Tag Archives: 爱丁堡格大学代写

高级物理化学|CHEM10056/CHEM11066 Advanced Physical Chemistry代写

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高级物理化学|CHEM10056/CHEM11066 Advanced Physical Chemistry代写

obtained as:
$$
K_{f}=\langle\sigma g\rangle \int_{\varepsilon_{a}}^{\infty} P_{c}\left(\varepsilon_{c}\right) \int_{\varepsilon_{v}=0}^{\varepsilon_{e}} P_{v}\left(\varepsilon_{v}\right) f_{B}\left(\varepsilon_{v}\right) f_{B}\left(\varepsilon_{c}-\varepsilon_{v}\right) d \varepsilon_{v} d \varepsilon_{c}
$$
For this model, the overall dissociation probability is:
$$
P_{V F D}=P_{c}\left(\varepsilon_{c}\right) P_{v}\left(\varepsilon_{v}\right)=A^{\prime}\left(\varepsilon_{v} / \varepsilon_{c}\right)^{\phi} \frac{\left(\varepsilon_{c}-\varepsilon_{a}\right)^{\psi}}{\left(\varepsilon_{c}\right)^{x}}
$$
where
$$
A^{\prime}=B \frac{\Gamma\left(\frac{\zeta_{v}}{2}\right)}{\Gamma\left(\frac{\zeta_{\mathrm{v}}}{2}+\phi\right)} \frac{\Gamma(\zeta+2-\omega+\phi)}{\Gamma(2-\omega) \Gamma\left(\zeta+b+\frac{3}{2}\right)}
$$



英国论文代写Viking Essay为您提供实分析作业代写Real anlysis代考服务

CHEM10056/CHEM11066 COURSE NOTES ：

$$
\mathrm{I}{2}+\mathrm{F} \rightarrow \mathrm{I}+\mathrm{IF} $$ With no $\mathrm{Cl}$ present, emission of $\mathrm{I}^{}$ from reaction (32) was detected. When $\mathrm{Cl}$ atoms were introduced via the reaction $$ \mathrm{F}+\mathrm{H} / \mathrm{DCl} \rightarrow \mathrm{H} / \mathrm{DF}+\mathrm{Cl} $$ strong enhancement of the $I^{}$ emission was observed. The proposed mechanism for enhancement was the sequence
$$
\begin{aligned}
\mathrm{Cl}+\mathrm{N}{3} & \rightarrow \mathrm{NCl}(a)+\mathrm{N}_{2} \
\mathrm{NCl}(a)+\mathrm{I} & \rightarrow \mathrm{NCl}(X)+\mathrm{I}^{*}
\end{aligned}
$$



风险中立的资产定价|MATH11157Risk-Neutral Asset Pricing代写

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风险中立的资产定价|MATH11157Risk-Neutral Asset Pricing代写

In the local volatility the risk-neutral dynamics of the underlying index or stock value is described by the SDE
$$
d S_{t}=\left(r_{t}-\delta_{t}\right) S_{t} d t+\sigma\left(t, S_{t}\right) S_{t} d W_{t},
$$
where $r_{t}$ is the deterministic instantaneous risk-free interest rate, $\delta_{t}$ is the deterministic instantaneous dividend yield and the volatility $\sigma\left(t, S_{t}\right)$ is a deterministic function dependent on the current index value $S_{t}$. Discretizing the SDE (1.3), yields the following difference equation
$$
S_{t_{i}}-S_{t_{i-1}}=\left(r_{t_{i-1}}-\delta_{t_{i-1}}\right) S_{t_{i-1}} \Delta_{t}+\sigma\left(t_{i-1}, S_{t_{i-1}}\right) S_{t_{i-1}} \Delta W_{t_{i-1}},
$$
where $\Delta W_{t_{i-1}}=W_{t_{i}}-W_{t_{i-1}} \sim N\left(\Delta_{t}\right)$. Hence we have the following conditional distribution of $S_{t_{i}}$
$$
S_{t_{i}} \mid S_{t_{i-1}} \sim N\left(S_{t_{i-1}}+\left(r_{t_{i-1}}-\delta_{t_{i-1}}\right) S_{t_{i-1}} \Delta_{t}, \sigma\left(t_{i-1}, S_{t_{i-1}}\right) S_{t_{i-1}} \Delta_{t}\right) .
$$


英国论文代写Viking Essay为您提供实分析作业代写Real anlysis代考服务

MATH11157 COURSE NOTES ：

A portfolio is a vector $\theta=\left(\theta_{0}, \theta_{1}, \ldots, \theta_{m}\right) \in \mathbb{R}^{m+1}$ $\theta_{j}$ is the number of units held in asset $A_{j}$ for all $j=0,1, \ldots, r$ Spot Value $($ at $t=0)$ of portfolio $\theta$ denoted $V_{\theta}^{(0)}$ is:
$$
V_{\theta}^{(0)}=\sum_{j=0}^{m} \theta_{j} \cdot S_{j}^{(0)}
$$
Value of portfolio $\theta$ in state $\omega_{i}($ at $t=1)$ denoted $V_{\theta}^{(i)}$ is:
$$
V_{\theta}^{(i)}=\sum_{j=0}^{m} \theta_{j} \cdot S_{j}^{(i)} \text { for all } i=1, \ldots, n
$$



金融中的随机分析|MATH11154 Stochastic Analysis in Finance代写

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这是一份ed.ac爱丁堡格大学MATH11154作业代写的成功案

金融中的随机分析|MATH11154 Stochastic Analysis in Finance代写

$$
\mathbb{P}(X+Y \leq a)=\int_{-\infty}^{\infty} F_{X}(a-y) f_{Y}(y) \mathrm{d} y
$$
Taking the derivatives of both sides with respect to $a$, we have
$$
f_{X+Y}(a)=\int_{-\infty}^{\infty} f_{X}(a-y) f_{Y}(y) \mathrm{d} y=f_{X} * f_{Y}(a)
$$
We used the fact that $F_{X}$ is sufficiently smooth.
A more mathematically rigorous proof goes as follows: Since the joint probability density function of $(X, Y)$ is equal to $f_{X}(x) f_{Y}(y)$, we have
$$
\mathbb{P}(X+Y \leq a)=\int_{-\infty}^{\infty} \int_{-\infty}^{a-y} f_{X}(x) f_{Y}(y) \mathrm{d} x \mathrm{dy}
$$
Differentiating both sides with respect to $a$, we obtain
$$
f_{X+Y}(a)=\int_{-\infty}^{\infty} f_{X}(a-y) f_{Y}(y) \mathrm{dy}
$$


英国论文代写Viking Essay为您提供实分析作业代写Real anlysis代考服务

MATH11154 COURSE NOTES ：

First, consider the case when $h$ is monotonically increasing. Since
$$
\operatorname{Pr}(a \leq X \leq x)=\operatorname{Pr}(h(a) \leq Y \leq h(x)),
$$
we have
$$
\int_{a}^{x} f_{X}(x) \mathrm{d} x=\int_{h(a)}^{h(x)} f_{Y}(y) \mathrm{d} y
$$
By differentiating with respect to $x$, we have
$$
f_{X}(x)=f_{Y}(h(x)) h^{\prime}(x)
$$
Now substitute $x=h^{-1}(y)$. For the case when $h$ is decreasing, note that
$$
\int_{a}^{x} f_{X}(x) \mathrm{d} x=\int_{h(x)}^{h(a)} f_{Y}(y) \mathrm{d} y=-\int_{h(a)}^{h(x)} f_{Y}(y) \mathrm{d} y
$$



随机控制和动态资产配置MATH11150 Stochastic Control and Dynamic Asset Allocation代写

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这是一份ed.ac爱丁堡格大学MATH11150作业代写的成功案

随机控制和动态资产配置MATH11150 Stochastic Control and Dynamic Asset Allocation代写



and is called the optimal value of the problem. When an optimal policy $\pi^{}$ exists, we write
$$
J^{}\left(x_{0}\right)=J_{\pi^{}}\left(x_{0}\right)=\min {n \in \Pi} J{n}\left(x_{0}\right) .
$$
When there does not exist an optimal policy we may be interested in finding an $\varepsilon$-optimal policy, i.e., a policy $\bar{\pi}$ such that
$$
J^{}\left(x_{0}\right) \leqslant J_{\bar{}}\left(x_{0}\right) \leqslant J^{*}\left(x_{0}\right)+\varepsilon,
$$
where $\varepsilon$ is some small number implying an acceptable deviation from optimality.


英国论文代写Viking Essay为您提供实分析作业代写Real anlysis代考服务

MATH11150 COURSE NOTES ：

In the above equations the minimizations indicated are over all functions $\mu_{k}$ such that $\mu_{k}\left(x_{k}\right) \in U_{k}\left(x_{k}\right)$ for all $x_{k}$ and $k$. In addition, the minimization is subject to the ever-present system equation constraint
$$
x_{k+1}=f_{k}\left[x_{k}, \mu_{k}\left(x_{k}\right), w_{k}\right] .
$$
Now we use the fact that for any function $F$ of $x, u$, we have
$$
\inf {\mu \in M} F[x, \mu(x)]=\inf {u \in U(x)} F(x, u)
$$
where $M$ is the set of all functions $\mu(x)$ such that $\mu(x) \in U(x)$ for all $x$.



广义相对论的几何学|MATH11138 Geometry of General Relativity代写

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这是一份ed.ac爱丁堡格大学MATH11138作业代写的成功案

广义相对论的几何学|MATH11138 Geometry of General Relativity代写

We can represent $V$ as a sum,
$$
\mathrm{V}=\pi V+V^{\perp},
$$
where $V^{\perp}$ is the component of $V$ normal to $T_{m}$. Now write $\partial / \partial x^{k}$ as $e_{k}$, and write
$$
\pi V=a^{1} e_{1}+\ldots+a^{n} e_{n},
$$
where the $a^{i}$ are the desired local coordinates. Then
$$
\begin{aligned}
V &=\pi V+V^{\perp} \
&=a^{1} e_{1}+\ldots+a^{n} e_{n}+V^{\perp}
\end{aligned}
$$


英国论文代写Viking Essay为您提供实分析作业代写Real anlysis代考服务

MATH11138 COURSE NOTES ：

which we can write in matrix form as
$$
\left[V e_{i}\right]=\left[a^{i}\right] g_{\text {事 }}
$$
whence
$$
\left[a^{i}\right]=\left[V \cdot e_{i}\right] g^{* } . $$ Finally, since $g^{ }$ is symmetric, we can transpose everything in sight to get $$ \left[a^{i}\right]=g^{ *}\left[V \cdot e_{i}\right],
$$



数据科学的大规模优化|MATH11147 Large Scale Optimization for Data Science代写

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这是一份ed.ac爱丁堡格大学MATH11147作业代写的成功案

数据科学的大规模优化|MATH11147 Large Scale Optimization for Data Science代写

$$
R_{t_{3}}=\frac{d c_{l}}{d u_{j}}=\frac{\partial c_{i}}{\partial u_{j}}+\sum_{k=1}^{n} \frac{\partial c_{i}}{\partial X_{k}} \frac{d X_{k}}{d u_{j}}
$$
We can now write in the form $R^{T} R \delta u=-R^{T} c$. This is cquivilent to the least squares problem MIN $|R \mathcal{x}+c|_{2}$ which can be solved by applying a constrained linear least squares solver.

The test coses discussed in this section are pressure matching cases inwolving a target pressure distribution obtained by analyzing an airfoll section similar to the ONERA M6 section. The initial airfoal was the NACA0012. The objective function is
$$
I=\frac{1}{2} \sum_{l=1}^{q} W_{k}\left[\left(c_{\mathrm{p}}\right){\ell}-\left(c{p}\right)_{1}\right]^{2}
$$


英国论文代写Viking Essay为您提供实分析作业代写Real anlysis代考服务

MATH11147  COURSE NOTES ：

$$
\beta(x)=\frac{\phi_{h}(x)}{\phi_{j o}(x)}
$$
and onstruct the linear approximation
$$
\beta_{k}(x)=\beta\left(x_{k}\right)+\nabla \beta\left(x_{k}\right)^{T}\left(x-x_{k}\right) .
$$
Then
$$
\bar{\phi}(x)=\beta_{k}(x) \phi_{l o}(x)
$$



应用数学的高级方法|MATH11138 Geometry of General Relativity代写

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这是一份ed.ac爱丁堡格大学MATH11138作业代写的成功案

应用数学的高级方法|MATH10086 Advanced Methods of Applied Mathematics代写

We can represent $V$ as a sum,
$$
\mathrm{V}=\pi V+V^{\perp}
$$
where $V^{\perp}$ is the component of $V$ normal to $T_{m}$. Now write $\partial / \partial x^{k}$ as $e_{k}$, and write
$$
\pi V=a^{1} e_{1}+\ldots+a^{n} e_{n},
$$
where the $a^{i}$ are the desired local coordinates. Then
$$
\begin{aligned}
V &=\pi V+V \perp \
&=a^{1} e_{1}+\ldots+a^{n} e_{n}+V^{\perp}
\end{aligned}
$$


英国论文代写Viking Essay为您提供实分析作业代写Real anlysis代考服务

MATH11138 COURSE NOTES ：

which we can write in matrix form as
$$
\left[V e_{i}\right]=\left[a^{i}\right] g_{\text {央 }}
$$
whence
$$
\left[a^{i}\right]=\left[V \cdot e_{i}\right] g^{* } $$ Finally, since $g^{ }$ is symmetric, we can transpose everything in sight to get $$ \left[a^{i}\right]=g^{ *}\left[V e_{i}\right],
$$



金融风险理论|MATH11132 Financial Risk Theory代写

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这是一份ed.ac爱丁堡格大学MATH11132作业代写的成功案

金融风险理论|MATH11132 Financial Risk Theory代写

fact take place, estimate the probabilities (which will clearly sum to 1) for each outcome from no injury up to death. In an obvious notation we can now obtain the valuc of risk as:
$$
R=\sum_{n=1}^{2} p(n) \sum_{3=1}^{6} p(n, s) w(s)
$$
where $p(1)$ and $p(2)$ are the respective probabilities of a bad fall and avalanche is, for example, the probability that an avalanche, if it occurs, will lead to death. The generalisation to a larger number of adverse occurrences is obvious. This measure of risk $R$ can be called, for obvious reasons, the “equivalent probability of death”, with the minimum value of 0 corresponding to no possibility of injury or death and the maximum value of 1 corresponding to imminent death with certainty.


英国论文代写Viking Essay为您提供实分析作业代写Real anlysis代考服务

MATH11132 COURSE NOTES ：

A Gaussian of mean $m$ and root mean square $\sigma$ is defined as:
$$
P_{G}(x) \equiv \frac{1}{\sqrt{2 \pi \sigma^{2}}} \exp \left(-\frac{(x-m)^{2}}{2 \sigma^{2}}\right) \text {. }
$$
The median and most probable value are in this case equal to $m$, while the MAD (or any other definition of the width) is proportional to the RMS (for example, $E_{a h m}=\sigma \sqrt{2 / \pi}$ ). For $m=0$, all the odd moments are zero while the even moments are given by $m_{2 n}=(2 n-1)(2 n-3) \ldots \sigma^{2 n}=$ $(2 n-1) ! ! \sigma^{2 n}$.

All the cumulants of order greater than two are zero for a Ganssian. This can be realised by examining its characteristic function:
$$
\hat{P}_{C}(z)=\exp \left(-\frac{\sigma^{2} z^{2}}{2}+i m z\right)
$$



应用数学的高级方法|MATH10086 Advanced Methods of Applied Mathematics代写

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这是一份ed.ac爱丁堡格大学MATH10086作业代写的成功案

应用数学的高级方法|MATH10086 Advanced Methods of Applied Mathematics代写

$$
\sigma_{x}=\frac{\partial^{2} U}{\partial x^{2}}, \quad \tau_{x y}=-\frac{\partial^{2} U}{\partial x \partial y}, \quad \sigma_{y}=\frac{\partial^{2} U}{\partial y^{2}}=0
$$
Keeping them in mind, we are able to combin into a single biharmonic equation defined by:
$$
\frac{\partial^{4} U}{\partial x^{4}}+2 \frac{\partial^{4} U}{\partial x^{2} \partial y^{2}}+\frac{\partial^{4} U}{\partial y^{4}}=0
$$
The solution to the above equation is that biharmonic function $U$ must be of the following form
$$
U(x, y)=\operatorname{Re}[\bar{z} \varphi(z)+\chi(z)]
$$


英国论文代写Viking Essay为您提供实分析作业代写Real anlysis代考服务

MATH10086 COURSE NOTES ：

$$
\alpha(z) \varphi(z)+\beta(z) z \overline{\varphi^{\prime}(z)}+\beta(z) \overline{\psi(z)}=F \text { for } z \in \Gamma
$$
where $\alpha, \beta$ and $F$ are associated with the boundary conditions. For force boundary conditions, we have
$$
\alpha=1, \quad \beta=1, \quad F=i \int_{z_{0}}^{z}\left(X_{n}+i Y_{n}\right) d s
$$
whereas for displacement boundary conditions,
$$
\alpha=\kappa, \quad \beta=-1, \quad F=2 \mu(\tilde{u}+i \tilde{v}) .
$$



线性分析|MATH10082 Linear Analysis代写

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线性分析|MATH10082 Linear Analysis代写

$$
\eta_{2} \equiv\left||M|^{-1 / 2} F|M|^{-1 / 2} \mid\right|_{2}<1
$$
then
$$
|\sin \Theta|_{F} \leqslant|Z| \frac{\eta_{\mathrm{F}}}{\sqrt{1-\eta_{2}}} / \min {\hat{\lambda} \in A, \hat{i} \in i} \frac{|\lambda-\hat{\lambda}|}{\sqrt{\hat{\lambda} \hat{\lambda}}} \text {, } $$ where $\eta{\mathrm{F}} \equiv\left||M|^{-1 / 2} F|M|^{-1 / 2} \mid\right|_{\mathrm{F}}$, and $Z$ is $J$-unitary.$$
|\sin \Theta|_{F}=|S|_{F} \leqslant|W|_{F} / \min _{\lambda \in A, \hat{i} \in i} \frac{|\hat{\lambda}-\hat{\lambda}|}{\sqrt{|\lambda \hat{\lambda}|}} .
$$


英国论文代写Viking Essay为您提供实分析作业代写Real anlysis代考服务

MATH10082 COURSE NOTES ：

To derive the absolute bound, multiply $r=(A-\hat{\lambda} I) \hat{x}$ by $Y^{}$ and use $Y^{} A=Y^{} A$, $$ Y^{} \hat{x}=(A-\hat{\lambda} I)^{-1} Y^{} r . $$ With $P$ being the orthogonal projector onto $\mathscr{S}^{\perp}=\operatorname{range}(Y)$ one gets $$ P \hat{x}=\left(Y^{\dagger}\right)^{} Y^{} \hat{x}=\left(Y^{\dagger}\right)^{}(\Lambda-\hat{\lambda} I)^{-1} Y^{*} r .
$$
From $|\hat{x}|_{2}=1$ follows
$$
|\sin \Theta|_{2}=|P \hat{x}|_{2} \leqslant \kappa(Y)\left|(\Lambda-\hat{\lambda} I)^{-1}\right|_{2}|r|_{2} .
$$
To derive the relative bound, we will use the absolute bound. Multiply $\left(D_{1} A D_{2}\right) \hat{x}=\hat{\hat{x}}$ by $D_{1}^{-1}$ and set $z \equiv D_{2} \hat{x} /\left|D_{2} \hat{x}\right|$,
$$
A z=\hat{\lambda} D_{1}^{-1} D_{2}^{-1} z .
$$
The residual for $\hat{\lambda}$ and $z$ is
$$
f \equiv A z-\hat{\lambda} z=\hat{\lambda}\left(D_{1}^{-1} D_{2}^{-1}-I\right) z=\hat{\lambda}\left(D_{1}^{-1}-D_{2}\right) \hat{x} /\left|D_{2} \hat{x}\right|_{2} .
$$
Hence
$$
|f|_{2} \leqslant|\hat{\lambda}| \alpha_{2}, \quad|f|_{2} \leqslant|\hat{\lambda}| \alpha_{1} /\left|D_{2} \hat{x}\right|_{2} .
$$
The idea is to first apply the absolute bound to the residual $f$ and then make an adjustment from $z$ to $\hat{x}$. Since $f$ contains $\hat{\lambda}$ as a factor we will end up with a relative bound.