Tag Archives: KCL代写

复分析|Complex Analysis代写5CCM212A

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这是一份kcl伦敦大学学院 5CCM212A作业代写的成功案

复分析|Complex Analysis代写5CCM212A
问题 1.

Let $S=S^{(n)}$ be a positive, even, unimodular matrix. Then $n \equiv 0 \bmod 8$.
Hint. Use the relation
$$
w:=1-\frac{1}{z}=\left(\frac{1}{1-z}-1\right)^{-1}
$$
and transform $\vartheta(S ; w)$ corresponding to these relations, by applying the formulas
$$
\vartheta(S ; z+1)=\vartheta(S ; z), \quad \vartheta(S ;-1 / z)=\sqrt{\frac{z}{i}}^{n} \vartheta(S ; z)
$$


证明 .

This gives the formula
$$
\sqrt{z / i}^{n}=\sqrt{z /(i(1-z))}^{n} \sqrt{(z-1) / i}^{n} .
$$
Now specialize $z=\mathrm{i}$ in it to infer
$$
1=e^{2 \pi \operatorname{in} / 8} \text {, i.e. } n \equiv 0 \quad \bmod 8
$$

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55CCM212A COURSE NOTES ：

The derivative of $s \mapsto 1-p^{-s}$ is $(\log p) p^{-s}$, the logarithmic derivative being thus
$$
\frac{(\log p) p^{-s}}{1-p^{-s}}=(\log p) \sum_{\nu=1}^{\infty} p^{-\nu s}
$$
This implies
$$
-\frac{\zeta^{\prime}(s)}{\zeta(s)}=\sum_{p}(\log p) \sum_{v=1}^{\infty} p^{-\nu s}
$$




古典动力学|Classical Dynamics代写5CCM231A

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这是一份kcl伦敦大学学院  5CCM231A作业代写的成功案

古典动力学|Classical Dynamics代写5CCM231A
问题 1.

Suppose that $f(r)=-\gamma / r^{2}$ where $\gamma>0$. Then $f(1 / u)=-\gamma u^{2}$ and the path equation becomes
$$
\frac{d^{2} u}{d \theta^{2}}+u=\frac{\gamma}{L^{2}}
$$
where $L$ is the angular momentum of the orbit. This has the form of the SHM equation with a constant on the right. The general solution is


证明 .

$$
u=A \cos \theta+B \sin \theta+\frac{\gamma}{L^{2}}
$$
which can be written in the form
$$
\frac{1}{r}=\frac{\gamma}{L^{2}}(1+e \cos (\theta-\alpha)),
$$

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5CCM231A COURSE NOTES ：

In order to express these reactions in terms of $\theta$ alone, we need to know $\dot{\theta}^{2}$ and $\ddot{\theta}$ as functions of $\theta$. From the previously derived equation of motion, we already have
$$
\dot{\theta}^{2}=\frac{3 g}{4 a}(1-2 \cos \theta)
$$
and, if we differentiate this equation with respect to $t$ (and cancel by $\dot{\theta}$ ), we obtain
$$
\ddot{\theta}=\frac{3 g}{4 a} \sin \theta .
$$
On making use of the above expressions for $\dot{\theta}^{2}$ and $\ddot{\theta}$, the required reactions are found to be
$$
N^{F}=\frac{M g}{4}\left(1-3 \cos \theta+9 \cos ^{2} \theta\right), \quad N^{W}=\frac{3 M g}{4} \sin \theta(3 \cos \theta-1)
$$




微积分II|Calculus II代写  4CCM112A

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这是一份kcl伦敦大学学院 4CCM112A作业代写的成功案

微积分II|Calculus II代写  4CCM112A
问题 1.

A famous distribution function is given by the standard normal distribution, whose probability density function $f$ is
$$
f(x)=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2}, \quad \text { for }-\infty<x<\infty
$$


证明 .

This is often called a “bell curve”, and is used widely in statistics. Since we are claiming that $f$ is a p.d.f., we should have
$$
\int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2} d x=1
$$
by formula , which is equivalent to
$$
\int_{-\infty}^{\infty} e^{-x^{2} / 2} d x=\sqrt{2 \pi}
$$

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4CCM112A COURSE NOTES ：

The expected value $E X$ of a random variable $X$ can be thought of as the “average” value of $X$ as it varies over its sample space. If $X$ is a discrete random variable, then
$$
E X=\sum_{x} x P(X=x) \text {, }
$$
with the sum being taken over all elements $x$ of the sample space. For example, if $X$ represents the number rolled on a six-sided die, then
$$
E X=\sum_{x=1}^{6} x P(X=x)=\sum_{x=1}^{6} x \frac{1}{6}=3.5
$$
is the expected value of $X$, which is the average of the integers $1-6$.




微积分I|Calculus I代写   4CCM111A

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这是一份kcl伦敦大学学院 4CCM111A作业代写的成功案

微积分I|Calculus I代写   4CCM111A
问题 1.

so
$e^{y}-2 x-e^{-y}=0$
or, multiplying by $e^{y}$,
$$
e^{2 y}-2 x e^{y}-1=0
$$
This is really a quadratic equation in $e^{y}$ :
$$
\left(e^{y}\right)^{2}-2 x\left(e^{y}\right)-1=0
$$


证明 .

Solving by the quadratic formula, we get
$$
e^{y}=\frac{2 x \pm \sqrt{4 x^{2}+4}}{2}=x \pm \sqrt{x^{2}+1}
$$
Note that $e^{y}>0$, but $x-\sqrt{x^{2}+1}<0$ (because $x<\sqrt{x^{2}+1}$ ). Thus the minus sign is inadmissible and we have
$$
e^{y}=x+\sqrt{x^{2}+1}
$$
Therefore
$$
y=\ln \left(e^{y}\right)=\ln \left(x+\sqrt{x^{2}+1}\right)
$$

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4CCM111A COURSE NOTES ：

Using Table 6 and the Chain Rule, we have
$$
\begin{aligned}
\frac{d}{d x}\left[\tanh ^{-1}(\sin x)\right] &=\frac{1}{1-(\sin x)^{2}} \frac{d}{d x}(\sin x) \
&=\frac{1}{1-\sin ^{2} x} \cos x=\frac{\cos x}{\cos ^{2} x}=\sec x
\end{aligned}
$$




应用微分方程|Applied Differential Equations代写   5CCM211A

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这是一份kcl伦敦大学学院 5CCM211A作业代写的成功案

应用微分方程|Applied Differential Equations代写   5CCM211A
问题 1.

$$
\mathrm{e}^{p t}\left(\frac{\mathrm{d} x}{\mathrm{~d} t}+p x\right)=q \mathrm{e}^{p t}
$$
and using this is simply
$$
\frac{\mathrm{d}}{\mathrm{d} t}\left(x \mathrm{e}^{p t}\right)=q \mathrm{e}^{p t} .
$$
For the general solution we integrate both sides to give
$$
x(t) \mathrm{e}^{p t}=\frac{q}{p} \mathrm{e}^{p t}+C,
$$


证明 .

so that
$$
x(t)=\frac{q}{p}+C \mathrm{e}^{-p t} .
$$
(It follows that if $p>0$ then $x(t) \rightarrow q / p$ as $t \rightarrow \infty$, independent of any initial condition.)
If we want the solution that has $x(a)=x_{a}$ then we need
$$
x_{a}=\frac{q}{p}+C \mathrm{e}^{-p a} \quad \Rightarrow \quad C=\left(x_{a}-\frac{q}{p}\right) \mathrm{e}^{p a}
$$
and so this solution is
$$
x(t)=\frac{q}{p}+\left(x_{a}-\frac{q}{p}\right) \mathrm{e}^{-p(t-a)} .
$$

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5CCM211A COURSE NOTES ：

$$
\frac{1}{I} \mathrm{~d} I=p(t) \mathrm{d} t
$$
and then by integration we get
$$
\ln |I(t)|=\int p(t) \mathrm{d} t .
$$
Finally we exponentiate both sides and choose $I(t)$ to be positive to give
$$
I(t)=\exp \left(\int p(t) \mathrm{d} t\right) .
$$
Given this integrating factor we should now be able to solve our general linear equation
$$
\frac{\mathrm{d} x}{\mathrm{~d} t}+p(t) x=q(t)
$$




代数数论|Algebraic Number Theory代写  7CCMMS03

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这是一份kcl伦敦大学学院  7CCMMS20作业代写的成功案

代数数论|Algebraic Number Theory代写  7CCMMS03
问题 1.

$$
|y|+\sum_{i=1}^{r_{1}}\left|y_{i}\right|+2 \sum_{j=1}^{r_{2}}\left|z_{j}\right| \leq t
$$
equivalently by
$$
\sum_{i=1}^{r_{1}}\left|y_{i}\right|+2 \sum_{j=1}^{r_{2}}\left|z_{j}\right| \leq t-|y|
$$


证明 .

Now if $|y|>t$, then $B_{t}$ is empty. For smaller values of $|y|$, suppose we change $y$ to $y+d y$. This creates a box in $(n+1)$-space with $d y$ as one of the dimensions. The volume of the box is $V\left(r_{1}, r_{2}, t-y\right) d y$. Thus
$$
V\left(r_{1}+1, r_{2}, t\right)=\int_{-t}^{t} V\left(r_{1}, r_{2}, t-|y|\right) d y
$$
which by the induction hypothesis is $2 \int_{0}^{t} 2^{r_{1}}(\pi / 2)^{r_{2}}\left[(t-y)^{n} / n !\right] d y$. Evaluating the integral, we obtain $2^{r+1}(\pi / 2)^{r_{2}} t^{n+1} /(n+1)$ !, as desired.
Finally, $V\left(r_{1}, r_{2}+1, t\right)$ is the volume of the set described by
$$
\sum^{r_{1}}\left|y_{i}\right|+2 \sum^{r_{2}}\left|z_{j}\right|+2|z| \leq t
$$

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7CCMMS03 COURSE NOTES ：

Proof. If $x \in B^{*}$, then by (6.1.1) and (2.1.6),
$$
\pm 1=N(x)=\prod_{i=1}^{n} \sigma_{i}(x)=\prod_{i=1}^{r_{1}} \sigma_{i}(x) \prod_{j=r_{1}+1}^{r_{1}+r_{2}} \sigma_{j}(x) \overline{\sigma_{j}(x)}
$$
Take absolute values and apply the logarithmic embedding to conclude that $\lambda(x)=$ $\left(y_{1}, \ldots, y_{r_{1}+r_{2}}\right)$ lies in the hyperplane $W$ whose equation is
$$
\sum_{i=1}^{r_{1}} y_{i}+2 \sum_{j=r_{1}+1}^{r_{1}+r_{2}} y_{j}=0
$$




代数几何学|Algebraic Geometry代写  7CCMMS20

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这是一份kcl伦敦大学学院  7CCMMS20作业代写的成功案

代数几何学|Algebraic Geometry代写  7CCMMS20
问题 1.

Let $R=K\left[T_{0}, \ldots, T_{n}\right]$. Then, for $k \geq 0$
$$
\operatorname{dim} R_{k}=\text { number of monomials of degree } k \text { in } T_{0}, \ldots, T_{n}=\left(\begin{array}{c}
k+n \
n
\end{array}\right)
$$


证明 .

Hence $P_{\mathrm{p} n}=\frac{1}{n !}(T+n) \cdots \ldots \cdot(T+1)=\frac{T^{n}}{n !}+\ldots$
If $X \subset p^{n}$ is a hypersurface given by a homogeneous polynomial $F$ of degree $d$ then, for $k \geq d$,
$$
\operatorname{dim}\left(K\left[T_{0}, \ldots, T_{n}\right] /(F)\right){k}=\left(\begin{array}{c} k+n \ n \end{array}\right)-\left(\begin{array}{c} k+n-d \ n \end{array}\right) $$ whence $P{X}=\frac{d}{(n-1) !} T^{n-1}+\ldots .$

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7CCMMS20 COURSE NOTES ：

Suppose, for simplicity, that $X$ is a smooth $n$-dimensional variety, and let $Y$ and $Z$ be two subvarieties of $X$. As was shown in Chap. $2, \S 6$, we have
$$
\operatorname{dim}(Y \cap Z) \geq \operatorname{dim} Y+\operatorname{dim} Z-n
$$
If equality holds, that is, if the varieties $Y$ and $Z$ meet properly then, to each component $W$ of $Y \cap Z$, intersection theory assigns some multiplicity $i(W ; Y, Z)$, and defines the intersection of $Y$ and $Z$ to be the cycle
$$
Y \cdot Z=\sum_{W} i(W ; Y, Z) \cdot[W]
$$




热物理和物质属性|Thermal Physics & Properties of Matter代写    5CCP4000代考

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这是一份KCL伦敦大学学院5CCP4000作业代写的成功案

纳米系统的理论处理|Theoretical Treatments Of Nano-systems代写   7CCP4473代考
问题 1.

The correct argumentation realises, that the particles are indistinguishable!
Before removing the separator:
$$
Z_{\mathrm{a}}=\frac{Z_{1}^{N}}{N !} \cdot \frac{Z_{1}^{N}}{N !}
$$
After removing the separator:
$$
Z_{\mathrm{b}}=\frac{\left(2 Z_{1}\right)^{2 N}}{(2 N) !}
$$


证明 .

Before removing the separator: $F_{\mathrm{a}}=-k_{\mathrm{B}} T \ln \left(\frac{Z_{1}^{N}}{N !} \cdot \frac{Z_{1}^{N}}{N !}\right)=-k_{\mathrm{B}} T 2\left[N \ln Z_{1}-N \ln N+N\right]$ After removing the separator: $F_{\mathrm{b}}=-k_{\mathrm{B}} T \ln \left(\frac{\left(2 Z_{1}\right)^{2 N}}{(2 N) !}\right)=-k_{\mathrm{B}} T\left[2 N \ln 2 Z_{1}-2 N \ln (2 N)+2 N\right]$ and $\Delta F=-k_{\mathrm{B}} T 2 N\left[\ln 2+\ln Z_{1}-\ln (N)-\ln 2+1-\left(\ln Z_{1}-\ln N+1\right)\right]=0$ hence $\Delta S=0$, which is correct!

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5CCP4000 COURSE NOTES ：


\begin{gathered}
P(p) \mathrm{d} p=\frac{1}{N} \quad \frac{V}{h^{3}} 4 \pi p^{2} \mathrm{~d} p \frac{N h^{3}}{V}\left(2 \pi m k_{\mathrm{B}} T\right)^{-\frac{3}{2}} \exp \left(-\frac{p^{2}}{2 m k_{\mathrm{B}} T}\right) \
P(p) \mathrm{d} p=4 \pi p^{2}\left(2 \pi m k_{\mathrm{B}} T\right)^{-\frac{3}{2}} \exp \left(-\frac{p^{2}}{2 m k_{\mathrm{B}} T}\right) \mathrm{d} p
\end{gathered}




纳米系统的理论处理|Theoretical Treatments Of Nano-systems代写   7CCP4473代考

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这是一份KCL伦敦大学学院 7CCP4473作业代写的成功案

纳米系统的理论处理|Theoretical Treatments Of Nano-systems代写   7CCP4473代考
问题 1.

$$
\frac{\partial T(t, x)}{\partial t}=k^{2} \frac{\partial^{2} T(t, x)}{\partial x^{2}}
$$
with initial and boundary conditions
$$
T\left(t_{0}, x\right)=T_{t}(x), T\left(t, x_{0}\right)=T_{0} \text { and } T\left(t, x_{f}\right)=T_{f} \text {. }
$$
A large number of analytical and numerical methods are available to solve the heat equation.
The analytic solution if
$$
T\left(t, x_{0}\right)=0 \text { and } T\left(t, x_{f}\right)=0
$$


证明 .

is given as
$$
\begin{aligned}
&T(t, x)=\sum_{i=1}^{\infty} B_{i} \sin \frac{i \pi x}{x_{f}} e^{-i^{2} \frac{k^{2} \pi^{2}}{x_{f}^{2}} t} \
&B_{i}=\frac{2}{x_{f}} \int_{x_{0}}^{x_{f}} T_{t}(x) \sin \frac{i \pi x}{x_{f}} d x
\end{aligned}
$$

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  7CCP4473 COURSE NOTES ：


$M={E, O, I} .$
The second entry is a letter chosen from the geometric set
$$
G={P, S, T, N, C, A} \text {. }
$$
That is, the actuator/sensor set is give as
$M G={(E, P),(E, S),(E, T), \cdots,(I, N),(I, C),(I, A)} .$
In general, we have
$M G={(m, g): m \quad M$ and $g \quad G} .$




物理学中的对称性|Symmetry In Physics代写  5CCP2332代考

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这是一份KCL伦敦大学学院 5CCP2332作业代写的成功案

物理学中的对称性|Symmetry In Physics代写  5CCP2332代考
问题 1.

$$
D_{\ell}=c_{D}(\ell) \exp [(i / h) \mathbf{P} \cdot \ell]
$$
where $c_{D}(\ell)$ is a constant number which may depend on $\ell$, and $\mathbf{P}$ is the total linear momentum operator of the system:
$$
\mathbf{P}=\sum_{i}^{N} \mathbf{p}_{i}
$$


证明 .

As is well known, the scalar product of an $f$ and a $\varphi$ gives
$$
\left(f_{\mathbf{r}{i}}, \varphi{\mathbf{p}{i}}\right)=\frac{1}{(2 \pi)^{3 / 2}} \exp \left(\frac{i}{\hbar} \mathbf{r}{i}^{\prime} \cdot \mathbf{p}{i}^{\prime}\right) \equiv \varphi{\mathbf{p}{i}}\left(\mathbf{r}{i}{ }^{\prime}\right)
$$
Using Eqs. (2.12) and $(2.16)$ for $D_{t}$ we get
$$
\begin{aligned}
\left(\varphi_{v^{\prime}}, D, f_{r^{\prime}}\right) &=c_{D}(l) \exp \left[(i / \hbar) \mathbf{P}^{\prime} \cdot l\right]\left(\varphi_{p^{\prime}}, f_{r^{\prime}}\right) \
&=c_{D}(l)\left(\varphi_{p^{\prime}}, f_{r^{\prime}-l}\right) .
\end{aligned}
$$

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  5CCP2332 COURSE NOTES ：


$$
D_{t} f_{r^{\prime}}=c_{D}(\ell) f_{r^{\prime}-t}
$$
We now fix the arbitrary phase factor free in $D_{\ell}$ by choosing $c_{D}(\ell)=1$. We then get
$$
D_{t}=\exp [(i / \hbar) \mathbf{P} \cdot \ell]
$$
and
$$
D_{l} f_{r^{\prime}}=f_{\mathrm{r}^{\prime}-l} .
$$
Equation (2.18) implies
$$
\psi_{O}(\mathbf{r})=\left(f_{x}, \psi_{O}\right)=\left(f_{\mathrm{r}}, D_{\ell} \psi_{o}\right)=\psi_{o}(\mathbf{r}+\ell) .
$$
Note that with this choice for $c_{D}(\ell)$, we also have
$$
D_{C \Psi_{D^{\prime}}}=\exp \left[(i / \hbar) \mathbf{P}^{\prime} \cdot \ell\right] \varphi_{\mathbf{D}^{\prime}} \cdot
$$