Tag Archives: KCL代写

实分析|Real Analysis  5CCM221A

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这是一份KCL伦敦大学 5CCM221A作业代写的成功案例

实分析|Real Analysis  5CCM221A
问题 1.


Proof. Let $h_{n} \rightarrow 0, h_{n} \neq 0$. Since $E$ is continuous, $E\left(h_{n}\right) \rightarrow E(0)=$ 1. Let $x_{n}=E\left(h_{n}\right)$; then $x_{n}>0, x_{n} \rightarrow 1$ and $x_{n} \neq 1$ (because $\left.h_{n} \neq 0\right)$, so by $9.5 .10$
$$
\frac{L\left(x_{n}\right)}{x_{n}-1} \rightarrow 1
$$


证明 .

That is,
$$
\frac{h_{n}}{E\left(h_{n}\right)-1} \rightarrow 1
$$
whence the lemmn.

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5CCM221A COURSE NOTES ：


(i) Show that $A$ is an odd function: $A(-x)=-A(x)$ for all $x \in \mathbb{R}$.
{Hint: $f$ is an even function.}
(ii) $A$ is strictly increasing. {Hint: $\left.A^{\prime}=f .\right}$
(iii) For every positive integer $k$,
$$
\frac{1}{1+k^{2}} \leq A(k)-A(k-1) \leq \frac{1}{1+(k-1)^{2}} .
$$
{Hint: Integrate $f$ over the interval
(iv) Let
$$
s_{n}=\sum_{k=1}^{n} \frac{1}{1+k^{2}} .
$$





量子场理论|Quantum Field Theory  7CCMMS32

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这是一份KCL伦敦大学 7CCMMS32作业代写的成功案例

量子场理论|Quantum Field Theory  7CCMMS32
问题 1.


$$
m \sim L^{-\bar{\beta} / \nu} \sim V^{-\bar{\beta} / \nu d}
$$
where we have used the notation $\tilde{\beta}$ for the critical exponent of the magnetization and $L$ denotes the linear extension of the spin system. The total magnetization at $\beta_{c}$ will be
$$
M=m V \sim V^{1-\tilde{\beta} / \nu d} .
$$


证明 .

$$
p \sim m, \quad p V \sim M
$$
The largest spin clusters at the critical point will be fractal. In peculation theory one defines the fractal dimension $D$ of a cluster by
$$
p L^{d}=L^{D}, \quad \text { i.e. } \quad p V=V^{D / d} .
$$
We conclude that $D$ is related to $\tilde{\beta}$ by
$$
\frac{D}{d}=1-\frac{\tilde{\beta}}{\nu d} \quad\left(=\frac{15}{16} \quad \text { for the Ising model }\right)
$$

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7CCMMS32 COURSE NOTES ：


$$
\left[-\frac{\hbar^{2}}{\mathcal{E}} \kappa^{2} G^{a b} \frac{\delta^{2}}{\delta q^{a} \delta q^{b}}+\sqrt{g} U\right] \Phi_{\mathcal{E}}[q]=0
$$
associated with the parameter $\varepsilon$. Finally, if we also impose the mass-shell constraint
$$
\boxplus \Psi=-\mathcal{M}^{2} \Psi
$$
then the only physical states are those with $\varepsilon=M^{2}$, and the (classically indeterminate) constant $M$ can be absorbed, via
$$
\hbar_{e f f}=\frac{\hbar}{\mathcal{M}}
$$
into a rescaling of Planck’s constant.





概率与统计|Probability And Statistics I4CCM141A

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这是一份KCL伦敦大学 4CCM141A作业代写的成功案例

概率与统计|Probability And Statistics I4CCM141A
问题 1.


The pf for the Poisson distribution is
$$
p_{k}=\frac{e^{-\lambda} \lambda^{k}}{k !}, \quad k=0,1,2, \ldots
$$
The probability generating function from Example $3.6$ is
$$
P(z)=e^{\lambda(z-1)}, \quad \lambda>0 .
$$


证明 .

The mean and variance can be computed from the probability generating function as follows:
$$
\begin{aligned}
\mathrm{E}(N) &=P^{\prime}(1)=\lambda \
\mathrm{E}[N(N-1)] &=P^{\prime \prime}(1)=\lambda^{2} \
\operatorname{Var}(N) &=\mathrm{E}[N(N-1)]+\mathrm{E}(N)-[\mathrm{E}(N)]^{2} \
&=\lambda^{2}+\lambda-\lambda^{2} \
&=\lambda
\end{aligned}
$$

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4CCM141A COURSE NOTES ：


It is not difficult to show that the probability generating function for the negative binomial distribution is
$$
P(z)=[1-\beta(z-1)]^{-r} .
$$
From this it follows that the mean and variance of the negative binomial distribution are
$$
\mathrm{E}(N)=r \beta \text { and } \operatorname{Var}(N)=r \beta(1+\beta)
$$





公制空间和拓扑学|Metric Spaces and Topology 5CCM226A

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这是一份KCL伦敦大学 5CCM226A作业代写的成功案例

公制空间和拓扑学|Metric Spaces and Topology 5CCM226A
问题 1.


Proof. This proof imitates that of Theorem.
Let $\left(x_{n}\right)$ be a Cauchy sequence in $\mathbb{R}$. Let $\delta>0$ be arbitrary. There exists a positive integer $N=N(\delta)$ such that for all $m \geq N$ and $n \geq N$, we have $\left|x_{n}-x_{m}\right|<\delta / 2$. In particular we have $\left|x_{n}-x_{N}\right|<\delta / 2$. Or, equivalently,
$$
x_{n} \in\left(x_{N}-\delta / 2, x_{N}+\delta / 2\right) \quad \text { for all } n \geq N .
$$


证明 .

From this we make the following observations:
(i) For all $n \geq N$, we have $x_{n}>x_{N}-\delta / 2$.
(ii) If $x_{n} \geq x_{N}+\delta / 2$, then $n \in{1,2, \ldots, N-1}$. Thus the set of $n$ such that $x_{n} \geq x_{N}+\delta / 2$ is finite.
We shall apply these two observations below for $\delta=1$ and $\delta=\varepsilon$.
We claim that $S$ is nonempty, bounded above and that $\sup S$ is the limit of the given sequence.
From (i), we see that $x_{N}-1 \in S$. Hence $S$ is nonempty.

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5CCM226A COURSE NOTES ：


$$
\left|x_{N}-\ell\right| \leq \varepsilon / 2
$$
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 $\lim {n \rightarrow \infty} x{n}=\ell$.





数学金融II：连续时间|Mathematical Finance II: Continuous Time 6CCM338A

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这是一份KCL伦敦大学 6CCM338A作业代写的成功案例

数学金融II：连续时间|Mathematical Finance II: Continuous Time 6CCM338A
问题 1.

$$
w_{t} \sim \mathcal{N}\left(0, \sigma^{2} t\right)
$$
Since the number of tiny little events which can happen between time $s$ and time $t$ is proportional to $t-s$ we assume
$$
w_{t}-w_{s} \sim \mathcal{N}\left(0, \sigma^{2}(t-s)\right), \forall s<t
$$


证明 .

Also we assume once we know the stock price $S_{s}^{1}$ at some time $s>0$ the future development $S_{t}^{1}$ for $t>s$ does not depend on the stock prices $S_{u}^{1}$ for $u<s$ before $s$. This translates to
$w_{t}-w_{s}$ is independent of $w_{u}, \forall u<s .$

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6CCM338A COURSE NOTES ：


$$
V_{t_{0}}(\varphi)=\varphi_{t_{0}} \cdot X_{t_{0}}
$$
Now he chooses not to change anything with his portfolio until time $t_{1}$. Then his portfolio at time $t_{1}$ still consists of $\varphi_{t_{0}}$ and hence has worth
$$
V_{t_{1}}(\varphi)=\varphi_{t_{0}} \cdot X_{t_{1}} .
$$
At time $t_{1}$ though he chooses to rearrange his portfolio and reinvest all the money from $V_{t_{1}}(\varphi)$ according to $\varphi_{t_{1}}$. After this rearrangement the worth of his portfolio calculates as
$$
V_{t_{1}}(\varphi)=\varphi_{t_{1}} \cdot X_{t_{1}} .
$$





数学金融 I 离散时间|Mathematical Finance I Discrete Time  6CCM388A

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这是一份KCL伦敦大学  6CCM388A作业代写的成功案例

数学金融 I 离散时间|Mathematical Finance I Discrete Time  6CCM388A
问题 1.

$$
\operatorname{cov}(R, L)=\frac{V_{0}^{\prime}}{b} \operatorname{cov}\left(R, R^{\prime}\right)
$$
( $R$ still corresponds to an arbitrary trading strategy). Hence can be rewritten as
$$
\bar{R}-r=-\frac{V_{0}^{\prime}}{b} \operatorname{cov}\left(R, R^{\prime}\right)
$$


证明 .

In particular, in the special case where you choose $H=H^{\prime}, says that
$$
\bar{R}^{\prime}-r=-\frac{V_{0}^{\prime}}{b} \operatorname{cov}\left(R^{\prime}, R^{\prime}\right)=-\frac{V_{0}^{\prime}}{b} \operatorname{var}\left(R^{\prime}\right)
$$
Using this to substitute for $V_{0}^{\prime} / b$ in where now we are back to an arbitrary trading strategy $H$, we obtain the following:

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6CCM388A COURSE NOTES ：


(recall $E\left[C_{1} L / B_{1}\right]=E_{Q}\left[C_{1} / B_{1}\right]$ ). With suitable assumptions about the utility function $u$ to ensure the optimal solution of $(2.22)$ will feature strictly positive consumption values, the following first order necessary conditions must be satisfied:
$$
u^{\prime}\left(C_{0}\right)=\lambda \quad \text { and } \quad u^{\prime}\left(C_{1}(\omega)\right)=\lambda L / B_{1}
$$
Hence
$$
C_{0}=I(\lambda) \text { and } C_{1}(\omega)=I\left(\lambda L / B_{1}\right)
$$





预微积分|Precalculus代写 MATH 1120

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这是一份northeastern东北大学(美国)  MATH 1120作业代写的成功案

预微积分|Precalculus代写 MATH 1120
问题 1.

$$
a_{k+1}>a_{k}
$$
so
$$
a_{k+1}+6>a_{k}+6
$$

证明 .

and
$$
\frac{1}{2}\left(a_{k+1}+6\right)>\frac{1}{2}\left(a_{k}+6\right)
$$
Thus
$$
a_{k+2}>a_{k+1}
$$

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MATH 1120COURSE NOTES ：

Let $a$ and $b$ be positive numbers with $a>b$. Let $a_{1}$ be their arithmetic mean and $b_{1}$ their geometric mean:
$$
a_{1}=\frac{a+b}{2} \quad b_{1}=\sqrt{a b}
$$
Repeat this process so that, in general,
$$
a_{n+1}=\frac{a_{n}+b_{n}}{2} \quad b_{n+1}=\sqrt{a_{n} b_{n}}
$$
(a) Use mathematical induction to show that
$$
a_{n}>a_{n+1}>b_{n+1}>b_{n}
$$(c) Show that $\lim {n \rightarrow \infty} a{n}=\lim {n \rightarrow \infty} b{n}$. Gauss called the common value of these limits the arithmetic-geometric mean of the numbers $a$ and $b$.




动态系统介绍|Introduction to Dynamical Systems for Joint Honours代写 5CCM131B

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这是一份kcl伦敦大学学院 5CCM131B作业代写的成功案

动态系统介绍|Introduction to Dynamical Systems for Joint Honours代写 5CCM131B
问题 1.

This can be separated to give
$$
\frac{1-3 y}{y} \mathrm{~d} y=\frac{5 x-2}{x} \mathrm{~d} x
$$
Integrating both sides we have
$$
\int\left(\frac{1}{y}-3\right) \mathrm{d} y=\int\left(5-\frac{2}{x}\right) \mathrm{d} x
$$

证明 .

taking $x$ and $y$ positive, because they represent the size of a population, we have no need of modulus signs in the logarithms arising from the integration,
$$
\ln y-3 y=5 x-2 \ln x+c .
$$
We can do no better than this implicit solution relating $x$ and $y$. However, we can represent the curves defined by
$$
F(x, y)=\ln y+2 \ln x-3 y-5 x=\text { constant }
$$

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5CCM131B COURSE NOTES ：

For this equation $x(t)=\mathrm{e}^{k t}$ is a solution if $k$ satisfies
$$
k^{2}+2 k+5=0
$$
which has roots
$$
k=\frac{-2 \pm \sqrt{4-20}}{2}=-1 \pm \sqrt{-4}=-1 \pm 2 \mathrm{i}
$$
So the general solution of $(12.10)$ is
$$
x(t)=\mathrm{e}^{-t}(A \cos 2 t+B \sin 2 t),
$$
showing that the origin is stable. Since
$$
\dot{x}(t)=e^{-t}((2 B-A) \cos 2 t-(2 A+B) \sin 2 t)
$$
the initial conditions pick out the solution with
$$
A=1 \quad 2 B-A=0,
$$
i.e. $A=1$ and $B=\frac{1}{2}$, so that
$$
x(t)=\mathrm{e}^{-t}\left(\cos 2 t+\frac{1}{2} \sin 2 t\right) .
$$




代数概论|Introduction To Algebra代写5CCM121B

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这是一份kcl伦敦大学学院 5CCM121B作业代写的成功案

代数概论|Introduction To Algebra代写5CCM121B
问题 1.

Proof. Let $\varphi: C(B) \longrightarrow C(A)$ be a homomorphism of $K$-algebras. Let $q_{j}$ : $B \longrightarrow K,\left(y_{1}, \ldots, y_{n}\right) \longmapsto y_{j}$ be the $j$ th coordinate function of $B$. The polynomial function $B \longrightarrow \bar{K}$ induced by $g=\sum_{k} c_{k} Y_{1}^{k_{1}} \cdots Y_{n}^{k_{n}} \in K\left[Y_{1}, \ldots, Y_{n}\right]$ sends $y=\left(y_{1}, \ldots, y_{n}\right) \in B$ to

证明 .

\begin{aligned}
g\left(y_{1}, \ldots, y_{n}\right) &=g\left(q_{1}(y), \ldots, q_{n}(y)\right) \
&=\sum_{k} c_{k} q_{1}(y)^{k_{1}} \ldots q_{n}(y)^{k_{n}}=\left(\sum_{k} c_{k} q_{1}^{k_{1}} \cdots q_{n}^{k_{n}}\right)(y)
\end{aligned}

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5CCM121B COURSE NOTES ：

If $\varphi: A \longrightarrow B$ is a homomorphism of left R-modules, then $\varphi_{}=\operatorname{Hom}{R}(M, \varphi): \operatorname{Hom}{R}(M, A) \longrightarrow \operatorname{Hom}{R}(M, B)$ is a homomorphism of abelian groups. Moreover: (1) if $\varphi$ is the identity on $A$, then $\varphi{}$ is the identity on $\operatorname{Hom}{R}(M, A)$; (2) $(\psi \circ \varphi){}=\psi_{} \circ \varphi_{}$ whenever $\varphi: A \longrightarrow B$ and $\psi: B \longrightarrow C$; (3) $(\varphi+\psi){}=\varphi{}+\psi_{}$ whenever $\varphi, \psi: A \longrightarrow B$.
Similarly, when $M, N$, and $A$ are left $R$-modules, every module homomorphism $\varphi: M \longrightarrow N$ induces a mapping




无限维矢量空间|Infinite Dimensional Vector Spaces代写7CCMMS05

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这是一份kcl伦敦大学学院  7CCMMS05作业代写的成功案

无限维矢量空间|Infinite Dimensional Vector Spaces代写7CCMMS05
问题 1.

Many problems in classical mathematical physics can be handled by reformulating them in terms of integral equations. A famous example is the Dirichlet problem discussed at the end of this section. Consider the simple operator $K$, defined in $C[0,1]$ by
$$
(K \varphi)(x)=\int_{0}^{1} K(x, y) \varphi(y) d y
$$

证明 .

where the function $K(x, y)$ is continuous on the square $0 \leq x, y \leq 1 . K(x, y)$ is called the kernel of the integral operator $K$. Since
$$
|(K \varphi)(x)| \leq\left(\sup {0 \leq x, y \leq 1}|K(x, y)|\right)\left(\sup {0 \leq y \leq 1}|\varphi(y)|\right)
$$
we see that
$$
|K \varphi|_{\infty} \leq\left(\sup {0 \leq x, y<1}|K(x, y)|\right)|\varphi|{\infty}
$$

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7CCMMS05 COURSE NOTES ：

Let $\langle M, \mu\rangle$ be a measure space and $\mathscr{H}=L^{2}(M, d \mu$ ). Then $A \in \mathscr{L}(\mathscr{H})$ is Hilbert-Schmidt if and only if there is a function
$$
K \in L^{2}(M \times M, d \mu \otimes d \mu)
$$
$$
(A f)(x)=\int K(x, y) f(y) d \mu(y)
$$
Moreover,
$$
|A|_{2}^{2}=\int|K(x, y)|^{2} d \mu(x) d \mu(y)
$$