diff --git a/pages/CSE442T/CSE442T_L24.md b/pages/CSE442T/CSE442T_L24.md index 792ca34..054f20f 100644 --- a/pages/CSE442T/CSE442T_L24.md +++ b/pages/CSE442T/CSE442T_L24.md @@ -169,7 +169,7 @@ One-way function is a class of functions that are easy to compute but hard to in ##### Weak one-way function -A weak one-way function is +A weak one-way function is $$ f:\{0,1\}^n\to \{0,1\}^* diff --git a/pages/Math4111/Math4111_L25.md b/pages/Math4111/Math4111_L25.md index 391d6ac..9adf247 100644 --- a/pages/Math4111/Math4111_L25.md +++ b/pages/Math4111/Math4111_L25.md @@ -1 +1,164 @@ -# Lecture 25 \ No newline at end of file +# Lecture 25 + +## Review + +Are the following statements true or false? You do not need to give a rigorous proof. + +1. $\exists N\in \mathbb{N}$, $\forall n\geq N$, $\left(\frac{1}{2}\right)^n < 0.001$ + - True, let $N = 10$ +2. $\forall \epsilon > 0$, $\exists N\in \mathbb{N}$, $\forall n\geq N$, $\left(\frac{1}{2}\right)^n < \epsilon$ + - True, let $N = \lceil \log_{\frac{1}{2}} \epsilon \rceil$ +3. $\forall x\in [0, 1)$, $\forall \epsilon > 0$, $\exists N\in \mathbb{N}$, $\forall n\geq N$, $x^n < \epsilon$ + - True, let $N = \lceil \log_x \epsilon \rceil$ +4. $\forall \epsilon > 0$, $\exists N\in \mathbb{N}$, $\forall n\geq N$, $\forall x\in [0, 1)$, $x^n < \epsilon$ + - False, $x$ can be arbitrarily close to 1 +5. $\forall \epsilon > 0$, $\forall x\in \left[0, \frac{1}{2}\right]$, $\exists N\in \mathbb{N}$, $\forall n\geq N$, $x^n < \epsilon$ + - True, let $N = \lceil \log_{\frac{1}{2}} \epsilon \rceil$ + +## New Materials + +### Sequences and series of functions + +#### Definition 7.1 + +Let $(f_n)$ be a sequence of functions $E\to \mathbb{R}$. Let $f:E\to \mathbb{R}$ be a function. We say $(f_n)$ converges pointwise to $f$ if $\forall x\in E$, $\lim_{n\to \infty} f_n(x) = f(x)$. + +i.e. $\forall x\in E$, $\forall \epsilon > 0$, $\exists N\in \mathbb{N}$, $\forall n\geq N$, $|f_n(x) - f(x)| < \epsilon$. + +Example: + +The sequence from the warm up exercise converges pointwise to $f(x) = \begin{cases} 0 & x\in [0, 1) \\ 1 \text{ otherwise} \end{cases}$. + +To check if a series of functions converges pointwise, we can take the limit of the series as $n\to \infty$. + +Example: + +Let $g_n(x):\mathbb{R}\to \mathbb{R}$ be defined by $g_n(x) = \sum_{k=0}^{n} \frac{x^2}{(1+x^2)^n}$. + +$$ +g_n(x) = \sum_{k=0}^{n} \left(\frac{x^2}{(1+x^2)^n}\right) +$$ + +And this is a geometric series with first term $\frac{x^2}{1+x^2}$ and common ratio $\frac{x^2}{1+x^2}$. + +Then $\left(g_n\right)$ converges pointwise to $g(x) =\begin{cases} 0 & x = 0 \\ 1+x^2 & \text{otherwise} \end{cases}$. + +This example shows that pointwise convergence does not preserve continuity. But if $(f_n)$ converges uniformly to $f$, then $f$ is continuous. + +#### Definition 7.7 + +Let $(f_n)$ be a sequence of functions $E\to \mathbb{R}$. We say $(f_n)$ converges uniformly to $f$ if $\forall \epsilon > 0$, $\exists N\in \mathbb{N}$, $\forall n\geq N$, $\forall x\in E$, $|f_n(x) - f(x)| < \epsilon$. + +i.e. $\forall \epsilon > 0$, $\exists N\in \mathbb{N}$, such that $\forall n\geq N$, $\forall x\in E$, $|f_n(x) - f(x)| < \epsilon$. + +> $f_n$ must always be within $[f-\epsilon, f+\epsilon]$ for all $n\geq N$. + +Example: + +Let $f_n(x) = \frac{1}{n}\sin(n^2 x)$. + +Ideas of proof: + +Given $\epsilon > 0$, let $N = \lceil \frac{1}{\epsilon} \rceil$. (This choice of $N$ does not depend on $x$.) + +Then $\forall n\geq N$, $\forall x\in \mathbb{R}$, $|\frac{1}{n}\sin(n^2 x)| < \epsilon$. + +Example: + +Let $f_n:[0, 1]\to \mathbb{R}$, $f_n(x) = x^n$. + +Then $(f_n)$ does not converge uniformly to $f(x) = \begin{cases} 0 & x<1 \\ 1 & x = 0 \end{cases}$. + +$f_n$ does not lie in the region $[0, 1]$ for all $n$. + +#### Theorem 7.12 (Corollary of Theorem 7.11) (Uniform limit theorem) + +Let $(f_n)$ be a sequence of continuous functions $E\to \mathbb{R}$. If $(f_n)$ converges uniformly to $f$ on $E$, then $f$ is continuous. + +Proof: + +Suppose $f_n\to f$ uniformly and $\forall n$, $f_n$ is continuous. + +Let $p\in E$ and $\epsilon > 0$. + +Since $f_n\to f$ uniformly, $\exists N\in \mathbb{N}$, $\forall n\geq N$, $\forall x\in E, |f_n(x) - f(x)| < \epsilon$. + +Since $f_N$ is continuous at $p$, $\exists \delta > 0$, such that $\forall x\in E$, if $|f_N(x) - f_N(p)| < \epsilon$. + +Suppose $x\in B_\delta(p)$. Then + +$$ +\begin{aligned} +|f(x) - f(p)| &= |f(x) - f_N(x) + f_N(x) - f_N(p) + f_N(p) - f(p)| \\ +&\leq |f(x) - f_N(x)| + |f_N(x) - f_N(p)| + |f_N(p) - f(p)| \\ +&< 3\epsilon +\end{aligned} +$$ + +_The $\epsilon$ bound would not hold if we only had pointwise convergence._ + +$|f_N(x) - f_N(p)| < 3\epsilon$. + +EOP + +> Recall: If $(s_n)$ is a sequence in $\mathbb{R}$, then $(s_n)$ converges to $s$ if and only if it is Cauchy. +> i.e. $\forall \epsilon > 0$, $\exists N\in \mathbb{N}$, $\forall n, m\geq N$, $|s_n - s_m| < \epsilon$. + +#### Theorem 7.9 (Cauchy criterion for uniform convergence) + +Let $(f_n)$ be a sequence of functions $E\to \mathbb{R}$. $(f_n)$ converges uniformly to $f$ on $E$ if and only if $(f_n)$ is uniformly Cauchy. + +i.e. $\forall \epsilon > 0$, $\exists N\in \mathbb{N}$, $\forall n, m\geq N$, $\forall x\in E$, $|f_n(x) - f_m(x)| < \epsilon$. + +Proof: + +Exercise. + +#### Theorem 7.10 (Weierstrass M-test) + +Let $(f_n)$ be a sequence of functions $E\to \mathbb{R}$ (or $\mathbb{C}$). Suppose + +- $\forall n$, $\exists M_n\in \mathbb{R}_{\geq 0}$, such that $\forall x\in E$, $|f_n(x)| \leq M_n$. +- $\sum_{n=1}^{\infty} M_n$ converges. + +Then $\sum_{n=1}^{\infty} f_n(x)$ converges uniformly on $E$. + +i.e. The sequence of partial sums $s_n(x) = \sum_{k=1}^{n} f_k(x)$ converges uniformly on $E$. + +Remark: + +The proof is nearly identical to the proof of the comparison test in Chapter 3. + +Proof: + +By Theorem 7.8, it's enough to show that $(s_n)$ is uniformly Cauchy. + +i.e. $\forall \epsilon > 0$, $\exists N\in \mathbb{N}$, $\forall m\geq n\geq N$, $\forall x\in E$, $|\sum_{k=n}^{m} f_k(x)| < \epsilon$. + +Let $\epsilon > 0$. Since $\sum_{n=1}^{\infty} M_n$ converges, $\exists N\in \mathbb{N}$, $\forall m\geq n\geq N$, $\sum_{k=n}^{m} M_k < \epsilon$. + +Suppose $m\geq n\geq N$ and $x\in E$. + +$$ +\begin{aligned} +|\sum_{k=n}^{m} f_k(x)| &\leq \sum_{k=n}^{m} |f_k(x)| \\ +&\leq \sum_{k=n}^{m} M_k \\ +&< \epsilon +\end{aligned} +$$ + +EOP + +Example: + +Let $f_n(x) = \sum_{n=0}^{\infty} \frac{1}{2^n} \cos(3^n x)$. + +$$ +\left|\frac{1}{2^n} \cos(3^n x)\right| \leq \frac{1}{2^n}=M_n +$$ + +By Weierstrass M-test, $f_n(x)$ converges uniformly on $\mathbb{R}$. + +By **Theorem 7.12**, $f(x) = \sum_{n=0}^{\infty} \frac{1}{2^n} \cos(3^n x)$ is continuous on $\mathbb{R}$. + +Fun fact: $f(x)$ is not differentiable at any $x\in \mathbb{R}$. \ No newline at end of file