diff --git a/pages/Math4111/Math4111_L22.md b/pages/Math4111/Math4111_L22.md index e0a3957..5986481 100644 --- a/pages/Math4111/Math4111_L22.md +++ b/pages/Math4111/Math4111_L22.md @@ -1 +1,175 @@ -# Lecture 22 \ No newline at end of file +# Lecture 22 + +## Review + +Let $(a_n)$ be a sequence, and let $b_n = a_{f(n)}$ be a rearrangement, where $f$ is given by the following: + +| $n$ | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | $\dots$ | +| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | +| $f(n)$ | 1 | 2 | 4 | 3 | 6 | 8 | 5 | 10 | 12 | $\dots$ | + +(The pattern is "odd,even,even,") Defined the partial sums $s_n = \sum_{k=1}^n a_k$ and $t_n = \sum_{k=1}^n b_k$. + +1. In terms of $a_1,a_2,\ldots$, determine $s_n-t_n$ for $n=1,2,3,4,5,6,7$. (For example, $s_3-t_3 = a_3-a_4$.) + $s_1 - t_1 = a_1 - a_1 = 0$ + $s_2 - t_2 = a_2 - a_2 = 0$ + $s_3 - t_3 = a_3 - a_4$ + $s_4 - t_4 = a_4 - a_4 = 0$ + $s_5 - t_5 = a_5 - a_6$ + $s_6 - t_6 = a_6 - a_8$ + $s_7 - t_7 = a_7 - a_8$ + +2. What is the smallest $n$ so that $s_n - t_n$ does not contain any of the terms $a_1,\dots, a_5$? + $n=7$ +3. What is the smallest $n$ so that $s_n - t_n$ does not contain any of the terms $a_1,\dots, a_{10}$? + $n=13$ + +## New Material + +### Absolute Convergence + +#### Theorem 3.55 + +Let $(a_n)$ be a sequence in $\mathbb{C}$ such that $\sum_{n=1}^\infty |a_n|$ converges. If $(b_n)$ is a rearrangement of $(a_n)$, then $\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty b_n$. + +Proof: + +Let $f:\mathbb{N}\to\mathbb{N}$ be a bijection and let $b_n = a_{f(n)}$. + +Let $I(n)=\{1,2,\dots,n\}$, $J(n)=\{f(1),f(2),\dots,f(n)\}$. + +Then $s_n = \sum_{k\in I(n)} a_k$ and $t_n = \sum_{k\in J(n)} a_k$. + +$$ +\begin{aligned} +s_n - t_n &= \sum_{k=1}^n a_k - \sum_{k=1}^n b_k \\ +&= \sum_{k\in I(n)} a_k - \sum_{k\in J(n)} a_{f(k)} \\ +&= \sum_{k\in I(n)\backslash J(n)} a_k - \sum_{k\in J(n)\backslash I(n)} a_{f(k)} \\ +|s_n - t_n|&\leq \sum_{i\in (I(n)\backslash J(n))\cup(J(n)\backslash I(n))} |a_i| +\end{aligned} +$$ + +We will show that $\lim_{n\to\infty} |s_n - t_n| = 0$. + +Let $\epsilon > 0$. + +By the Cauchy criterion applied to $\sum_{n=1}^\infty |a_n|$, there exists $N$ such that if $m,n\geq N$, then $\sum_{k=m+1}^n |a_k| < \epsilon$. + +Then we choose $p\in\mathbb{N}$ such that $I(n)\subset J(p)$ ($\{1,2,\dots,N\}\subset\{f(1),f(2),\dots,f(p)\}$). $p=\max\{f^{-1}(1),f^{-1}(2),\dots,f^{-1}(N)\}$. + +Note: This implies that $p$ is at least $N$. + +If $n\geq p$, then $I(n)\subset J(p)\subset I(n)\cap J(n)$ so $s_n = t_n$. + +So, + +$$ +|s_n - t_n| \leq \sum_{i=N+1}^{\max J(n)} |a_i| < \epsilon +$$ + +> Back to the example of the review question. +> +> $I(9)=\{1,2,\dots,9\}$, $J(9)=\{1,2,4,3,6,8,5,10,12\}$, $I(9)\backslash J(9)=\{7,9\}$, $J(9)\backslash I(9)=\{10,12\}$. +> +>$$ +|s_9 - t_9| \leq |a_7|+|a_9|+|a_{10}|+|a_{12}| \leq \sum_{k=7}^{12} |a_k| < \epsilon +$$ + +This proves that $\lim_{n\to\infty} |s_n - t_n| = 0$. + +Since $\lim_{n\to\infty} s_n$ exists, $\lim_{n\to\infty} s_n = \lim_{n\to\infty} t_n$. + +EOP + +#### Theorem 3.54 + +Special case of Riemann rearrangement theorem + +Suppose $a_n\in \mathbb{R}$, $\sum_{n=1}^\infty a_n$ converges conditionally. (i.e. $\sum_{n=1}^\infty a_n$ converges but $\sum_{n=1}^\infty |a_n|$ diverges.) Then $\forall \alpha\in\mathbb{R}$, there exists a rearrangement $(b_n)$ of $(a_n)$ such that $\sum_{n=1}^\infty b_n = \alpha$. + +## Chapter 4 Continuity + +### Limits of Functions + +#### Definition 4.1 + +Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces and $E\subset X$, $p\in E'$, $q\in Y$. Let $f:E\to Y$. We say that $\lim_{x\to p} f(x) = q$ if $\forall \epsilon > 0$, $\exists \delta > 0$ such that $\forall x\in E$, if $0 < d_X(x,p) < \delta$, then $d_Y(f(x),q) < \epsilon$. + +This is same as: + +If $\lim_{x\to p} f(x)=q$, then + +$$ +\forall \epsilon > 0, \exists \delta > 0, \forall x\in E, 0 < d_X(x,p) < \delta \implies d_Y(f(x),q) < \epsilon +$$ + +In many definitions, $E=X$ + +#### Theorem 4.2 + +$\lim_{x\to p} f(x) = q \iff$ forall sequence $(p_n)$ in $E\backslash\{p\}$ with $p_n\to p$, $f(p_n)\to q$. + +Proof: + +$\implies$ + +Suppose $\lim_{x\to p} f(x) = q$. + +Let $(p_n)$ be a sequence in $E\backslash\{p\}$ with $p_n\to p$. + +Let $\epsilon > 0$. + +By definition of limit of function, $\exists \delta > 0$ such that if $0 < d_X(x,p) < \delta$, then $d_Y(f(x),q) < \epsilon$. + +Since $p_n\to p$, $\exists N$ such that if $n\geq N$, then $0 < d_X(p_n,p) < \delta$. So $f(p_n)\in B_\epsilon(q)$. + +Thus, we showed $\forall \epsilon > 0$, $\exists N$ such that if $n\geq N$, then $f(p_n)\in B_\epsilon(q)$. + +$\impliedby$ + +We proceed by contrapositive. + +Suppose $\lim_{x\to p} f(x) \neq q$, then $\exists$ sequence $(p_n)$ in $E\backslash\{p\}$ with $p_n\to p$ such that $f(p_n)\cancel{\to} q$. + +Suppose $\lim_{n\to\infty} f(p_n) \neq q$, then $\exists \epsilon > 0$ such that for all $\delta > 0$, there exists $x\in E\cap B_\delta(p)\backslash\{p\}$ such that $f(x)\notin B_\epsilon(q)$. + +For $n\in\mathbb{N}$, if we apply this with $\delta = \frac{1}{n}$, we get $p_n\in E\cap B_{\frac{1}{n}}(p)\backslash\{p\}$ such that $f(p_n)\notin B_\epsilon(q)$. + +Then: $(p_n)$ is a sequence in $E\backslash\{p\}$ with $d_X(p_n,p) = \frac{1}{n}\to 0$ so that as $n\to\infty$, $f(p_n)\notin B_\epsilon(q)$. + +So $\lim_{n\to\infty} f(p_n) \neq q$. + +EOP + +With this theorem, we can use the properties of limit of sequences to study limits of functions. + +To prove things about limits of functions, we can use sequences. + +- If $\lim_{x\to p} f(x) = q$, and $\lim_{x\to p} f(x)=q'$, then $q=q'$. +- If $\lim_{x\to p} f(x) = A$ and $\lim_{x\to p} g(x) = B$, then $\lim_{x\to p} f(x) + g(x) = A+B$. + +### Continuous functions + +#### Definition 4.5 + +Let $(X,d_X)$ and $(Y,d_Y)$ be metric spaces and $E\subset X$, $p\in E$. Let $f:E\to Y$. We say that $f$ is continuous at $p$ if $\forall \epsilon > 0$, $\exists \delta > 0$ such that $f(E\cap B_\delta(p))\subset B_\epsilon(f(p))$. + +- We say that $f$ is continuous on $E$ if $f$ is continuous at every $p\in E$. + +> Remark: For $p\in E$, there are two possibilities. +> +> - $p$ is an isolated point of $E$. +> - $p$ is a limit point of $E$. +> +> In the first case, $f$ is automatically continuous at $p$. ($E\cap B_\delta(p)=\{p\}$ for all $\delta > 0$.) +> +> In the second case, $f$ is continuous at $p$ if and only if $\lim_{x\to p} f(x) = f(p)$. + +#### Theorem 4.8 + +A function $f:E\to Y$ is continuous at $p\in E$ if the pre-image of every open set is open. + +Two consequences if $f:E\to Y$ is continuous: + +- Image of compact set is compact. (Implies Extreme Value Theorem) +- Image of connected set is connected. (Implies Intermediate Value Theorem) diff --git a/pages/Math4111/_meta.js b/pages/Math4111/_meta.js index b088f16..7dfe1d6 100644 --- a/pages/Math4111/_meta.js +++ b/pages/Math4111/_meta.js @@ -21,9 +21,7 @@ export default { Math4111_L19: "Lecture 19", Math4111_L20: "Lecture 20", Math4111_L21: "Lecture 21", - Math4111_L22: { - display: 'hidden' - }, + Math4111_L22: "Lecture 22", Math4111_L23: { display: 'hidden' },