188 lines
6.8 KiB
TeX
188 lines
6.8 KiB
TeX
\documentclass[../../script.tex]{subfiles}
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%! TEX root = ../../script.tex
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\begin{document}
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\section{Sequences, Series and Limits}
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\begin{defi}[Sequences and Convergence]
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Let $\metric$ be a metric space. A sequence is a mapping $\natn \rightarrow X$. We write $\seq{x}_{n \in \natn}$ or $\seq{x}$.
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The sequence $\seq{x}$ is said to be convergent to $x \in X$ if
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\[
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\forall \epsilon > 0 ~\exists N \in \natn ~\forall n \ge N: ~~d(x_n, x) < \epsilon
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\]
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$x$ is said to be the limit, and sequences that aren't convergent are called divergent.
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\end{defi}
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\begin{rem}
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On $\realn$ the metric is the Euclidian metric $|\cdot|$, therefore this new definition of convergence is merely a generalization of the old one.
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\end{rem}
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\begin{thm}
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Let $\seq{x}$ be a sequence in the metric space $\metric$ and $x \in X$. Then the following statements are equivalent:
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\begin{enumerate}[(i)]
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\item $\seq{x}$ converges to $x$
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\item $\forall \epsilon > 0$ $\oball(x)$ contains all but finitely many elements of the sequence (almost every (a.e.) element)
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\item $(d(x, x_n))$ is a null sequence
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\end{enumerate}
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\end{thm}
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\begin{proof}
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(ii) is merely a reformulation of (i), and $(ii) \iff (iii)$ follows from
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\begin{equation}
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d(x_n, x) = |d(x_n, x) - 0|
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\end{equation}
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\end{proof}
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\begin{thm}
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Let $\left(x^{(n)}\right) = (x_1^{(n)}, x_2^{(n)}, \cdots, x_d^{(n)}) \subset \realn^d$ and
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\[
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x = (x_1, \cdots, x_d) \in \realn^d
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\]
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$\left(x^{(n)}\right)$ is said to converge to $x$ if and only if $x_i^{(n)}$ converges to $x_i$ for all $i$ in $\set{1, \cdots, d}$
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\end{thm}
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\begin{proof}
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For $y = (y_1, \cdots, y_d) \in \realn^d$ we have
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\begin{equation}
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\norm{y_i} < \norm{y} ~~\forall i \in \set{1, \cdots, d}
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\end{equation}
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If $\left(x^{(n)}\right)$ converges to $x$, then
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\begin{equation}
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\abs{x_i^{(n)} - x_i} \le \norm{x^{(n)} - x} \conv{} 0
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\end{equation}
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If $(x_i^{(n)})$ converges to $x_i ~~\forall i \in \set{1, \cdots d}$, then
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\begin{equation}
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\forall \epsilon > 0 ~\exists N \in \natn ~\forall n > N: ~~\abs{x_i^{(n)} - x_i} < \frac{\epsilon}{\sqrt{d}} ~~\forall i \in \set{1, \cdots d}
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\end{equation}
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Thus
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\begin{equation}
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\begin{split}
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\norm{x^{(n)} - x} &= \sqrt{(x_1^{(n)} - x_1)^2 + (x_2^{(n)} - x_2)^2 + \cdots + (x_d^{(n)} - x_d)^2} \\
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&\le \sqrt{\frac{\epsilon^2}{d} + \frac{\epsilon^2}{d} + \cdots + \frac{\epsilon}{2}} \\
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&= \epsilon
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\end{split}
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\end{equation}
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So $\left(x^{(n)}\right)$ converges to $x$.
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\end{proof}
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\begin{thm}
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Every convergent sequence has exactly one limit and is bounded.
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\end{thm}
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\begin{proof}
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Assume that $x, y$ are limits of $(x_n)$ with $x \ne y$. Then $d(x, y) > 0$. There exists $N_1, N_2 \in \natn$, such that
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\begin{subequations}
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\begin{align}
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d(x_n, x) &< \frac{d(x, y)}{2} ~~\forall n \ge N_1 \\
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d(x_n, x) &< \frac{d(x, y)}{2} ~~\forall n \ge N_2
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\end{align}
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\end{subequations}
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From this follows that
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\begin{equation}
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d(x, y) \le d(x, x_n) + d(x_n, y) < d(x, y) ~~\forall \max\set{N_1, N_2}
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\end{equation}
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which is a contradiction, thus sequences can have only one limit.
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Now if $\seq{x}$ converges to $x$, then
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\begin{equation}
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\exists N \in \natn ~\forall n \ge N: ~~d(x_n, x) < 1
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\end{equation}
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Then
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\begin{equation}
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d(x_n, x) \le \max\set{d(x_1, x), d(x_2, x), \cdots, d(x_{N-1}, x), 1}
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\end{equation}
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\end{proof}
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\begin{thm}
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Let $\normed$ be a normed space over $\field$. Let $\seq{x}, \seq{y} \subset V$ be sequences with limits $x, y \in V$
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and $\anyseqdef[\lambda]{\field}$ a sequence with limit $\lambda \in \field$. Then
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\begin{align*}
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x_n + y_n \longrightarrow x + y && \lambda_n x_n \longrightarrow \lambda x
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\end{align*}
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\end{thm}
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\begin{proof}
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\reader
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\end{proof}
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\begin{defi}[Cauchy sequences and completeness]
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A sequence $\seq{x}$ in a metric space $\metric$ is called Cauchy sequence if
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\[
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\forall \epsilon > 0 ~\exists N \in \natn: ~~d(x_n, x_m) < \epsilon ~~\forall m, n \ge N
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\]
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A metric space is complete if every Cauchy sequence converges. A complete normed space is called Banach space.
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\end{defi}
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\begin{eg}
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\item $(\realn, \abs{\cdot})$ and $(\cmpln, \abs{\cdot})$ are complete
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\item $(\ratn, \abs{\cdot})$ is not complete
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\end{eg}
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\begin{thm}
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Every convering series is a Cauchy sequence
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\end{thm}
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\begin{proof}
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Let $\seq{x} \conv{} x$. This means that
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\begin{equation}
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\forall \epsilon > 0 ~\epsilon N \in \natn: ~~d(x_n, x) < \frac{\epsilon}{2} ~~\forall n \ge N
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\end{equation}
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Then
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\begin{equation}
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d(x_n, x_m) \le d(x_n, x) + d(x, x_m) < \epsilon ~~\forall m, n \ge N
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\end{equation}
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\end{proof}
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\begin{thm}
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$\realn^n$ with the Euclidian norm is complete.
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\end{thm}
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\begin{proof}
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Let $\left(x^{(n)}\right) \subset \realn^n$ be a Cauchy sequence. We know that
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\begin{equation}
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\forall y \in \realn^n: ~~\abs{y_i} \le \norm{y} ~~\forall i \in \set{1, \cdots, n}
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\end{equation}
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We also know that $(x_i^{(n)})$ are Cauchy sequences because
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\begin{equation}
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\abs{(x_i^{(n)} - x_i^{m})} \le \norm{x^{(n)} - x^{(m)}} ~~\forall i \in \set{1, \dots, n}
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\end{equation}
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Thus $x_i^{(n)} \conv{} x_i$ and therefore $\left(x^{(n)}\right) \conv{} x$.
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\end{proof}
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\begin{defi}[Series and (absolute) convergence]
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Let $\normed$ be a normed space and $\anyseqdef{V}$. The series
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\[
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\series{k} x_k
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\]
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is the sequence of partial sums
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\[
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s_n = \series[n]{k} x_k
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\]
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If the series converges then $\series{k} x_k$ also denotes the limit.
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The series is said to absolutely convergent if
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\[
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\series{k} \norm{x_k} < \infty
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\]
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\end{defi}
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\begin{thm}
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In Banach spaces every absolutely convergent series is convergent.
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\end{thm}
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\begin{proof}
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Let $\normed$, $\anyseqdef{V}$ and require $\series{n} \normed{x_n} < \infty$. We need to show that $s_n = \series[n]{k} x_k$ is a Cauchy sequence.
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Let $\epsilon > 0$ and $t_n = \series[n]{k} \norm{x_k}$. $\seq{t}$ is convergent in $\realn$, and thus a Cauchy sequence.
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I.e.
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\begin{equation}
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\exists N \in \natn: ~~|t_n - t| < \epsilon ~~\forall m, n \ge N
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\end{equation}
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For $n > m > N$:
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\begin{equation}
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\norm{s_n - s_m} = \norm{\sum_{k=m+1}^n x_k} \le \sum_{k=m+1}^n \norm{x_k} = t_n - t_m = |t_n - t_m| < \epsilon
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\end{equation}
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\end{proof}
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\begin{thm}
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Let $\normed$ be a Banach space, $\series{k} x_k$ absolutely convergent and let $\sigma: \natn \rightarrow \natn$ be a bijective mapping. Then
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\[
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\series{k} x_k = \series{k} x_{\sigma(k)}
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\]
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\end{thm}
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\begin{proof}
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Analogous to $\Cref{259}$
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\end{proof}
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\end{document} |