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MATH4604课程简介
Calculus is the mathematical study of continuous change, in the same way that geometry is the study of shape, and algebra is the study of generalizations of arithmetic operations.
Sequel to MATH 4603. Topology of n-dimensional Euclidean space. Rigorous treatment of multivariable differentiation and integration, including chain rule, Taylor’s Theorem, implicit function theorem, Fubini’s Theorem, change of variables, Stokes’ Theorem.
prereq: 4603 or 5615 or instr consent
Prerequisites
It has two major branches, differential calculus and integral calculus; the former concerns instantaneous rates of change, and the slopes of curves, while the latter concerns accumulation of quantities, and areas under or between curves. These two branches are related to each other by the fundamental theorem of calculus, and they make use of the fundamental notions of convergence of infinite sequences and infinite series to a well-defined limit.
MATH4604 Calculus HELP（EXAM HELP， ONLINE TUTOR）
Show: Let $V:=\mathbb{R}^2, W \in \mathrm{TS}, f: V \rightarrow W$.
Assume $0_V \in \mathbb{D}f^{\prime \prime}$. Then $\left(\partial_1 \partial_2 f\right){0_V}=\left(\partial_2 \partial_1 f\right)_{0_V}$.
Proof: Want: $\forall k \in \mathcal{I}W, \pi_k^W\left(\left(\partial_1 \partial_2 f\right){0_V}\right)=\pi_k^W\left(\left(\partial_2 \partial_1 f\right){0_V}\right)$. Given $k \in \mathcal{I}_W$. Want: $\pi_k^W\left(\left(\partial_1 \partial_2 f\right){0_V}\right)=\pi_k^W\left(\left(\partial_2 \partial_1 f\right){0_V}\right)$. We have $\partial_1 \partial_2 f: V \rightarrow W$, so $\mathbb{I}{\partial_1 \partial_2 f} \subseteq W$.
We have $\partial_2 \partial_1 f: V \rightarrow W$, so $\mathbb{I}{\partial_2 \partial_1 f} \subseteq W$. By a class theorem $\mathbb{D}_f^{\prime \prime} \subseteq \mathbb{D}{\partial_1 \partial_2 f}$ and $\mathbb{D}f^{\prime \prime} \subseteq \mathbb{D}{\partial_2 \partial_1 f}$.
Let $L:=\pi_k^W$. Then $L \in \mathcal{L}{\mathbb{R}}^W$. Then $\mathbb{D}_L=W$ and $\mathbb{D}_L^{\prime \prime}=W$. Want: $L\left(\left(\partial_1 \partial_2 f\right){0_V}\right)=L\left(\left(\partial_2 \partial_1 f\right)_{0_V}\right)$.
By $\mathrm{HW} # 12-2$, we have: $\partial_2(L \circ f) \supseteq L \circ\left(\partial_2 f\right)$.
Then $\partial_1 \partial_2(L \circ f) \supseteq \partial_1\left(L \circ\left(\partial_2 f\right)\right)$.
By HW#12-2, we have: $\partial_1\left(L \circ\left(\partial_2 f\right)\right) \supseteq L \circ\left(\partial_1 \partial_2 f\right)$
Then: $\partial_1 \partial_2(L \circ f) \supseteq \partial_1\left(L \circ\left(\partial_2 f\right)\right) \supseteq L \circ\left(\partial_1 \partial_2 f\right)$.
Then $\left(\partial_1 \partial_2(L \circ f)\right){0_V} \quad=\left(L \circ\left(\partial_1 \partial_2 f\right)\right){0_V}$.
We have $0_V \in \mathbb{D}f^{\prime \prime} \subseteq \mathbb{D}{\partial_1 \partial_2 f}$, so $\left(\partial_1 \partial_2 f\right){0_V} \in \mathbb{I}{\partial_1 \partial_2 f}$.
Since $\left(\partial_1 \partial_2 f\right){0_V} \in \mathbb{I}{\partial_1 \partial_2 f} \subseteq W=\mathbb{D}L$, we get $\left(L \circ\left(\partial_1 \partial_2 f\right)\right){0_V}=L\left(\left(\partial_1 \partial_2 f\right){0_V}\right)$. Then $\left(\partial_1 \partial_2(L \circ f)\right){0_V}==^\left(L \circ\left(\partial_1 \partial_2 f\right)\right){0_V}=L\left(\left(\partial_1 \partial_2 f\right){0_V}\right)$, so $\left(\partial_1 \partial_2(L \circ f)\right){0_V}=L\left(\left(\partial_1 \partial_2 f\right){0_V}\right)$
By HW#12-2, we have: $\partial_1(L \circ f) \supseteq L \circ\left(\partial_1 f\right)$.
Then $\partial_2 \partial_1(L \circ f) \supseteq \partial_2\left(L \circ\left(\partial_1 f\right)\right)$.
By HW#12-2, we have: $\partial_2\left(L \circ\left(\partial_1 f\right)\right) \supseteq L \circ\left(\partial_2 \partial_1 f\right)$.
Then: $\partial_2 \partial_1(L \circ f) \supseteq \partial_2\left(L \circ\left(\partial_1 f\right)\right) \supseteq L \circ\left(\partial_2 \partial_1 f\right)$.
Then $\left(\partial_2 \partial_1(L \circ f)\right){0_V} \quad=^\left(L \circ\left(\partial_2 \partial_1 f\right)\right){0_V}$.
We have $0_V \in \mathbb{D}f^{\prime \prime} \subseteq \mathbb{D}{\partial_2 \partial_1 f}$, so $\left(\partial_2 \partial_1 f\right){0_V} \in \mathbb{I}{\partial_2 \partial_1 f}$.
Since $\left(\partial_2 \partial_1 f\right){0_V} \in \mathbb{I}{\partial_2 \partial_1 f} \subseteq W=\mathbb{D}L$, we get $\left(L \circ\left(\partial_2 \partial_1 f\right)\right){0_V}=L\left(\left(\partial_2 \partial_1 f\right){0_V}\right)$. Then $\left(\partial_2 \partial_1(L \circ f)\right){0_V}==^\left(L \circ\left(\partial_2 \partial_1 f\right)\right){0_V}=L\left(\left(\partial_2 \partial_1 f\right){0_V}\right)$, so $\left(\partial_2 \partial_1(L \circ f)\right){0_V}=L\left(\left(\partial_2 \partial_1 f\right){0_V}\right)$,
Since $0_V \in \mathbb{D}f^{\prime \prime}$ and $f{0_V} \in W=\mathbb{D}L^{\prime \prime}$, we get: $0_V \in \mathbb{D}{f \circ L}^{\prime \prime}$.
Then, by HW#11-5, we have: $\left(\partial_1 \partial_2(L \circ f)\right){0_V}=\left(\partial_1 \partial_2(L \circ f)\right){0_V}$.
Then $L\left(\left(\partial_1 \partial_2 f\right){0_V}\right)=\left(\partial_1 \partial_2(L \circ f)\right){0_V}$
$=\left(\partial_2 \partial_1(L \circ f)\right){0_V}=L\left(\left(\partial_2 \partial_1 f\right){0_V}\right) . \quad$ QED
Show: Let $V, W \in \mathrm{TS}, q, u, v \in V, f: V \rightarrow W, \phi:=f \circ \alpha_q^{u v}$. Then $\left(\nabla_u \nabla_v f\right)q$ 兰 $\left(\partial_1 \partial_2 \phi\right){(0,0)}$.
Proof: Want: $\left(\left(\nabla_v f\right) \circ \alpha_q^u\right)0^{\prime} \quad =\left(\left(\partial_2 \phi\right) \circ \alpha{(0,0)}^{(1,0)}\right)0^{\prime}$. Want: $\left(\left(\nabla_v f\right) \circ \alpha_q^u\right)^{\prime}=\left(\left(\partial_2 \phi\right) \circ \alpha{(0,0)}^{(1,0)}\right)^{\prime}$.
Want: $\left(\nabla_v f\right) \circ \alpha_q^u=\left(\partial_2 \phi\right) \circ \alpha_{(0,0)}^{(1,0)}$.
Given $s \in \mathbb{R}$. Want: $\left(\left(\nabla_v f\right) \circ \alpha_q^u\right)s \quad \underline{x} \quad\left(\left(\partial_2 \phi\right) \circ \alpha{(0,0)}^{(1,0)}\right)s$. We have $\left(\alpha_q^u\right)_s=q+s u$. We have $\left(\alpha{(0,0)}^{(1,0)}\right)s=(s, 0)$. Want: $\left(f \circ \alpha{q+s u}^v\right)0^{\prime}$ 关 $\left(\phi \circ \alpha{(s, 0)}^{(0,1)}\right)0^{\prime}$. Want: $\left(f \circ \alpha{q+s u}^v\right)^{\prime}=\left(\phi \circ \alpha_{(s, 0)}^{(0,1)}\right)^{\prime}$.
Want: $f \circ \alpha_{q+s u}^v=\phi \circ \alpha_{(s, 0)}^{(0,1)}$.
By definition of $\phi$, we have: $\phi=f \circ \alpha_q^{u v}$.
Want: $f \circ \alpha_{q+s u}^v=f \circ \alpha_q^{u v} \circ \alpha_{(s, 0)}^{(0,1)}$.
Want: $\alpha_{q+s u}^v=\alpha_q^{u v} \circ \alpha_{(s, 0)}^{(0,1)}$.
Want: $\quad \forall t \in \mathbb{R}, \quad\left(\alpha_{q+s u}^v\right)t=\left(\alpha_q^{u v} \circ \alpha{(s, 0)}^{(0,1)}\right)t$. Given $t \in \mathbb{R}$. Want: $\left(\alpha{q+s u}^v\right)t=\left(\alpha_q^{u v} \circ \alpha{(s, 0)}^{(0,1)}\right)t$. We have $\left(\alpha{q+s u}^v\right)t=q+s u+t v$. We have $\left(\alpha{(s, 0)}^{(0,1)}\right)t=(s, t)$. Also, $\left(\alpha_q^{u v}\right){(s, t)}=q+s u+t v$.
Then $\left(\alpha_q^{u v} \circ \alpha_{(s, 0)}^{(0,1)}\right)t=\left(\alpha_q^{u v}\right){(s, t)}=q+s u+t v$.
Then $\left(\alpha_{q+s u}^v\right)t^v=q+s u+t v=\left(\alpha_q^{u v} \circ \alpha{(s, 0)}^{(0,1)}\right)_t . \quad$ QED
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