140 lines
3.1 KiB
Markdown
140 lines
3.1 KiB
Markdown
# Math4121 Lecture 4
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## Chapter 5. Differentiation
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### The continuity of the derivative
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#### Theorem 5.12
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Suppose $f$ is differentiable on $[a,b]$, Then $f'$ attains intermediate values between $f'(a)$ and $f'(b)$.
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Proof:
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Let $\lambda\in (f'(a),f'(b))$. We need to show that there exists $x\in (a,b)$ such that $f'(x)=\lambda$.
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Let $g(x)=f(x)-\lambda x$. Then $g$ is differentiable on $(a,b)$ and
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$$
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g'(x)=f'(x)-\lambda.
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$$
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So $g'(a)=f'(a)-\lambda<0$ and $g'(b)=f'(b)-\lambda>0$.
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We need to show that $g'(x)=0$ for some $x\in (a,b)$.
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Since $g'(a)<0$, $\exists t_1\in (a,b)$ such that $g'(t_1)<g(a)$.
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If not, then $g(t)\geq g(a)$ for all $t\in (a,b)$. But then $g'(a)\gets \frac{g(t)-g(a)}{t-a}\geq 0$, which contradicts $g'(a)<0$.
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With the loss of generality, since $g'(b)>0$, $\exists t_2\in (a,b)$ such that $g'(t_2)<g(b)$.
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Hence, $g$ attains its infimum on $[a,b]$ at some $x\in (a,b)$. Then this $x$ is a local minimum of $g$ on $(a,b)$.
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So $g'(x)=0$ and $f'(x)=\lambda$.
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QED
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### L'Hôpital's Rule
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#### Theorem 5.13
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Suppose $f$ and $g$ are differentiable on $(a,b)$ and $g'(x)\neq 0$ for all $x\in (a,b)$, where $-\infty\leq a<b\leq \infty$. Suppose
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$$
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\frac{f'(x)}{g'(x)}\to A \text{ as } x\to a\dots
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$$
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If
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$$
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f(x)\to 0 \text{ and } g(x)\to 0 \text{ as } x\to a,
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$$
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or
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$$
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g(x)\to \infty \text{ as } x\to a,
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$$
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then
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$$
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\frac{f(x)}{g(x)}\to A \text{ as } x\to a.
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$$
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Note that all these numbers $A$ can be $\infty$ or $-\infty$ (on extended real line).
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We're using the open neighborhood definition of $\to$ here. An open neighborhood of $\infty$ is an interval of the form $(c,\infty)$ for some $c\in \mathbb{R}$.
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> Recall the [Definition 3.1](https://notenextra.trance-0.com/Math4111/Math4111_L13#definition-31).
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Proof:
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Main step:
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Suppose $-\infty\leq A\leq \infty$, and let $q>A$ with neighborhood $(-,\infty,q)$. Then $\exists c\in \mathbb{R}$ such that $\frac{f(x)}{g(x)}<q,\forall x\in (a,c)$.
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Proof of the main step:
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Fix $A<r<q$. Then $\exists c\in (a,b)$ such that $\frac{f'(x)}{g'(x)}<r,\forall x\in (a,c)$.
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Now, for any $a<x<y<c$, by generalized mean value theorem, $\exists t\in (x,y)$ such that
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$$
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\frac{f(x)-f(y)}{g(x)-g(y)}=\frac{f'(t)}{g'(t)}
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$$
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Since $t\in (a,c)$, $\frac{f'(t)}{g'(t)}<r$.
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Case 1: $f(x)\to 0$ and $g(x)\to 0$ as $x\to a$.
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As $x\to a$, $f(x)\to 0$ and $g(x)\to 0$. So
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$$
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\begin{aligned}
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\lim_{x\to a}\frac{f(x)-f(y)}{g(x)-g(y)}&=\lim_{x\to a}\frac{0-f(y)}{0-g(y)}\\
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&=\lim_{x\to a}\frac{f(y)}{g(y)}\\
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&=\frac{f'(y)}{g'(y)}\\
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&\leq r<q
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\end{aligned}
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$$
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$\forall y\in (a,c)$, $\frac{f(y)}{g(y)}<q$.
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Case 2: $g(x)\to \infty$ as $x\to a$.
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We can find $c_1\in (a,y)$ such that $g(x)>g(y)$ for all $x\in (a,c_1)$.
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Therefore,
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$$
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\begin{aligned}
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\frac{f(x)-f(y)}{g(x)}&<\frac{r[g(x)-g(y)]}{g(x)}\\
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\frac{f(x)}{g(x)}&<r-\frac{rg(y)}{g(x)}+\frac{f(y)}{g(x)}
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\end{aligned}
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$$
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To make the right side less than $q$, we need
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$$
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\frac{|rg(y)|+|f(y)|}{|g(x)|}<q-r
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$$
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so,
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$$
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|g(x)|>\frac{|rg(y)|+|f(y)|}{q-r}
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$$
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There exists $c_2\in (a,c_1)$ such that $|g(x)|>\frac{|rg(y)|+|f(y)|}{q-r},\forall x\in (a,c_2)$.
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So $\forall x\in (a,c_2)$,
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$$
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\frac{f(x)}{g(x)}<\frac{rg(y)+f(y)}{g(x)}<r+(q-r)=q
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$$
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$\forall x\in (a,c_2)$, $\frac{f(x)}{g(x)}<q$.
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QED
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