# 高级衍射几何学与拓扑学 5M:Adv Diff Geom & Topology MATHS5039_1 5M

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Proof of (b) is easy:
$$d\left(y_{n}, x\right) \leq d\left(y_{n}, x_{n}\right)+d\left(x_{n}, x\right) \rightarrow 0 .$$
(c) is also easy. Look at
$$d\left(x_{n}, y_{n}\right) \leq d\left(x_{n}, x_{n}^{\prime}\right)+d\left(x_{n}^{\prime}, y_{n}^{\prime}\right)+d\left(y_{n}^{\prime}, y_{n}\right) .$$
Taking limits, we get $\lim {n} d\left(x{n}, y_{n}\right) \leq \lim {n} d\left(x{n}^{\prime}, y_{n}^{\prime}\right)$. Similar argument shows the other way inequality and hence the proof.

## MATHS5039_1 COURSE NOTES ：

For, if $\varepsilon>0$ is given, by the Cauchy nature of $\left(x_{n}\right)$, there exists $N \in \mathbb{N}$ such that $d\left(x_{m}, x_{n}\right)<\varepsilon$ for $m, n \geq N$. We have, for $n \geq N$,
$$d\left(\tilde{x}{n}, \xi\right):=\lim {m \rightarrow \infty} d\left(x_{n}, x_{m}\right)<\varepsilon .$$
It is clear that the map $\varphi: X \rightarrow \tilde{X}$ given by $\varphi(x)=\tilde{x}$ is an isometry. $d(\varphi(x), \varphi(y)):=\lim _{n} d(x, y)$, the limit of a constant sequence.
We claim that the image $\varphi(X)$ of $X$ under the map $\varphi: x \mapsto \tilde{x}$ is dense

# 数学 MATH MATHS2025_1/MATHS3016_1

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Let $f$ be a function from a nonempty open subset $E$ of $\mathbb{R}$ to $\mathbb{R}$. The function $f$ is said to be differentiable at $c \in E$ if
$$\lim {x \rightarrow c} \frac{f(x)-f(c)}{x-c}$$ or, equivalently, $$\lim {h \rightarrow 0} \frac{f(c+h)-f(c)}{h}$$
exists. This limit (if it exists) is called the derivative of $f$ at $c$. If the derivative of $f$ exists at every $c \in E$, then $f$ is said to be differentiable on $E$ (or just differentiable). The derivative of $f$ as a function from $E$ to $\mathbb{R}$ is denoted by
$$f^{\prime} \text { or } \frac{d f}{d x}$$
Note that the limit in Eq. (7.1) is understood as the limit of the function
$$g(x)=\frac{f(x)-f(c)}{x-c}, \quad x \in E \backslash{c}$$

## MATHS2025_1/MATHS3016_1COURSE NOTES ：

From the definition of $f^{\prime}(c)$, it follows that for every $\varepsilon>0$, there exists $\delta>0$ such that $x \in E,|x-c|<\delta$, and $x \neq c$ imply
$$\left|\frac{f(x)-f(c)}{x-c}-f^{\prime}(c)\right|<\varepsilon .$$
Thus, for every $x \in E$ with $|x-c|<\delta$,
$$|f(x)-\varphi(x)| \leq \varepsilon|x-c|,$$
where $\varphi$ is the linear function defined by
$$\varphi(x)=f(c)+f^{\prime}(c)(x-c), \quad x \in \mathbb{R}$$

# 复杂分析的方法 Methods in Complex Analysis MATHS4076_1

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To find the best probability assignment subject to the constraint of given mean, we use the MaxEnt method. I.e. maximse
$$S[p]=-k_{\mathrm{B}} \sum_{k \geq 0} p(k) \ln p(k)$$
subject to the constraint
$$\sum_{k \geq 0} k p(k)=\mu .$$
using the method of Lagrangian multipliers to take constraints into account.
To simplify notation we can assume $k_{\mathrm{B}}=1$; this amounts to redefining Lagrangian multipliers in units of $k_{\mathrm{B}}$ (can you see why?). Thus define
$$\mathcal{L}[p]=-\sum_{k \geq 0} p(k) \ln p(k)+\lambda_{0}\left(\sum_{k \geq 0} p(k)-1\right)+\lambda_{1}\left(\sum_{k \geq 0} k p(k)-\mu\right)$$

## MATHS4076_1COURSE NOTES ：

$$\mu=\sum_{k \geq 0} k p(k)=\left(1-\mathrm{e}^{\lambda_{1}}\right) \sum_{k=1}^{\infty} k \mathrm{e}^{\lambda_{1} k}=\frac{\mathrm{e}^{\lambda_{1}}}{1-\mathrm{e}^{\lambda_{1}}}$$
required by the constraint of the given mean. Hence
$$\mathrm{e}^{\lambda_{1}}=\frac{\mu}{1+\mu} \quad \Longrightarrow \quad p(k)=\frac{1}{1+\mu}\left(\frac{\mu}{1+\mu}\right)^{k}$$
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# 随机过程 Stochastic Processes STATS4024_1 /STATS5026_1

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from which it follows that, by monotonicity,
$$\forall n \in \mathbb{N}: \quad E\left[Y_{n+1} \mid X=x\right] \geq E\left[Y_{n} \mid X=x\right], \quad P_{X} \text {-a.s. }$$
Moreover,
$$\forall B \in \mathcal{B}: \quad \int_{[X \in B]} Y_{n} d P=\int_{B} E\left[Y_{n} \mid X=x\right] d P_{X}(x)$$
and
$$\forall B \in \mathcal{B}: \quad \int_{[X \in B]} Y d P \geq \int_{[X \in B]} Y_{n} d P$$
Thus
$$E[Y \mid X=x] \geq E\left[Y_{n} \mid X=x\right], \quad P_{X} \text {-a.s. }$$

## STATS4024_1 /STATS5026_1COURSE NOTES ：

Therefore,
$$P([Y \in A] \mid \cdot)=P([Y \in A]), \quad P_{X} \text {-a.s. }$$
and if $Y$ is a real-valued integrable random variable, then
$$E[Y \mid \cdot]=E[Y], \quad P_{X}-a . s .$$
Proof: Independence of $X$ and $Y$ is equivalent to
$$P([X \in B] \cap[Y \in A])=P([X \in B]) P([Y \in A]) \quad \forall A \in \mathcal{B}{1}, B \in \mathcal{B}$$ or \begin{aligned} \int{[X \in B]} I_{[Y \in A]}(\omega) P(d \omega) &=P([Y \in A]) \int I_{B}(x) d P_{X}(x) \ &=\int_{D} P([Y \in A]) d P_{X}(x) \end{aligned}

# 进一步的复杂分析 Further Complex Analysis MATHS5070_1 /MATHS4104_1

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$$l_{\mathrm{p}}(\psi)=c-\frac{n}{2} \log (\mathrm{SSB}+\lambda \mathrm{SSW})+\frac{n-m}{2} \log \lambda,$$
where $\lambda=1+k \psi$. The maximum of $l_{\mathrm{p}}$ is given by
$$\hat{\lambda}=\left(1-\frac{1}{m}\right) \frac{\mathrm{MSB}}{\mathrm{MSW}},$$
where MSB $=\operatorname{SSB} /(m-1)$ and MSW $=\operatorname{SSW} /(n-m)$, or
$$## MATHS5070_1 /MATHS4104_1COURSE NOTES ：$$
\hat{F}{r}(x)=\frac{1}{m{r}} \sum_{u=1}^{m_{r}} 1_{\left(\hat{\alpha}{r}, u \leq x\right)} \stackrel{P}{\longrightarrow} F{r}(x), \quad x \in C\left(F_{r}\right),
$$where \hat{\alpha}{r, u} is the u th component of \hat{\alpha}{r}, 1 \leq r \leq s, and$$
\hat{F}{0}(x)=\frac{1}{n} \sum{u=1}^{n} 1_{\left(\tilde{\epsilon}{u} \leq x\right)} \stackrel{P}{\longrightarrow} F{0}(x), \quad x \in C\left(F_{0}\right),
$$# 线性混合模型 Linear Mixed Models STATS5054_1/STATS4045_1 0 这是一份GLA格拉斯哥大MATHS5077_1作业代写的成功案例 The methods discussed above for the full parameter vector now directly carry over to calculate local influences on the geometric surface defined by \mathrm{LD}{1}(\omega). We now partition \ddot{L} as$$ \ddot{L}=\left(\begin{array}{ll} \ddot{L}{11} & \ddot{L}{12} \ \ddot{L}{21} & \ddot{L}{22} \end{array}\right), $$according to the dimensions of \theta{1} and \theta_{2}. Cook (1986) has then shown that the local influence on the estimation of \theta_{1}, of perturbing the model in the direction of a normalized vector \boldsymbol{h}, is given by$$
C_{h}\left(\boldsymbol{\theta}{1}\right)=2\left|\boldsymbol{h}^{\prime} \Delta^{\prime}\left[\ddot{L}^{-1}-\left(\begin{array}{cc} 0 & 0 \ 0 & \ddot{L}{22}^{-1}
\end{array}\right)\right] \Delta \boldsymbol{h}\right|
$$Because all eigenvalues of the malrix$$
\left(\begin{array}{ll}
\ddot{L}{11} & \ddot{L}{12} \
\ddot{L}{21} & \ddot{L}{22}
\end{array}\right)\left(\begin{array}{cc}
0 & 0 \
0 & \ddot{L}{22}^{-1} \end{array}\right)=\left(\begin{array}{cc} 0 & \ddot{L}{12} \ddot{L}_{22}^{-1} \
0 & I
\end{array}\right)
$$## MATHS5077_1COURSE NOTES ： A measure of influence, proposed , is then$$
\rho_{i}=-\left(\hat{\boldsymbol{\theta}}-\hat{\boldsymbol{\theta}}{(i)}^{1}\right)^{\prime} \ddot{L}\left(\hat{\boldsymbol{\theta}}-\hat{\boldsymbol{\theta}}{(i)}^{1}\right)

\boldsymbol{\Delta}{i}=-\sum{j \neq i} \boldsymbol{\Delta}{j}=\ddot{L}{(i)}(\widehat{\boldsymbol{\theta}})\left(\hat{\boldsymbol{\theta}}{(i)}^{1}-\widehat{\boldsymbol{\theta}}\right) $$such that expression (11.4) becomes$$ C{i}=-2\left(\hat{\boldsymbol{\theta}}-\hat{\boldsymbol{\theta}}{(i)}^{1}\right)^{\prime} \ddot{L}{(i)} \ddot{L}^{-1} \ddot{L}{(i)}\left(\hat{\boldsymbol{\theta}}-\hat{\boldsymbol{\theta}}{(i)}^{1}\right)
$$# 代数专题 : Topics in Algebra MATHS5077_1 0 这是一份GLA格拉斯哥大MATHS5077_1作业代写的成功案例 Although the inverse of a homomorphism may not exist, a homomorphism does share some properties in common with isomorphisms. For example, if \theta: G \rightarrow G^{\prime} is a homomorphism, and if e and e^{\prime} are the identities in G and G^{\prime}, respectively, then \theta(e)=\theta\left(e^{2}\right)=\theta(e) \theta(e), so that \theta(e)=e^{\prime}. Thus a homomorphism maps the identity in G to the identity in G^{\prime}. Similarly, as e^{\prime}=\theta(e)=\theta\left(g g^{-1}\right)= \theta(g) \theta\left(g^{-1}\right), we see that for all g in G$$
\theta(g)^{-1}=\theta\left(g^{-1}\right) .
$$The most obvious example of a homomorphism is a linear map between vector spaces. Indeed, a vector space is a group with respect to addition, and as any linear map \alpha: V \rightarrow W satisfies$$
\alpha(u+v)=\alpha(u)+\alpha(v)
$$## MATHS5077_1COURSE NOTES ： Proof First, every element of x_{0} K is a solution to the given equation, for the general element of x_{0} K is x_{0} k, where k \in K, and \theta\left(x_{0} k\right)=\theta\left(x_{0}\right) \theta(k)=y e^{\prime}= y. Next suppose that \theta(x)=y, and consider the element x_{0}^{-1} x in G. Then$$
\theta\left(x_{0}^{-1} x\right)=\theta\left(x_{0}^{-1}\right) \theta(x)=\left(\theta\left(x_{0}\right)\right)^{-1} y=y^{-1} y=e^{\prime},
$$so that x_{0}^{-1} x \in K. This means that x \in x_{0} K and so the set of solutions of \theta(x)=y is the coset x_{0} K. # 通用线性模型 Generalised Linear Models STATS4043_1 0 这是一份GLA格拉斯哥大STATS4043_1作业代写的成功案例 warning message, Clearly missing value cannot be allowed in certain contexts and wil1 be faulted, for instance an array cannot be given a shape containing a wissing value. In order to allow the user to detect missing values and replace them, if required, three special functions are supplied;$$
\begin{aligned}
\text { \&EQMN }(X) &=1 \text { (true) if } x=* \
&=0 \text { (false) otherwise } \
\text { xMYV }(X ; Y) &=* \text { if } y=1 \quad \text { (true) } \
&=x \text { if } y=0 \quad \text { (false) } \
\not{\gamma V M}(X ; Y) &=x \text { if } x \neq * \
&=y \text { if } x=*
\end{aligned}
$$## STATS4043_1COURSE NOTES ： The likelihood function can be taken to be$$
L(u)=\exp \left(-\frac{1}{2} \frac{\Sigma\left(y_{i}-\mu\right)^{2}}{a^{2}}\right)
$$with 10 g-1 ikelihood function$$
\ell(\mu)=-\frac{1}{2} \frac{\varepsilon\left(\gamma_{i}-\mu\right)^{2}}{\sigma^{2}}
$$Following through the usual maximam 1 ikelihood calculations, we have$$
\begin{aligned}
\mathbb{E}^{\prime}(u) &=E\left(y_{i}-\mu\right) / d^{2} \
Q^{*}(\mu) &=n / o^{2}
\end{aligned}
$$# 数学物理 Mathematical Physics MATHS5073_1 /MATHS4107_1 0 这是一份GLA格拉斯哥大学MATHS5073_1 /MATHS4107_1作业代写的成功案例 Now approaching this problem using vector notation, we note that the torque about the fulcrum due to F_{1} is \tau_{1}=r_{1} \times F_{1}. Here r_{1} is a vector from the fulcrum to the point of application of \mathbf{F}{1}, and$$ \tau{1}=\mathbf{r}{1} \times \mathbf{F}{1}=r_{1} F_{1} \sin \theta_{1} \hat{\mathbf{n}}=d_{1} F_{1} \hat{\mathbf{n}}
$$where \hat{\mathbf{n}} is a unit vector normal to, and out of the plane of the figure. A similar analysis of the torque \tau_{2} reveals that it is$$
\tau_{2}=\mathbf{r}{2} \times \mathbf{F}{2}=-r_{2} F_{2} \sin \theta_{2} \hat{\mathbf{n}}=-d_{2} F_{2} \hat{\mathbf{n}}
$$the minus sign occurring because \hat{\mathbf{n}} has the same meaning as before but this torque is directed into the plane of the figure. We now observe that \theta_{2}=\pi-\theta_{1}, and therefore \sin \theta_{2}=\sin \theta_{1}. At equilibrium the sum of these torques is zero. ## MATHS5073_1 /MATHS4107_1 COURSE NOTES ： Sometimes the determination of the gradient is straightforward. Given$$
f(x, y, z)=\frac{x y}{z}
$$we easily find$$
\frac{\partial f}{\partial x}=\frac{y}{z}, \quad \frac{\partial f}{\partial y}=\frac{x}{z}, \quad \frac{\partial f}{\partial z}=-\frac{x y}{z^{2}}
$$leading to \nabla f=\frac{y}{z} \hat{\mathbf{e}}{x}+\frac{x}{z} \hat{\mathbf{e}}{y}-\frac{x y}{z^{2}} \hat{\mathbf{e}}_{z} # 弹性回归 Flexible Regression STATS5052_1/STATS4040_1 0 这是一份GLA格拉斯哥大学MATHS3021_1作业代写的成功案例 where G_{0} is a standard exponential distribution. The distribution of \epsilon_{t} is therefore$$
f_{v, \xi}\left(\epsilon_{t}\right)=\sum_{j=1}^{\infty} w_{j} v\left(-\xi_{j} e^{-\lambda}, \xi_{j} e^{\lambda}\right)
$$and the conditional return distribution is$$
f_{G, \lambda}\left(\gamma_{t} \mid \sigma_{t}\right)=\sum_{j=1}^{\infty} w_{j} v\left(\gamma_{t} \mid-\xi_{j} \sigma_{t} e^{-\lambda}, \xi_{j} \sigma_{t} e^{\lambda}\right)
$$## STATS5052_1/STATS4040_1 COURSE NOTES ： Utilising the latent variable, the copula yields$$
P\left(Y_{1}=0, Y_{2} \leq Y_{2}\right)=P\left(Y_{1}^{} \leq 0, Y_{2} \leq Y_{2}\right)=C\left(F_{1}^{}(0), F_{2}\left(\gamma_{2}\right)\right)
$$and$$
P\left(Y_{1}=1, Y_{2} \leq \gamma_{2}\right)=P\left(Y_{1}^{}>0, Y_{2} \leq \gamma_{2}\right)=F_{2}\left(\gamma_{2}\right)-C\left(F_{1}^{}(0), F_{2}\left(\gamma_{2}\right)\right) \text {, }
$$leading to the mixed binary-continuous density$$
p\left(\gamma_{1}, \gamma_{2}\right)=\left(\frac{\partial C\left(F_{1}^{}(0), F_{2}\left(\gamma_{2}\right)\right)}{\partial F_{2}\left(\gamma_{2}\right)}\right)^{1-\gamma_{1}} \cdot\left(1-\frac{\partial C\left(F_{1}^{}(0), F_{2}\left(\gamma_{2}\right)\right)}{\partial F_{2}\left(\gamma_{2}\right)}\right)^{\gamma_{1}} \cdot p_{2}\left(\gamma_{2}\right),