数学代写|Math444 Mathematical Analysis

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Math444课程简介

This course is an introduction to ε – δ analysis on real numbers, which makes what the students have learned from calculus courses rigorous. This course is for students who do not plan to do graduate study in Mathematics (those students should take Math 447). Topics covered by Math 444 include the real number system, limits, continuity, derivatives, the Darboux integral, the Riemann integral, and sequences of functions.

Prerequisites 

Exams: This course has three 90-minute midterm tests and a 3-hour final exam.

Proctorship Information: Exams in this course may be taken online.

Students with a course start date prior to January 1st, 2023 should use proctor information on this page to take their exams.

Students with a course start date on or after January 1st, 2023 should refer to this page for exam taking information.

Once registered, students will find relevant exam taking information and all other course requirements within their Nexus student dashboard.

Math444 Mathematical Analysis HELP（EXAM HELP， ONLINE TUTOR）

First, note that $S$ is nonempty, since $0$ is a lower bound of $S$ (since $x < \frac{1}{x}$ for all $x>1$) and $S$ is bounded above by $-1$ (since $x < \frac{1}{x}$ implies $x^2 < 1$ and hence $|x|<1$, so $x\leq -1$ cannot hold). Thus, by the completeness axiom, $S$ has an infimum and a supremum.

To find the infimum and supremum, we consider the function $f(x) = x – \frac{1}{x}$ defined on $(0,\infty)$. Note that $x < \frac{1}{x}$ if and only if $f(x) < 0$, so $S$ is precisely the set of positive real numbers $x$ such that $f(x) < 0$. We will use calculus to find the minimum and maximum values of $f(x)$ on $(0,\infty)$, which will enable us to determine the infimum and supremum of $S$.

Taking the derivative of $\$ f(x) \$$ with respect to $\$ x \$$, we find that
$$
f^{\prime}(x)=1+\frac{1}{x^2} .
$$

Setting $f'(x) = 0$, we see that $x = 1$ is the only critical point of $f(x)$ on $(0,\infty)$. Moreover, $f”(x) = -\frac{2}{x^3} < 0$ for all $x>0$, so $x=1$ is a global maximum of $f(x)$ on $(0,\infty)$.

Thus, $f(x)$ is strictly decreasing on $(0,1]$ and strictly increasing on $[1,\infty)$. In particular, $f(x) < f(1) = 0$ for all $x \in (0,1]$ and $f(x) < f(1) = 0$ for all $x \in [1,\infty)$. Therefore, the supremum of $S$ is $1$ and the infimum of $S$ is $-\infty$.

To summarize: we have shown that $S$ is nonempty and bounded above by $-1$, so by the completeness axiom, $S$ has a supremum. We have also shown that $f(x) = x – \frac{1}{x}$ is a strictly decreasing function on $(0,1]$ and a strictly increasing function on $[1,\infty)$, with a global maximum of $0$ at $x=1$. Thus, $S$ consists precisely of the positive real numbers less than $1$, and hence the supremum of $S$ is $1$. Since $S$ has no lower bound, the infimum of $S$ is $-\infty$.

Since $S$ is non-empty and bounded from above, it follows that $\sup S$ exists. Let $T = {-s : s \in S}$, which is also non-empty since $S$ is non-empty. We claim that $\inf T$ exists and is equal to $-\sup S$.

First, we will show that $-\sup S$ is a lower bound of $T$. Let $t \in T$, so that $t = -s$ for some $s \in S$. Since $s \leq \sup S$, we have $-s \geq -\sup S$, which implies that $t \geq -\sup S$. Therefore, $-\sup S$ is a lower bound of $T$.

Next, we will show that $-\sup S$ is the greatest lower bound of $T$. Let $l$ be any lower bound of $T$. Then $-l$ is an upper bound of $S$, since $-s \leq -l$ implies $s \geq l$ for all $s \in S$. Therefore, $\sup S \leq -l$, or equivalently, $l \leq -\sup S$. Since $l$ was an arbitrary lower bound of $T$, we conclude that $-\sup S$ is the greatest lower bound of $T$.

Hence, $\inf T = -\sup S$. Finally, note that $s \leq \sup S$ for all $s \in S$, which implies that $-s \geq -\sup S$ for all $s \in S$, or equivalently, $-s \in T$ for all $s \in S$. Therefore, $-\sup S$ is a upper bound of $T$, and since it is the least upper bound of $T$, we have $\sup T = -\inf T = -\sup S$.

In summary, we have shown that $\inf T = -\sup S$ and $\sup T = -\inf T = -\sup S$, which implies that $\sup S = -\inf T$.

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