# 微积分网课代修|常微分方程代写Ordinary Differential Equation代考|MATH2410 Review of Calculus

• 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|常微分方程代写Ordinary Differential Equation代考|Review of Calculus

The basic definitions of the calculus extend easily to multidimensional spaces. In fact, these definitions are essentially the same when extended to infinite dimensional spaces. Thus, we will begin our review with the definition of differentiation in a Banach space.

Definition 1.150. Let $U$ be an open subset of a Banach space $X$, let $Y$ denote a Banach space, and let the symbol || denote the norm in both Banach spaces. A function $f: U \rightarrow Y$ is called (Fréchet) differentiable at $a \in U$ if there is a bounded linear operator $\operatorname{Df}(a): X \rightarrow Y$, called the derivative of $f$, such that
$$\lim _{h \rightarrow 0} \frac{1}{|h|}|f(a+h)-f(a)-D f(a) h|=0 .$$
If $f$ is differentiable at each point in $U$, then function $f$ is called differentiable.

Using the notation of Definition $1.150$, let $L(X, Y)$ denote the Banach space of bounded linear transformations from $X$ to $Y$, and note that the derivative of $f: U \rightarrow Y$ is the function $D f: U \rightarrow L(X, Y)$ given by $x \mapsto D f(x)$.
The following proposition is a special case of the chain rule.

## 微积分网课代修|常微分方程代写Ordinary Differential Equation代考|The Mean Value Theorem

The mean value theorem for functions of several variables is very important. However, the proof is somewhat more delicate than the usual proof for the case of a scalar function of one variable. Let us begin with a special case.
Theorem 1.156. Suppose that $[a, b]$ is a closed interval, $Y$ is a Banach space, and $f:[a, b] \rightarrow Y$ is a continuous function. If $f$ is differentiable on the open interval $(a, b)$ and there is some number $M>0$ such that $\left|f^{\prime}(t)\right| \leq M$ for all $t \in(a, b)$, then
$$|f(b)-f(a)| \leq M(b-a) .$$
Proof. Let $\epsilon>0$ be given and define $\phi:[a, b] \rightarrow \mathbb{R}$ by
$$\phi(t)=|f(t)-f(a)|-(M+\epsilon)(t-a) .$$
Clearly, $\phi$ is a continuous function such that $\phi(a)=0$. We will show that $\phi(b) \leq \epsilon$

## 微积分网课代修|常微分方程代写Ordinary Differential Equation代 考|Review of Calculus

$$\lim _{h \rightarrow 0} \frac{1}{|h|}|f(a+h)-f(a)-D f(a) h|=0 .$$

## 微积分网课代修|常微分方程代写Ordinary Differential Equation代考|The Mean Value Theorem

$$|f(b)-f(a)| \leq M(b-a) .$$

$$\phi(t)=|f(t)-f(a)|-(M+\epsilon)(t-a) .$$