Calculus Note

This note is for my reference only.

Number Set

Bounded set

To prove $\beta$ is the supremum of $E$, we could show that for any $\epsilon > 0$, $\beta - \epsilon$ is not the upper bound of $E$. that is, find a element of $E$ that is greater or equal to $\beta - \epsilon$.

The extended real number system

$\overline{\R} = \R \ \cup \{-\infin, \infin\}$.

Complex number

$$\overline{z + w} = \overline{z} + \overline{w}\\ \overline{z\cdot w} = \overline{z} \cdot \overline{w}\\ \mid zw\mid = \mid z\mid \cdot \mid w \mid \\ \mid z + w \mid \leq \mid z\mid + \mid w \mid \\ \mid z \mid = \mid z\overline z\mid^{\frac{1}{2}} \Rightarrow (1+a\overline b)^2 = (1+a\overline b)(1+\overline ab)$$

Euler’s Formula

$$e^{ix} = \cos x + i \sin x$$

$$(e^{ix})^n = \cos (nx) + i \sin(nx)$$

$$\cos x = \frac{e^{ix} + e^{-ix}}{2}, \sin x = \frac{e^{ix} - e^{-ix}}{2}$$

Sequence Convergence

1. $\epsilon-N$ Definition.
2. Squeeze Theorem
3. Monotone Convergence Theorem
4. Odd/Even Subsequence: if they are convergence to the same number.
5. Cauchy Criterion (by definition or by proving $\mid x_{n+1}-x_n\mid\leq r\mid x_n-x_{n-1}\mid, 0<r<1$).

Tricks for sequence

有些名字是自己编的。

1. To show equality such as $a = b$, try to show both $a \leq b$ and $a \geq b$ hold.

2. To show a number $a$ is irrational, usually prove by contradiction, assume $a = \frac{x}{y}$.

3. To prove the equivalent ($\Leftrightarrow$), prove both $\Leftarrow$ and $\Rightarrow$ hold.

$1+\epsilon$ Trick

Example 1

show that $\lim_{n\to\infin} \sqrt[n]{n} = 1$.

​ Pf. Let $\sqrt[n]{n} = 1 + x_n$, for $x_n > 0$. we get

$$n = (1+x_n)^n\geq \tbinom{n}{2} x_n^2 = \frac{n(n-1)}{2}x_n^2$$

​ It implies that. $x_n^2 \leq \frac{2}{n - 1}, n\geq 2$. Thus $\lim_{n\to\infin} x_n = 0$.

Example 2

show that $\lim_{n\to\infin} a^n = 0$, for $0 < a < 1$.

​ Pf. Let $a = \frac{1}{\frac{1}{a}} = \frac{1}{b + 1}$, for $b > 0$. Then $a^n = \frac{1}{(b+1)^n} < \frac{1}{nb}$

Example 3

show that $\lim_{n\to\infin} \frac{n^k}{a^n} = 0$, for $a>0, k\in \N_+$.

​ Pf. to show $\lim_{n\to\infin} \frac{n}{(1+b)^{\frac{n}{k}}} = 0$

​ $0 < \frac{n}{(1+b)^{\frac{n}{k}}} < \frac{n}{\frac{\frac{n}{k}\cdot(\frac{n}{k}-1)}{2}b^2}\to 0$

Order rule & squeeze

Example 1

Find the limit $\lim_{n\to\infin}\sqrt[n]{n^2-5n+(-1)^n}$.

Sol. It’s obvious that $\lim_{n\to\infin} \frac{n^2-5n+(-1)^n}{n^2}=1$. By the order rule, $0.5 < \frac{n^2-5n+(-1)^n}{n^2} < 1.5$. then we get

$$\sqrt[n]{0.5}\cdot(\sqrt[n]{n})^2 < \sqrt[n]{n^2-5n+(-1)^n}<\sqrt[n]{1.5}\cdot(\sqrt[n]{n})^2$$

Example 2

Suppose $\{a_n\}$ is a sequence of positive number s.t. $\lim_{n\to\infin} \frac{a_{n+1}}{a_n} = L$, where $0\leq L < 1$. Prove that $\lim_{n\to\infin} a_n = 0$.

Since $L < \frac{L + 1}{2} < 1$, by the order rule, we can get $\frac{a_{n+1}}{a_n} < \frac{L + 1}{2}$, for $n > N$. Then we can compute

$$0 < a_n = \frac{a_n}{a_{n-1}}\cdot\frac{a_{n-1}}{a_{n-2}}\cdots\frac{a_{N+1}}{a_N}\cdot a_N < (\frac{L + 1}{2})^{n-N}\cdot a_N$$

Because $\lim_{n\to\infin}(\frac{L+1}{2})^{n-N} = 0$, so $\lim_{n\to\infin} a_n = 0$

AM-GM Inequality

$$\prod_{i=1}^n x_i^{\lambda_i} \leq \sum_{i=1}^n \lambda_ix_i, \lambda_1+\cdots+\lambda_n=1$$

To show $(1+\frac{1}{n})^n \leq (1+\frac{1}{n+1})^{n+1}$.

$$(1+\frac{1}{n})^n = \underbrace{\frac{n+1}{n}\cdot\frac{n+1}{n}\cdots\frac{n+1}{n}}_{n\ \text{terms}}\cdot 1$$

$$\begin{aligned} (1+\frac{1}{n})^{\frac{n}{n+1}} &= \underbrace{(\frac{n+1}{n})^{\frac{1}{n+1}}\cdot(\frac{n+1}{n})^{\frac{1}{n+1}}\cdots(\frac{n+1}{n})^{\frac{1}{n+1}}}_{n\ \text{terms}}\cdot1^{\frac{1}{n+1}}\\ &\leq \underbrace{\frac{1}{n+1}\cdot\frac{n+1}{n} +\cdots+ \frac{1}{n+1}\cdot\frac{n+1}{n}}_{n\ \text{terms}} + \frac{1}{n+1}\cdot 1\\ &= \frac{n+2}{n+1} = (1+\frac{1}{n+1}) \end{aligned}$$

Therefore, $(1+\frac{1}{n})^n \leq (1+\frac{1}{n+1})^{n+1}$

Binomial Formula

The most common usage method has already been stated.

Prove that $x_n = (1+\frac1n)^n$ is bounded above.

$$\begin{aligned} x_n &= \sum_{k = 0}^n\tbinom{n}{k}\frac{1}{n^k}\\ &= 2 + \sum_{k=2}^n\frac{n(n-1)\cdots(n-k+1)}{k!}\cdot\frac1{n^k}\\ &= 2 + \sum_{k=2}^n\frac{n(n-1)\cdots(n-k+1)}{n^k}\cdot\frac1{k!}\\ &< 2 + \sum_{k=2}^n \frac{1}{k!}\\ &< 2 + (1-\frac1n)\\ &< 3 \end{aligned}$$

Re-grouping

To show $\sum_{k=1}^\infin \frac1k$ diverges, for any $N>0$, we can have any $n$ s.t. $2^n > N$, and we can get,

$$\mid x_{2^{n+1}}-x_{2^n}\mid = \frac1{2^n+1}+\frac1{2^n+2}+\cdots+\frac1{2^{n+1}} > 2^n\cdot\frac1{2^{n+1}} = \frac12$$

By taking $0 <\epsilon < \frac12$, $\mid x_{2^{n+1}}-x_{2^n}\mid> \epsilon$. So it diverges.

Common formula

$$x^3-y^3 = (x-y)(x^2+xy+y^2)$$

$$\mid \sin x \mid \leq \mid x\mid$$

$$0 < 1-e^{-t} \leq t, \ \text{for}\ t\geq 0$$

三角恒等式

Change Variable

Function Convergence

1. $\epsilon-\delta$ Definition.
2. Squeeze Theorem.
3. Arithmetic rules
4. Composition rule
5. One sided limit

Prop. 2.18

The limit of the function $f(x)$ as $x$ approaches $a$ is $L$ if and only if for every sequence $\{x_n\}$ that converges to $a$, the sequence of function values $\{f(x_n)\}$ converges to L.

The proposition could be used to prove that a function limit DNE by construct two sequence $\{x_n\}$ and $\{y_n\}$ that both converges to $a$, but the limit of $\{f(x_n)\}$ and $\{f(y_n)\}$ are different.

Derivative

$$(f^{-1})'(a) = \frac1{f'(f^{-1}(a))}$$