# 微积分代写Calculus II|MATH 15200 University of Chicago Assignment

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The Department of Mathematics at this institution offers a variety of programs in mathematics and applied mathematics at both the undergraduate and graduate levels. At the undergraduate level, students can pursue either a Bachelor of Arts (BA) or a Bachelor of Science (BS) degree in Mathematics. Additionally, the department offers two BS degree programs: one in Applied Mathematics and one in Mathematics with a specialization in Economics.

The BA program in Mathematics provides students with a broad foundation in mathematics, while the BS program is designed for students who wish to focus on a more rigorous and specialized mathematical education. The BS program in Applied Mathematics combines coursework in mathematics with applied fields such as physics, engineering, or biology. The BS program in Mathematics with a specialization in Economics is tailored for students who wish to apply mathematical concepts to economic analysis and decision-making.

In addition to the major programs, students in other fields of study can also complete a minor in Mathematics. The minor program is designed to provide students with a basic understanding of mathematics and its applications, which can enhance their academic and career opportunities.

Overall, the Department of Mathematics at this institution aims to provide students with a comprehensive education in mathematics, preparing them for careers in a variety of fields that require mathematical skills and knowledge. The department also actively engages in research activities, contributing to the advancement of mathematical knowledge and its applications in various fields.

Let $a<b$ in $\mathbb{R}$. Every continuous map of the interval $[a, b]$ into itself has a fixed point.

Let $f:[a, b] \rightarrow[a, b]$ be continuous. If $f(a)=a$ or $f(b)=b$, then the result is obvious; so we suppose that $f(a)>a$ and $f(b)0$. Since $g(a)<0<g(b)$, we may conclude from the intermediate value theorem that $0 \in \operatorname{ran} g$. That is, there exists $z$ in $(a, b)$ such that $g(z)=z-f(z)=0$. Thus $z$ is a fixed point of $f$.

Let $f: A \rightarrow \mathbb{R}$ where $A \subseteq \mathbb{R}$, and let $a$ be an accumulation point of $A$. If $f(x) \rightarrow b$ as $x \rightarrow a$, and if $f(x) \rightarrow c$ as $x \rightarrow a$, then $b=c$.

Argue by contradiction. If $b \neq c$, then $\epsilon:=|b-c|>0$. Thus there exists $\delta_1>0$ such that $|f(x)-b|<\epsilon / 2$ whenever $x \in A$ and $0<|x-a|<\delta_1$ and there exists $\delta_2>0$ such that $|f(x)-c|<\epsilon / 2$ whenever $x \in A$ and $0<|x-a|<\delta_2$. Let $\delta=\min \left{\delta_1, \delta_2\right}$. Since $a \in A^{\prime}$, the set $A \cap J_\delta(a)$ is nonempty. Choose a point $x$ in this set. Then
$$\epsilon=|b-c| \leq|b-f(x)|+|f(x)-c|<\epsilon / 2+\epsilon / 2=\epsilon$$
It is worth noticing that the preceding proof cannot be made to work if $a$ is not required to be an accumulation point of $A$. To obtain a contradiction we must know that the condition $0<|x-a|<\delta$ is satisfied for at least one $x$ in the domain of $f$.

If $f: A \rightarrow \mathbb{R}$ where $A \subseteq \mathbb{R}$, and $a \in A^{\prime}$, then
$$\lim {h \rightarrow 0} f(a+h)=\lim {x \rightarrow a} f(x)$$
in the sense that if either limit exists, then so does the other and the two limits are equal.

Let $g$ : $h \mapsto f(a+h)$. Notice that $h \in \operatorname{dom} g$ if and only if $a+h \in \operatorname{dom} f$; $\operatorname{dom} f=a+\operatorname{dom} g$.
That is, $\operatorname{dom} f={a+h: h \in \operatorname{dom} g}$.
First we suppose that
$$l:=\lim {h \rightarrow 0} g(h)=\lim {h \rightarrow 0} f(a+h) \text { exists. }$$
We show that $\lim {x \rightarrow a} f(x)$ exists and equals $l$. Given $\epsilon>0$ there exists (by (Q.4)) a number $\delta>0$ such that $$|g(h)-l|<\epsilon \text { whenever } h \in \operatorname{dom} g \text { and } 0<|h|<\delta .$$ Now suppose that $x \in \operatorname{dom} f$ and $0<|x-a|<\delta$. Then by (Q.3) $$x-a \in \operatorname{dom} g$$ and by (Q.5) $|f(x)-l|=|g(x-a)-l|<\epsilon$. Thus given $\epsilon>0$ we have found $\delta>0$ such that $|f(x)-l|<\epsilon$ whenever $x \in \operatorname{dom} f$ and $0<$ $|x-a|<\delta$. That is, $\lim {x \rightarrow a} f(x)=l$.

The converse argument is similar. Suppose $l:=\lim {x \rightarrow a} f(x)$ exists. Given $\epsilon>0$ there exists $\delta>0$ such that $|f(x)-l|<\epsilon$ whenever $x \in \operatorname{dom} f$ and $0<|x-a|<\delta$. If $h \in \operatorname{dom} g$ and $0<|h|<\delta$, then $a+h \in \operatorname{dom} f$ and $0<|(a+h)-a|<\delta$. Therefore $|g(h)-l|=|f(a+h)-l|<\epsilon$, which shows that $\lim {h \rightarrow 0} g(h)=l$.