这是一份uwa西澳大学MATH3033的成功案例

几何学|MATH3033 Geometry代写 UWA代写

Concretely parametrize the sphere in the usual way:
$$
x_{1}=\sin \theta \sin \phi, \quad x_{2}=\sin \theta \cos \phi, \quad x_{3}=\cos \theta
$$
then with the poles removed the range of values is $0<\theta<\pi, 0 \leq \phi<2 \pi$. The antipodal map is
$$
\theta \mapsto \pi-\theta, \quad \phi \mapsto \phi+\pi .
$$
We can therefore identify the space of lines in $\mathbf{R}^{2}$ as the pairs
$$
(\theta, \phi) \in(0, \pi) \times[0, \pi]
$$

MATH3033 COURSE NOTES ：

We can add symmetric bilinear forms: $(B+C)(v, w)=B(v, w)+C(v, w)$ and multiply by a scalar $(\lambda B)(v, w)=\lambda B(v, w)$ so they form a vector space isomorphic to the space of symmetric $n \times n$ matrices which has dimension $n(n+1) / 2$. If we take a different basis
$$
w_{i}=\sum_{j} P_{j i} v_{j}
$$
then
$$
B\left(w_{i}, w_{j}\right)=B\left(\sum_{k} P_{k i} v_{k}, \sum_{\ell} P_{\ell j} v_{\ell}\right)=\sum_{k, \ell} P_{k i} B\left(v_{k}, v_{\ell}\right) P_{\ell j}
$$
so that the matrix $\beta_{i j}=B\left(v_{i}, v_{j}\right)$ changes under a change of basis to
$$
\beta^{\prime}=P^{T} \beta P .
$$

这是一份uwa西澳大学MATH3032的成功案例

拓扑结构和分析|MATH3032 Topology and Analysis代写 UWA代写

By Fubini’s theorem, it follows that for $k \neq j$ that
$$
\int_{\Omega_{n}}\left(X_{k}-x\right)\left(X_{j}-x\right) d \mu_{z}^{n}=\left[\int_{{0,1}}(\omega-x) d \mu_{z}(\omega)\right]^{2}=0
$$
and
$$
\int_{\Omega_{n}}\left(X_{k}-x\right)^{2} d \mu_{z}^{n}=\int_{{0,1}}(\omega-x)^{2} d \mu_{z}(\omega)=(1-x)^{2} x+x^{2}(1-x) \leq 2
$$
Combining the last three displayed equations shows that
$$
\mu_{z}^{n}\left(\left|S_{n}-x\right|>\epsilon\right) \leq \frac{1}{n^{2} \epsilon^{2}} 2 n=\frac{2}{n \epsilon^{2}}
$$
which combined with $\mathrm{Eq}$. (5.4) implies that
$$
\sup {z \in[0,1]}\left|f(x)-p{n}(x)\right| \leq \frac{4 M}{n \epsilon^{2}}+\delta_{c}
$$
and therefore
$$
\lim {n \rightarrow \infty} \sup \sup {z \in[0,1]}\left|f(x)-p_{n}(x)\right| \leq \delta_{c} \rightarrow 0 \text { as } \epsilon \rightarrow 0
$$

MATH3032 COURSE NOTES ：

To prove the claim, let $V_{y}$ be an open neighborhood of $y$ such that $\left|f-g_{z y}\right|<\epsilon$ on $V_{y}$ so in particular $f<\epsilon+g_{z y}$ on $V_{y}$. By compactness, there exists $\Lambda \subset \subset X$ such that $X=\bigcup_{y \in \Lambda} V_{y}$. Set
$$
g_{z}(z)=\max \left{g_{z y}(z): y \in \Lambda\right},
$$
then for any $y \in \Lambda, f<\epsilon+g_{z y}<\epsilon+g_{z}$ on $V_{y}$ and therefore $f<\epsilon+g_{z}$ on $X$. Moreover, by construction $f(x)=g_{z}(x)$.

We now will finish the proof of the theorem. For each $x \in X$, let $U_{z}$ be a neighborhood of $x$ such that $\left|f-g_{z}\right|<\epsilon$ on $U_{z}$. Choose $\Gamma \subset \subset X$ such that $X=\bigcup_{z \in \Gamma} U_{z}$ and define $$ g=\min \left{g_{z}: x \in \Gamma\right} \in \mathcal{A} . $$ Then $f0$ is arbitrary it follows that $f \in \overline{\mathcal{A}}=\mathcal{A}$.