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这是美国康涅狄格大学微积分课程的代写成功案例。
Math1131课程简介
Course communication: Much of the communication for this course will occur online, via HuskyCT, email, and/or this page. Please monitor this page, course HuskyCT announcements, and your UConn email regularly for updates related to MATH 1131Q. However, you are also expected to be in class every day to receive important communications there, and to contact a classmate, or your instructor or TA, if you miss class to make sure you have not missed important course information.
Prerequisites
Time Zone: All times for due dates, online and in-person classes, office hours, Supplemental Instruction, and Q Center hours in this course are given in US Eastern Time.
Description: This course is an introduction to differential and integral calculus, which is the mathematical language used in any science concerned with dynamically changing quantities. The main topics it covers are limits, derivatives, integrals, the Fundamental Theorem of Calculus, and some basic applications of these ideas.
Prerequisites: A qualifying score of 22 on the mathematics placement exam (MPE). Students who fail to achieve this minimum score are required to spend time on the preparatory and learning modules before re-taking the MPE, or to register for a lower level Mathematics course. Not open for credit to students who have passed MATH 1132Q or 1152Q.
Math1131 Calculus HELP（EXAM HELP， ONLINE TUTOR）
Show: Let $V, W \in \operatorname{TS}, f: V \rightarrow W, q \in \mathbb{D}_f^{\prime \prime}, u, v \in V$.
Then: $\quad\left(\nabla_u \nabla_v f\right)_q=\left(\nabla_v \nabla_u f\right)_q$.
Show: Let $V, W \in \mathrm{TS}, f: V \cdots W, q \in \mathbb{D}f^{\prime \prime}, u, v \in V$. Then: $\quad\left(\nabla_u \nabla_v f\right)_q=\left(\nabla_v \nabla_u f\right)_q$. Proof: By HW#12-4, we have: $\left(\nabla_u \nabla_v f\right)_q$ 生 $\left(\partial_1 \partial_2 \phi\right){(0,0)}$.
By a class theorem, we have: $\left(\nabla_v \nabla_u f\right)q \stackrel{*}{=}\left(\partial_2 \partial_1 \phi\right){(0,0)}$.
Let $U:=\mathbb{R}^2$. Then $(0,0)=0_U$.
Then: $\left(\nabla_u \nabla_v f\right)q$ 兰 $\left(\partial_1 \partial_2 \phi\right){0_U}$ and $\left(\nabla_v \nabla_u f\right)q$ 兰 $\left(\partial_2 \partial_1 \phi\right){0_U}$.
By HW#12-3 (with $V$ replaced by $U)$, we have: $\left(\partial_1 \partial_2 \phi\right){0_U}=\left(\partial_2 \partial_1 \phi\right){0_U}$.
Then: $\quad\left(\nabla_u \nabla_v f\right)q=\left(\partial_1 \partial_2 \phi\right){0_U}=\left(\partial_2 \partial_1 \phi\right)_{0_U}=\left(\nabla_v \nabla_u f\right)_q$. QED
Show: Let $V:=\mathbb{R}^2, f: V \rightsquigarrow \mathbb{R}$.
Assume: $0=\left(\partial_2 f\right){0_V}=\left(\partial_1 \partial_2 f\right){0_V}=\left(\partial_2 \partial_2 f\right){0_V}$. Assume: $0_V \in \mathbb{D}{\partial_2 f}^{\prime}$. Then: $\triangle^2 f \in \mathcal{O}_{\mathbb{R}}^V(2)$
Show: Let $V:=\mathbb{R}^2, f: V \rightarrow \mathbb{R}$.
Assume: $0=\left(\partial_2 f\right){0_V}=\left(\partial_1 \partial_2 f\right){0_V}=\left(\partial_2 \partial_2 f\right){0_V}$ Assume: $0_V \in \mathbb{D}{\partial_2 f}^{\prime}$. Then: $\triangle^2 f \in \mathcal{O}{\mathbb{R}}^V(2)$. Proof: Let $g:=\partial_2 f$. Then $0=g{0_V}=\left(\partial_1 g\right){0_V}=\left(\partial_2 g\right){0_V}$.
Then, by a class theorem $g \in \mathcal{O}{\mathbb{R}}^V(1)$. Since $\partial_2 f=g \in \mathcal{O}{\mathbb{R}}^V(1)$
by a class theorem, we get: $\triangle^2 f \in \mathcal{O}_{\mathbb{R}}^V(2)$. QED
Show: Let $V:=\mathbb{R}^2, f: V \rightarrow \mathbb{R}$.
Assume: $0=f_{0_V}=\left(\partial_1 f\right){0_V}=\left(\partial_1 \partial_1 f\right){0_V}$.
Then: $\triangle^1 f \in \mathcal{O}_{\mathbb{R}}^V(2)$
Show: Let $V:=\mathbb{R}^2, f: V \longrightarrow \mathbb{R}$.
Assume: $0=f_{0_V}=\left(\partial_1 f\right){0_V}=\left(\partial_1 \partial_1 f\right){0_V}$. Then: $\triangle^1 f \in \mathcal{O}{\mathbb{R}}^V(2)$. Proof: Let $z:=0_V, u:=\varepsilon_1^V$. Then $0=f_z=\left(\nabla_u f\right)_z=\left(\nabla_u \nabla_u f\right)_z$. Then $0=\left(f \circ \alpha_z^u\right)_0=\left(\left(\nabla_u f\right) \circ \alpha_z^u\right)_0=\left(\left(\nabla_u \nabla_u f\right) \circ \alpha_z^u\right)_0$. Since $z=0_V$, we get: $\quad \forall t \in \mathbb{R},\left(\alpha_z^u\right)_t=t \cdot u$. Let $g:=f \circ \alpha_z^u$. Then: $\forall t \in \mathbb{R}, g_t$ 兰 $f{t \cdot u}$. So, since $f_{0 \cdot u}=f_z=0$, we get $g_0=0$. Since $g=f \circ \alpha_z^u$, by a class theorem, we get:
$$
g^{\prime}=\left(\nabla_u f\right) \circ \alpha_z^u \quad \text { and } g^{\prime \prime}=\left(\nabla_u \nabla_u f\right) \circ \alpha_z^u .
$$
Then $g_0^{\prime}$ 生 $\left(\nabla_u f\right)z=0$, so $g_0^{\prime}=0$. Since $0=g_0=g_0^{\prime}=g_0^{\prime \prime}$, by a class theorem, we get: $g \in \mathcal{O}{\mathbb{R}}^{\mathbb{R}}(2)$.
Since $\pi_1^V \in \mathcal{L}{\mathbb{R}}^V=\mathcal{H}{\mathbb{R}}^V(1) \subseteq \widehat{\mathcal{O}}{\mathbb{R}}^V(1)$ and since $g \in \mathcal{O}{\mathbb{R}}^{\mathbb{R}}(2)$
we get: $g \circ \pi_1^V \in \mathcal{O}{\mathbb{R}}^V(2)$. Want: $\triangle^1 f=g \circ \pi_1^V$. Want: $\forall v \in V, \quad\left(\triangle^1 f\right)_v$ 是 $\left(g \circ \pi_1^V\right)_v$. Given $v \in V$. Want: $\left(\triangle^1 f\right)_v \quad \stackrel{}{=}\left(g \circ \pi_1^V\right)_v$. Let $x:=v_1$. Then $\pi_1^V(v)=x$ and $K^1 C_v^V=(x, 0)$. We have $\left(g \circ \pi_1^V\right)_v$ 兰 $g_x$ 兰 $f{x \cdot u}$. Want: $\left(\triangle^1 f\right)v$ 兰 $f{x \cdot u \cdot}$.
We have $K^1 C_v^V=(x, 0)=x \cdot(1,0)=x \cdot \varepsilon_1^V=x \cdot u$,
We have $K^0 C_v^V=0_V=z, \quad$ so $\left(f \circ K^0 C^V\right)_v \quad$ 恙 $f_z$.
So, since $f_z=0$, we get: $\left(f \circ K^0 C^V\right)_v=0$.
Then $\left(\triangle^1 f\right)_v \quad \stackrel{}{=}\left(\left(f \circ K^1 C^V\right)-\left(f \circ K^0 C^V\right)\right)_v$
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