Tag Archives: 英国代写

组合学 Combinatorics MATH314301

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组合学 Combinatorics MATH314301
问题 1.

Let $y \in G$, and write $y$ in the form (1). We can write
$$
\varphi(x)-\varphi(x y)=\left[\varphi(x 8)-\varphi\left(x s_{1}\right)\right]+\ldots+\left[\varphi\left(x s_{1} \ldots s_{\ell-1}-\varphi(x y)\right] .\right.
$$
It follows, for example, by the Canchy-Schwarz inequality that
$$
(\varphi(x)-\varphi(x y))^{2} \leq \ell^{*} \sum^{\ell}\left(\varphi\left(x s_{1} \ldots s_{i-1}\right)-\varphi\left(x s_{1} \ldots s_{i}\right)\right)^{2}
$$

证明 .

where $\ell^{*}$ is the number of nonzero terms in the sum, and is bounded above by $d=\operatorname{diam}(\bar{C})$, since $\gamma$ is geodesic. Summing this inequality over $x \in G$ we get
$$
\sum_{x \in G}(\varphi(x)-\varphi(x y))^{2} \leq d \sum_{t \in C, s \in S} N_{\gamma}(s, \bar{C})(\varphi(z)-\varphi(z s))^{2}
$$
Since this holds for all $y \in G$, we may average the left hand side with respect to $y$ with weights $\approx(y)$ to get
$$
\sum_{x, y \in C}(\varphi(x)-\varphi(x y))^{2} \tilde{\pi}(y) \leq d \sum_{z \in G, s \in S} N_{\gamma}(s, \bar{C})(\varphi(z)-\varphi(z s))^{2}
$$

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MATH314301 COURSE NOTES ：

Last time, we proved
Theorem 1 Let $C$ be a subset of the group $G, S=S^{-1}$ a symmetric generating set, $\pi=U(S)$ the uniform distribution on $S$, and $p_{\pi}$ the “one-step evolution” of the random walk (i.e. $p_{\pi} \varphi=U(S) * \varphi$ ). Then for any probability distribution $\varphi$,
$$
\left|p_{n} \varphi\right|^{2} \leq\left(1-\frac{|G \backslash C|}{2 \cdot A \cdot|G|}\right) \cdot|\varphi|^{2}
$$
where $A=d \cdot|S| \cdot \max {s \in S} \max {g \in C} \mu_{s}(g), d=\operatorname{diam}(\bar{C}, G)$.
We will use this theorem to bound the escape time of a random walk $X_{\mathrm{t}}$ generated by $S$. For a subset $C$ of $G$, set
$$
\varphi_{t}(g)=\operatorname{Pr}\left[X_{t}=g \text { and } X_{i} \in C \forall i=1 \ldots t\right]
$$
Obviously, supp $\varphi_{t} \subset C,\left|\varphi_{0}\right| \leq 1$ ( 1 if $C$ contains $1_{G}, 0$ otherwise) and








变量微积分 Calculus of Variations MATH265001

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这是一份leeds利兹大学MATH265001作业代写的成功案例

变量微积分 Calculus of Variations MATH265001
问题 1.

Let $\epsilon \in \mathbb{R}, \varphi \in C_{0}^{\infty}(a, b), \lambda=\left(2\left|\varphi^{\prime}\right|_{L^{\infty}}\right)^{-1}$ and
$$
\xi(x, \epsilon)=x+\epsilon \lambda \varphi(x)=y \text {. }
$$
Observe that for $|\epsilon| \leq 1$, then $\xi(., \epsilon):[a, b] \rightarrow[a, b]$ is a diffeomorphism with $\xi(a, \epsilon)=a, \xi(b, \epsilon)=b$ and $\xi_{x}(x, \epsilon)>0$. Let $\eta(., \epsilon):[a, b] \rightarrow[a, b]$ be its inverse, i.e.
$$
\xi(\eta(y, \epsilon), \epsilon)=y .
$$

证明 .

Since
$$
\xi_{x}(\eta(y, \epsilon), \epsilon) \eta_{y}(y, \epsilon)=1 \text { and } \xi_{x}(\eta(y, \epsilon), \epsilon) \eta_{e}(y, \epsilon)+\xi_{e}(\eta(y, \epsilon), \epsilon)=0,
$$
we find $(O(t)$ stands for a function $f$ so that $|f(t) / t|$ is bounded in a neighborhood of $t=0$ )
$$
\begin{aligned}
&\eta_{y}(y, \epsilon)=1-\epsilon \lambda \varphi^{\prime}(y)+O\left(\epsilon^{2}\right) \
&\eta_{\epsilon}(y, \epsilon)=-\lambda \varphi(y)+O(\epsilon)
\end{aligned}
$$

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MATH265001 COURSE NOTES ：

$$
f(x, u, \xi)=f(u, \xi)=\frac{1}{2} \xi^{2}-u .
$$
The second form of the Euler-Lagrange equation is
$$
0=\frac{d}{d x}\left[f\left(u(x), u^{\prime}(x)\right)-u^{\prime}(x) f_{\xi}\left(u(x), u^{\prime}(x)\right)\right]=-u^{\prime}(x)\left[u^{\prime \prime}(x)+1\right],
$$
and it is satisfied by $u \equiv 1$. However, $u \equiv 1$ does not verify the Euler-Lagrange equation, which is in the present case
$$
u^{\prime \prime}(x)=-1 \text {. }
$$








流体动力学 Fluid Dynamics MATH262009/MATH262501

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这是一份leeds利兹大学MATH262009/MATH262501作业代写的成功案例

流体动力学 Fluid Dynamics MATH262009/MATH262501
问题 1.

For $\beta \gg 1$ it seems logical to look for solutions in the form
$$
\begin{aligned}
\psi(x, y, \tilde{t}, \beta)=& \psi_{0}(x, y, \tilde{t})+\frac{1}{\beta} \psi_{1}(x, y, \tilde{t}) \
&+\frac{1}{\beta^{2}} \psi_{2}(x, y, \tilde{t})+\cdots
\end{aligned}
$$

证明 .

whose substitution in will yield a sequence of problems for $\psi_{0}, \psi_{1}$, etc. by comparing like orders in $\beta^{-1}$. The lowest order, or $\mathrm{O}(1)$, problem for $\psi_{0}$ is
$$
\frac{\partial}{\partial t}\left(\nabla^{2} \psi_{0}-F \psi_{0}\right)+\frac{\partial \psi_{0}}{\partial x}=0,
$$
whose solution can be written as a general superposition of Rossby waves. e.g.,
$$
\psi_{0}=\sum_{j} a_{j} \cos \theta_{j},
$$
where the range of the sum over the integral index $j$ is formally infinite, but where in fact only a finite number of the amplitudes $a_{j}$ may be different from zero. The phase of each wave is
$$
\theta_{j}=k_{j} x+l_{j} y-\sigma_{j} \tilde{t}+\phi_{j},
$$

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MATH262009/MATH262501 COURSE NOTES ：

the solution forced by the term $a_{m} a_{n} B\left(K_{m}, K_{n}\right) \cos \left(\theta_{m}+\theta_{n}\right)$. A periodic forced solution for $\psi_{1}$ may be sought in the form
$$
\psi_{1}=A_{1 m n} \sin \left(\theta_{m}+\theta_{n}\right)
$$
where $A_{1 m n}$ is determined by $(3.26 .11 \mathrm{~b})$ to be
$$
A_{1 m n}=\frac{a_{m} a_{n} B\left(K_{m}, K_{n}\right)}{\left(K_{m n}^{2}+F\right)\left(\omega_{m n}-\sigma_{m n}\right)} .
$$








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

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这是一份exeter埃克塞特大学MTH2003-JD作业代写的成功案

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

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.

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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

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这是一份exeter埃克塞特大学MTH1004作业代写的成功案

概率、统计和数据|MTH1004 Probability, Statistics and Data代写

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
$$
\mathrm{P}(V>80)=1-\mathrm{P}(V \leq 80)=1-\left(3-\frac{180}{80}\right)=\frac{1}{4}
$$

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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

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这是一份exeter埃克塞特大学MTH1003作业代写的成功案

数学建模|MTH1003 Mathematical Modelling代写

$$
y^{+}=\frac{1}{K} \frac{\mu+\mu_{t}}{\mu} .
$$
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:
$$
y_{r}^{+}=\frac{1}{K} \ln \left(\varepsilon y_{r}^{+}\right)
$$

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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
$$
\Psi=\int_{\mathrm{V}} \rho B_{\phi} \mathrm{d} V \approx\left(\rho B_{\phi} V\right){\mathrm{P}{0}} \text { and } \phi=\phi(\mathbf{r}, t)
$$



傅里叶分析|MATH10051 Fourier Analysis代写

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Understand the statements and proofs of important theorems and be able to explain the key steps in proofs, sometimes with variation.

这是一份drps.ed爱丁堡个大学MATH10051作业代写的成功案

傅里叶分析|MATH10051 Fourier Analysis代写

To do this we introduce complex numbers by adding $i$ times the second equation to the first to obtain the single equation
$$
\frac{d}{d t}(U+i V)=B(V-i U) .
$$
If we set $\Phi=U+i V$ this equation takes the form
$$
\dot{\Phi}=-i B \Phi
$$
This is an equation that we can solve to obtain
$$
\Phi(t)=A e^{-i B t}
$$
where $A$ is a fixed complex number.
How does this solution fit in with our original problem. recall that we can write
$$
A=r \exp i \alpha
$$

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MATH10051 COURSE NOTES ：

If $x=R \cos \theta, y=R \sin \theta$ where $R$ is very large then to a very good approximation
$$
S(R \cos \theta, R \sin \theta, t)=R^{-2} \exp (i(\omega t-\lambda R) Q(u)
$$
where
$$
Q(u)=\sum_{k=-N}^{N} A_{k} \exp \left(i\left(k u-\phi_{k}\right)\right)
$$
and $u=\lambda^{-1} l \sin \theta$.
In the discussion that follows we use the notation of Lemma $4.1$ and the discussion that preceded it. We set
$$
P(u)=\sum_{k=-N}^{N} A_{k} \exp (i k u)
$$



金融数学 |MATH10003 Financial Mathematics代写

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The next block focuses on continuous time finance and contains an introduction to the basic ideas of Stochastic calculus. The last chapter is an overview of Actuarial Finance. This course is a great introduction to finance theory and its purpose is to give students a broad perspective on the topic.”

这是一份drps.ed爱丁堡个大学MATH10003作业代写的成功案

金融数学 |MATH10003 Financial Mathematics代写

Once again, differentiating in $(b, c)$, we find that the random variable $\left(X_{T}, M_{T}^{X}\right)$ has the bivariate density
$$
f^{X}(T, b, c)=\frac{2(2 c-b)}{T \sqrt{T}} \phi\left(\frac{2 c-b}{\sqrt{T}}\right) \cdot e^{\mu b-\frac{1}{2} \mu^{2} T} .
$$
Note that the processes $\left(\mu t+\sigma B_{t}\right)$ and $\left(\mu t-\sigma B_{t}\right)$ have the same law. Hence we consider the process
$$
Y_{t}=\mu t+\sigma B_{t} \text { for } \sigma>0 .
$$
Write $F^{Y}(T, b, c)=P\left(Y_{T}<b, M_{T}^{Y}<c\right)$. Consider
$$
\widehat{X}{t}=\sigma^{-1} Y{t}=\frac{\mu}{\sigma} t+B_{t}
$$

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MATH10003  COURSE NOTES ：

Consider again the situation with two assets, the riskless bond
$$
S_{t}^{0}=e^{r t}
$$
and a risky asset $S^{1}$ with dynamics
$$
d S_{t}^{1}=S_{t}^{1}\left(\mu d t+\sigma d B_{t}\right) .
$$
$\left(B_{t}\right)$ is a standard Brownian motion on a probability space $(\Omega, \mathcal{F}, P)$. Consider the risk-neutral probability $P^{\theta}$ and the $P^{\theta}$-Brownian motion $W^{\theta}$ given by
$$
d W_{t}^{\theta}=\theta d t+\sigma d B_{t}
$$
Here $\theta=\frac{r-\mu}{\sigma}$. Under $P^{\theta}$,
$$
d S_{t}^{1}=S_{t}^{1}\left(r d t+\sigma d W_{t}^{\theta}\right),
$$



数学 Mathematics 1 MATHS1017_1

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这是一份GLA格拉斯哥大学MATHS1017_1作业代写的成功案例

数学 Mathematics 1 MATHS1017_1
问题 1.

$2 x+3$ and $x-1$ are factors of $2 x^{4}+a x^{3}-3 x^{2}+b x+3$.
Find $a$ and $b$ and all zeros of the polynomial.
Since $2 x+3$ and $x-1$ are factors,
$2 x^{4}+a x^{3}-3 x^{2}+b x+3=(2 x+3)(x-1)($ a quadratic) $=\left(2 x^{2}+x-3\right)\left(x^{2}+c x-1\right)$ for some $c$
Equating coefficients of $x^{2}$ gives: $\quad-3=-2+c-3$
$\therefore c=2$

证明 .

Equating coefficients of $x^{3}: \quad a=2 c+1=4+1=5$
Equating coefficients of $x$ :
$$
\begin{aligned}
b &=-1-3 c \
\therefore \quad b &=-1-6=-7
\end{aligned}
$$
$\therefore \quad P(x)=(2 x+3)(x-1)\left(x^{2}+2 x-1\right)$ which has zeros of: $\quad-\frac{3}{2}, 1$ and $\frac{-2 \pm \sqrt{4-4(1)(-1)}}{2}=\frac{-2 \pm 2 \sqrt{2}}{2}=-1 \pm \sqrt{2}$
$\therefore$ the zeros are $-\frac{3}{2}, 1$ and $-1 \pm \sqrt{2}$.

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MATHS1017_1 COURSE NOTES ：

Find $k$ given that $x-2$ is a factor of $x^{3}+k x^{2}-3 x+6$. Hence, fully factorise $x^{3}+k x^{2}-3 x+6$.
Let $P(x)=x^{3}+k x^{2}-3 x+6$
By the Factor theorem, as $x-2$ is a factor then $P(2)=0$
$$
\begin{aligned}
\therefore \quad(2)^{3}+k(2)^{2}-3(2)+6 &=0 \
\therefore 8+4 k &=0 \quad \text { and so } k=-2
\end{aligned}
$$
Now $x^{3}-2 x^{2}-3 x+6=(x-2)\left(x^{2}+a x-3\right)$ for some constant $a$.
Equating coefficients of $x^{2}$ gives: $\quad-2=-2+a \quad$ i.e., $a=0$
Equating coefficients of $x$ gives: $\quad-3=-2 a-3 \quad$ i.e., $\quad a=0$
$$
\begin{aligned}
\therefore x^{3}-2 x^{2}-3 x+6 &=(x-2)\left(x^{2}-3\right) \
&=(x-2)(x+\sqrt{3})(x-\sqrt{3})
\end{aligned}
$$
$$
\therefore P(2)=4 k+8 \quad \text { and since } \quad P(2)=0, \quad k=-2
$$
Now $P(x)=(x-2)\left(x^{2}+[k+2] x+[2 k+1]\right)$
$$
\begin{aligned}
&=(x-2)\left(x^{2}-3\right) \
&=(x-2)(x+\sqrt{3})(x-\sqrt{3})
\end{aligned}
$$









多变量方法（M级) Multivariate Methods (Level M) STATS5021_1

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这是一份GLA格拉斯哥大学STATS5021_1作业代写的成功案例

多变量方法（M级) Multivariate Methods (Level M) STATS5021_1
问题 1.

If every $y$ in the population is multiplied by a constant $a$, the expected value is also multiplied by $a$ :
$$
E(a y)=a E(y)=a \mu .
$$
The sample mean has a similar property. If $z_{i}=a y_{i}$ for $i=1,2, \ldots, n$, then
$$
\bar{z}=a \bar{y}
$$

证明 .

The variance of the population is defined as $\operatorname{var}(y)=\sigma^{2}=E(y-\mu)^{2}$. This is the average squared deviation from the mean and is thus an indication of the extent to which the values of $y$ are spread or scattered. It can be shown that $\sigma^{2}=E\left(y^{2}\right)-\mu^{2}$.
The sample variance is defined as
$$
s^{2}=\frac{\sum_{i=1}^{n}\left(y_{i}-\bar{y}\right)^{2}}{n-1}
$$
which can be shown to be equal to
$$
s^{2}=\frac{\sum_{i=1}^{n} y_{i}^{2}-n \bar{y}^{2}}{n-1} .
$$

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STATS5021_1 COURSE NOTES ：

$$
\begin{aligned}
z_{i} &=a_{1} y_{i 1}+a_{2} y_{i 2}+\cdots+a_{p} y_{i p} \
&=\mathbf{a}^{\prime} \mathbf{y}{i}, \quad i=1,2, \ldots, n \end{aligned} $$ The sample mean of $z$ can be found either by averaging the $n$ values $z{1}=\mathbf{a}^{\prime} \mathbf{y}{1}, z{2}=$ $\mathbf{a}^{\prime} \mathbf{y}{2}, \ldots, z{n}=\mathbf{a}^{\prime} \mathbf{y}{n}$ or as a linear combination of $\overline{\mathbf{y}}$, the sample mean vector of $\mathbf{y}{1}$, $\mathbf{y}{2}, \ldots, \mathbf{y}{n}$ :
$$
\bar{z}=\frac{1}{n} \sum_{i=1}^{n} z_{i}=\mathbf{a}^{\prime} \bar{y}
$$