Honors Calculus Notes for [3 Differentiation]

Definition

Def. 3.1 Derivative

Let $f:D\rightarrow \mathbb R$ be a function, and $a\in D$. We denote:

\[f^{\prime}(a)=\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h} \]

and it is called the derivative of $f$ at $a$.

Notation

\[f^{\prime}(x)=\frac{d}{dx}f(x)=\frac{df}{dx} \]

Def. 3.2 Differentiable Functions

We say $f$ is differentiable at $a$ if the following condition holds:

$\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}$
exists.

Furthermore, if $f$ is differentiable at every point $a$ in an interval $I$, then we say that $f$ is differentiable on $I$. If $f$ is differentiable on $\mathbb R$, we may also say $f$ is differentiable everywhere.

Def. 3.3 One-Sided Derivatives

Given a function $f:\mathbb R\rightarrow\mathbb R$, we define:

\[f^{\prime}_{+}(a)=\lim_{h\rightarrow 0^{+}}\frac{f(a+h)-f(a)}{h} \]

\[f^{\prime}_{-}(a)=\lim_{h\rightarrow 0^{-}}\frac{f(a+h)-f(a)}{h} \]

which is called the right/left-hand derivatives respectively.

$f$ is differentiable at $a$ if and only if both $f^{\prime}_{+}(a)$ and $f^{\prime}_{-}(a)$ exist and they are equal.

Def. 3.4 Little-o, Big-O

Given two functions $f(x)$ and $g(x)$, we say $f(x)\in o(g(x))$ as $x\rightarrow a$ if $\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=0$.

Given two functions $f(x)$ and $g(x)$, we say $f(x)\in O(g(x))$ as $x\rightarrow a$ if there exist constants $C>0$ and $\delta>0$ such that if $0<|x-a|<\delta$, then $|f(x)|\leq C|g(x)|$.

Def. 3.5 Monotonicity

Let $f:[a, b]\rightarrow\mathbb R$ be a differentiable function. We say:

$f$ is monotone increasing on $[a,b]$, if $\forall x,y\in[a,b] \text{ with } x<y$, we have $f(x)\leq f(y)$.
$f$ is monotone decreasing on $[a,b]$, if $\forall x,y\in[a,b] \text{ with } x<y$, we have $f(x)\geq f(y)$.

Def. 3.6 Global/Local Maximum/Minimum.

Given a function $f:D\rightarrow\mathbb R$, we say:

$f$ has an global maximum at $c$ if $f(c)\geq f(x)$ for any $x\in D$.
$f$ has an global minimum at $c$ if $f(c)\leq f(x)$ for any $x\in D$.
$f$ has an local maximum at $c$ if there exists $\delta>0$ such that $f(c)\geq f(x)$ for any $x\in (c-\delta, c+\delta)\cap D$
$f$ has an local minimum at $c$ if there exists $\delta>0$ such that $f(c)\leq f(x)$ for any $x\in (c-\delta, c+\delta)\cap D$

If $f$ has an global maximum at $c$, then we say $(c, f(c))$ is an global maximum point and $f(c)$ is the global maximum value.

Def. 3.7 Real analytic functions

A function $f$ is said to be real analytic at $x=a$ if there exists $\delta>0$ such that

\[f(x)=\sum_{k=0}^{\infty}\frac{f^{(k)}(a)}{k!}(x-a)^k \]

for any $x\in(a-\delta, a+\delta)$. If $f$ is real analytic at any $a$ in an open interval $I$, we may say it is real analytic on $I$.

Def. 3.7 Convex/Concave Function

A function $f$ is convex on an interval if for any $x, y$ in the interval, the straight line $L_{x, y}$ connecting the points $(x, f(x))$ and $(y, f(y))$ lies above the part of the graph of $f$ over $[x, y]$. This means

\[L_{x,y}(z):=f(x)+\frac{f(y)-f(x)}{y-x}(z-x)\geq f(z) \text{ for any } x\leq z\leq y \]

A function $f$ is convex on an interval if for any $x, y$ in the interval, the straight line $L_{x, y}$ connecting the points $(x, f(x))$ and $(y, f(y))$ lies below the part of the graph of $f$ over $[x, y]$. This means

\[L_{x,y}(z):=f(x)+\frac{f(y)-f(x)}{y-x}(z-x)\leq f(z) \text{ for any } x\leq z\leq y \]

Proposition & Theorem

Theo. 3.1

If $f$ is differentiable at $a$, then $f$ is also continuous at $a$.

Proof Theo. 3.1

Let $x=a+h$ (regarding $a$ is fixed and $h$ is the variable), we have:

\[f^{\prime}(a)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a} \]

Then:

\[\begin{aligned} \lim_{x\rightarrow a}(f(x)-f(a))&=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{(x-a)}(x-a)\\\\ &=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{(x-a)}\times\lim_{x\rightarrow a}(x-a)\\\\ &=f^{\prime}(a)\times 0=0 \end{aligned} \]

This shows $\lim_{x\rightarrow a}=f(a)$. Therefore, $f$ is continuous at $a$.

Pro. 3.2

A function $f(x)$ is differentiable at $x=a$ if and only if there exists a constant $m\in\mathbb R$ such that

\[f(x)=f(a)+m(x-a)+o(x-a) \]

as $x\rightarrow a$. In such case, it is necessary that $m=f^{\prime}(a)$.

Proof Pro. 3.2

Necessity:

If $f(x)$ is differentiable at $x=a$, then $f^\prime(a)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$ exists, which means $\lim_{x\rightarrow a}[\frac{f(x)-f(a)}{x-a}-f^\prime(a)]=\lim_{x\rightarrow a}\frac{f(x)-[f(a)+f^\prime(a)(x-a)]}{x-a}=0$.

Let $g(x-a)=f(x)-[f(a)+f^\prime(a)(x-a)]$, with Def. 3.4, $g(x-a)\in o(x-a)$.

Rewrite the above equation, we have $f(x)=f(a)+f^\prime(a)(x-a)+g(x-a)$, which means $f(x)=f(a)+f^\prime(a)(x-a)+o(x-a)$.

Thus we find a constant $m=f^\prime(a)$, which meets the above conditions.

Sufficiency:

If the condition:

$f(x)=f(a)+m(x-a)+o(x-a)$ as $x\rightarrow a$

holds, then we rewrite this condition:

\[o(x-a)=f(x)-f(a)-m(x-a) \]

Divide both sides of the equation by $(x-a)$ simultaneously:

\[\frac{o(x-a)}{x-a}=\frac{f(x)-f(a)}{x-a}-m \]

With Def. 3.4, we have $\lim_{(x-a)\rightarrow 0}\frac{o(x-a)}{x-a}=0$.

$\therefore \lim_{x\rightarrow a}(\frac{f(x)-f(a)}{x-a}-m)=0$, which means $m=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}=f^\prime(a)$.

As the constant $m$ exists, we can conclude that $f^\prime(a)$ exists, which means $f(x)$ is differentiable at $x=a$.

Pro. 3.3 Product rule & quetient rule

If $f$ and $g$ are differentiable at $a$, then so does $f\pm g$ and $fg$, and if $g(a)\neq 0$, then so do $\frac{f}{g}$. Also, we have:

\[\begin{aligned} (f\pm g)^\prime(a)&=f^\prime(a)\pm g^\prime(a) \\\\ (fg)^\prime(a)&=f^\prime(a)g(a)+f(a)g^\prime(a) \\\\ \left(\frac{f}{g}\right)^\prime(a)&=\frac{f^\prime(a)g(a)-f(a)g^\prime(a)}{g^2(a)} \end{aligned} \]

Proof Pro. 3.3.1

We have

\[f(x)=f(a)+f^\prime(a)(x-a)+o(x-a), g(x)=g(a)+g^\prime(a)(x-a)+o(x-a) \]

as $x\rightarrow a$. It is clear that

\[f(x)+g(x)=f(a)+g(a)+(x-a)(f^\prime(a)+g^\prime(a))+o(x-a)+o(x-a) \]

If $p(x), q(x)\in o(x-a)$, then clearly $p(x)+q(x)\in o(x-a)$. Therefore, we have

\[f(x)+g(x)=f(a)+g(a)+(x-a)(f^\prime(a)+g^\prime(a))+o(x-a) \]

which is in linear approximation form. Therefore, $f(x)+g(x)$ is differentiable at $x=a$, with $(f+g)^\prime(a)=f^\prime(a)+g^\prime(a)$ which is the coefficient of $x-a$.

The proof of $f-g$ is silimar with the above proof.

So we can conclude that:

\[(f\pm g)^\prime(a)=f^\prime(a)\pm g^\prime(a) \]

Proof Pro. 3.3.2

We have

\[f(x)=f(a)+f^\prime(a)(x-a)+o(x-a), g(x)=g(a)+g^\prime(a)(x-a)+o(x-a) \]

as $x\rightarrow a$. It is clear that

\[f(x)g(x)=f(a)g(a)+(x-a)(f^\prime(a)g(a)+f(a)g^\prime(a))+o^2(x-a)+o(x-a)[f(a)+g(a)+f^\prime(a)(x-a)+g^\prime(a)(x-a)]+f^\prime(a)g^\prime(a)(x-a)^2 \]

Clearly $(x-a)^2\in o(x-a)$.

If $p(x)\in o(x-a)$, then clearly $p(x)q(x)\in o(x-a)$. Therefore, we have

\[f(x)g(x)=f(a)g(a)+(x-a)(f^\prime(a)g(a)+f(a)g^\prime(a))+o(x-a) \]

which is in linear approximation form. Therefore, $f(x)g(x)$ is differentiable at $x=a$, with $(fg)^\prime(a)=f^\prime(a)g(a)+f(a)g^\prime(a)$ which is the coefficient of $x-a$.

So we can conclude that:

\[(fg)^\prime(a)=f^\prime(a)g(a)+f(a)g^\prime(a) \]

Pro. 3.4 Chain rule

Suppose $g(x)$ is differentiable at $x=a$, and $f(y)$ is differentiable at $y=g(a)$, then $(f\circ g)(x):=f(g(x))$ is differentiable at $x=a$ with $(f\circ g)^\prime(a)=f^\prime(g(a))g^\prime(a)$.

Proof Pro. 3.4

Given that

\[\begin{aligned} g(x)&=g(a)+g^\prime(a)(x-a)+h(x),&\text{where $h(x)\in (x-a)$ as $x\rightarrow a$} \\\\ f(y)&=f(g(a))+f^\prime(g(a))(y-g(a))+k(y),&\text{where $k(y)\in o(y-g(a))$ as $y\rightarrow g(a)$} \end{aligned} \]

Given any $\epsilon>0$, with $k(y)\in o(y-g(a))$ there exists $\delta>0$ such that if $0<|y-g(a)|<\delta$, then

\[\left|\frac{k(y)}{y-g(a)}\right|<\epsilon \Rightarrow |k(y)|<\epsilon|y-g(a)| \]

Since $k(y)$ is nothing but just $f(y)-f(g(a))-f^\prime(g(a))(y-g(a))$, so $k(y=g(a))=0$. Therefore, in fact we have:

\[|y-g(a)|<\delta\Rightarrow |k(y)|\leq \epsilon|y-g(a)| \]

(It is this subtle modification that allows us to let $y=g(x)$ without worrying whether $y$ could be equal to $g(a)$ or not when $x$ is near $a$.)

\[|k(g(x))|\leq \epsilon|g(x)-g(a)|. \]

Now, consider $f(y)=f(g(a))+f^\prime(g(a))(y-g(a))+k(y)$, which holds even when $y=g(a)$, we can substitute $y=g(x)$ and get:

\[\begin{aligned} f(g(x))&=f(g(a))+f^\prime(g(a))(g(x)-g(a))+k(g(x)) \\\\ &=f(g(a))+f^\prime(g(a))(g^\prime(a)(x-a)+h(x))+k(g(x)) \\\\ &=f(g(a))+f^\prime(g(a))g^\prime(a)(x-a)+f^\prime(g(a))h(x)+k(g(x)) \end{aligned} \]

It is clear that $f^\prime(g(a))h(x)\in o(x-a)$. As for $k(g(x))$, as long as $0<|x-a|<\eta$, we have

\[\left|\frac{k(g(x))}{x-a}\right|\leq\epsilon\left|\frac{g(x)-g(a)}{x-a}\right|. \]

As $\left|\frac{g(x)-g(a)}{x-a}\right|\rightarrow|g^\prime(a)|$ as $x\rightarrow a$, by comparison rule $\exists \eta^{\prime}>0$ such that if $0<|x-a|<\eta^{\prime}$, then

\[\left|\frac{g(x)-g(a)}{x-a}\right|<|g^\prime(a)|+c\text{, where $c>0$ is a constant.} \]

Hence, for any $\epsilon>0$, by taking $\theta=\min\lbrace\eta, \eta^\prime\rbrace$, we have $0<|x-a|<\theta$ implies

\[\left|\frac{g(x)-g(a)}{x-a}\right|\leq\epsilon(|g^\prime(a)|+c) \]

By replacing $\epsilon$ by $\frac{\epsilon}{|g^\prime(a)|+c}$, we have proved that $k(g(k))\in o(x-a)$ as $x\rightarrow a$. Therefore, we finished the proof that

\[f(g(x))=f(g(a))+f^\prime(g(a))g^\prime(a)(x-a)+o(x-a). \]

Pro. 3.5

Let $f:(a, b)\rightarrow \mathbb R$ be a differentiable function whose inverse $f^{-1}:f(\mathbb R)\rightarrow (a,b)$ exists. Then, for any $y_0\in f(\mathbb R)$ such that $f^\prime(x_0)\neq 0$ where $x_0=f^{-1}(y_0)$, then $f^{-1}$ is differentiable at $y_0$, and

\[(f^{-1})^\prime(y_0)=\frac{1}{f^\prime(x_0)}=\frac{1}{f^\prime(f^{-1}(y_0))}. \]

Proof Pro. 3.5

\[\begin{aligned} \lim_{y\rightarrow y_0}\frac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0}&=\lim_{x\rightarrow x_0}\frac{f^{-1}(f(x))-f^{-1}(y_0)}{f(x)-y_0} \\\\ &=\lim_{x\rightarrow x_0}\frac{x-x_0}{f(x)-f(x_0)}\\\\ &=\lim_{x\rightarrow x_0}\frac{1}{\frac{f(x)-f(x_0)}{x-x_0}}\\\\ &=\frac{1}{f^\prime(x_0)} \end{aligned} \]

Pro. 3.6

Let $f:[a, b]\rightarrow\mathbb R$ be a differentiable function. We say:

If $f$ is monotone increasing on $[a,b]$, then $f^\prime(x)\geq 0$ for any $x\in(a, b)$.
If $f$ is monotone decreasing on $[a,b]$, then $f^\prime(x)\leq 0$ for any $x\in(a, b)$.

Pro. 3.7

If $c\in(a,b)$ is an interior local maximum or minimum of $f:[a, b]\to \mathbb R$ and $f(x)$ is differentiable at $x=c$, then $f^\prime(c)=0$.

Local extrema must be in the following cases:

end points of the interval.
$f'(x)$ is not defined.
$f'(x) = 0$.

Second or Higher Derivative Test

Theorem: Suppose $f(x) = f(a) + \Delta x^n + o(\Delta x^n)$,

If $n$ is odd, $c \neq 0$, then $f(x)$ is not a local extreme.
If $n$ is even, $c > 0$, then $f(x)$ is a local minimum.
If $n$ is even, $c < 0$, then $f(x)$ is local maximum.

where $f'(a) = f''(a) = \cdots = f^{(n-1)}(a) = 0$ and $c \triangleq f^{(n)}(a) \neq 0$.

Theo. 3.8 Mean Value Theorem (MVT)

Suppose $f:[a,b]\rightarrow \mathbb R$ is a function such that:

$f$ is continuous on $[a,b]$, and
$f$ is differentiable on $(a,b)$

Then, there exists at least one $c$ in $(a, b)$ such that:

\[\frac{f(b)-f(a)}{b-a}=f^\prime(c), \text{ or equivalently, } f(b)-f(a)=f^\prime(c)(b-a) \]

Proof Theo. 3.8

Given a function $f:[a, b]\rightarrow \mathbb R$ which is continuous on $[a, b]$ and differentiable on $(a, b)$, we define a new function $g:[a, b]\rightarrow\mathbb R$ by

\[g(x):=f(x)-\left(\frac{b-x}{b-a}f(a)+\frac{x-a}{b-a}f(b)\right)\text{.} \]

The graph

\[y=\frac{b-x}{b-a}f(a)+\frac{x-a}{b-a}f(b) \]

is a straight line joining the points $(a, f(a))$ and $(b, f(b))$, so $g(x)$ is measuring the gap between $f(x)$ and this straightline.

It is geometrically clear that $g(a)=g(b)$.

By Coro 3.11 (Rolle's Theorem), there exists $c\in(a, b)$ such that $g^\prime(c)=0$. Note that

$g^\prime(x)=f^\prime(x)-\frac{f(b)-f(a)}{b-a}$

Therefore $g^\prime(c)=0$ implies that $f^\prime(c)=\frac{f(b)-f(a)}{b-a}$.

Pro. 3.9

Given a differentiable function $f:[a,b]\rightarrow \mathbb R$, then:

If $f^\prime(x)>0$ on $(a, b)$, then $f(x)$ is strictly increasing on $[a, b]$.
If $f^\prime(x)<0$ on $(a, b)$, then $f(x)$ is strictly decreasing on $[a, b]$.
$f^\prime(x)\geq 0$ on $(a, b)$ if and only if $f(x)$ is monotone increasing on $[a, b]$.
$f^\prime(x)\leq 0$ on $(a, b)$ if and only if $f(x)$ is monotone decreasing on $[a, b]$.
$f^\prime(x)= 0$ on $(a, b)$ if and only if $f(x)$ is a constant function on $[a, b]$.

Proof Pro. 3.9

Only prove the last one.

Assuming that $f^\prime(x)= 0$ on $(a, b)$.

According to MVT, we have $\forall x\in[a, b], \exists c\in(a, x) \text{ such that }$

\[f(x)-f(a)=f^\prime(c)(x-a) \]

Since $c\in [a, b]$, $f^\prime(c)=0$, which means

\[f(x)-f(a)=0 \]

$\therefore \forall x\in[a, b], f(x)=f(a)$.

$\therefore f(x)$ is a constant function on $[a, b]$

Theo. 3.10 Lagrange's Extreme Value Theorem (EVT)

Suppose $f:[a, b]\rightarrow \mathbb R$ is continuous on the closed and bounded interval $[a, b]$, then $f$ must have a global maximum and global minimum.

i.e., there exist $c_1, c_2\in[a,b]$ such that

\[f(c_1)\leq f(x)\leq f(c_2) \text{, }\forall x\in[a, b]\text{.} \]

Proof Pro. 3.10

Assuming there exists such a function that is not bounded from above, then for any $n\in \mathbb N, \exists x_n\in[a, b]$ such that $f(x_n)>N$. Although $\lbrace x_n \rbrace$
may not have a limit, Bolzano-Weiestrass Theorem shows the interval $[a, b]$ being bounded implies there exists a converging subsequence $\lbrace x_{n_k}\rbrace^{\infty}_{k=1}$. Suppose $x_{n_k}\rightarrow L$ as $k\rightarrow \infty$, then by $a\leq x_{n_k}\leq b$ for any $k$ we see that $L\in[a, b]$.

However, by continuity of $f$, we have

\[\lim_{k\rightarrow \infty}f(x_{n_k})=f(L)\in\mathbb R \]

However, by $f(x_{n_k})>n_k$ and $n_k\rightarrow \infty$, we get that

\[\lim_{k\rightarrow \infty}f(x_{n_k})=+\infty \]

which is a contradiction. This shows $f$ is bounded from above.

Now by the Completeness Axiom, the set $\lbrace f(x):x\in[a, b]\rbrace$ which is bounded from above has a supremum. Denote

$M:=\sup\lbrace f(x):x\in[a, b]\rbrace\text{.}$

Then, for any $n\in\mathbb N$, by considering $M-\frac{1}{n}$ which is strictly less than $M$, there must exist $y_n\in[a, b]$ such that $M-\frac{1}{n}<f(y_n)\leq M$, otherwise $M$ would not be the least upper bound for $\lbrace f(x):x\in[a, b]\rbrace$. Applying Bolzano-Weiestrass Theorem again, there exists a converging subsequence $\lbrace y_{n_k}\rbrace$ such that $y_{n_k}\rightarrow c$ for some $c\in[a, b]$. Applying squeeze theorem on $M-\frac{1}{n}<f(y_n)\leq M$, we conclude that

\[\lim_{k\rightarrow \infty}f(y_{n_k})=M\text{,} \]

and by continuity of $f$ it shows $f(c)=M$. Recall that $M$ is an upper bound for $f(x)$, so we conclude that $f(x)\leq f(c)$ for any $x\in[a, b]$.

Similar argument shows there exists $c^\prime\in[a, b]$ such that $f(c^\prime)\leq f(x)$ for any $x\in[a, b]$.

Corollary 3.11 Rolle's Theorem

Suppose $f$ is continuous on $[a, b]$, differentiable on $(a, b)$ and $f(a)=f(b)$, then there exists $c\in(a, b)$ such that $f^\prime(c)=0$.

Proof Coro. 3.11

By EVT, there exist $c_1, c_2$ such that

\[f(c_1)\leq f(x)\leq f(c_2)\text{, } \forall x\in[a, b] \]

If $f(a)=f(b)=f(c_1)=f(c_2)$, we can see that $f$ is a constant function. Therefore $\forall x\in(a, b), f^\prime(x)=0$.

If not, $c_1$ or $c_2$ is not equal to neither $a$ nor $b$. Thus $c_1\in(a, b)$ or $c_2\in(a, b)$.

By Pro. 3.7, we have $f^\prime(c_1)=0$ or $f^\prime(c_2)=0$.

Theo. 3.12 Cauthy's Mean Value Theorem

Let $f$ and $g$ be real-valued functions continuous on $[a, b]$ and differentiable on $(a, b)$. And for any $x\in(a, b)$, $g^\prime(x)\neq 0$.

Then there exists $c\in (a, b)$ such that $\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f^\prime(c)}{g^\prime(c)}$.

Proof Theo. 3.12

$F(x):=[f(b)-f(a)][g(x)-g(a)]-[g(b)-g(a)][f(x)-f(a)]$

Obviously $F(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$. And $F(a)=F(b)$.

According to Rolle's Theorem, there exists $c\in(a, b)$ such that $F^\prime(c)=[f(b)-f(a)]g^\prime(c)-[g(b)-g(a)]f^\prime(c)=0$.

$\because g^\prime(c)\neq 0$

$\therefore \frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f^\prime(c)}{g^\prime(c)}$

Theo. 3.13 L'Hospital's Rule

Suppose there exists $\delta >0$ such that $f(x)$ and $g(x)$ are differentiable on $(a-\delta, a+\delta)\setminus\lbrace a\rbrace$, and they satisfy:

\[\begin{aligned} \lim_{x\rightarrow a}f(x)=0&\text{ and }\lim_{x\rightarrow a}g(x)=0 \\\\ \text{or}: \lim_{x\rightarrow a}f(x)=\pm \infty&\text{ and }\lim_{x\rightarrow a}g(x)=\pm\infty \end{aligned} \]

If $\lim_{x\rightarrow a}\frac{f^\prime(x)}{g^\prime(x)}$ exists, then we have $\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\frac{f^\prime(x)}{g^\prime(x)}$.

Proof Theo. 3.13

Part one

Now given

\[\lim_{x\rightarrow a}f(x)=\lim_{x\rightarrow a} g(x)=0\text{ and } \lim_{x\rightarrow a}\frac{f^\prime(x)}{g^\prime(x)}=L, \]

we need to show $\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a}\frac{f^\prime(x)}{g^\prime(x)}=L.$

First note that the limit of a function as $x\rightarrow a$ is not relevant to the value of the function at $x=a$, and in view of the assumption that $\lim_{x\rightarrow a}f(x)=\lim_{x\rightarrow a} g(x)=0$, it is not a problem to assume without loss of generality that $f(a)=g(a)=0$ so that $f$ and $g$ would be continuous at $x=a$. By the result of Cauthy's Mean Value Theorem, there exists $c\in(a, x)$ such that

\[\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f^\prime(c)}{g^\prime(c)}\Rightarrow f^\prime(c)g(x)=g^\prime(c)f(x)\Rightarrow\frac{f(x)}{g(x)}=\frac{f^\prime(x)}{g^\prime(x)} \]

Note that $a<c<x$, so if $x\rightarrow a^+$, by squeeze theorem, $c\rightarrow a^+$ as well. Thereofore,

\[\lim_{x\rightarrow a^+}\frac{f^\prime(x)}{g^\prime(x)}=L\Rightarrow\lim_{x\rightarrow a^+}\frac{f^\prime(c)}{g^\prime(c)}=L \]

and hence $\lim_{x\rightarrow a^+}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a^+}\frac{f^\prime(x)}{g^\prime(x)}=L$.

Similar argument can prove that $\lim_{x\rightarrow a^-}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a^-}\frac{f^\prime(x)}{g^\prime(x)}=L$

Part two

Select $x_0\in(a, a+\epsilon)$, $x\in(a, x_0)$, where $\epsilon\rightarrow 0$.

Then $f(x),g(x)$ is continuous on $[x, x_0]$ and differentiable on $(x, x_0)$.

According to Cauthy's Mean Value Theorem, there exists $c\in(x, x_0)$ such that $\frac{f(x_0)-f(x)}{g(x_0)-g(x)}=\frac{f^\prime(c)}{g^\prime(c)}$

$\therefore \frac{f(x)}{g(x)}=\frac{f^\prime(c)}{g^\prime(c)}\cdot\frac{g(x)-g(x_0)}{g(x)}+\frac{f(x_0)}{g(x)}$

$\because x_0\rightarrow a^+, x\rightarrow a^+$

$\therefore c\rightarrow a^+$

$\because \lim_{x\rightarrow a^+}g(x)=\infty$

$\therefore \lim_{x\rightarrow a^+}\frac{f(x)}{g(x)}=\lim_{x\rightarrow a^+}\left(\frac{f^\prime(c)}{g^\prime(c)}-\frac{g(x_0)\cdot\frac{f^\prime(c)}{g^\prime(c)}}{g(x)}+\frac{f(x_0)}{g(x)}\right)=\lim_{x\rightarrow a^+}\frac{f^\prime(c)}{g^\prime(c)}=L$

Pro. 3.14

Given that $f$ is $n$-times differentiable at $x=a$, then

\[f(x)=\sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a)^k+o((x-a)^n) \text{ as } x\rightarrow a\text{ .} \]

The polynomial $\sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a)^k$ is called the $n$-th degree Taylor polynomial/approximation of $f$ near $x=a$.

Theo. 3.15 Lagrange Remainder

Suppose $f$ be $(n+1)$-times differentiable function on an open interval $I$ containing $a$, then for each $x\in I\setminus\lbrace a\rbrace$, there exists $c\in(a, x)$ or $c\in(x, a)$, which may depend on $x, a$ and $ n$, such that

\[f(x)=\sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a)^k+\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}\text{ .} \]

Theo. 3.16 Cauchy's remainder

Given $f$ is $(n+1)$-times differentiable on an open interval $I$ containing $a$. Then, for any $x\in I\setminus \lbrace a\rbrace$, there exists $c\in(a, x)$ or $(x, a)$ such that

\[f(x)=\sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a)^k+\frac{f^{(n+1)}(c)}{n!}(x-c)^{n}(x-a)\text{ .} \]

Theo. 3.17 Convex and Concave (differentiable function)

A differentiable function $f(x)$ on an interval is convex if and only if $f^\prime(x)$ is increasing. If $f(x)$ has second order derivative, then this is equivalent to $f^{\prime\prime}(x)\geq 0$.

A differentiable function $f(x)$ on an interval is concave if and only if $f^\prime(x)$ is decreasing. If $f(x)$ has second order derivative, then this is equivalent to $f^{\prime\prime}(x)\leq 0$

Theo. 3.17 Leibniz formula

\[[f(x)\cdot g(x)]^{(n)}=\sum_{k=0}^nC_{n}^kf^{(k)}(x)g^{(n-k)}(x) \]

Proof Theo. 3.17

This clearly holds when $n=1$.

Assuming that this holds when $n=t$.

When $n=t+1$, we have

\[\begin{aligned} [f(x)\cdot g(x)]^{(t+1)}&=\lbrace[f(x)\cdot g(x)]^{(t)}\rbrace^\prime \\\\ &=[\sum_{k=0}^tC_{t}^kf^{(k)}(x)g^{(t-k)}(x)]^\prime \\\\ &=\sum_{k=0}^tC_{t}^k[f^{(k)}(x)g^{(t-k)}(x)]^\prime \\\\ &=\sum_{k=0}^tC_{t}^k[f^{(k+1)}(x)g^{(t-k)}(x)+f^{(k)}(x)g^{(t+1-k)}(x)] \\\\ &=f^{(t+1)}(x)g(x)+\sum_{k=1}^{t}[(C_{t}^{k-1}+C_{t}^{k})f^{(k)}(x)g^{(t+1-k)}(x)]+f(x)g^{(t+1)}(x) \\\\ &=\sum_{k=0}^{t+1}C_{t+1}^kf^{(k)}(x)g^{(t+1-k)}(x) \end{aligned} \]

Samples

Smp. 3.1 Derivative of Inverse Trigonometric Functions

\[\begin{aligned} \frac{d}{dx}\sin^{-1}x&=\frac{1}{\sqrt{1-x^2}} \\\\ \frac{d}{dx}\cos^{-1}x&=-\frac{1}{\sqrt{1-x^2}} \\\\ \frac{d}{dx}\tan^{-1}x&=\frac{1}{1+x^2} \\\\ \frac{d}{dx}\cot^{-1}x&=-\frac{1}{1+x^2} \end{aligned} \]

Proof Smp. 3.1

\[\begin{aligned} \frac{d}{dx}\sin y&=\frac{d}{dx}x \\\\ \cos y\cdot\frac{dy}{dx}&=1 \\\\ \frac{dy}{dx}&=\frac{1}{\cos y} \\\\ &=\frac{1}{\cos(\sin^{-1}x)} \end{aligned} \]

Since $\cos(\sin^{-1} x)=\sqrt{1-x^2}$. Therefore we have

\[\frac{d}{dx}\sin^{-1}x=\frac{1}{\sqrt{1-x^2}} \]

Other proofs are similar to those mentioned above.

Smp. 3.2

Let $g(x)$ be a function which is continuous on $[a, b]$, and differentiable on $(a,b)$, and given that $g^\prime(x)$ is strictly decreasing on $[a, b]$, and $g(a)=g(b)=0$. There exists $c\in(a, b)$ such that $g$ is strictly decreasing on $[a, c]$, and strictly decreasing on $[c,b]$. Hence, $g(x)\geq 0$ for all $x\in [a, b]$.

Proof Smp. 3.2.1

According to MVT, there exists $c_0\in(a, b)$ such that $g(b)-g(a)=g^\prime(c_0)(b-a)$.

Since $g(a)=g(b)=0$ and $a\neq b$, we have $g^\prime(c_0)=0$

Since $g^\prime(x)$ is strictly decreasing, we have $g^\prime(x)>0$ for all $x\in (a, c_0)$ and $g^\prime(x)<0$ for all $x\in (c_0, b)$.

According to Pro. 3.9, we have $g(x)$ is strictly increasing on $[a, c_0]$ and strictly decreasing on $[c_0, b]$.

Let $f(x)$ be a twice differentiable function such that $f^{\prime\prime}(x)>0$ on $[a, b]$. Consider the function $$h(x)=(b-x)f(a)+(x-a)f(b)-(b-a)f(x), x\in [a, b]\text{,}$$ show that $$f(x)\leq\frac{b-x}{b-a}f(a)+\frac{x-a}{b-a}f(b) \text{, } \forall x\in[a, b]\text{.}$$

Sol. Smp. 3.2.2

$h(a)=h(b)=0$.

$h^\prime(x)=f(b)-f(a)-(b-a)-f^\prime(x)$.

Since $f^{\prime\prime}(x)>0$ on $[a, b]$, according to Pro. 3.9, we have $f^{\prime}(x)$ is strictly increasing on $[a, b]$.

$\therefore h^\prime(x)$ is strictly decreasing on $[a, b]$.

Since $f(x)$ is differentiable on $[a, b]$, $h(x)$ is differentiable on $[a, b]$.

According to Smp 3.1.1, we have $h(x)\geq 0$ for all $x\in[a,b]$.

$\therefore f(x)=\frac{b-x}{b-a}f(a)+\frac{x-a}{b-a}f(b)-\frac{h(x)}{b-a}\leq \frac{b-x}{b-a}f(a)+\frac{x-a}{b-a}f(b)$.

Smp. 3.3

\[\begin{aligned} \sin x&=\sum_{i=0}^\infty(-1)^{i}\frac{x^{2i+1}}{(2i+1)!}\\\\ \cos x&=\sum_{i=0}^\infty(-1)^{i}\frac{x^{2i}}{(2i)!}\\\\ \ln(1+x)&=\sum_{i=1}^\infty (-1)^{i-1}\frac{x^i}{i} \end{aligned} \]

as $x\rightarrow 0$

posted @ 2024-12-11 22:09 Displace 阅读(29) 评论(0) 收藏举报

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