121 lines
2.6 KiB
Markdown
121 lines
2.6 KiB
Markdown
# Math4501 Lecture 3
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## Review from last lecture
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### Bisection method for finding root
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P1. Let $f$ be a continuous function on $[a,b]\to \mathbb{R}$, find $\xi \in [a,b]$ such that $f(\xi)=0$.
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P2. Let $g$ be a continuous function on $[a,b]\to \mathbb{R}$, find $\xi \in [a,b]$ such that $g(\xi)=\xi$.
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#### Theorem 1:
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solution to P1 exists if $f(a)f(b)<0$.
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#### Theorem 2:
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Fixed point theorem:
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If solution to P2 exists, then $g:[a,b]\to [a,b]$.
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#### Bijection method
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Obtain two sequence $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$.
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Initially, we set $a_0=a$ and $b_0=b$.
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If $|a_n-b_n|<2^{-n}|a_0-b_0|$, then $a_n$ and $b_n$ are Cauchy sequence. So their limit exists.
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$\lim_{n\to\infty} a_n=\lim_{n\to\infty} b_n=\xi$.
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### Simple iteration method
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#### Definition of Simple Iteration
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Given $x_0\in [a,b]$, we define a sequence $(x_n)_{n=0}^\infty$ by
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$$
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x_{n+1}=g(x_n)
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$$
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If a simple iteration converges, the it converges to a fixed point of
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<details>
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<summary>Proof</summary>
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Let $c\coloneqq \lim_{n\to\infty} x_n=g(c)$.
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$g:[a,b]\to \mathbb{R}$ is continuous if and only if $g$ is continuous at $x_0\in [a,b]$.
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</details>
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#### Definition of Lipschitz continuous
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A function $g:[a,b]\to \mathbb{R}$ is Lipschitz continuous if for all $x,y\in [a,b]$, there exists a constant $L>0$ such that
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$$
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|g(x)-g(y)|\leq L|x-y|
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$$
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for some $L>0$.
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> [!NOTE]
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>
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> Lipschitz continuous is a stronger condition than continuous.
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>
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> If a function is Lipschitz continuous, then it is continuous.
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>
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> However, the converse is not true.
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#### Definition of contraction mapping
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A function $g:[a,b]\to \mathbb{R}$ is a contraction mapping if it is Lipschitz continuous with $L<1$.
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#### Theorem of simple iteration
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Let $g:[a,b]\to [a,b]$ is a contraction mapping (Lipschitz continuous with $L<1$, with pointwise ontinuous $g$).
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Then
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- $g$ has a unique fixed point $\xi \in [a,b]$
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- Simple iteration $x_{n+1}=g(x_n)$ converges to $\xi$ for any $x_0\in [a,b]$.
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<details>
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<summary>Proof</summary>
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**Uniqueness**:
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Suppose $\xi_1$ and $\xi_2$ are two fixed points of $g$.
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Then $|x_1-x_2|=|g(\xi_1)-g(\xi_2)|\leq L|\xi_1-\xi_2|$.
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Thus, $(1-L)|\xi_1-\xi_2|\leq 0$, which implies $|\xi_1-\xi_2|=0$.
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A more general result:
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Brouwer's fixed point theorem
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**Convergence**:
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Let $\xi\in [a,b]$ be the unique fixed point of $g$.
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Then,
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$$
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\begin{aligned}
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|x_n-\xi|&=|g(x_{n-1})-g(\xi)|\\
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&\leq L|x_{n-1}-\xi|\\
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&=L|g(x_{n-2})-g(\xi)|\\
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&\leq L^2|x_{n-2}-\xi|\\
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&\vdots\\
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&\leq L^n|x_0-\xi|
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$$
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Thus, we can always find $N$ such that $L^N|x_0-\xi|<\epsilon$ for any $\epsilon>0$. Choose $N=\log(\frac{\epsilon}{|x_0-\xi|})/\log(L)$.
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Therefore, $x_n$ converges to $\xi$.
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</details>
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