{-1},z_c)$.
If $b(z) \le 3$ then for $\beta$ sufficiently small (not depending on $z$) it is in fact
the case that $b(z) \le 2$.
\end{prop}

The next proposition is a replacement for the bound
on $ C_{\mu_z}*  E_z$ in \cite[Proposition~1.9]{HHS03} which required a delicate
Fourier analysis of $C_{\mu_z}$, and of the bound of \cite[Lemma~4]{BHK18} which used
the expansion $ C_{1/\degree}(x) = a|x|^{-(d-2)} + b|x|^{-d} + O(|x|^{-(d+2)})$ from \cite{Uchi98}
which was also proved by careful Fourier analysis.

\begin{prop}
\label{prop:h}
Let $d>4$ and let $\beta$ be sufficiently small.
Let $z\in [\frac{1}{\degree},z_c]$.
Under the assumption that $b(z) \le 3$, the derivatives $\nabla^\alpha \hat f_z(k)$ obey
\begin{equation}
\label{e:nablafbd}
    \|\nabla^\alpha \hat f_z  \|_{L^1(\T^d )} \le O(\beta)
\end{equation}
provided $|\alpha|\le d-2$ when $d=5$ and $|\alpha| \le d-1$ for $d \ge 6$,
with constant depending only on $d$ (not on $z$).
\end{prop}

The importance of \eqref{e:nablafbd} resides in the fact that the smoothness
of a function on the torus implies bounds on the decay of its (inverse) Fourier transform.
More precisely, it is proved in \cite[Corollary~3.3.10]{Graf10} using integration by parts that
there is a constant $\kappa_{d,s}$, depending only on the dimension $d$ and the
maximal order $s$
of differentiation,
such that if $\nabla^\alpha \hat\psi
\in L^1(\T^d)$ for all multi-indices $\alpha$ with $|\alpha| \le s$ then
\begin{equation}
\label{e:Graf}
    |\psi(x)|
    \le
    \kappa_{d,s} \frac{1}{1\vee |x|^s}
    \max_{|\alpha| \in \{0,s\}}\|\nabla^\alpha \hat\psi\|_{L^1(\T^d)}.
\end{equation}
In addition to the quantitative estimate \eqref{e:Graf}, it follows from
the integrability of
$\nabla^\alpha \hat\psi$ together with the Riemann--Lebesgue Lemma (e.g.,
\cite[Proposition~3.3.1]{Graf10}) that the inverse Fourier transform of
$\nabla^\alpha \hat\psi$ vanishes at infinity, so
$|x|^s \psi(x) \to 0$ as $|x|\to\infty$.

\begin{proof}[Proof of Proposition~\ref{prop:boot}]
Suppose that $b(z) \le 3$.
Let $n_d=d-2$ for $d=5$ and $n_d=d-1$ for $d \ge 6$.
By \eqref{e:nablafbd}, as discussed around \eqref{e:Graf},
$|f_z(x)|= o(|x|^{-n_d})$ and
$|f_z(x)| \le  O(\beta |x|^{-n_d})$,
with $z$-independent constant in the latter bound.
Therefore,
\begin{align}
\label{e:GCasy}
    G_z(x) & = \lambda_z C_{\mu_z}(x)  + o(|x|^{-n_d}),
    \\
\label{e:GCbd}
    G_z(x) & = \lambda_z C_{\mu_z}(x)  + O(\beta |x|^{-n_d}),
\end{align}
with the constant in \eqref{e:GCbd} independent of $x$ and $z$.
As is well known (e.g., \cite{Uchi98,LL12}),
$C_{1/\degree}(x) \sim a_d  |x|^{-(d-2)}$.
From \eqref{e:GCbd},
we see that by taking $\beta$ sufficiently small we can obtain
\begin{equation}
    G_z(x) \le (1+O(\beta))C_{1/\degree}(x) + O(\beta)C_{1/\degree}(x) \le 2 C_{1/\degree}(x),
\end{equation}
i.e., $b(z) \le 2$.
This completes the proof.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:wsaw}]
Since $b(z) \le b(\degree^{-1})\le 1$ for $z \le \degree^{-1}$ by definition,
it follows from Proposition~\ref{prop:boot} and the continuity of the function $b$
that the interval $(2,3]$ is forbidden for values of $b(z)$,
so $b(z) \le 2$ for all $z\in [0,z_c)$.  By monotone convergence, also $b(z_c) \le 2$.  Thus $\lambda_z$ approaches a limit
$\lambda_{z_c}=1+O(\beta)$ as $z \to z_c^-$.
Since $b(z_c) \le 2$, \eqref{e:GCasy} holds for $z=z_c$ so
$G_{z_c}(x)=\lambda_{z_c}C_{1/\degree}(x) +o(|x|^{-n_d})$.
Thus Theorem~\ref{thm:wsaw} is proved subject to Proposition~\ref{prop:h}.
\end{proof}

\subsection{Proof of Proposition~\ref{prop:h}}

It remains only to prove Proposition~\ref{prop:h}.
The next lemma is closely related to \cite[Lemma~7.2]{HHS03}.  It
reflects our choice of $\lambda_z,\mu_z$ to achieve \eqref{e:Emoments}.

\begin{lemma}
\label{lem:Ebd}
Let $d>4$ and let $\beta$ be sufficiently small.
Let $z \in [\frac{1}{\degree},z_c]$ and suppose that $b(z) \le 3$.
There is a $c_0>0$ (independent of $z$)
such that $|\nabla^\alpha \hat E_z(k)| \le c_0 \beta$
for $|\alpha| < 2d-6$, and moreover
\begin{equation}
    |\nabla^\alpha \hat E_z(k)| \le
    c_0  \beta \times
    \begin{cases}
    |k|^{4-|\alpha|} & (d>5)
    \\
    |k|^{4-|\alpha|} \log |k|^{-1} & (d=5)
    \end{cases}
    \qquad (|\alpha| \le 3).
\end{equation}
\end{lemma}

\begin{proof}
We first prove that $|\nabla^\alpha \hat E_z(k)| \le c_0 \beta$
for $|\alpha| < 2d-6$, via
the inequality $|\nabla^\alpha \hat E_z(k)| \le \sum_x |x^\alpha E_z(x)|$ together with
term-by-term estimation in the formula for $E_z$ in \eqref{e:Ezdef}.
Indeed, since $\lambda_z=1+O(\beta)$, the contribution from all moments of the term
$(1-\lambda_z)(\delta - D)$ is $O(\beta)$, and since $|\hat\Pi_z(0)| \le O(\beta)$
by \eqref{e:Pibd}, all moments of $\lambda_z \hat\Pi_z(0)D$ are also $O(\beta)$.
Finally, the moments of $\lambda_z|\Pi_z|$ with order less than $2d-6$
are bounded by $O(\beta)$ using \eqref{e:Pibd}.
In the remainder of the proof, we can therefore restrict to $|\alpha|\le 3$.

Let $g_x(k) = \cos (k\cdot x) - 1 + \frac{(k\cdot x)^2}{2!}$.
By symmetry and by \eqref{e:Emoments},
\begin{align}
    \hat E_z(k) & = \sum_{x\in\Z^d}  E_z(x) \cos (k\cdot x)
    =  \sum_{x\in\Z^d}  E_z(x) g_x(k)
    .
\end{align}
By explicit computation of the derivatives and elementary properties of sine and cosine,
\begin{gather}
     |g_x(k)| \le c(|k|^4|x|^4 \wedge (1 + |k|^2|x|^2)),
    \qquad
    |\nabla_i g_x(k)| \le c(|k|^3|x|^4 \wedge (|x|+|k||x|^2)),
    \\
     |\nabla_{ij}^2 g_x(k)| \le c(|k|^2|x|^4 \wedge |x|^2),
    \qquad
    |\nabla_{ijl}^3 g_x(k)| \le c(|k||x|^4 \wedge |x|^3),
    \qquad
\end{gather}
where $\wedge$ denotes minimum.  In the upper bound $|\nabla^\alpha \hat E_z(k)|
\le \sum_x |E_z(x)|\, |\nabla^{\alpha}g_x(k)|$,
we estimate the sum over $|x|\le |k|^{-1}$ using the $|x|^4$ bound on $\nabla^\alpha g_x(k)$,
and we estimate the sum over $|x|> |k|^{-1}$ using the other alternative in the minimum.
By a term-by-term estimate using \eqref{e:Ezdef} with \eqref{e:Pibd} (as in the previous
paragraph), for $|\alpha| \le 3$ and $d \ge 5$, and for small $|k|$,
\begin{align}
    \sum_{|x| \le |k|^{-1}} |x|^4 |E_z(x)|
    & \le
    O(\beta) \int_1^{|k|^{-1}} \frac{r^{d-1+4}}{r^{3(d-2)}}dr
    =
    O(\beta) \int_1^{|k|^{-1}} \frac{1}{r^{2d-9}}dr
    =
    \begin{cases}
        O(\beta) & (d>5)
        \\
        O(\beta \log |k|^{-1}) & (d=5),
    \end{cases}
\\
    \sum_{|x| > |k|^{-1}} |x|^{|\alpha|} |E_z(x)|
    & \le
    O(\beta) \int_{|k|^{-1}}^\infty \frac{r^{d-1+|\alpha|}}{r^{3(d-2)}}dr
    =
    O(\beta) \int_{|k|^{-1}}^\infty\frac{1}{r^{2d-5-|\alpha|}}dr
    =
        O(\beta |k|^{2d-6-|\alpha|}) ,
\end{align}
and the desired result then follows after some bookkeeping.
\end{proof}

\begin{proof}[Proof of Proposition~\ref{prop:h}]
Let $n_d=d-2$ for $d=5$ and $n_d=d-1$ for $d \ge 6$; then $n_d < 2d-6$.
Our goal is to prove the bound \eqref{e:nablafbd} on derivatives of $\hat f_z$ of order
up to $n_d$.
By the infrared bound and Lemma~\ref{lem:Ebd},  for $d>5$ we have
\begin{equation}
    |\hat f_z(k)| \le  |k|^{-2} O( \beta) |k|^4 |k|^{-2} = O(\beta),
\end{equation}
and a similar estimate holds for $d=5$ with an additional factor $\log |k|^{-1}$ in
the upper bound.

Estimation of the $L^1$ norm of derivatives of $\hat f_z(k)$
is an exercise in power counting, as follows.
For $|\alpha| \le n_d$, by the product rule for differentiation
$\nabla^\alpha \hat f_z(k)$ involves terms
\begin{equation}
    \nabla^{\alpha_1}\hat C_{\mu_z}(k)
    \nabla^{\alpha_2}\hat E_{z}(k) \nabla^{\alpha_3}\hat G_{z}(k) ,
    \qquad
    (|\alpha_1|+|\alpha_2|+|\alpha_3| =|\alpha|).
\end{equation}
By Lemma~\ref{lem:Ebd},
each derivative on $\hat E$ up to the fourth order reduces by $1$ the original power $|k|^4$ for that factor (an unimportant factor $\log |k|^{-1}$ is present for $d=5$),
and subsequent derivatives do not cause further reduction; the net effect
is therefore reduction by $\min\{|\alpha_2|,4\}$.
Similarly, each derivative on $\hat C_{\mu_z}$ or $\hat G_z$
reduces (worsens)
its power $|k|^{-2}$ by $1$; we illustrate the idea for $d=5$, for which $n_d=d-2=3$:
\begin{align}
    \left| \nabla_i \frac{1}{ \hat F} \right|
    &
    = \left| \frac{\nabla_i\hat F }{\hat F^2} \right|
    \le c_1 \frac{|k_i|}{ |k|^{4}} \le c_1 \frac{1}{ |k|^{3}},
    \qquad
    \left| \nabla_i^2 \frac{1}{\hat F} \right|
    \le \left| \frac{\nabla_i^2\hat F }{\hat F^2} \right|
    +
    \left| \frac{2( \nabla_i\hat F)^2 }{ \hat F^3} \right|
    \le c_2 \frac{k_i^0}{|k|^4} + c_2 \frac{k_i^2}{|k|^6} ,
     \\
    \left| \nabla_i^3 \frac{1}{\hat F} \right|
    &
    \le \left| \frac{\nabla_i^3\hat F }{\hat F^2} \right|
    +
    \left| \frac{6( \nabla_i\hat F)(\nabla_i^2 \hat F) }{ \hat F^3} \right|
    +
    \left| \frac{6( \nabla_i\hat F)^3 }{ \hat F^4} \right|
    \le c_3 \frac{k_i^0}{|k|^4} + c_3 \frac{k_i^{1+0}}{|k|^6} +c_3 \frac{k_i^3}{|k|^8}   .
\end{align}
A detail in the above calculation is that $|\nabla_i\hat F|$ can be bounded by a multiple
of $|\nabla_i\hat D| + |\nabla_i\hat \Pi|$, with the first term of order $k_i$ by explicit
calculation and the second also of order $k_i$ by Taylor's Theorem, symmetry,
and the boundedness of $\nabla_i^2\Pi$.

In the general case, in advancing from one derivative to the next, when the derivative acts
on the numerator it either maintains the same power of $|k|$ or reduces it
by $1$, and if it acts on the denominator then it increases the power of the denominator
by $|k|^2$ and increases the power of the numerator by $1$; the net result is
reduction of the overall power of $|k|$ by at most $1$.
For $|\alpha| \le n_d$, the total resulting power is (for small $|k|$) at worst
\begin{equation}
    \frac{|k|^{4-\min\{|\alpha_2|,4\}}}{|k|^{2+ |\alpha_1|} |k|^{2+|\alpha_3|}}
    =
    \frac{1}{|k|^{ |\alpha_1|+\min\{|\alpha_2|,4\} +|\alpha_3|}}
    \le
    \frac{1}{|k|^{|\alpha|}}
    \le
    \frac{1}{|k|^{n_d}}.
\end{equation}
This is in $L^1(\T^d)$ (also with the additional logarithmic factor when $d=5$)
and the norm in $L^1(\T^d)$
is $O(\beta)$ due to the factor $\beta$ in the bound on $\hat E_z$ in Lemma~\ref{lem:Ebd}.
This completes the proof.\footnote{An earlier proof of Proposition~\ref{prop:h}
was based on Kotani's Theorem \cite{Slad20_Kotani}
rather than via the direct proof given here.}
\end{proof}

\subsection{Concluding remarks}

\begin{enumerate}[label=(\roman*)]
\item
The main difference between the above proof and the proofs in \cite{HHS03,Hara08,BHK18}
is our avoidance of any need to convert the decay in $x$-space of factors in a convolution
$C_{\mu_z}*E_z*G_z$ into decay of the convolution (which is delicate
since, e.g., the convolution of two factors each with decay $|x|^{-(d-2)}$ has worse
decay $|x|^{-(d-4)}$ when $d>4$).  In the above proof, we encounter
instead the Fourier transform $\hat C_{\mu_z}\hat E_z \hat G_z$ and the corresponding
step is handled simply via the product rule for differentiation.

\item
To control the lace expansion,
the analysis in any of \cite{HHS03,Hara08,BHK18}
only requires a bound of the form $|\Pi(x)|\le O(|x|^{-(d+2+\epsilon)})$ for $\epsilon>0$
(with $\epsilon=2$ in \cite{BHK18}), whereas
the above proof
requires the more demanding bound $|\Pi(x)| \le  O(|x|^{-(2d-2+\epsilon)})$
in order for $\hat\Pi(k)$ to have derivatives of order $d-2$.
For self-avoiding walk, the upper bound \eqref{e:Pibd} has power $3(d-2) = 2d-2 + (d-4)$ which is sufficient.
For the Ising and $\varphi^4$ models, $\Pi(x)$  also
obeys an upper bound $|x|^{-3(d-2)}$ \cite{BHH19,Saka07,Saka15}.  However,
the above proof appears not to apply to percolation or to lattice trees and
lattice animals, where
the bound on $\Pi(x)$ is $|x|^{-2(d-2)}$ for percolation and $|x|^{-3(d-2)+d}$ for
lattice trees and lattice animals \cite{HHS03}.  It would be of interest to understand
better why this breakdown occurs and whether there is any possiblity to overcome it in these settings with upper critical dimension equal to $6$ or $8$.

\item
With further effort, it may be possible to extend our approach to spread-out
models of strictly self-avoiding walk or the Ising or $\varphi^4$ models in dimensions $d>4$
by proving a version of \eqref{e:nablafbd} in those settings.
Possibly this could simplify aspects of the analysis in \cite{HHS03,Saka07,Saka15,BHH19}.
However, this is beyond our current scope and we do not draw a conclusion about
this question here.

\item
It is natural to ask whether our approach could be used for the nearest-neighbour
strictly self-avoiding walk in dimensions $d \ge 5$, to give an alternate proof
of the $|x|^{-(d-2)}$ decay proved in \cite{Hara08}.
This certainly could not be done, at present, without
a portion of the very sizeable and computer-assisted input from \cite{HS92b} listed
in \cite[Proposition~1.3]{Hara08}; in fact results beyond those of \cite{HS92b} may be needed
to deal with the higher derivatives of $\hat\Pi(k)$ encountered here.
A further and serious obstacle, as pointed out and overcome
with a different method in \cite{Hara08}, is that
the amplitude $a_d = \frac d2 \pi^{-d/2} \Gamma((d-2)/2)$ in the asymptotic formula for the critical simple random walk two-point
function (and appearing in Theorem~\ref{thm:wsaw})
grows rapidly with $d$, so the small parameter that facilitates the bootstrap argument is
more hidden and more delicate to exploit.
\end{enumerate}

\section*{Acknowledgement}

This work was supported in part by NSERC of Canada.

config: arxiv


TITLE: RSA cipher Encryption with $n=210757$ and $a=3$ and Decryption with $n=14659$ and $a=3$
QUESTION [0 upvotes]: I think I understand correctly how to encrypt something with an RSA cipher, but I am a little lost on how to find the decryption key...(also I apologize for the formatting errors)

An RSA cipher is set up with modulus $n=210757$ and encryption key $a=3$. The plaintext is "$54321$". Encrypt it.

$$E_{N,e}(x)=x^e\pmod N$$
$$E_{210757,3}(54321)=54321^3\pmod{210757}$$
$$E_{210757,3}(54321)=(54321^2*54321)\pmod{210757}$$
$$E_{210757,3}(54321)=(54321^2*54321)\pmod{210757}$$
$$E_{210757,3}(54321)=(2950771041*54321)\pmod{210757}$$
$$E_{210757,3}(54321)=(173041*54321)\pmod{210757}$$
$$E_{210757,3}(54321)=9399760161\pmod{210757}$$
$$E_{210757,3}(54321)=208718\pmod{210757}$$

An RSA cipher is set up with modulus $n=14659$ and encryption key $a=3$. Find the decryption key.

$$E_{N,e}(x)=x^e\pmod N$$
$$E_{14659,3}(x)=x^3\pmod{14659}$$
$$D_{N,d}(y)=y^d\pmod N$$
$$D_{14659,d}(y)=y^d\pmod{14659}$$

REPLY [1 votes]: We have the following for 5)$$n=14659, a=3$$
Here we can find the prime factorisation for $n$, e.g. $n=107 * 137$
Take one from both of these prime factors, and multiply.$$(107-1)(137-1)=14416$$
We will now use this number as the modulo for $a^{-1}$, this will give us our required $b$ for decryption $\text{e}^b\pmod{n}$
$$b= a^{-1} \pmod {14416}$$$$3^{-1}\pmod {14416}=9611$$
Now we can take the cipher text and take it to the power of $9611\pmod{14659}$

Example: We want to decrypt the cipher 33.
Since we know $b$, this simply requires us to compute $33^{9611} \pmod {14659}=11282 \pmod{14659}$


TITLE: $\lim_{x \to 0} {1-x\cot x\over x^2}$
QUESTION [0 upvotes]: What's the value of this limit? I keep getting half, but the answer's 1/3. I think there might be a problem with writing ${x\cot x \over x^2 }$  as $ {\cos x \over x^2}$. Is there ?
I get (1-cosx)/x^2 on using this substitution, and if I use L'hôpital's Rule on this, I get the answer as sinx/2x = 1/2.
Edit: I can only use results like $\sin x=x$ and L'Hospital's rule. No series expansion.
Edit 2: Here's my attempt:
1/x^2 - cosx/(x*sinx) = 1/x^2 - (cosx/x^2)(x/sinx) = 1/x^2 - (cosx/x^2)(1)
{Since sinx/x=1}
= (1-cosx)/x^2
Being a 0/0 case, L'hôpital's Rule:
sinx/2x = (1/2)(sinx/x) = 1/2, since sinx/x=1

REPLY [0 votes]: By L'Hospital and using $\dfrac{\sin x}x\to1$,
$$\lim_{x\to0}\frac{\sin x-x\cos x}{x^3}=\lim_{x\to0}\frac{x\sin x}{3x^2}=\frac13.$$


TITLE: Integrals over closed contour
QUESTION [3 upvotes]: Calculate integral
$$\oint_{\gamma} \frac{z+1}{z^4+2iz^3} dz$$
where $\gamma$ is parameterization of circle $B(0,1)$ along one positive rotation.
I did something like this with Cauchy.
\begin{align}
\oint_{\gamma} \frac{z+1}{z^4+2iz^3}dz&=\oint_{\gamma} \frac{z+1}{z^3(z+2i)} =\oint_{\gamma} \frac{\dfrac{z+1}{z+2i}}{(z-0)^{2+1}}dz \\
&=\frac{2i\pi}{n!}f^{''}(z_0) \\
&=\frac{2i \pi}{2!}\left(\frac{2-4i}{(0+2i)^3}\right) \\
&=\frac 12 \pi i-\frac 14 \pi
\end{align}
with this one I'm not that sure
Calculate integral
$$\oint_{\gamma} \frac{z^3+3}{z(z-i)^2}dz $$
Where $\gamma$ is 8-like curve where the rotation is positive around $(0,i)$ and negative around $(0,0)$.
Isn't the integral zero over $\oint_{\gamma_1}(\text{positive part})$ + $\oint_{\gamma_2}(\text{negative part})=0$. So is it two integrals were $\oint_{\gamma_2}=-\oint_{\gamma_1}$.
And the integral would be
\begin{align}\oint_{\gamma_1}&=\frac{\dfrac{z^3+3}{z}}{(z-i)^2}dz \\
&=\frac{2\pi i}{1!}\left( \frac{2(-i)^3 -3}{(-i)^2}\right) \\
&=2\pi i (3-2i)\\
&=4\pi+6\pi i \\
\oint_{\gamma_2}&= -4\pi-6\pi i
 \end{align}
Did I get these question correct? I'm not 100% on the second question.

REPLY [1 votes]: The second one, assuming I guessed correctly what the path is, I'd do as follows: let $\;\gamma_1\;$ be a circle (with radius $\;\le 1/2)\;$) around $\;i=(0,1)\;$ in the positive direction, and $\;\gamma_2\;$ a circle (with radius $\;\le1/2\;$ in the negative direction around the original, and the union of both paths above is the whole $\;\gamma\;$ ,  then
$$\int\limits_\gamma\frac{z^3+3}{z(z-i)^2}dz=\oint\limits_{\gamma_1}\frac{\frac{z^3+3}z}{(z-i)^2}dz-\oint\limits_{\gamma_2}\frac{\frac{z^3+3}{(z-i)^2}}zdz=$$
$$2\pi i\left[ \frac d{dz}\overbrace{\left(\frac{z^3+3}z\right)}^{z^2+\frac3z}{}_{z=i}-\left(\frac{z^3+3}{(z-i)^2}\right)_{z=0}\right]=2\pi i\left[\left(2z-\frac3{z^2}\right)_{z=i}-\frac3{(-i)^2}\right]=$$
$$2\pi i(2i+3+3)=-4\pi+12\pi i$$
Check the above: either you or I have a mistake somewhere.


TITLE: Find the general solution to non homogeneous linear equation
QUESTION [2 upvotes]: The question:
State the general solutuon of:
$$5x_1-2x_2+4x_3=5$$
My attempt: I tried following closely an example provided for homogeneous equations without any luck:
$$
x=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=
\begin{bmatrix}2x_2-4x_3+5\\x_2\\x_3\end{bmatrix}=
x_2\begin{bmatrix}2\\1\\0\end{bmatrix}
+x_3\begin{bmatrix}-4\\0\\1\end{bmatrix}
+\begin{bmatrix}5\\1\\1\end{bmatrix}
$$
And after this step I get stuck since I don't know whether is correct and should proceed with the solution space (and also how to find the solution space for non homogenous equations)
As for who might be concerned, the question is homework related.
It would be nice to know why I got downvoted so I can improve my question
Based on the feedback received from @Isham the system should have been written as follows:
$$
x=\begin{bmatrix}5x_1\\x_2\\x_3\end{bmatrix}=
\begin{bmatrix}2x_2-4x_3+5\\x_2\\x_3\end{bmatrix}=
\begin{bmatrix}2/5x_2-4/5x_3+1\\x_2\\x_3\end{bmatrix}=
x_2\begin{bmatrix}2/5\\1\\0\end{bmatrix}
+x_3\begin{bmatrix}-4/5\\0\\1\end{bmatrix}
+\begin{bmatrix}1\\0\\0\end{bmatrix}
$$

REPLY [2 votes]: According to your equation Lucian, it should be rather this:
$x=\begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix}=
\begin{bmatrix}2/5x_2-4/5x_3+1\\x_2\\x_3\end{bmatrix}=
x_2\begin{bmatrix}2/5\\1\\0\end{bmatrix}
+x_3\begin{bmatrix}-4/5\\0\\1\end{bmatrix}
+\begin{bmatrix}1\\0\\0\end{bmatrix}$


TITLE: Choosing Senators
QUESTION [0 upvotes]: Just working my way through the W.Feller introduction to probability theory and its applications volume 1. An example in the chapter on combinatorial analysis asks the question:
"Each of the 50 states has 2 senators. If we choose 50 senators at random, what is the probability a given state is represented?"
I understand the answer given which uses the complement of the event but was curious whether the method where you force the desired outcome to occur, then work out how many ways the remaining cells can be chosen, would work here too?
AJ

REPLY [1 votes]: Yep, that alternative method would work fine.  But you need to keep an inclusion/exclusion effect in mind.
Suppose that the senators in the state of interest are $A$ and $B$. There are $\pmatrix{100\\50}$ ways of choosing the fifty senators.  Of these, $\pmatrix{99\\49}$  will include $A$ and $\pmatrix{99\\49}$ will include $B$.  However, $\pmatrix{98\\48}$ will include both!  The probability is therefore:
$$
\frac{2\pmatrix{99\\49}-\pmatrix{98\\48}}{\pmatrix{100\\50}}.$$

The easiest way I can think of to use a method based on complements is this:
There are $\pmatrix{100\\50}$ ways of choosing the 50 senators, and $\pmatrix{98\\ 50}$ ways to choose the 50 without including either $A$ or $B$.  The probability of excluding both $A$ and $B$ is therefore:
$$
1-\frac{\pmatrix{98\\50}}{\pmatrix{100\\50}}.$$
It's easy to see numerically that these methods agree, but you should try an analytic (algebraic) proof to verify that these methods agree for the more general case of choosing $k$ senators from a population of $n$ when excluding 2 (or more, although then the inclusion/exclusion calculations are trickier).


TITLE: For a Turing machine and input $w$- Is "Does M stop or visit the same configuration twice" a decidable question?
QUESTION [2 upvotes]: I have the following question out of an old exam that I'm solving:
Input: a Turing machine and input w
Question: Does on running of M on w, at least one of the following things happen
          -M stops of w
          -M visits the same configuration at least twice

First I thought that it's clearly in $RE$, it is a recursive enumerable question since we can simulate running of $M$ on $w$ and list the configuration or wait until it stops,
but then I thought to myself:"If it visits the same configuration more than twice, it must be in an infinite loop", because, as I understand, if it reached the same configuration it's gonna copy the same transitions, over and over again, so the problem might be in $R$, it's decidable, since it's the same question as "It stops for $w$ or it doesn't"?
What do you think? Thank you!

REPLY [4 votes]: The $A = \{\langle M, w \rangle : M \text{ halts on } w \text{ or } M \text{ visits the same configuration twice }\}$ is r.e.
The algorithm that witnesses this property would just run $M$ on $w$ until (if ever) it halts or visits a configuration twice.
$A$ is not computable. You can reduce the Halting Problem $K = \{\langle M, w \rangle : M \text{ halts on } w\}$ to $A$. Given $\langle M, w \rangle$, ask if $\langle M, w \rangle \in A$. If it is then run $M$ on $w$. Since $\langle M, w \rangle$ is in $A$, you are certain that $M$ will halt on $w$ or $M$ will reach a configuration twice. Keep running $M$ until one of the two happens.  If $M$ halts on $w$, then $\langle M, w \rangle \in K$. If $M$ on $w$ reaches a configuration twice, then it loops so $\langle M, w \rangle \notin K$. If $\langle M, w \rangle \notin A$, then clearly $\langle M, w \rangle \notin K$. Thus using $A$, you can computably know $K$. Since $K$, the halting problem, is know to be R.E. but not computable, $A$ can not be computable. If fact since $K$ is a complete c.e. set with respect to $m$-reductions, $A$ is actually $m$-equivalent to $K$, the halting problem. $m$-reduction is stronger is stronger that Turing reduction. So in a very strong sense $A$ and $K$ are the same.
Also your claim "since it's the same question as "It stops for w or it doesn't"? is not true. It is possible that you not halt but do not repeat any configuration. For example a Turing machine that just keeps writing $1$ on the tape and never halts is a Turing that never halts but never comes back to the same configuration since the tape will never be the same as it was in any previous stage.

General hint. To prove that something is recursive (decidable) or r.e., you need to produce a Turing Machine that witnesses that fact that something is decidable or r.e. However to show that something is not recursive or r.e., you need to the reduce something known to not recursive or r.e. to your set. The good set that is is r.e. but not computable is the halting problem $K$. Since $K$ is not computable, the complement of $K$, $\bar{K}$, is not r.e.
Of course this will not always work. As mentioned above $K$ is the most computationally power r.e. set. There are r.e. that are not computable but strictly weaker than $K$.

REPLY [2 votes]: We can modify a Turing machine $T$ by replacing every computation step of $T$ by a procedure in which we go to the left end of the (used portion of) the tape, and one more step left, print a special symbol $U$, and then hustle back to do the intended step. So in the modified machine, a configuration never repeats. Thus if we can solve your problem, we can solve the Halting Problem. The conclusion is that your problem is not decidable.


TITLE: I've read that abelian categories can naturally be enriched in $\mathbf{Ab}.$ How does this actually work?
QUESTION [4 upvotes]: Wikipedia defines the notion of an abelian category as follows (link).

A category is abelian iff

it has a zero object,
it has all binary products and binary coproducts, and
it has all kernels and cokernels.
all monomorphisms and epimorphisms are normal.

It later explains that an abelian category can naturally be enriched in $\mathbf{Ab},$ as a result of the first three axioms above.
How does this actually work?

REPLY [8 votes]: The claim is not trivial and requires a bit of ingenuity. For a guided solution, see Q6 here. The four steps are as follows:

Show that the category has finite limits and finite colimits.
Show that a morphism $f$ is monic (resp. epic) if and only if $\ker f = 0$ (resp. $\operatorname{coker} f = 0$).
Show that the canonical morphisms $A + B \to A \times B$ are isomorphisms, and then deduce that the category is enriched over commutative monoids.
Show that every morphism has an additive inverse (by inverting an appropriate matrix).

A complete proof is given as Theorem 1.6.4 in [Borceux, Handbook of categorical algebra, Vol. 2].


TITLE: Sliding a semicircle within a rectangle
QUESTION [1 upvotes]: Let ABCD be a rectangle with sides sizes of $1$ and $CD=AB=\sqrt{2}$. And let a semicircle with center $O$ of radius $1$ be such that its diameter is on the $AB$-produced so $CD$ will be tangent to the circle at point $Q$. $AB$ is always within the diameter of the semicircle. So semicircle cuts AC and BD at points P and R respectively. How to prove that the angle $POR$ always  is at least (or maybe exactly) $\pi/2$ ?
PS - I am trying to prove a theorem relating "lattice point covering property" and I reduced the proof to the question I asked above; so $POR$-angle to be $\ge \pi/2$ will finish the proof. Please help!

REPLY [1 votes]: As stated in TonyK's comment, the rectangle should be $ABDC$. (The result does not hold for $ABCD$.)
Note that $PR\ge\sqrt{2}$ and $\displaystyle \cos\angle POR=\frac{1^2+1^2-PR^2}{2(1)(1)}$.
We can conclude that $\cos\angle POR\le 0$ and hence $\angle POR\ge\dfrac\pi2$.