Tag Archives: imperial代写

商业评估 Business Valuation FINANCE 3442

0
Posted on by

这是一份 Imperial帝国理工大学 FINANCE 3442作业代写的成功案例

商业评估 Business Valuation FINANCE 3442
问题 1.

The arithmetic mean is the simple average of the numbers in each series. We use equation to calculate it.
$$
r_{A}=\frac{1}{n} \sum_{t=1}^{n} r_{t}
$$
The geometric mean return is the compound rate of return over the period. Its formula is in equation
$$
r_{G}=\left[\frac{V_{n}}{V_{0}}\right]^{\frac{1}{n}}-1 .
$$

证明 .

Another way to represent the geometric mean retum is in equation .
$$
r_{G}=\left[\prod_{t=1}^{n}\left(1+r_{t}\right)\right]^{\frac{1}{n}}-1 .
$$


英国论文代写Viking Essay为您提供作业代写代考服务

FINANCE 3442 COURSE NOTES ：

  1. We determine eamings before interest but after taxes (EBIBAT) as the income measure. ${ }^{7}$ This should be normalized EBIBAT. 8
  2. We discount EBIBAT using the WACC.
  3. We must subtract the market value of debt from the calculated market value of invested capital to get the market value of equity.
  4. We must calculate a new WACC for each new iteration of FMV of equity.
  5. We do not show the calculation of unlevered beta but will assume that it has already been calculated to be $1.05$.








企业融资 Corporate Finance FINANCE 1122

0
Posted on by

这是一份 Imperial帝国理工大学 FINANCE 1122作业代写的成功案例

企业融资 Corporate Finance FINANCE 1122
问题 1.

The 6 percent coupon bond with maturity 2002 starts with 3 years left until maturity and sells for $\$ 1,010.77$. At the end of the year, the bond has only 2 years to maturity and investors demand an interest rate of 7 percent. Therefore, the value of the bond becomes
$$
\mathrm{PV} \text { at } 7 \%=\frac{\$ 60}{(1.07)}+\frac{\$ 1,060}{(1.07)^{2}}=\$ 981.92
$$

证明 .

You invested $\$ 1,010.77$. At the end of the year you receive a coupon payment of $\$ 60$ and have a bond worth $\$ 981.92$. Your rate of return is therefore
Rate of return $=\frac{\$ 60+(\$ 981.92-\$ 1,010.77)}{\$ 1,010.77}=.0308$, or $3.08 \%$
The yield to maturity at the start of the year was $5.6$ percent. However, because interest rates rose during the year, the bond price fell and the rate of return was below the yield to maturity.


英国论文代写Viking Essay为您提供作业代写代考服务

FINANCE 1122 COURSE NOTES ：

You are now in a position to determine the value of shares in United. If investors demand a return of $r=10$ percent, then price today should be
$P_{0}=\mathrm{PV}$ (dividends from Years 1 to 3 ) $+\mathrm{PV}$ (forecast stock price in Year 3)
$$
\begin{aligned}
\mathrm{PV}(\text { dividends }) &=\frac{\$ 1.00}{1.10}+\frac{\$ 1.20}{1.10^{2}}+\frac{\$ 1.44}{1.10^{3}}=\$ 2.98 \
\mathrm{PV}\left(P_{H}\right) &=\frac{\$ 30.24}{(1.10)^{3}}=\$ 22.72 \
P_{0} &=\$ 2.98+\$ 22.72=\$ 25.70
\end{aligned}
$$








财务建模 Financial modelling FINANCE 101

0
Posted on by

这是一份 Imperial帝国理工大学 FINANCE 101作业代写的成功案例

财务建模 Financial modelling FINANCE 101
问题 1.

Note that in this connection if $q$ is a positive decreasing function then
$$
\int_{0}^{\infty}\left|\chi_{\delta}(v)\right| \mathrm{d} v=2 \int_{0}^{\delta} q(v) \mathrm{d} v .
$$
Suppose that $\pi \delta$ converges weakly, as $\delta \rightarrow 0$, to a probability measure $\pi$ on $[0, l]$, i.e.
$$
\pi \delta \stackrel{w}{\rightarrow} \pi .
$$

证明 .

Suppose that $l<\infty$ and let
$$
\Psi_{\delta}(u)=\int_{0}^{u} \psi_{\delta}(v) \mathrm{d} v \quad \text { and } \quad \bar{\Psi}{\delta}(u)=\int{l+\delta-u}^{l+\delta} \psi_{\delta}(v) \mathrm{d} v,
$$
so that $c(\delta)^{-1} \Psi_{\delta}$ is the distribution function, say $\Pi_{\delta}$, of $\pi_{\delta}$.
Next, for $k=1,2, \ldots$, let
$$
\begin{aligned}
c_{k}(\delta) &=\int_{(k-1) \delta}^{k \delta} \psi_{\delta}(u) \mathrm{d} u \
&=\int_{0}^{\delta} \psi_{\delta}((k-1) \delta+u) \mathrm{d} u \
&=\delta \int_{0}^{1} \psi_{\delta}((k-1+u) \delta) \mathrm{d} u
\end{aligned}
$$


英国论文代写Viking Essay为您提供作业代写代考服务

FINANCE 101 COURSE NOTES ：

which tends to 0 at the order of $\delta^{2}$. To obtain $\pi_{\delta} \stackrel{w}{\rightarrow} \delta_{0}$ we therefore only need to add the assumption that
$$
\Psi_{\delta}(l+\delta)-\Psi_{\delta}(t+\delta)=o(c(\delta))
$$
Now,
$$
\Psi_{\delta}(l+\delta)-\Psi_{\delta}(t+\delta)=\sum_{k=n+1}^{\infty} c_{k}(\delta)
$$
Thus, letting
$$
\bar{c}{k}(\delta)=\frac{c{k}(\delta)}{c(\delta)}
$$







统计学的概率 PROBABILITY FOR STATISTICS MATH 647

0
Posted on by

这是一份 Imperial帝国理工大学 MATH 647作业代写的成功案例

统计学的概率 PROBABILITY FOR STATISTICS MATH 647
问题 1.

(Extending independence on $\pi$-systems)
(a) Suppose the $\pi$-system $\mathcal{C}$ and a class $\mathcal{D}$ are independent. Then $\sigma[\mathcal{C}]$ and $\mathcal{D}$ are independent.
(b) Suppose the $\pi$-systems $\mathcal{C}$ and $\mathcal{D}$ are independent. Then $\sigma[\mathcal{C}]$ and $\sigma[\mathcal{D}]$ are independent.

证明 .

(c) If $\mathcal{C}{1}, \ldots, \mathcal{C}{n}$ are independent $\pi$-systems (see (2)), then $\sigma\left[\mathcal{C}{1}\right], \ldots, \sigma\left[\mathcal{C}{n}\right] \quad$ are independent $\sigma$-fields.


英国论文代写Viking Essay为您提供作业代写代考服务

MATH 647 COURSE NOTES ：

$\quad \sum_{1}^{n} a_{i} X_{i} \cong\left(\sum_{1}^{n} a_{i} \mu_{i}, \sum_{1}^{n} a_{i}^{2} \sigma_{i}^{2}\right)$.
In particular, suppose $X_{1}, \ldots, X_{n}$ are uncorrelated $\left(\mu, \sigma^{2}\right)$. Then
$\bar{X}{n} \equiv \frac{1}{n} \sum{1}^{n} X_{i} \cong\left(\mu, \frac{\sigma^{2}}{n}\right)$, while $\quad \sqrt{n}\left(\bar{X}{n}-\mu\right) / \sigma \cong(0,1)$, provided that $0<\sigma<\infty$. Moreover $$ \operatorname{Cov}\left[\sum{i=1}^{m} a_{i} X_{i}, \sum_{j=1}^{n} b_{j} Y_{j}\right]=\operatorname{Cov}\left[\sum_{i=1}^{m} a_{i}\left(X_{i}-\mu_{X_{i}}\right), \sum_{j=1}^{n} b_{j}\left(Y_{j}-\mu_{Y_{j}}\right)\right]
$$
$$
=\sum_{i=1}^{m} \sum_{j=1}^{n} a_{i} b_{j} \operatorname{Cov}\left[X_{i}, Y_{j}\right]
$$







微积分I Calculus I MATH 192

0
Posted on by

这是一份 Imperial帝国理工大学 MATH 192作业代写的成功案例

微积分I Calculus I MATH 192
问题 1.

This looks like Greens Theorem with only the dimension of $U$ changed from 2 to 3 . This is right. The proof is the same too.
Recall that we had for
$$
\begin{aligned}
g: & I^{2} \longrightarrow U \subset \mathbb{R}^{2} \
& {\left[\begin{array}{l}
u \
v
\end{array}\right] \rightsquigarrow\left[\begin{array}{l}
x \
y
\end{array}\right] }
\end{aligned}
$$

证明 .

and if $\omega$ is a $1-$ form on $U, \omega=P d x+Q d y, g^{} \omega$ was the $1-$ form on $I^{2}$ defined by $$ \left(g^{} \omega\right)\left[\begin{array}{l}
u \
v
\end{array}\right]=P\left[g\left[\begin{array}{l}
u \
v
\end{array}\right]\right] d x+Q\left[g\left[\begin{array}{l}
u \
v
\end{array}\right]\right] d y
$$
and
$$
\left[\begin{array}{l}
d x \
d y
\end{array}\right]=g^{\prime}\left[\begin{array}{l}
d u \
d v
\end{array}\right]=\left[\begin{array}{ll}
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \
\frac{\partial y}{\partial u} & \frac{\partial u}{\partial v}
\end{array}\right]\left[\begin{array}{l}
d u \
d v
\end{array}\right]
$$


英国论文代写Viking Essay为您提供作业代写代考服务

MATH 192 COURSE NOTES ：

If $\omega$ is a smooth differential 1-form on $U \subseteq \mathbb{R}^{3}$ and $g: I^{2} \longrightarrow U$ is a smooth embedding a.e.,
$$
\int_{\partial g\left(I^{2}\right)} \omega=\int_{\partial I^{2}} g^{} \omega $$ Proof We have for each edge of $\left(I^{2}\right), g$ defines a parametric curve in $U \subseteq \mathbb{R}^{3}$ which is part of the boundary of $g\left(I^{2}\right)$ and $g^{} \omega$ is constructed to take care of the length stretch automatically:
$$
\begin{aligned}
\int_{g(I)} \omega=\int_{0}^{1}\left(P \frac{d x}{d t}+Q \frac{d y}{d t}+R \frac{d z}{d t}\right) d t \
&=\int_{I} g^{*} \omega
\end{aligned}
$$







高级微积分 Advanced Calculus  MATH411/MATH410

0
Posted on by

这是一份imperial帝国理工大学 MATH411/MATH410作业代写的成功案例

高级微积分 Advanced Calculus  MATH411/MATH410
问题 1.

Suppose $f(x, y)=x y^{2}$. Then the rate of change of $f(x, y)$, at the point $(2,1)$, in the direction $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right\rangle$ is
$$
\frac{\sqrt{2}}{2} \frac{\partial f}{\partial x}(2,1)+\frac{\sqrt{2}}{2} \frac{\partial f}{\partial y}(2,1)=\frac{\sqrt{2}}{2} 1+\frac{\sqrt{2}}{2} 4=\frac{5 \sqrt{2}}{2}
$$



证明 .

The rate of change of $f(x, y)$, in the direction of $V$, is called a directional derivative, and is denoted as $\nabla_{V} f(x, y)$. Hence, we have the formula
$$
\nabla_{V} f(x, y)=a \frac{\partial f}{\partial x}+b \frac{\partial f}{\partial y}
$$
But notice that the right side of this equation can be rewritten as a dot product:
$$
\nabla_{V} f(x, y)=\langle a, b\rangle \cdot\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right\rangle
$$


英国论文代写Viking Essay为您提供作业代写代考服务

MMATH411/MATH410 COURSE NOTES ：

We may safely assume that neither $a$ nor $b$ are zero. Dividing the first equation by $2 b^{2}$ and the second by $2 a^{2}$ then gives
$$
\begin{aligned}
&0=-a^{2}+100-2 a b \
&0=-b^{2}+100-2 a b
\end{aligned}
$$
Solving both equations for $2 a b$ and setting them equal to each other gives us
$$
-a^{2}+100=-b^{2}+100
$$
and thus (since we are only interested in positive values of $a$ and $b$ ) we may conclude $a=b$. Combining this with Equation 7-1 tells us
$$
0=-a^{2}+100-2 a^{2}
$$








生命科学的微积分II Calculus II for the Life Sciences MATH131/MATH130

0
Posted on by

这是一份imperial帝国理工大学 MATH131作业代写的成功案例

生命科学的微积分II Calculus II for the Life Sciences MATH131
问题 1.

Putting it all together,
$$
C=3 \pi r^{2}+3 \pi r^{2}+4 \pi r h=6 \pi r^{2}+4 \pi r h
$$
Since the goal is to express the cost as a function of the radius alone, you must find a way to express the height $h$ in terms of $r$. To do this, use the fact that the volume $V=\pi r^{2} h$ is to be $24 \pi$. That is, set $\pi r^{2} h$ equal to $24 \pi$ and solve for $h$ to get
$$
\pi r^{2} h=24 \pi \quad \text { or } \quad h=\frac{24}{r^{2}}
$$



证明 .

Now substitute this expression for $h$ into the formula for $C$ :
$$
\begin{aligned}
&C(r)=6 \pi r^{2}+4 \pi r\left(\frac{24}{r^{2}}\right) \
&C(r)=6 \pi r^{2}+\frac{96 \pi}{r}
\end{aligned}
$$
A graph of the relevant portion of this cost function is sketched in Figure 1.34. Notice that there is some radius $r$ for which the cost is minimal. you will learn how to find this optimal radius using calculus.


英国论文代写Viking Essay为您提供作业代写代考服务

MATH131/MATH130 COURSE NOTES ：

If $x$ is the number of units manufactured and sold, the total revenue is given by $R(x)=110 x$ and the total cost by $C(x)=7,500+60 x$.
a. To find the break-even point, set $R(x)$ equal to $C(x)$ and solve:
$$
\begin{aligned}
110 x &=7,500+60 x \
50 x &=7,500 \
x &=150
\end{aligned}
$$
It follows that the manufacturer will have to sell 150 units to break even








随机微分方程 Stochastic Differential Equations M45A51

0
Posted on by

这是一份imperial帝国理工大学 M45A51作业代写的成功案例

随机微分方程 Stochastic Differential Equations M45A51
问题 1.

$$
d q_{t}=\beta q_{t} d t+\theta \gamma(\theta) d W_{t}, q_{0}=0 .
$$
It is easy to verify that
$$
q_{t}=\theta \gamma(\theta) \int_{0}^{t} e^{-\sqrt{\theta^{2}+\beta^{2}(t-s)}} d X_{s} .
$$



证明 .

In this case the LAMN property holds with
$$
m_{T}(\theta)=e^{\theta T}, \quad \zeta(\theta)=\left(\int_{0}^{\infty} e^{-\sqrt{\theta^{2}+\beta^{2}}(t-s)} d W_{s}\right)^{2} .
$$


英国论文代写Viking Essay为您提供作业代写代考服务

M45A51 COURSE NOTES ：

The indefinite integral is defined as
$$
\int_{0}^{t} \psi(s) d W_{s}^{H}=\int_{0}^{1} \psi(t) I_{[0, t]} d W_{t}^{H}
$$
This integral has a continuous version and a Gaussian process. However,
$$
E\left(\int_{0}^{t} \psi(s) d W_{s}^{H}\right) \neq 0 .
$$
To overcome this situation, Duncan, Hu and Pasik-Duncan (2000) introduced an integral using Wick calculus for which
$$
E\left(\int_{0}^{t} f(s) d W_{s}^{H}\right)=0 .
$$








偏微分方程简介 Introduction to Partial Differential Equations MATH96018/97027/97104

0
Posted on by

这是一份imperial帝国理工大学 MATH96018/97027/97104作业代写的成功案例

偏微分方程简介 Introduction to Partial Differential Equations MATH96018/97027/97104
问题 1.

uniformly on each compact set in the $(\zeta, \varepsilon)$ space as $y$ tends to infinity in any closed subsector of the open sector,
$$
|\arg y|<\frac{3 \pi}{5}
$$
moreover
$$
E_{3}(y ; \zeta, \varepsilon)=\frac{2}{5} y^{5 / 2}+\zeta y^{1 / 2}
$$



证明 .

and $B_{3, N}$ are polynomials in $(\zeta, \varepsilon)$.
We note that, setting
$$
\omega=\exp \left[i \frac{2}{5} \pi\right]
$$
and
$$
\mathscr{Y}{3, k}(y ; \zeta, \varepsilon)=\mathscr{Y}{3}\left(\omega^{-k} y ; \omega^{-2 k} \zeta, \omega^{-3 k} \varepsilon\right),
$$


英国论文代写Viking Essay为您提供作业代写代考服务

MATH96018/97027/97104 COURSE NOTES ：

Let us now consider the above equation in a neighborhood of $\left(0, \zeta_{0}\right)$. There exists a positive $\delta$ such that each root of the equation
$$
C_{0}(\zeta, \varepsilon)=0
$$
for some positive integer $p$, is a holomorphic function of $\varepsilon^{1 / p}$, for $0<|\varepsilon|<\delta$, that is
$$
\zeta(\varepsilon)=\zeta_{0}+\sum_{j=0}^{\infty} c_{j}\left(\varepsilon^{1 / p}\right)^{j}=g\left(\varepsilon^{1 / p}\right)
$$
This is actually a consequence of Theorem $3.2 .6$ in $[6]$, observing that $\gamma(\zeta, \varepsilon) \neq 0$ implies that the function $g$ has no polar singularity at the origin.

As a matter of fact the function $g$ is holomorphic in a full neighborhood of the origin so that $\zeta\left(\eta^{p}\right)=g(\eta)$, which is a well-defined holomorphic function of $\eta$.








数学0D2基础 Mathematics 0D2 Foundation MATH19872

0
Posted on by

这是一份imperial帝国理工大学 MATH19872作业代写的成功案例

数学0D2基础 Mathematics 0D2 Foundation MATH19872
问题 1.

(i) $\phi^{(k)}(0)=0$ for $k2 n$;
(ii) $\phi^{(k)}(0)$ and $\phi^{(k)}(1)$ are integers for all $k \in \mathbf{N}$.
Suppose that $\pi^{2}=p / q$ for some positive integers $p, q$, and define
$$
F=q^{n} \sum_{k=0}^{n}(-1)^{2 n-k} \pi^{2(n-k)} \phi^{(2 k)}
$$



证明 .

Show that $F(0)$ and $F(1)$ are integers, and that
$$
F^{\prime \prime}(x)+\pi^{2} F(x)=\pi^{2} p^{n} \phi(x) .
$$
Setting
$$
G(x)=F^{\prime}(x) \sin \pi x-\pi F(x) \cos \pi x
$$
show that
$$
G^{\prime}(x)=\pi^{2} p^{n} \phi(x) \sin \pi x
$$
and hence that
$$
\pi \int_{0}^{1} p^{n} \phi(x) \sin \pi x \mathrm{~d} x
$$


英国论文代写Viking Essay为您提供作业代写代考服务

MATH19872 COURSE NOTES ：

Proof. Let $F$ be a Lebesgue primitive of $f$. Then $F \in \mathcal{P}{f}$, , $$ F{}(\eta)-F_{}(\xi) \leq F(\eta)-F(\xi) \quad(\xi \leq \eta) .
$$
Since a finite union of sets of measure zero has measure zero, it follows from this inequality and Theorem (2.1.10) that
$$
f \leq F_{}^{\prime} \leq F^{\prime}=f $$ almost everywhere. Hence $F_{}^{\prime}=f$ almost everywhere, and $F_{*}$ is a Lebesgue primitive of $f$.