Tag Archives: 华威大学代写

高级线性代数/ 群与环 Advanced Linear Algebra/Groups and Rings MA251-12/MA249-12

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这是一份warwick华威大学MA251-12/MA249-12的成功案例

高级线性代数/ 群与环 Advanced Linear Algebra/Groups and Rings MA251-12/MA249-12


Let $A \in \mathbb{C}^{m \times n}$. A yields
(a) $\sigma_{i}\left(A^{T}\right)=\sigma_{i}\left(A^{}\right)=\sigma_{i}(\bar{A})=\sigma_{i}(A)$, for $i=1,2, \ldots, q$. (b) Let $k=\operatorname{rank}(A)$. Then $\sigma_{i}\left(A^{\dagger}\right)=\sigma_{k-i+1}^{-1}(A)$ for $i=1, \ldots, k$, and $\sigma_{i}\left(A^{\dagger}\right)=0$ for $i=$ $k+1, \ldots, q$. In particular, if $m=n$ and $A$ is invertible, then $$ \sigma_{i}\left(A^{-1}\right)=\sigma_{n-i+1}^{-1}(A), \quad i=1, \ldots, n . $$ (c) For any $j \in \mathbb{N}$ $$ \begin{gathered} \sigma_{i}\left(\left(A^{} A\right)^{j}\right)=\sigma_{i}^{2 j}(A), \quad i=1, \ldots, q ; \
\sigma_{i}\left(\left(A^{} A\right)^{j} A^{}\right)=\sigma_{i}\left(A\left(A^{*} A\right)^{j}\right)=\sigma_{i}^{2 j+1}(A) \quad i=1, \ldots, q .
\end{gathered}
$$

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MA251-12/MA249-12 COURSE NOTES ：

(Submatrices) Take $A \in \mathbb{C}^{m \times n}$ and let $B$ denote $A$ with one of its rows or columns deleted. Then $\sigma_{i+1}(A) \leq \sigma_{i}(B) \leq \sigma_{i}(A), \quad i=1, \ldots, q-1 .$
Take $A \in \mathbb{C}^{m \times n}$ and let $B$ be $A$ with a row and a column deleted. Then
$$
\sigma_{\mathrm{i}+2}(A) \leq \sigma_{\mathrm{i}}(B) \leq \sigma_{\mathrm{i}}(A), \quad i=1, \ldots, q-2 .
$$
The $i+2$ cannot be replaced by $i+1$. (Example 2 )
Take $A \in \mathbb{C}^{m \times n}$ and let $B$ be an $(m-k) \times(n-l)$ submatrix of $A$. Then
$$
\sigma_{i+k+l}(A) \leq \sigma_{i}(B) \leq \sigma_{i}(A), \quad i=1, \ldots, q-(k+l)
$$
Take $A \in \mathbb{C}^{m \times n}$ and let $B$ be $A$ with some of its rows and/or columns set to zero. Then $\sigma_{i}(B) \leq$ $\sigma_{i}(A), \quad i=1, \ldots, q$.
Let $B$ be a pinching of $A$. Then $\operatorname{sv}(B) \preceq_{w} \operatorname{sv}(A)$. The inequalities $\prod_{i=1}^{k} \sigma_{i}(B) \leq \prod_{i=1}^{k} \sigma_{i}(A)$ and $\sigma_{k}(B) \leq \sigma_{k}(A)$ are not necessarily true for $k>1$.










数学建模的方法 Methods of Mathematical Modelling MA146-10/MA144-10

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这是一份warwick华威大学MA146-10/MA144-10的成功案例

数学建模的方法 Methods of Mathematical Modelling MA146-10/MA144-10


where $\sigma_{0}$ is the surface tension and $|\mathbf{v}|$ is the absolute value of the fluid velocity. The convective heat flux then becomes:
$$
\dot{Q}{\text {can }}=\pi d{o} \kappa \mathrm{Nu}\left(T-T_{o}\right),
$$
where, $T$ and $T_{0}$ are temperatures of the continuous and dispersed phases respectively and the Nusselt number is given by:
$$
\mathrm{Nu}=2+0.6 \operatorname{Re}^{0.5} \operatorname{Pr}^{0.33}
$$
In the previous equation the Prandtl number is defined as:
$$
\operatorname{Pr}=\frac{\mu C_{p}}{\kappa}
$$
Reynolds number is:
$$
\operatorname{Re}=\frac{\rho\left|\mathbf{v}{o}-\mathbf{v}\right| d{o}}{\mu}
$$

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MA140-10/MA152-15 COURSE NOTES ：

$$
W(s)=1+\sum_{i=1}^{l} b^{i} f^{j}(s)
$$
where $i$ is the number of variables which influence the adaptation, $b^{t}$ are constants and $f(s)$ are adaptation functions or their first derivatives. The adaptation function, which appears in the same equation, is integrated along the length of the grid as:
$$
F^{l}(s)=\int_{0}^{s} f^{l}(s) d s
$$
together give:
$$
\xi(s)=\frac{s+\sum_{i=1}^{l} b^{i} F^{l}(s)}{S_{\max }+\sum_{i=1}^{l} b^{i} F^{i}\left(S_{\max }\right)} .
$$










数学分析 Mathematical Analysis  MA140-10/MA152-15 

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这是一份warwick华威大学MA140-10/MA152-15的成功案例

数学分析 Mathematical Analysis  MA140-10/MA152-15 


Proof As is shown in linear algebra, the matrix $A$ that represents $T$ is a product of elementary matrices
$$
A=E_{1} \cdots E_{k} .
$$
Each elementary $2 \times 2$ matrix is one of the following types:
$$
\left[\begin{array}{ll}
\lambda & 0 \
0 & 1
\end{array}\right] \quad\left[\begin{array}{ll}
1 & 0 \
0 & \lambda
\end{array}\right] \quad\left[\begin{array}{ll}
0 & 1 \
1 & 0
\end{array}\right] \quad\left[\begin{array}{ll}
1 & \sigma \
0 & 1
\end{array}\right]
$$
where $\lambda>0$. The first three matrices represent isomorphisms whose effect on $I^{2}$ is obvious: $I^{2}$ is converted to the rectangles $\lambda I \times I, I \times \lambda I, I^{2}$. In each case, the area agrees with the magnitude of the determinant. The fourth isomorphism converts $I^{2}$ to the parallelogram$\Pi$ is Riemann measurable since its boundary is a zero set. By Fubini’s Theorem, we get
$$
|\Pi|=\int \chi_{\Pi}=\int_{0}^{1}\left[\int_{x=\sigma x}^{x=1+\sigma y} 1 d x\right] d y=1=\operatorname{det} E .
$$

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MA140-10/MA152-15 COURSE NOTES ：

$$
d x_{I}: \varphi \mapsto \int_{l^{k}} \frac{\partial \varphi_{I}}{\partial u} d u
$$
where this integral notation is shorthand for
$$
\int_{0}^{1} \ldots \int_{0}^{1} \frac{\partial\left(\varphi_{i_{1}}, \ldots, \varphi_{i_{k}}\right)}{\partial\left(u_{1}, \ldots, u_{k}\right)} d u_{1 \ldots} . d v_{k}
$$
If $f$ is a smooth function on $\mathbb{R}^{n}$ then $f d x_{l}$ is the functional
$$
f d x_{I}: \varphi \mapsto \int_{I^{k}} f(\varphi(u)) \frac{\partial \varphi_{I}}{\partial u} d u
$$










数学基础知识 Foundations MA132-10

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这是一份warwick华威大学MA132-10的成功案例

数学基础知识 Foundations MA132-10 


Let $F \subset \mathbb{R}^{n}$ and suppose that $f: F \rightarrow \mathbb{R}^{m}$ satisfies a Hölder condition
$$
|f(x)-f(y)| \leqslant c|x-y|^{\alpha} \quad(x, y \in F) .
$$
Then $\operatorname{dim}{H} f(F) \leqslant(1 / \alpha) \operatorname{dim}{H} F$.
Proof. If $s>\operatorname{dim}{H} F$ then by Proposition $2.2 \mathcal{H}^{s / \alpha}(f(F)) \leqslant c^{s / \alpha} \mathcal{H}^{s}(F)=0$, implying that $\operatorname{dim}{\mathrm{H}} f(F) \leqslant s / \alpha$ for all $s>\operatorname{dim}_{\mathrm{H}} F$.

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MA132-10 COURSE NOTES ：

Fundamental to most definitions of dimension is the idea of ‘measurement at scale $\delta$. For each $\delta$, we measure a set in a way that ignores irregularities of size less than $\delta$, and we see how these measurements behave as $\delta \rightarrow 0$. For example, if $F$ is a plane curve, then our measurement, $M_{\delta}(F)$, might be the number of steps required by a pair of dividers set at length $\delta$ to traverse $F$. A dimension of $F$ is then determined by the power law (if any) obeyed by $M_{\delta}(F)$ as $\delta \rightarrow 0$. If
$$
M_{\delta}(F) \sim c \delta^{-s}
$$
for constants $c$ and $s$, we might say that $F$ has ‘divider dimension’ $s$, with $c$ regarded as the ‘ $s$-dimensional length’ of $F$. Taking logarithms
$$
\log M_{\delta}(F) \simeq \log c-s \log \delta
$$
in the sense that the difference of the two sides tends to 0 with $\delta$, and
$$
s=\lim {\delta \rightarrow 0} \frac{\log M{\delta}(F)}{-\log \delta}
$$










用R进行线性统计建模 Linear Statistical Modelling with R ST231-10  

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这是一份warwick华威大学ST231-10的成功案例

用R进行线性统计建模 Linear Statistical Modelling with R ST231-10  


Suppose $Y(k)=m(k)+X(k)$ with a deterministic function $m(k)$ and $X(k)$ is a stochastic process with all moments $\mathbb{E}\left[|X(k)|^{p}\right]$ existing for $k \in \mathbb{N}$ and distributions $F_{k}(x):=$ $\operatorname{Prob}({\omega \in \Omega: X(k, \omega) \leq x})$ symmetric to the origin. Then the following are equivalent:

  1. For each $p \in \mathbb{N}$ it holds:
    $$
    \mathbb{E}\left[|Y(k)-\mathbb{E}[Y(k)]|^{p}\right]=c(p) \cdot \sigma^{p}|k|^{p H}
    $$
  2. For each $k$ the following functional scaling law holds on $\operatorname{Sym}{0}^{0}(\mathbb{R})$ : $$ F{k}(x)=F_{1}\left(k^{-H} x\right)
    $$
    where $S y m C_{0}^{0}(\mathbb{R})$ is the set of symmetric (with respect to the $y$-axis) continuous functions with compact support.

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ST231-10  COURSE NOTES ：

$$
\begin{aligned}
\sigma_{\log (x+)} &=\frac{1}{x_{l}} \cdot\left(\frac{q \cdot(1-q)}{n \cdot\left(f\left(x_{l}\right)\right)^{2}}\right)^{\frac{1}{2}} \
&=\sqrt{\frac{q \cdot(1-q)}{n}} \cdot \frac{1}{x_{l} \cdot f\left(x_{l}\right)} .
\end{aligned}
$$
For example, if $X \sim \mathcal{N}\left(0, \sigma^{2}\right)$, the propagation of the error can be written as
$$
\begin{aligned}
\sigma_{\log (x)} &=\sqrt{\frac{q \cdot(1-q)}{n}} \cdot \frac{\sqrt{2 \pi} \cdot \sigma}{x \cdot \exp \left(-\frac{x^{2}}{2 \sigma^{2}}\right)} \
&=\sqrt{\frac{q \cdot(1-q)}{n}} \cdot \frac{\sqrt{2 \pi}}{y \cdot \exp \left(-\frac{y^{2}}{2}\right)}
\end{aligned}
$$










数学统计 Mathematical Statistics ST230-10 

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这是一份warwick华威大学ST230-10的成功案例

数学统计 Mathematical Statistics ST230-10 


If the value of $\mu$ is lnown, then
$$
\left[\frac{\sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2}}{\chi_{n}^{2}(\alpha / 2)}, \frac{\sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2}}{\chi_{n}^{2}(1-\alpha / 2)}\right]
$$
is a $100(1-\alpha) \%$ confidence interval for $\sigma^{2}$, where $\chi_{n}^{2}(p)$ is the $100(1-p) \%$ point of the chi-square distribution with $n$ degrees of freedom.

If the value of $\mu$ is estimated from the data, then Theorem $6.1$ can be used to demonstrate that
$$
\left[\frac{\sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}}{\chi_{n-1}^{2}(\alpha / 2)}, \frac{\sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}}{\chi_{n-1}^{2}(1-\alpha / 2)}\right]
$$
is a $100(1-\alpha) \%$ confidence interval for $\sigma^{2}$, where $\chi_{n-1}^{2}(p)$ is the $100(1-p) \%$ point of the chi-square distribution with $(n-1)$ degrees of freedom.

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ST230-10 COURSE NOTES ：

If $X_{1}, X_{2}, \ldots, X_{n}$ is a random sample from a distribution with parameter $\theta$, and $\mu_{k}=E\left(X^{k}\right)$ is a function of $\theta$ for some $k$, then a method of moments estimator (or MOM estimator) of $\theta$ is obtained using the following procedure:
Solve $\mu_{k}=\widehat{\mu_{k}}$ for the parameter $\theta$.
For example, let $X$ be a uniform random variable on the interval $[0, b]$, and assume that $b>0$ is unknown. Since $E(X)=b / 2$, a MOM estimator is obtained as follows:
$$
\mu_{1}=\widehat{\mu_{1}} \Longrightarrow E(X)=\frac{1}{n} \sum_{i=1}^{n} X_{i} \Longrightarrow \frac{b}{2}=\bar{X} \Longrightarrow \widehat{b}=2 \bar{X}
$$










数学统计学的概率 Probability for Mathematical Statistics ST229-10

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这是一份warwick华威大学ST227-10 的成功案例


For a given $k$-tuple, $\left(x_{1}, x_{2}, \ldots, x_{k}\right)$, let
$$
\mathbf{x}{\text {obs }}^{2}=\sum{i=1}^{k} \frac{\left(x_{i}-n p_{i}\right)^{2}}{n p_{i}}
$$
be the observed value of Pearson’s statistic. Use the chi-square approximation to the distribution of $\mathbf{X}^{2}$ to compute $P\left(\mathbf{X}^{2} \geq \mathbf{X}{\mathrm{obs}}^{2}\right)$. Then the following hold: (i) If $P\left(\mathbf{X}^{2} \geq \mathbf{X}{\text {obs }}^{2}\right)>0.10$, the fit is judged to be good (the observed data are judged to be consistent with the multinomial model).
(ii) If $0.05<P\left(\mathbf{X}^{2} \geq \mathbf{x}{\mathrm{obs}}^{2}\right)<0.10$, the fit is judged to be fair (the observed data are judged to be marginally consistent with the model). (iii) If $P\left(\mathbf{X}^{2} \geq \mathbf{X}{\text {obs }}^{2}\right)<0.05$, the fit is judged to be poor (the observed data are judged to be not consistent with the model).

The probability $P\left(\mathbf{X}^{2} \geq \mathbf{x}_{\text {obs }}^{2}\right)$ is called the $p$ value of the test. The p value measures the strength of the evidence against the given multinomial model.

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ST229-10 COURSE NOTES ：

The bias of the estimator $\widehat{\theta}$ is the difference between the expected value of the estimator and the true parameter:
$$
B \operatorname{BIS}(\widehat{\theta})=E(\widehat{\theta})-\theta
$$
If $E(\widehat{\theta})=\theta$, then $\widehat{\theta}$ is said to be an unbiased estimator of $\theta$; otherwise, $\widehat{\theta}$ is said to be a biased estimator of $\theta$.

For example, let $\bar{X}, S^{2}$, and $S$ be the sample mean, sample variance, and sample standard deviation of a random sample of size $n$ from a normal distribution. Since $E(\bar{X})=\mu, \bar{X}$ is an unbiased estimator of $\mu$. Since $E\left(S^{2}\right)=\sigma^{2}, S^{2}$ is an unbiased estimator of $\sigma^{2}$. Since
$$
E(S)=\sigma \sqrt{\frac{2}{n-1}} \frac{\Gamma(n / 2)}{\Gamma((n-1) / 2)} \neq \sigma
$$
$S$ is a biased estimator of $\sigma .$










统计和概率的数学方法 Mathematical Methods for Statistics and Probability ST228-10

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这是一份warwick华威大学ST228-10的成功案例

随机过程 Stochastic Processes ST227-10


Knowledge of the distribution of a sample summary is important in statistical applications. For example, suppose that the sample mean and sample variance of a random sample of size $n$ from a normal distribution are used to estimate the unlnown $\mu$ and $\sigma^{2}$. Let $\chi_{p}^{2}$ and $\chi_{1-p}^{2}$ be the $p^{t h}$ and $(1-p)^{t h}$ quantiles of the chi-square distribution with $(n-1)$ degrees of freedom, respectively. Then
$$
1-2 p=P\left(\chi_{p}^{2} \leq \frac{(n-1) S^{2}}{\sigma^{2}} \leq \chi_{1-p}^{2}\right)=P\left(\frac{(n-1) S^{2}}{\chi_{1-p}^{2}} \leq \sigma^{2} \leq \frac{(n-1) S^{2}}{\chi_{p}^{2}}\right)
$$
If the observed value of the sample variance is $s^{2}=8.72, n=12$ and $p=0.05$, then the interval
$$
\left[\frac{11(8.72)}{19.68}, \frac{11(8.72)}{4.57}\right]=[4.87,20.99]
$$
is an estimate of an interval containing $\sigma^{2}$ with probability $0.90$.

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ST228-10 COURSE NOTES ：

The distribution of the ratio of $S_{x}^{2} / S_{y}^{2}$ to $\sigma_{x}^{2} / \sigma_{y}^{2}$ is important in statistical applications. For example, suppose that all four parameters $\left(\mu_{x}, \sigma_{x}, \mu_{y}, \sigma_{y}\right)$ are unknown. Let $f_{p}$ and $f_{1-p}$ be the $p^{t h}$ and $(1-p)^{t h}$ quantiles of the $f$ ratio distribution with $(n-1)$ and $(m-1)$ degrees of freedom, respectively. Then
$$
1-2 p=P\left(f_{p} \leq \frac{S_{x}^{2} / S_{y}^{2}}{\sigma_{x}^{2} / \sigma_{y}^{2}} \leq f_{1-p}\right)=P\left(\frac{S_{x}^{2} / S_{y}^{2}}{f_{1-p}} \leq \frac{\sigma_{x}^{2}}{\sigma_{y}^{2}} \leq \frac{S_{x}^{2} / S_{y}^{2}}{f_{p}}\right)
$$
If the observed sample variances are $s_{x}^{2}=18.75$ and $s_{y}^{2}=3.45, n=8, m=10$, and $p=0.05$, then the interval
$$
\left[\frac{18.75 / 3.45}{3.29}, \frac{18.75 / 3.45}{0.27}\right]=[1.65,20.13]
$$
is an estimate of an interval containing $\sigma_{x}^{2} / \sigma_{y}^{2}$ with probability $0.90$.










随机过程 Stochastic Processes ST227-10

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这是一份warwick华威大学ST227-10 的成功案例

随机过程 Stochastic Processes ST227-10


Let $\left(\mathbf{u}{t}\right){0 \leq t \leq T}$ be an $m$-dimensional process and
$$
\begin{aligned}
&\mathbf{a}:[0, T] \times \Omega \rightarrow \mathbb{R}^{m}, \mathbf{a} \in \mathcal{C}{1 \mathbf{w}}([0, T]) \ &b:[0, T] \times \Omega \rightarrow \mathbb{R}^{m n}, b \in \mathcal{C}{1 \mathbf{w}}([0, T])
\end{aligned}
$$
The stochastic differential $d \mathbf{u}(t)$ of $\mathbf{u}(t)$ is given by
$$
d \mathbf{u}(t)=\mathbf{a}(t) d t+b(t) d \mathbf{W}(t)
$$
if, for all $0 \leq t_{1}<t_{2} \leq T$
$$
\mathbf{u}\left(t_{2}\right)-\mathbf{u}\left(t_{1}\right)=\int_{t_{1}}^{t_{2}} \mathbf{a}(t) d t+\int_{t_{1}}^{t_{2}} b(t) d \mathbf{W}(t)
$$

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ST227-10  COURSE NOTES ：

Let $A$ be a nonempty subset of a metric space $(X, d)$. Define
$$
d_{A}(x):=\inf {d(x, a): a \in A}, \quad x \in X .
$$
Then $d_{A}$ is continuous. (Geometrically, we think of $d_{A}(x)$ as the distance of $x$ to $A$.)

We give a proof even though it is easy, because of the importance of this result. Let $x, y \in X$ and $a \in A$ be arbitrary. We have, from the triangle inequality $d(a, x) \leq d(a, y)+d(y, x)$,
$$
\begin{aligned}
&d(a, y) \geq d(a, x)-d(y, x) \
&d(a, y) \geq d_{A}(x)-d(y, x)
\end{aligned}
$$
The inequality says that $d_{A}(x)-d(y, x)$ is a lower bound for the set ${d(a, y): a \in A}$. Hence the greatest lower bound of this set, namely, $\inf {d(a, y)=a \in \backslash A} n d$ is greater than or equal to this lower bound, that is,
$$
d_{A}(y) \geq d_{A}(x)-d(y, x)
$$










度量空间 Metric Spaces MA222-10

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这是一份warwick华威大学MA222-10 的成功案例

度量空间 Metric Spaces MA222-10


By the LUB axiom, there exists $\ell \in \mathbb{R}$ which is sup $s$. We clamm that $\lim x_{n}=\ell$. Let $\varepsilon>0$ be given. As $\ell$ is an upper bound for $S$ and $x_{N}-\varepsilon / 2 \in S$ (by (i)) we infer that $x_{N}-\varepsilon / 2 \leq \ell$. Since $\ell$ is the least upper bound for $S$ and $x_{N}+\varepsilon / 2$ is an upper bound for $S$ (from (ii)) we see that $\ell \leq x_{N}+\varepsilon / 2$. Thus we have $x_{N}-\varepsilon / 2 \leq \ell \leq x_{N}+\varepsilon / 2$ or
$$
\left|x_{N}-\ell\right| \leq \varepsilon / 2
$$
For $n \geq N$ we have
$$
\begin{aligned}
\left|x_{n}-\ell\right| & \leq\left|x_{n}-x_{N}\right|+\left|x_{N}-\ell\right| \
&<\varepsilon / 2+\varepsilon / 2=\varepsilon
\end{aligned}
$$
We have thus shown that $\lim {n \rightarrow \infty} x{n}=\ell$.

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MA222-10  COURSE NOTES ：

Let $A$ be a nonempty subset of a metric space $(X, d)$. Define
$$
d_{A}(x):=\inf {d(x, a): a \in A}, \quad x \in X .
$$
Then $d_{A}$ is continuous. (Geometrically, we think of $d_{A}(x)$ as the distance of $x$ to $A$.)

We give a proof even though it is easy, because of the importance of this result. Let $x, y \in X$ and $a \in A$ be arbitrary. We have, from the triangle inequality $d(a, x) \leq d(a, y)+d(y, x)$,
$$
\begin{aligned}
&d(a, y) \geq d(a, x)-d(y, x) \
&d(a, y) \geq d_{A}(x)-d(y, x)
\end{aligned}
$$
since $d(a, x) \geq \inf \left{d\left(a^{\prime}, x\right): a^{\prime} \in A\right}:=d_{A}(x)$. The inequality says that $d_{A}(x)-d(y, x)$ is a lower bound for the set ${d(a, y): a \in A}$. Hence the greatest lower bound of this set, namely, $\inf {d(a, y)=a \in \backslash A} n d$ is greater than or equal to this lower bound, that is,
$$
d_{A}(y) \geq d_{A}(x)-d(y, x)
$$