# 数学|MATH1141/MATH1131 Mathematics代写 unsw代写

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MATH1131, Mathematics 1A, and MATH1141, Higher Mathematics 1A, are first year courses
taught by the School of Mathematics and Statistics in semester 1, and are each worth six units
of credit. MATH1131 is also taught in semester 2. Students who pass MATH1131 in semester
1 usually continue to study MATH1231, Mathematics 1B, in semester 2. Those students who
pass MATH1141 with a Credit usually continue to study MATH1241, Higher Mathematics 1B,
in semester 2. MATH1231 is also taught in Summer Session. MATH1131 and MATH1231 (or
MATH1141 and MATH1241) are generally specified in Engineering programs, as well as many
Science programs.

Next, if $\alpha>\omega$ is a limit ordinal, then according to what we have said before, $\sum_{\xi<\omega^{\alpha}} \xi=\sup {\beta<\alpha} \sum{\xi<\omega^{\sigma}} \xi$, and here in the supremum we can take the supremum for successor ordinals $\beta$ smaller than $\alpha$. Thus, according to what we have just proved, in this case
$$\sum_{\xi<\omega^{\alpha}} \xi=\sup {\beta<\alpha} \omega^{\beta \cdot 2}=\omega^{\sigma},$$ where $\sigma=\sup {\beta<\alpha} \beta \cdot 2$. Here if $\alpha$ equals one of the powers of $\omega$, say $\alpha=\omega^{\tau}$, then
$$\omega^{\tau}=\sup {\beta<\alpha} \beta \leq \sup {\beta<\alpha} \beta \cdot 2 \leq \sup _{\beta<\alpha} \beta \cdot \omega=\omega^{\tau},$$

$$\beta=\omega^{\xi_{n}} \cdot a_{n}+\cdots+\omega^{\xi_{0}} \cdot\left(a_{0}-1\right)+\delta,$$
where $\delta<\omega^{\xi_{0}}$, and here
$$\beta \cdot 2=\omega^{\varepsilon_{n}} \cdot\left(2 a_{n}\right)+\omega^{\xi_{n-1}} \cdot a_{n-1}+\cdots+\omega^{\varepsilon_{0}} \cdot\left(a_{0}-1\right)+\delta,$$
thus
$$\sigma=\sup {\beta<\alpha} \beta \cdot 2=\omega^{\xi{n}} \cdot\left(2 a_{n}\right)+\cdots+\omega^{\xi_{0}} \cdot a_{0}=\alpha \cdot 2 .$$
In summary, the sum in question is equal to $\omega^{2 \alpha-1}$ if $\alpha$ is finite, it equals $\omega^{\alpha}$ if $\alpha$ is a power of $\omega$, and in all other cases it equals $\omega^{\alpha \cdot 2}$.

## MATH1141/MATH1131 COURSE NOTES ：

Actually, the method we used for the existence can be easily extended to yield both the existence and unicity of the representation. In fact, if the normal form of $\alpha$ is $\alpha=\omega^{\xi_{n}} \cdot a_{n}+\cdots+\omega^{\varepsilon_{0}} \cdot a_{0}$ and set $\delta_{0}=\xi_{0}$ and choose $\delta_{i}$, $1 \leq i \leq n$, so that $\xi_{i}=\xi_{i-1}+\delta_{i}$, then $\delta_{0} \geq 0$ and $\delta_{i}>0$ for $1 \leq i \leq n$. Now
$$\alpha=\omega^{\delta_{0}} \cdot a_{0} \cdot\left(\omega^{\delta_{1}}+1\right) \cdot a_{1} \cdots a_{n-1} \cdot\left(\omega^{\delta_{n}}+1\right) \cdot a_{n}$$
and if $\delta_{0}=\omega^{\gamma_{m}}+\cdots+\omega^{\gamma_{0}}$ with $\gamma_{m} \geq \cdots \geq \gamma_{0}$, then
$$\alpha=\omega^{\omega^{\gamma m}} \cdots \omega^{\omega v 0} \cdot a_{0} \cdot\left(\omega^{\delta_{1}}+1\right) \cdot a_{1} \cdots a_{n-1} \cdot\left(\omega^{\delta_{n}}+1\right) \cdot a_{n}$$

# 数学 Maths 2 PHYS130001/PHYS238001

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Let $b_{\alpha \beta \mid \sigma}:=\partial_{\sigma} b_{\alpha \beta}-\Gamma_{\alpha \sigma}^{\tau} b_{\tau \beta}-\Gamma_{\beta \sigma}^{\tau} b_{\alpha \tau}$ denote the first-order covariant derivatives of the curvature tensor, defined here by means of its covariant components. Show that these covariant derivatives satisfy the Codazzi-Mainardi identities
$$b_{\alpha \beta \mid \sigma}=b_{\alpha \sigma \mid \beta}$$
which are themselves equivalent to the relations (Thm. 2.8-1)
$$\partial_{\sigma} b_{\alpha \beta}-\partial_{\beta} b_{\alpha \sigma}+\Gamma_{\alpha \beta}^{\tau} b_{\tau \sigma}-\Gamma_{\alpha \sigma}^{\tau} b_{\tau \beta}=0$$
Hint: The proof is analogous to that given in for establishing the relations $\left.b_{\beta}^{\tau}\right|{\alpha}=\left.b{\alpha}^{\tau}\right|_{\beta}$.

## PPHYS130001/PHYS238001COURSE NOTES ：

$u_{i}^{\varepsilon}\left(x^{\varepsilon}\right)=u_{i}(\varepsilon)(x)$ for all $x^{\varepsilon}=\pi^{\varepsilon} x \in \bar{\Omega}^{\varepsilon}$,
where $\pi^{c}\left(x_{1}, x_{2}, x_{3}\right)=\left(x_{1}, x_{2}, \varepsilon x_{3}\right)$. We then assume that there exist constants $\lambda>0, \mu>0$ and functions $f^{i}$ independent of $\varepsilon$ such that
$$\begin{gathered} \lambda^{\varepsilon}=\lambda \text { and } \mu^{\varepsilon}=\mu, \ f^{i, \varepsilon}\left(x^{\varepsilon}\right)=\varepsilon^{p} f^{i}(x) \text { for all } x^{\varepsilon}=\pi^{e} x \in \Omega^{\varepsilon}, \end{gathered}$$

# 工程原理 Engineering Principles GENG0005W1-01

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$$m=m_{0} \sqrt{1-\frac{v^{2}}{c^{2}}}$$
where
$m$ – mass of a moving body (also called the variable mass)
$m_{0}$ – rest mass of a body (velocity is zero)
$v-$ velocity of a moving body
$c$ – speed of light
The ratio, $v^{2} / c^{2}$ is usually denoted as $\beta^{2}$. Thus
$$m=m_{0} / \sqrt{1-\beta^{2}}$$
Similarly, the relativistic energy of a body moving with velocity $v$ is no more expressed as $m v^{2} / 2$, but
$$E=m_{0} c^{2} / \sqrt{1-\beta^{2}}$$

## GENG0005W1-01COURSE NOTES ：

$$E^{2}=(p c)^{2}+\left(m_{0} c^{2}\right)^{2}$$
By definition, momentum can be described as a function of the mass and velocity of a moving body:
$$p=m v=m_{0} v / \sqrt{1-\beta^{2}}$$
Squaring Eq. (3-6)
\begin{aligned} &E^{2}=\left(m c^{2}\right)^{2}=\left(m_{0} c^{2}\right)^{2} / 1-\frac{v^{2}}{c^{2}} \rightarrow m^{2} c^{4}\left(1-\frac{v^{2}}{c^{2}}\right)=m_{0}^{2} c^{4} \ &m^{2} c^{4}=E^{2}=m_{0}^{2} c^{4}+m^{2} c^{2} v^{2} \end{aligned}
where $p=m v$, thus giving
$$E^{2}=(p c)^{2}+\left(m_{0} c^{2}\right)^{2}$$
For a massless particle (like a photon) it follows that the total energy depends on its momentum and the speed of light: $E=p c$. This aspect will be discussed in greater detail in later sections.

# 机械科学 Mechanical Science GENG0003W1-01

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If final pressure of the gas is $p_{3}$, for a constant volume process $3-1$,
$$p_{3}=\frac{T_{3}}{T_{1}} p_{1}=\frac{323}{423} \times 10=7.6 \mathrm{bar}$$
Let us find the mass of the gas $m$ and
$$m=\frac{p_{1} \forall \forall_{1}}{R T_{1}}=\frac{10 \times 10^{5} \times 0.336}{293 \times 423}=2.7 \mathrm{~kg}$$
Change in internal energy
$$d U=U_{3}-U_{1}=m C_{7}\left(T_{3}-T_{1}\right)=2.7 \times 0.703(323-423)=-189.8 \mathrm{~kJ}$$
The negative sign indicates that there is a decrease in internal energy.

## GENG0003W1-01COURSE NOTES ：

The path followed by the system, $p_{1} \forall_{1}^{\prime}=p_{2} \forall_{2}$.
So,
$$\left(\frac{\forall_{2}}{\forall_{1}}\right)^{\gamma}=\frac{p_{1}}{p_{2}}$$
and
$$\forall_{2}=\left(\frac{p_{1}}{p_{2}}\right)^{1 / \gamma} \forall_{1}=\left(\frac{500}{100}\right)^{1 / L 4} \times 0.2=0.6313 \mathrm{~m}^{3}$$
Hence work done,
$$W_{1-2}=\frac{p_{1} \forall_{1}-p_{2} \forall_{2}}{\gamma-1}=\frac{(500 \times 0.2-100 \times 0.6313) 10^{3}}{(1.4-1) \times 10^{3}}=92.175 \mathrm{~kJ}$$

# 数学 Mathematics B GENG0002W1-01

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from which we see that $f+(-f)=0$. The additive inverse law follows. For the distributive laws note that for real numbers $a, b$ and continuous functions $f, g \in V$, we have that for all $0 \leq x \leq 1$,
$$a(f+g)(x)=a(f(x)+g(x))=a f(x)+a g(x)=(a f+a g)(x),$$
which proves the first distributive law. For the second distributive law, note that for all $0 \leq x \leq 1$,
$$((a+b) g)(x)=(a+b) g(x)=a g(x)+b g(x)=(a g+b g)(x),$$
and the second distributive law follows. For the scalar associative law, observe that for all $0 \leq x \leq 1$,
$$((a b) f)(x)=(a b) f(x)=a(b f(x))=(a(b f))(x),$$
so that $(a b) f=a(b f)$, as required. Finally, we see that
$$(1 f)(x)=1 f(x)=f(x),$$

## GENG0002W1-01COURSE NOTES ：

First let $f(x), g(x) \in V$ and let $c$ be a scalar. By definition of the set $V$ we have that $f(1 / 2)=0$ and $g(1 / 2)=0$. Add these equations together and we obtain
$$(f+g)(1 / 2)=f(1 / 2)+g(1 / 2)=0+0=0 .$$
It follows that $V$ is closed under addition with these operations. Furthermore, if we multiply the identity $f(1 / 2)=0$ by the real number $c$ we obtain that
$$(c f)(1 / 2)=c \cdot f(1 / 2)=c \cdot 0=0 .$$
It follows that $V$ is closed under scalar multiplication. Now certainly the zero function belongs to $V$, since this function has value 0 at any argument. Therefore, $V$ contains an additive identity element. Finally, we observe that the negative of a function $f(x) \in V$ is also an element of $V$, since
$$(-f)(1 / 2)=-1 \cdot f(1 / 2)=-1 \cdot 0=0 .$$

# 高级衍射几何学与拓扑学 5M:Adv Diff Geom & Topology MATHS5039_1 5M

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Proof of (b) is easy:
$$d\left(y_{n}, x\right) \leq d\left(y_{n}, x_{n}\right)+d\left(x_{n}, x\right) \rightarrow 0 .$$
(c) is also easy. Look at
$$d\left(x_{n}, y_{n}\right) \leq d\left(x_{n}, x_{n}^{\prime}\right)+d\left(x_{n}^{\prime}, y_{n}^{\prime}\right)+d\left(y_{n}^{\prime}, y_{n}\right) .$$
Taking limits, we get $\lim {n} d\left(x{n}, y_{n}\right) \leq \lim {n} d\left(x{n}^{\prime}, y_{n}^{\prime}\right)$. Similar argument shows the other way inequality and hence the proof.

## MATHS5039_1 COURSE NOTES ：

For, if $\varepsilon>0$ is given, by the Cauchy nature of $\left(x_{n}\right)$, there exists $N \in \mathbb{N}$ such that $d\left(x_{m}, x_{n}\right)<\varepsilon$ for $m, n \geq N$. We have, for $n \geq N$,
$$d\left(\tilde{x}{n}, \xi\right):=\lim {m \rightarrow \infty} d\left(x_{n}, x_{m}\right)<\varepsilon .$$
It is clear that the map $\varphi: X \rightarrow \tilde{X}$ given by $\varphi(x)=\tilde{x}$ is an isometry. $d(\varphi(x), \varphi(y)):=\lim _{n} d(x, y)$, the limit of a constant sequence.
We claim that the image $\varphi(X)$ of $X$ under the map $\varphi: x \mapsto \tilde{x}$ is dense

# 数学 MATH MATHS2025_1/MATHS3016_1

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Let $f$ be a function from a nonempty open subset $E$ of $\mathbb{R}$ to $\mathbb{R}$. The function $f$ is said to be differentiable at $c \in E$ if
$$\lim {x \rightarrow c} \frac{f(x)-f(c)}{x-c}$$ or, equivalently, $$\lim {h \rightarrow 0} \frac{f(c+h)-f(c)}{h}$$
exists. This limit (if it exists) is called the derivative of $f$ at $c$. If the derivative of $f$ exists at every $c \in E$, then $f$ is said to be differentiable on $E$ (or just differentiable). The derivative of $f$ as a function from $E$ to $\mathbb{R}$ is denoted by
$$f^{\prime} \text { or } \frac{d f}{d x}$$
Note that the limit in Eq. (7.1) is understood as the limit of the function
$$g(x)=\frac{f(x)-f(c)}{x-c}, \quad x \in E \backslash{c}$$

## MATHS2025_1/MATHS3016_1COURSE NOTES ：

From the definition of $f^{\prime}(c)$, it follows that for every $\varepsilon>0$, there exists $\delta>0$ such that $x \in E,|x-c|<\delta$, and $x \neq c$ imply
$$\left|\frac{f(x)-f(c)}{x-c}-f^{\prime}(c)\right|<\varepsilon .$$
Thus, for every $x \in E$ with $|x-c|<\delta$,
$$|f(x)-\varphi(x)| \leq \varepsilon|x-c|,$$
where $\varphi$ is the linear function defined by
$$\varphi(x)=f(c)+f^{\prime}(c)(x-c), \quad x \in \mathbb{R}$$

# 复杂分析的方法 Methods in Complex Analysis MATHS4076_1

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To find the best probability assignment subject to the constraint of given mean, we use the MaxEnt method. I.e. maximse
$$S[p]=-k_{\mathrm{B}} \sum_{k \geq 0} p(k) \ln p(k)$$
subject to the constraint
$$\sum_{k \geq 0} k p(k)=\mu .$$
using the method of Lagrangian multipliers to take constraints into account.
To simplify notation we can assume $k_{\mathrm{B}}=1$; this amounts to redefining Lagrangian multipliers in units of $k_{\mathrm{B}}$ (can you see why?). Thus define
$$\mathcal{L}[p]=-\sum_{k \geq 0} p(k) \ln p(k)+\lambda_{0}\left(\sum_{k \geq 0} p(k)-1\right)+\lambda_{1}\left(\sum_{k \geq 0} k p(k)-\mu\right)$$

## MATHS4076_1COURSE NOTES ：

$$\mu=\sum_{k \geq 0} k p(k)=\left(1-\mathrm{e}^{\lambda_{1}}\right) \sum_{k=1}^{\infty} k \mathrm{e}^{\lambda_{1} k}=\frac{\mathrm{e}^{\lambda_{1}}}{1-\mathrm{e}^{\lambda_{1}}}$$
required by the constraint of the given mean. Hence
$$\mathrm{e}^{\lambda_{1}}=\frac{\mu}{1+\mu} \quad \Longrightarrow \quad p(k)=\frac{1}{1+\mu}\left(\frac{\mu}{1+\mu}\right)^{k}$$
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# 随机过程 Stochastic Processes STATS4024_1 /STATS5026_1

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from which it follows that, by monotonicity,
$$\forall n \in \mathbb{N}: \quad E\left[Y_{n+1} \mid X=x\right] \geq E\left[Y_{n} \mid X=x\right], \quad P_{X} \text {-a.s. }$$
Moreover,
$$\forall B \in \mathcal{B}: \quad \int_{[X \in B]} Y_{n} d P=\int_{B} E\left[Y_{n} \mid X=x\right] d P_{X}(x)$$
and
$$\forall B \in \mathcal{B}: \quad \int_{[X \in B]} Y d P \geq \int_{[X \in B]} Y_{n} d P$$
Thus
$$E[Y \mid X=x] \geq E\left[Y_{n} \mid X=x\right], \quad P_{X} \text {-a.s. }$$

## STATS4024_1 /STATS5026_1COURSE NOTES ：

Therefore,
$$P([Y \in A] \mid \cdot)=P([Y \in A]), \quad P_{X} \text {-a.s. }$$
and if $Y$ is a real-valued integrable random variable, then
$$E[Y \mid \cdot]=E[Y], \quad P_{X}-a . s .$$
Proof: Independence of $X$ and $Y$ is equivalent to
$$P([X \in B] \cap[Y \in A])=P([X \in B]) P([Y \in A]) \quad \forall A \in \mathcal{B}{1}, B \in \mathcal{B}$$ or \begin{aligned} \int{[X \in B]} I_{[Y \in A]}(\omega) P(d \omega) &=P([Y \in A]) \int I_{B}(x) d P_{X}(x) \ &=\int_{D} P([Y \in A]) d P_{X}(x) \end{aligned}

# 进一步的复杂分析 Further Complex Analysis MATHS5070_1 /MATHS4104_1

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$$l_{\mathrm{p}}(\psi)=c-\frac{n}{2} \log (\mathrm{SSB}+\lambda \mathrm{SSW})+\frac{n-m}{2} \log \lambda,$$
where $\lambda=1+k \psi$. The maximum of $l_{\mathrm{p}}$ is given by
$$\hat{\lambda}=\left(1-\frac{1}{m}\right) \frac{\mathrm{MSB}}{\mathrm{MSW}},$$
where MSB $=\operatorname{SSB} /(m-1)$ and MSW $=\operatorname{SSW} /(n-m)$, or
$$ 英国论文代写Viking Essay为您提供作业代写代考服务 ## MATHS5070_1 /MATHS4104_1COURSE NOTES ：$$
$$where \hat{\alpha}{r, u} is the u th component of \hat{\alpha}{r}, 1 \leq r \leq s, and$$