# 微积分网课代修|微分学代写Differential calculus代考|MATH272 Theorems of Analysis

• 单变量微积分
• 多变量微积分
• 傅里叶级数
• 黎曼积分
• ODE
• 微分学

## 微积分网课代修|微分学代写Differential calculus代考|Theorems of Analysis

The theorems in this section will be used to prove new theorems. It can be skipped on the first reading. We accept the following fundamental result without proof:
Theorem 1.7.1: Monotone Convergence Theorem
Every bounded and monotonic sequence is convergent.
In other words, if a sequence $a_{n}$ is

either increasing, $a_{n} \leq a_{n+1}$ for all $n$, or decreasing, $a_{n} \geq a_{n+1}$ for all $n$, and

bounded, $\left|a_{n}\right| \leq Q$ for some number $Q$,
then it has a limit.

## 微积分网课代修|微分学代写Differential calculus代考|Uniqueness of Supremum and Infimum

• For a given set, there can be only one least upper bound.
• For a given set, there can be only one greatest lower bound.
Proof.
Thus, $M=\sup S$ means that:
1. $M$ is an upper bound of $S$.
2. If $M^{\prime}$ is another upper bound of $S$, then $M^{\prime} \geq M$.
Now, if we have another $M^{\prime}=\sup S$, then:
3. $M^{\prime}$ is an upper bound of $S$.
4. If $M$ is another upper bound of $S$, then $M \geq M^{\prime}$.
Therefore, $M=M^{\prime}$.

## 微积分网课代修|微分学代写Differential calculus代考|Uniqueness of Supremum and Infimum

• 对于给定的集合，只能有一个最小上界。
• 对于给定的集合，只能有一个最大的下限。
证明。
因此， $M=\sup S$ 意思是:
1. $M$ 是上界 $S$.
2. 如果 $M^{\prime}$ 是另一个上限 $S$ ， 然后 $M^{\prime} \geq M$.
现在，如果我们有另一个 $M^{\prime}=\sup S$ ，然后：
3. $M^{\prime}$ 是上界 $S$.
4. 如果 $M$ 是另一个上限 $S$ ， 然后 $M \geq M^{\prime}$.
所以， $M=M^{\prime}$.