进一步数论 Further Number Theory MATH3078W1-01

Posted on 2022年5月6日 by admin

这是一份southampton南安普敦大学MATH3078W1-011作业代写的成功案例

进一步数论 Further Number Theory MATH3078W1-01

问题 1.

If $p$ is a prime and $k \geq 0$ we have
$$
\sigma\left(p^{k}\right)=\frac{p^{k+1}-1}{p-1} .
$$

证明 .

Since $p$ is prime, the divisors of $p^{k}$ are $1, p, p^{2}, \ldots, p^{k}$. Hence
$$
\sigma\left(p^{k}\right)=1+p+p^{2}+\cdots+p^{k}=\frac{p^{k+1}-1}{p-1}
$$
as desired.
Let $n=p_{1}^{e_{1}} p_{2}^{e_{2}} \cdots p_{r}^{e_{r}}$. Our proof is by induction on $r$. If $r=1, n=p_{1}^{e_{1}}$ and the result follows. Suppose the result is true when $1 \leq r \leq k$. Consider now the case $r=k+1$. That is, let
$$
n=p_{1}^{e_{1}} \cdots p_{k}^{e_{k}} p_{k+1}^{e_{k+1}}
$$

MATH3078W1-01 COURSE NOTES ：

$$
\sigma(n)=\sigma\left(2^{k}\right) \sigma(q)=\left(2^{k+1}-1\right) \sigma(q)
$$
So we have
$$
2^{k+1} q=2 n=\sigma(n)=\left(2^{k+1}-1\right) \sigma(q),
$$
hence
() $2^{k+1} q=\left(2^{k+1}-1\right) \sigma(q) .$ Now $\sigma^{}(q)=\sigma(q)-q$, so $$ \sigma(q)=\sigma^{}(q)+q . $$ Putting this in () we get
$$
2^{k+1} q=\left(2^{k+1}-1\right)\left(\sigma^{*}(q)+q\right)
$$

精算数学 Actuarial Mathematics II MATH3066W1-01

Posted on 2022年5月6日 by admin

这是一份southampton南安普敦大学MATH3066W1-01作业代写的成功案例

精算数学 Actuarial Mathematics II MATH3066W1-01

问题 1.

$Z$ is the present-value random variable for an $n$-year endowment insurance. Hence
$$
\mathrm{E}[Y]=\bar{a}{x: \bar{n}]}=\mathrm{E}\left[\frac{1-Z}{\delta}\right]=\frac{1-\bar{A}{x: \bar{n}}}{\delta}
$$
and
$$
\operatorname{Var}(Y)=\frac{\operatorname{Var}(Z)}{\delta^{2}}=\frac{{ }^{2} \bar{A}{x: \bar{n}}-A{x:\left.\bar{n}\right|^{2}}}{\delta^{2}}
$$

证明 .

In terms of annuity values, becomes
$$
\begin{aligned}
\operatorname{Var}(Y) &=\frac{1-2 \delta^{2} \bar{a}{x: i n}-\left(1-\delta \bar{a}{x: i n}\right)^{2}}{\delta^{2}} \
&=\frac{2}{\delta}\left(\bar{a}{x: i n}-{ }^{2} \bar{a}{x: \bar{n}}\right)-\left(\bar{a}_{x: i n t}\right)^{2}
\end{aligned}
$$

MATH3066W1-01 COURSE NOTES ：

Since $Y=(1-Z) / d$, where
$$
Z= \begin{cases}v^{K+1} & 0 \leq K<n \ v^{n} & K \geq n\end{cases}
$$
is the present-value random variable for a unit of endowment insurance, payable at the end of the year of death or at maturity, we have
$$
\ddot{a}{x: \bar{n}}=\frac{1-\mathrm{E}[Z]}{d}=\frac{1-A{x \cdot \bar{n}}}{d} ;
$$

局部微分方程IPartial Differential Equations MATH2038W1-01

Posted on 2022年5月5日 by admin

这是一份southampton南安普敦大学MATH2038W1-01作业代写的成功案例

局部微分方程IPartial Differential Equations MATH2038W1-01

问题 1.

Now
$$
\mathbb{Q}^{+}=\int_{V_{1}}^{V_{2}} \Lambda_{V}\left(V, T_{2}\right) d V
$$
Furthermore
$$
W=\int_{\Gamma} P d V=\int_{T_{1}}^{T_{2}} \int_{V_{1}(T)}^{V_{2}(T)} \frac{\partial P}{\partial T} d V d T
$$

证明 .

by the Gauss-Green Theorem.
$$
\int_{V_{1}}^{V_{2}} \Lambda_{V}\left(V, T_{2}\right) d V=\frac{T_{2}}{T_{2}-T_{1}} \int_{T_{1}}^{T_{2}} \int_{V_{1}(T)}^{V_{2}(T)} \frac{\partial P}{\partial T} d V d T
$$
$\operatorname{Let} T_{1} \rightarrow T_{2}=T_{}$ $$ \int_{V_{1}}^{V_{2}} \Lambda_{V}\left(V, T_{}\right) d V=T_{} \int_{V_{1}}^{V_{2}} \frac{\partial P}{\partial T}\left(V, T_{}\right) d V
$$
Divide by $V_{2}-V_{1}$ and let $V_{2} \rightarrow V_{1}=V_{*}$, to deduce

MATH2038W1-01 COURSE NOTES ：

The quantities $n ! / k !(n-k) !$ are the famous binomial coefficients, and they are denoted by
$$
\left(\begin{array}{l}
n \
k
\end{array}\right)=\frac{n !}{k !(n-k) !} \quad(n \geq 0 ; 0 \leq k \leq n)
$$
Some of their special values are
$$
\begin{gathered}
\left(\begin{array}{l}
n \
0
\end{array}\right)=1 \quad(\forall n \geq 0) ; \quad\left(\begin{array}{l}
n \
1
\end{array}\right)=n \quad(\forall n \geq 0) ; \
\left(\begin{array}{l}
n \
2
\end{array}\right)=n(n-1) / 2 \quad(\forall n \geq 0) ; \quad\left(\begin{array}{l}
n \
n
\end{array}\right)=1 \quad(\forall n \geq 0) .
\end{gathered}
$$
It is convenient to define $\left(\begin{array}{l}n \ k\end{array}\right)$ to be 0 if $k<0$ or if $k>n$.

算法IAlgorithms MATH2014W1-01

Posted on 2022年5月5日 by admin

这是一份southampton南安普敦大学MATH2014W1-01作业代写的成功案例

问题 1.

$$
x_{n+1}=3 x_{n}+n \quad\left(n \geq 0 ; x_{0}=0\right) .
$$
The winning change of variable, is to let $x_{n}=3^{n} y_{n}$. After substituting and simplifying, we find
$$
y_{n+1}=y_{n}+n / 3^{n+1} \quad\left(n \geq 0 ; y_{n}=0\right)
$$

证明 .

Now by summation,
$$
y_{n}=\sum_{j=1}^{n-1} j / 3^{j+1} \quad(n \geq 0) .
$$
Finally, since $x_{n}=3^{n} y_{n}$ we obtain in the form
$$
x_{n}=3^{n} \sum_{j=1}^{n-1} j / 3^{j+1} \quad(n \geq 0) .
$$

MATH2014W1-01 COURSE NOTES ：

随机过程IStochastic Processes MATH2012W1-01

Posted on 2022年5月3日 by admin

这是一份southampton南安普敦大学MATH2012W1-01作业代写的成功案例

问题 1.

with measurement equation
$$
z(k)=\left[\begin{array}{ll}
1 & 0
\end{array}\right] x^{i}(k)+w(k)
$$
The models differ in the control gain parameter $b^{i}$. The process and measurement noises are mutually uncorrelated with zero mean and variances given by
$$
E[v(k) v(j)]=0.16 \delta_{k j}
$$
and
$$
E[w(k) w(j)]=\delta_{k j}
$$

证明 .

The control gain parameters were chosen to be $b^{1}=2$ and $b^{2}=0.5$.
The Markov transition matrix was selected to be
$$
\left[\begin{array}{ll}
0.8 & 0.2 \
0.1 & 0.9
\end{array}\right]
$$
For this example $N=7$, and the cost parameters $R(k)$ and $Q(k)$, were selected as
$$
R(k)=5.0 \quad k=1,2, \ldots, N-1
$$

MATH2012W1-01 COURSE NOTES ：

Proof: Since $f(d)$ is the minimal polynomial of $\boldsymbol{F}, p(d)$ can be factored as
$$
p(d)=g(d) . f(d)
$$
for some polynomial $g(d)=\sum_{i=0}^{s} b_{i} d^{s-i}$
Let $z_{k}$ and $\bar{z}{k}$ be linear combinations of $y{k}$ defined as in by using polynomials $f(d)$ and $p(d)$, respectively.
$\bar{z}{k} \quad$ can be expressed in terms of $z{k}$ as
$$
\bar{z}{k}=\sum{i=0}^{s} b_{i} z_{k-i}
$$

统计建模I Statistical Modelling I MATH2010W1-01

Posted on 2022年5月3日 by admin

这是一份southampton南安普敦大学MATH2010W1-011作业代写的成功案例

统计建模I Statistical Modelling I MATH2010W1-01

问题 1.

Our second test for normality.The observations $y_{1}, y_{2}, \ldots, y_{n}$ are ordered as $y_{(1)} \leq y_{(2)} \leq \cdots \leq y_{(n)}$, and we calculate
$$
\begin{aligned}
&D=\frac{\sum_{i=1}^{n}\left[i-\frac{1}{2}(n+1)\right] y_{(i)}}{\sqrt{n^{3} \sum_{i=1}^{n}\left(y_{i}-\bar{y}\right)^{2}}}, \
&Y=\frac{\sqrt{n}\left[D-(2 \sqrt{\pi})^{-1}\right]}{.02998598} .
\end{aligned}
$$
A table of percentiles for $Y$, given for $10 \leq n \leq 250$, is provided in Table A.4.
The final test we report is by Lin and Mudholkar (1980). The test statistic is
$$
z=\tanh ^{-1}(r)=\frac{1}{2} \ln \left(\frac{1+r}{1-r}\right)
$$

证明 .

If the $y$ ‘s are normal, $z$ is approximately $N(0,3 / n)$. A more accurate upper $100 \alpha$ percentile is given by
$$
z_{\alpha}=\sigma_{n}\left[u_{\alpha}+\frac{1}{24}\left(u_{\alpha}^{3}-3 u_{\alpha}\right) \gamma 2 n\right],
$$
with
$$
\sigma_{n}^{2}=\frac{3}{n}-\frac{7.324}{n^{2}}+\frac{53.005}{n^{3}}, \quad u_{\alpha}=\Phi^{-1}(\alpha), \quad \gamma_{n}=-\frac{11.70}{n}+\frac{55.06}{n^{2}}
$$
where $\Phi$ is the distribution function of the $N(0,1)$ distribution; that is, $\Phi(x)$ is the probability of an observation less. The inverse function $\Phi^{-1}$ is essentially a quantile. For example, $u_{.05}=-1.645$ and $u_{.95}=1.645$.

MATH2010W1-01 COURSE NOTES ：

has a beta distribution, which is related to the $F$ distribution. To obtain a $Q-Q$ plot, the values $u_{1}, u_{2}, \ldots, u_{n}$ are ranked to give $u_{(1)} \leq u_{(2)} \leq \cdots \leq u_{(n)}$, and we plot $\left(u_{(i)}, v_{i}\right)$, where the quantiles $v_{i}$ of the beta are given by the solution to
$$
\int_{0}^{v_{i}} \frac{\Gamma(a+b)}{\Gamma(a) \Gamma(b)} x^{a-1}(1-x)^{b-1} d x=\frac{i-\alpha}{n-\alpha-\beta+1}
$$
where $a=\frac{1}{2} p, b=\frac{1}{2}(n-p-1)$,
$$
\begin{aligned}
\alpha &=\frac{p-2}{2 p} \
\beta &=\frac{n-p-3}{2(n-p-1)}
\end{aligned}
$$

电子和电气工程的数学 Mathematics for Electronic and Electrical Engineering MATH1055W1-01

Posted on 2022年5月3日 by admin

这是一份southampton南安普敦大MATH1055W1-01作业代写的成功案例

电子和电气工程的数学 Mathematics for Electronic and Electrical Engineering MATH1055W1-01

问题 1.

If a square plane plate is considered as an example, we should remember that the potential in a finite and very thin plane plate can be evaluated by (Balanis, 1990):
$$
V(x, y, z=0)=\frac{1}{4 \pi \varepsilon} \int_{-a}^{a} \mathrm{~d} x^{\prime} \int_{-b}^{b} \mathrm{~d} y^{\prime} \frac{\rho\left(x^{\prime}, y^{\prime}\right)}{\left[\left(x-x^{\prime}\right)^{2}+\left(y-y^{\prime}\right)^{2}\right]^{1 / 2}}
$$
Thus, after applying the method of the moments, knowing the function of the approximated solution $f(x, y)$, the expansion function $g(x, y)$ and the weighed function $W(x, y)$, the potential in a square plane plate will be estimated by the inner product of these functions:

证明 .

$$
V(x, y)=\langle g, W, f\rangle \frac{1}{R}=\int_{-a}^{a} \frac{g(x, y) W(x, y) f(x, y)}{R(x, y)} \mathrm{d} x
$$
where
$$
R(x, y)=\sqrt{\left(x-x^{\prime}\right)^{2}+\left(y-y^{\prime}\right)^{2}}
$$
Dividing the plate in equal segments and applying the weighed function as being the Dirac delta function, we had that $W_{m}=\delta\left(x-x_{m}\right) \delta\left(y-y_{m}\right)$, being the inner product in the point given by:
$$
V(x, y, z=0)=\left\langle W_{m}, f, L g\right\rangle
$$

MATH1055W1-01 COURSE NOTES ：

$$
H_{z} \sim 1+k_{0,1} W_{\mathrm{TE}}
$$
where
$$
k_{0, \perp} \sim \omega\left[\varepsilon_{0} \varepsilon_{\mathrm{r}} \mu_{0} \mu_{\mathrm{r}}-\frac{1}{2}\left(\frac{\gamma}{\omega}\right)^{2}\right]
$$
is the quasi-static wavenumber.
On the other hand, in the T.M. case it is shown that
$$
E_{2} \sim 1+k_{0, \perp} W_{\mathrm{TM}}
$$
where
$$
\Delta W_{\mathrm{TM}}=0, \quad \text { in } Y \backslash C
$$

工程与环境的数学 Mathematics for Engineering and the Environment MATH1054W1-01

Posted on 2022年5月3日 by admin

这是一份southampton南安普敦大学MATH1054W1-01作业代写的成功案例

Industry and technology concept. INDUSTRY 4.0

问题 1.

As in the harmonic balance method, we neglect the effect of the higher harmonic, and see that since $-a \omega \sin \omega t=\dot{x}$ on the limit cycle, can be written as
$$
\varepsilon\left(x^{2}-1\right) \dot{x} \approx \varepsilon\left(\frac{1}{4} a^{2}-1\right) \dot{x}
$$
We now replace this in the differential equation to give the linear equation
$$
\ddot{x}+\varepsilon\left(\frac{1}{4} a^{2}-1\right) \dot{x}+x=0 .
$$

证明 .

The non-periodic solutions are spirals in the phase plane. Consider the motion for which $x(0)=a_{0}, \dot{x}(0)=0$ : for the next few ‘cycles’ $a_{0}$ will serve as the amplitude used in the above approximation, so may be written
$$
\ddot{x}+\varepsilon\left(\frac{1}{4} a_{0}^{2}-1\right) \dot{x}+x-0 .
$$
With the initial conditions given, the solution is
$$
x(t)=\frac{a_{0}}{\beta} \mathrm{e}^{\alpha t}[\beta \cos \beta t-\alpha \sin \beta t]
$$
where
$$
\left.\alpha=\frac{1}{2} \varepsilon\left(1-\frac{1}{4} a_{0}^{2}\right), \quad \beta=\frac{1}{2} \sqrt{4}-\varepsilon^{2}\left(1-\frac{1}{4} a_{0}^{2}\right)\right]
$$

MATH1054W1-01 COURSE NOTES ：

Therefore $\operatorname{sgn}(x)$ is replaced by $(4 / \pi) \cos \omega t$, which in turn is replaced by $(4 / \pi) x / a$. The equivalent linear equation is then
$$
\ddot{x}+\frac{4}{\pi a} x=0 .
$$
The solution, having any amplitude $a$, of the form $a \cos \omega t$ is
$$
x(t)=a \cos \left[\left(\frac{4}{\pi a}\right)^{1 / 2} t\right]
$$
Therefore
$$
\omega=\frac{2}{\sqrt{(\pi a)}}
$$

科学家的数学方法 1b 1b Math Methods for Scientist 1bMATH1009W1-01

Posted on 2022年4月28日 by admin

这是一份southampton南安普敦大学MATH1009W1-01作业代写的成功案例

科学家的数学方法 1b 1b Math Methods for Scientist 1bMATH1009W1-01

问题 1.

We examine a one-dimensional equation
$$
\frac{\partial^{2} u}{\partial t^{2}}=\frac{\partial^{2} u}{\partial x^{2}}+c^{2} u
$$
in the interval $0 \leq x \leq l$ with homogeneous initial conditions
$$
\left.u\right|{t=0}=\left.u{t}\right|{t=0}=0 $$ and boundary-value conditions $$ \left.u\right|{x=0}=\omega_{1}(t) ;\left.\quad u\right|{x=l}=\omega{2}(t) .
$$

证明 .

It should be mentioned that the initial condition may always be reduced to homogeneous. The Bessel function of the imaginary variable $I_{0}\left(c \sqrt{t^{2}-x^{2}}\right)$ is the fundamental solution.

Placing the continuously acting sources, corresponding to this solution, at the ends of the interval $[0, l]$, we obtain, as may easily be seen, the ‘simple layer’ potentials:
$$
\begin{gathered}
\int_{0}^{t-x} \varphi(\tau) I_{0}\left(c \sqrt{(t-\tau)^{2}-x^{2}}\right) d \tau, \
\int_{0}^{t-(l-x)} \psi(\tau) I_{0}\left(c \sqrt{(t-\tau)^{2}-(l-x)^{2}}\right) d \tau,
\end{gathered}
$$

MATH1009W1-01 COURSE NOTES ：

We now determine functions $T_{n}(t)$, corresponding to the form of the wave $X_{n}(x)$. For this purpose, the value $\lambda_{n}$ is substituted into the equation for $T$ :
$$
T_{n}^{\prime \prime}+\frac{a^{2} \pi^{2}}{l^{2}} n^{2} T_{n}=0 .
$$
The general integral of this equation has the form::
$$
T_{n}(t)=B_{n} \sin \frac{\pi a n}{l} t+C_{n} \cos \frac{\pi a n}{l} t=A_{n} \sin \left(\frac{\pi a n}{l} t+\varphi_{n}\right) \text {, }
$$
where $B_{n}$ and $C_{n}$ or $A_{n}$ and $\varphi_{n}$ are arbitrary constants.
Using $X_{n}$ and $T_{n}$, we can write the final expression for all possible standing waves:
$$
u_{n}(x, t)=A_{n} \sin \left(\frac{\pi a n}{l} t+\varphi_{n}\right) \sin \frac{\pi n x}{l}
$$

物理科学家的数学方法 1b Mathematical Methods For Physical Scientists 1b MATH1007W1-01

Posted on 2022年4月28日 by admin

这是一份southampton南安普敦大学 MATH1007W1-01作业代写的成功案例

物理科学家的数学方法 1b Mathematical Methods For Physical Scientists 1b MATH1007W1-01

问题 1.

Since $f(x)$ possesses a finite derivative at the point $x$, say $a$,
$$
\lim _{\delta x \rightarrow 0} \frac{f(x+\delta x)-f(x)}{\delta x}=a
$$
and so,
$$
|f(x+\delta x)-f(x)|=|\delta x|(|a|+\varepsilon)
$$
where $\varepsilon \rightarrow 0$ as $\delta x \rightarrow 0$. Since $a$ is finite
$$
\mid f(x)+\delta x)-f(x) \mid \rightarrow 0 \text { as }|\delta x| \rightarrow 0
$$
Hence $f(x)$ is continuous at the point $x$.
(2) If $f(x)$ is a constant, its derivative is zero, since

证明 .

Differentials
$$
\lim {\delta x \rightarrow 0} \frac{f(x+\delta x)-f(x)}{\delta x}=\lim \frac{0}{\delta x}=0 $$ (3) If $f(x)=x$, its derivative is unity, since $$ \lim {\delta x \rightarrow 0} \frac{(x+\delta x)-x}{\delta x}=\lim \frac{\delta x}{\delta x}=1
$$

MATH1007W1-01 COURSE NOTES ：

If $y=\sin ^{-1} x$ where $-1<x<1$ then $x=\sin y$ and is increasing steadily in the range $-\pi / 2<y<\pi / 2$. Since $d x / d y=\cos y$, it follows
$$
\frac{d y}{d x}=\frac{1}{\cos y}
$$
However, in the range $-\pi / 2<y<\pi / 2, \cos y$ is positive and given by
$$
\begin{gathered}
\cos y=\left(1-\sin ^{2} y\right)^{1 / 2}=\left(1-x^{2}\right)^{1 / 2} \
\frac{d y}{d x}=\frac{1}{\left(1-x^{2}\right)^{1 / 2}}
\end{gathered}
$$

英国论文代写Viking Essay

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Tag Archives: 南安普敦大学代写

进一步数论 Further Number Theory MATH3078W1-01

MATH3078W1-01 COURSE NOTES ：

精算数学 Actuarial Mathematics II MATH3066W1-01

MATH3066W1-01 COURSE NOTES ：

局部微分方程IPartial Differential Equations MATH2038W1-01

MATH2038W1-01 COURSE NOTES ：

算法IAlgorithms MATH2014W1-01

MATH2014W1-01 COURSE NOTES ：

随机过程IStochastic Processes MATH2012W1-01

MATH2012W1-01 COURSE NOTES ：

统计建模I Statistical Modelling I MATH2010W1-01

MATH2010W1-01 COURSE NOTES ：

电子和电气工程的数学 Mathematics for Electronic and Electrical Engineering MATH1055W1-01

MATH1055W1-01 COURSE NOTES ：

工程与环境的数学 Mathematics for Engineering and the Environment MATH1054W1-01

MATH1054W1-01 COURSE NOTES ：

科学家的数学方法 1b 1b Math Methods for Scientist 1bMATH1009W1-01

MATH1009W1-01 COURSE NOTES ：

物理科学家的数学方法 1b Mathematical Methods For Physical Scientists 1b MATH1007W1-01

MATH1007W1-01 COURSE NOTES ：