[Calculus] 8. Taylor Expansion

2024-02-24

Definition 8.1 (higher derivative) Given $f: S \to \R$, and $a \in S$. We say $f$ is k-times differentiable at $a$ if $f$ is differentiable at every $x$ near $a$, and $f'$ is (k-1)-times differentiable at $a$.

Proposition 8.2 Equivalently, here is another definition of higher derivatives.

Given $f: S \to \R$, and $a \in S$. We say $f$ is k-times differentiable at $a$ if $\exists \delta > 0[(a - \delta, a + \delta) \sube S]$ and $f', f'', \cdots, f^{(k-1)}$ exists on $(a - \delta, a + \delta)$, and $f^{(k-1)}$ is differentiable at $a$.

Proof ($\Rarr$)

For times less than $k-1$, we have:

\begin{align*} f \text{ is } k\text{-times diff.} &\Rarr f \text{ is diff. near } a{, and } f' \text{ is } (k-1)\text{-times diff. at } a\\ &\Darr\\ f' \text{ is } (k-1)\text{-times diff.} &\Rarr f' \text{ is diff. near } a{, and } f'' \text{ is } (k-2)\text{-times diff. at } a\\ &\Darr\\ \vdots\\ &\Darr\\ f^{(k-2)} \text{ is } k\text{-times diff.} &\Rarr f^{(k-2)} \text{ is diff. near } a{, and } f^{(k-1)} \text{ is } (k-1)\text{-times diff. at } a \end{align*}

($\Larr$)

$f^{(k-1)}$ is diff. at $a$, and $f^{(k-2)}$ is diff. near $a$, therefore $f^{(k-2)}$ is 1-times diff.

Repeat that, and we will obtain the answer □

Lemma 8.3 Given $f: I \to \R$, and $a,b \in I, a < b$.

If $f$ is convex and $\exists\delta > 0 (a - \delta, a + \delta) \in I$, then $f_{-}'(a)$ and $f_{+}'(a)$ both exist, and $f_{-}'(a) \le f_{+}'(a)$.
If $f$ is convex and $\exists \delta >0 (a - \delta, b + \delta) \in I$, then $f_{+}'(a) \le f_{-}'(b)$.

Proof (1)

By proposition 7.4, consider $a - \delta_1 < a < a + \delta_2$ where $\delta_1,\delta_2 \le \delta$, we have

\[ \frac{f(a) - f(a-\delta_1)}{\delta_1} \le \frac{f(a+\delta_2) - f(a)}{\delta_2} \]

Fix $\delta_1$, and let $delta_2$ be approaching to $0$, $\frac{f(a+\delta_2) - f(a)}{\delta_2}$ is decreasing in the process, and it is bounded. Therefore, $f_{+}'(a)=\lim_{\delta_2 \to 0}\frac{f(a+\delta_2) - f(a)}{\delta_2}$ exists.

Similarly, we can obtain that $f_{-}'(a)$ exists, too.

And if letting $\delta_1,\delta_2$ be approaching 0 at the same time, we can obtain that $f_{-}'(a) \le f_{+}'(b)$.

(2)

Let $x,y \in I$, such that $a < x < y < b$. Then

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

As $x \to a$ and $y \to b$, similarly as (1), we have $f_{-}'(a) \le f_{+}'(b)$. □

Proposition 8.4 (convex test by 2-times derivative) If $f: I \to \R$, where $I$ is an open interval, is 2-times differentiable everywhere, then $f$ is convex iff $f''(x) \ge 0$ for all $x \in I$.

Proposition 8.5 By lemmas, we can have that for $a < x (\in I)$,

\[ f_{-}'(a) \le f_{+}'(a) \le f_{-}'(x) \le f_{+}'(x) \]

Since $f$ is 2-times differentiable everywhere, i.e,. $f'(x)$ is differentiable everywhere, the left limit is equal to the right limit equaled to the limit itself, we have $f'(a) \le f'(x)$.

\[ \lim{x \to 0}\frac{ f'(x) - f'(a) }{ x - a } = f''(a) \le 0 \]

for any $a \in I$.

Definition 8.6 We say two maps $f(x), g(x)$ is k-times close as $x \to a$ if

\[ \lim_{x \to a}\frac{f(x) - g(x)}{(x-a)^k} = 0 \]

, denoted as $f(x) \underset{k}{\sim} g(x)$.

Proposition 8.7 If $P(x)$ is a k-times polynomial, then \[ P(x) = \sum_{i=0}^{k}\frac{P^{(i)}(a)}{i!}(x-a)^i \]

Proof Too hard for me. I'll give it a try another day. □

Proposition 8.8 If $f(x) \underset{k}{\sim} P(x)$ and $f(x) \underset{k}{\sim} Q(x)$ as $x \to a$, then $P(x) \underset{k}{\sim} Q(x)$ as $x \to a$, and $P(a) = Q($), $P^{(i)}(a) = Q^{(i)}(a)$ for $i = 1,2,\cdots, k$.

\begin{align*} \lim_{x \to a} \frac{P(x) - Q(x)}{(x - a)^k} &= \lim_{x \to a} \frac{P(x) - f(x) + f(x) - Q(x)}{(x - a)^k}\\ &= \lim_{x \to a} \frac{f(x) - Q(x)}{(x - a)^k} - \lim_{x \to a} \frac{f(x) - P(x)}{(x - a)^k}\\ &= 0 \end{align*}

Proof Let $ F(x) = P(x) - Q(x)$. Then $\lim_{x \to a}\frac{F(x)}{(x-a)^k} = 0 \Rarr \lim_{x \to a}(P(x) - Q(x)) = 0$, i.e., $P(a) = Q(a)$.

Then

\begin{align*} \lim_{x \to a}\frac{P(x) - Q(x)}{(x-a)^k} &= \lim_{x \to a}\frac{P(x) - P(a) + P(a) - Q(x)}{(x-a)^k}\\ &= \lim_{x \to a}\frac{P(x) - P(a) + Q(a) - Q(x)}{(x-a)^k}\\ &= \lim_{x \to a}\frac{\frac{P(x) - P(a)}{x-a} - \frac{Q(x) - Q(a)}{x-a}}{(x-a)^{k-1}}\\ &= \lim_{x \to a}\frac{P'(a) - Q'(a)}{(x-a)^{k-1}}\\ &= 0 \end{align*}

this makes sure that $P'(a) = Q'(a)$, else the limit will grow to infinity as $x \to a$.

Similarly, we can get the result we want. □

Definition 8.9 If $f$ is k-times differentiable at $a$, then we call

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

the k-th Taylor polynomial of $f$ at $a$.

Theorem 8.10 Theorem 8.9 $f: I \to \R$ is k-times differentiable at $a$, then $f(x) \underset{k}{\sim} T_k(x)$ as $x \to a$.

Furthermore, if $f$ is (k+1)-times differentiable near $a$, then for every $x \in I$, there exists $c$ between $x$ and $a$,

\[ f(x) = f(a) + \frac{f'(a)}{1!}(x-a) + \cdots + \frac{f^{(k)}}{k!}(x-a)^k + \frac{f^{(k+1)}(c)}{(k+1)!}(x-a)^{k+1} \]