# 统计计算|MTH3045 Statistical Computing代写

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A variation on least squares scaling is the so-called Sammon mapping which minimizes
$$\sum_{i \neq i^{\prime}} \frac{\left(d_{i i^{\prime}}-\left|z_{i}-z_{i^{\prime}}\right|\right)^{2}}{d_{i i^{\prime}}} .$$
Here more emphasis is put on preserving smaller pairwise distances.
In classical scaling, we instead start with similarities $s_{i i^{r}}:$ often we use the centered inner product $s_{i i^{\prime}}=\left\langle x_{i}-\bar{x}, x_{i^{\prime}}-\bar{x}\right\rangle$. The problem then is to minimize
$$\sum_{i, i^{\prime}}\left(s_{i i^{\prime}}-\left\langle z_{i}-\bar{z}, z_{i^{\prime}}-\bar{z}\right\rangle\right)^{2}$$

## MTH3045 COURSE NOTES ：

We approximate each point by an affine mixture of the points in its neighborhood:
$$\min {W{i k}}\left|x_{i}-\sum_{k \in \mathcal{N}(i)} w_{i k} x_{k}\right|^{2}$$
over weights $w_{i k}$ satisfying $w_{i k}=0, k \notin \mathcal{N}(i), \sum_{k=1}^{N} w_{i k}=1 . w_{i k}$ is the contribution of point $k$ to the reconstruction of point $i$. Note that for a hope of a unique solution, we must have $K<p$.

Finally, we find points $y_{i}$ in a space of dimension $d<p$ to minimize
$$\sum_{i=1}^{N}\left|y_{i}-\sum_{k=1}^{N} w_{i k} y_{k}\right|^{2}$$
with $w_{i k}$ fixed.

# 组合学|MTH3021RWA Combinatorics代写

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The definition of boxed products implies the coefficient relation
$$A_{n}=\sum_{k=1}^{n}\left(\begin{array}{l} n-1 \ k-1 \end{array}\right) B_{k} C_{n-k} .$$
The binomial coefficient that appears in the standard labelled product is now modified since only $n-1$ labels need to be distributed between the two components, $k-1$ going to the $\mathcal{B}$ component (that is constrained to contain the label 1 already) and $n-k$ to the $\mathcal{C}$ component. From the equivalent form
$$\frac{A_{n}}{n !}=\frac{1}{n} \sum_{k=0}^{n}\left(\begin{array}{l} n \ k \end{array}\right)\left(k B_{k}\right) C_{n-k}$$

## MTH3021RWA COURSE NOTES ：

A useful special case is the min-rooting operation,
$$\mathcal{A}={1}^{0} \star \mathcal{C} \text {. }$$
for which a variant definition goes as follows. Take in all possible ways elements $\gamma \in \mathcal{C}$, prepend an atom with a label smaller than the labels of $\gamma$, for instance 0 , and relabel in the canonical way over $[1 \ldots(n+1)]$ by shifting all label values by 1 . Clearly $A_{n+1}=C_{n}$ which yields
$$A(z)=\int_{0}^{z} C(t) d t$$
a result also consistent with the general formula of boxed products.
For some applications, it is easier to impose constraints on the maximal label rather than the minimum. The max-boxed product written
$$A=\left(B^{-} * \mathcal{C}\right) \text {. }$$

# 微分方程|MTH2003-JD Differential Equations – DEFERRAL代写

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that any $\lambda \in \mathrm{C}$ such that $\operatorname{Re}(\lambda)<-\mu$ is contained in $\rho(A)$ and in particular
$$R_{A}(\lambda)=\int_{0}^{\infty} e^{k t} T(t) d t$$
Hence it follows for every $\varpi \in \mathbb{R}$ that
$$\omega\left(R_{A}\left(\mu^{\prime}+i \varpi\right) \xi\right)=\frac{1}{\sqrt{2 \pi}} \int_{\mathbb{R}} e^{i \boxminus t} h(t) d t$$
where
$$h(t):=\sqrt{2 \pi} e^{\mu / t} \omega(T(t) \xi)$$
for all $t \in[0, \infty)$ and $h(t):=0$ for all $t \in(-\infty, 0)$. In particular,
$$|h(t)| \leqslant \sqrt{2 \pi} c\left|k u|\mid \xi \xi| \cdot e^{-|\mu+\mu| t}\right.$$
for every $t \geqslant 0$ and hence $h \in L_{\mathrm{C}}^{2}(\mathrm{R})$. Hence $\left(\mathrm{R} \rightarrow \mathrm{C}, \varpi \mapsto \omega\left(R_{A}\left(\mu^{\prime}+i \varpi\right) \xi\right)\right) \in$ $L_{\mathrm{C}}^{2}(\mathbb{R})$ as well as (4.3.1) follows by the Fourier inversion theorem.

## MMTH2003-JD COURSE NOTES ：

we have:
$$\left(A_{r}^{-1} f\right)(x)=\int_{0}^{x} f(y) d y$$
for all $x \in I$ and $f \in X$ and hence
$$\left(\left(A_{r}^{}\right)^{-1} f\right)(x)=\int_{x}^{a} f(y) d y$$ for all $x \in I$ and $f \in X$. Further, by the proof of Corollary 5.1.4 $$\bar{A}{l}^{-1}=U \bar{A}{r}^{-1} U$$
and hence
\begin{aligned} &\left(\bar{A}{l}^{-1} f\right)(x)=\left(\bar{A}{r}^{-1} U f\right)(a-x)=\int_{0}^{a-x} f(a-y) d y \ &=\int_{x}^{a} f\left(y^{\prime}\right) d y^{\prime}=\left(\left(A_{r}^{}\right)^{-1} f\right)(x) \end{aligned} # 概率、统计和数据|MTH1004 Probability, Statistics and Data代写 0 这是一份exeter埃克塞特大学MTH1004作业代写的成功案 for all speeds v between 60 and 90 . We can now obtain the probability density f_{V} of V by differentiating:
f_{V}(v)=\frac{\mathrm{d}}{\mathrm{d} v} F_{V}(v)=\frac{\mathrm{d}}{\mathrm{d} v}\left(3-\frac{180}{v}\right)=\frac{180}{v^{2}}
$$for 60 \leq v \leq 90. It is amusing to note that with the second model the traffic police write fewer speeding tickets because$$
\mathrm{P}(V>80)=1-\mathrm{P}(V \leq 80)=1-\left(3-\frac{180}{80}\right)=\frac{1}{4}
$$## MTH1004 COURSE NOTES ： As an example, let X be a random variable with an N\left(\mu, \sigma^{2}\right) distribution, and let Y=r X+s. Then this rule gives us$$
f_{Y}(y)=\frac{1}{r} f_{X}\left(\frac{y-s}{r}\right)=\frac{1}{r \sigma \sqrt{2 \pi}} \mathrm{e}^{-\frac{1}{2}((y-r \mu-s) / r \sigma)^{2}}
$$for -\infty<y<\infty. On the right-hand side we recognize the probability density of a normal distribution with parameters r \mu+s and r^{2} \sigma^{2}. This illustrates the following rule. # 数学建模|MTH1003 Mathematical Modelling代写 0 这是一份exeter埃克塞特大学MTH1003作业代写的成功案$$
y^{+}=\frac{1}{K} \frac{\mu+\mu_{t}}{\mu} .
$$For the k- \varepsilon model of turbulence this distance is:$$
y^{+}=\frac{\rho \sqrt[4]{C_{\mu}} \sqrt{k_{p}} \delta_{n p}}{\mu} .
$$In the previous equation, k_{P} is the kinetic energy of turbulence at the centre of the boundary cell, while \sigma_{n s} denotes the normal distance from the centre of the boundary cell to the wall. The viscous sub-layer thickness is defined as the larger root of the equation:$$
y_{r}^{+}=\frac{1}{K} \ln \left(\varepsilon y_{r}^{+}\right)
$$## MTH1003 COURSE NOTES ： \int_{V} \operatorname{grad} \psi \mathrm{d} V=\int_{S} \psi \mathrm{ds} \Rightarrow \quad(\operatorname{grad} \psi){\mathrm{P}{0}} \approx \frac{1}{V_{\mathrm{P}{0}}} \sum{j=1}^{n_{f}} \psi_{j} \mathbf{s}{j} Here, \psi{j} is the value of variable \psi at the cell face centre. The first term in the prototype equation is different to the others because it contains an integral with respect to time. If the equation is rearranged into the following form:$$
\frac{d \Psi}{d t}=F(\phi)
$$where$$
\Psi=\int_{\mathrm{V}} \rho B_{\phi} \mathrm{d} V \approx\left(\rho B_{\phi} V\right){\mathrm{P}{0}} \text { and } \phi=\phi(\mathbf{r}, t)
$$# 数学方法 Mathematical Methods MTH1002/MTH1002-JD 0 这是一份exeter埃克塞特大学MTH1002作业代写的成功案例 问题 1. Now we determine the curves for which L^{2}=4 \pi A. To do this we find conditions for which A is equal to the upper bound we obtained for it above. First note that$$
\sum_{n=1}^{\infty} n\left(a_{n}^{2}+b_{n}^{2}+c_{n}^{2}+d_{n}^{2}\right)=\sum_{n=1}^{\infty} n^{2}\left(a_{n}^{2}+b_{n}^{2}+c_{n}^{2}+d_{n}^{2}\right)
$$证明 . implies that all the coefficients except a_{0}, c_{0}, a_{1}, b_{1}, c_{1} and d_{1} are zero. The constraint,$$
\pi \sum_{n=1}^{\infty} n\left(a_{n} d_{n}-b_{n} c_{n}\right)=\frac{\pi}{2} \sum_{n=1}^{\infty} n\left(a_{n}^{2}+b_{n}^{2}+c_{n}^{2}+d_{n}^{2}\right)
$$then becomes$$
a_{1} d_{1}-b_{1} c_{1}=a_{1}^{2}+b_{1}^{2}+c_{1}^{2}+d_{1}^{2}
$$## MTH1002/MTH1002-JDCOURSE NOTES ： The Fourier sine series has the form$$
x(1-x)=\sum_{n=1}^{\infty} a_{n} \sin (n \pi x)
$$The norm of the eigenfunctions is$$
\int_{0}^{1} \sin ^{2}(n \pi x) d x=\frac{1}{2}
$$The coefficients in the expansion are$$
\begin{aligned}
a_{n} &=2 \int_{0}^{1} x(1-x) \sin (n \pi x) \mathrm{d} x \
&=\frac{2}{\pi^{3} n^{3}}(2-2 \cos (n \pi)-n \pi \sin (n \pi)) \
&=\frac{4}{\pi^{3} n^{3}}\left(1-(-1)^{n}\right)
\end{aligned}
$$# 数学结构 Mathematical Structures MTH1001/MTH1001-JD 0 这是一份exeter埃克塞特大学MTH1001/MTH1001-JD作业代写的成功案例 问题 1. Let$$
s=f(t)
$$be the distance as a function f of the time t. We just noticed:$$
f(n t)=n f(t) \quad \text { for } n \in \mathbf{N} .
$$Replace t with \frac{1}{n} t.Then$$
f(t)=n f\left(\frac{1}{n} t\right)
$$证明 . which read$$
f\left(\frac{1}{n} t\right)=\frac{1}{n} f(t)
$$yields in \frac{1}{n} of the time \frac{1}{n} of the distance is covered. If in the last formula t is replaced with m t(m \in N) one gets$$
f\left(\frac{m}{n} t\right)=\frac{1}{n} f(m t)=\frac{m}{n} f(t)
$$## MTH1001/MTH1001-JDCOURSE NOTES ： Take similar triangles with their areas as a function f of their – variable height h,$$
f(h)=\alpha h^{2} \quad \text { with fixed } \alpha ;
$$the difference$$
f(h+\delta)-f(h)
$$is the area of a strip of height \delta on the base, the difference quotient$$
\frac{f(h+\delta)-f(h)}{\delta}
$$approximately equals the base, and the differential quotient$$
\frac{\mathrm{d} f}{\mathrm{~d} h}=2 \alpha h
$$equals the base,$$
b=2 \alpha h .
$$Thus indeed$$
f(h)=\frac{1}{2} b h .