Assignment-daixieTM为您提供杜克大学Duke UniversityMATH 222 Advanced Multivariable Calculu高级多变量微积分代写代考和辅导服务！
Instructions:
This course offers a brief introduction to the multivariate calculus required to build many common machine learning techniques. We start at the very beginning with a refresher on the “rise over run” formulation of a slope, before converting this to the formal definition of the gradient of a function. We then start to build up a set of tools for making calculus easier and faster. Next, we learn how to calculate vectors that point up hill on multidimensional surfaces and even put this into action using an interactive game. We take a look at how we can use calculus to build approximations to functions, as well as helping us to quantify how accurate we should expect those approximations to be.
A function $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous if and only if $f^{\leftarrow}(U)$ is open whenever $U$ is open in $\mathbb{R}$.
Suppose $\mathrm{f}$ is continuous. Let $V$ be an open subset of $\mathbb{R}$. To show that $f^{\leftarrow}(V)$ is open it suffices to prove that each point of $f^{\leftarrow}(V)$ is an interior point of that set. (Notice that if $f^{\leftarrow}(V)$ is empty, then there is nothing to prove. The null set is open.) If $a \in f^{\leftarrow}(V)$, then $V$ is a neighborhood of $f(a)$. Since $f$ is continuous at $a$, the set $f^{\leftarrow}(V)$ contains a neighborhood of $a$, from which we infer that $a$ is an interior point of $f^{\leftarrow}(V)$.
Conversely, suppose that $f^{\leftarrow}(V) \subseteq \mathbb{R}$ whenever $V \subseteq \mathbb{R}$. To see that $f$ is continuous at an arbitrary point $a$ in $\mathbb{R}$, notice that if $V$ is a neighborhood of $f(a)$, then $a \in f^{\leftarrow}(V) \subseteq \mathbb{R}$. That is, $f^{\leftarrow}(V)$ is a neighborhood of $a$. So $f$ is continuous at $a$.
Find the set of all points on the real line that are within 5 units of the number −2.
Find those numbers x such that d(x, −2) ≤ 5. In other words, solve the
inequality
|x + 2| = |x − (−2)| ≤ 5.
This may be rewritten as
−5 ≤ x + 2 ≤ 5,
which is the same as
−7 ≤ x ≤ 3.
Thus the points in the closed interval [−7, 3] are those that lie within 5 units of −2.
Proposition. If $A$ and $B$ are sets of real numbers, then
$$
(A \cap B)^{\circ}=A^{\circ} \cap B^{\circ} .
$$
Hint. Show separately that $(A \cap B)^{\circ} \subseteq A^{\circ} \cap B^{\circ}$ and that $A^{\circ} \cap B^{\circ} \subseteq(A \cap B)^{\circ}$.
Since $A \cap B \subseteq A$ we have $(A \cap B)^{\circ} \subseteq A^{\circ}$ by 1.2.9. Similarly, $(A \cap B)^{\circ} \subseteq$ $B^{\circ}$. Thus $(A \cap B)^{\circ} \subseteq A^{\circ} \cap B^{\circ}$. To obtain the reverse inclusion take $x \in A^{\circ} \cap B^{\circ}$. Then there exist $\epsilon_1, \epsilon_2>0$ such that $J_{\epsilon_1}(x) \subseteq A$ and $J_{\epsilon_2}(x) \subseteq B$. Then $J_\epsilon(x) \subseteq A \cap B$ where $\epsilon=\min \left{\epsilon_1, \epsilon_2\right}$. This shows that $x \in(A \cap B)^{\circ}$.
