# 数学分析|MATH2023/MATH2923 Analysis代写 Sydney代写

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We have
$$\Phi_{p}(f)=|\Omega|^{-\frac{1}{p}}|f|_{L^{p}(\Omega)} .$$
By corollary $\Phi_{p}(f)$ viewed as a function of $p$ is increasing with
$$\Phi_{p}(f) \leq|f|_{L^{\infty}(\Omega)}$$ that
$$|f|_{L^{\infty}(\Omega)} \leq \lim {p \rightarrow \infty} \Phi{p}(f)$$
For $K \in \mathbf{R}$ let
$$A_{K}:={x \in \Omega|| f(x) \mid \geq K} .$$
The set $A_{k}$ is measureable since $f$ is and $\left|A_{K}\right|>0$ if $K<|f|_{L^{\infty}(\Omega)}$. Moreover,
$$\Phi_{p}(f) \geq|\Omega|^{-\frac{1}{p}}\left(\int_{A_{K}}|f(x)|^{p} d x\right)^{1 / p} \geq|\Omega|^{-\frac{1}{p}}\left|A_{K}\right|^{\frac{1}{p}} K .$$
Passing to the limit $p \rightarrow \infty$ we obtain
$$\lim {p \rightarrow \infty} \Phi{p}(f) \geq K$$
Because this holds for all $K<|f|_{L^{\infty}(\Omega)}$ we conclude
$$\lim {p \rightarrow \infty} \Phi{p}(f) \geq|f|_{L^{\infty}(\Omega)}$$

## MATH2023/MATH2923 COURSE NOTES ：

Let $\left(f_{k}\right){k \in \mathrm{N}} \subset L^{P}(\Omega)$ be a Cauchy sequence. It suffices to show that $\left(f{k}\right)$ has a convergent subsequence. For every $i \in \mathrm{N}$ there is an integer $N_{i}$ so that
$$\left|f_{n}-f_{m}\right|_{L P(\Omega)} \leq 2^{-i} \text { whenever } n, m \geq N_{i} .$$
We construct a subsequence $\left(f_{k_{i}}\right) \subset\left(f_{k}\right)$ so that
$$\left|f_{k_{i+1}}-f_{k_{i}}\right|_{L P(\Omega)} \leq 2^{-i}$$
by setting $k_{i}:=\max \left{i, N_{i}\right}$. In order to simplify notation we will from now on assume that
$$\left|f_{k+1}-f_{k}\right|_{L F(\Omega)} \leq 2^{-k}$$
so that
$$M:=\sum_{k \in \mathbf{N}}\left|f_{k+1}-f_{k}\right|_{L^{p}(\Omega)}<\infty$$

# 数学分析 Mathematical Analysis  MA140-10/MA152-15

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Proof As is shown in linear algebra, the matrix $A$ that represents $T$ is a product of elementary matrices
$$A=E_{1} \cdots E_{k} .$$
Each elementary $2 \times 2$ matrix is one of the following types:
$$\left[\begin{array}{ll} \lambda & 0 \ 0 & 1 \end{array}\right] \quad\left[\begin{array}{ll} 1 & 0 \ 0 & \lambda \end{array}\right] \quad\left[\begin{array}{ll} 0 & 1 \ 1 & 0 \end{array}\right] \quad\left[\begin{array}{ll} 1 & \sigma \ 0 & 1 \end{array}\right]$$
where $\lambda>0$. The first three matrices represent isomorphisms whose effect on $I^{2}$ is obvious: $I^{2}$ is converted to the rectangles $\lambda I \times I, I \times \lambda I, I^{2}$. In each case, the area agrees with the magnitude of the determinant. The fourth isomorphism converts $I^{2}$ to the parallelogram$\Pi$ is Riemann measurable since its boundary is a zero set. By Fubini’s Theorem, we get
$$|\Pi|=\int \chi_{\Pi}=\int_{0}^{1}\left[\int_{x=\sigma x}^{x=1+\sigma y} 1 d x\right] d y=1=\operatorname{det} E .$$

## MA140-10/MA152-15 COURSE NOTES ：

$$d x_{I}: \varphi \mapsto \int_{l^{k}} \frac{\partial \varphi_{I}}{\partial u} d u$$
where this integral notation is shorthand for
$$\int_{0}^{1} \ldots \int_{0}^{1} \frac{\partial\left(\varphi_{i_{1}}, \ldots, \varphi_{i_{k}}\right)}{\partial\left(u_{1}, \ldots, u_{k}\right)} d u_{1 \ldots} . d v_{k}$$
If $f$ is a smooth function on $\mathbb{R}^{n}$ then $f d x_{l}$ is the functional
$$f d x_{I}: \varphi \mapsto \int_{I^{k}} f(\varphi(u)) \frac{\partial \varphi_{I}}{\partial u} d u$$

# 数学分析 Analysis MATH41220-WE01/MATH1051-WE01/MATH3011-WE01

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Thus $\left(d\left(p_{n}, q_{n}\right)\right)$ is a Cauchy sequence in $\mathbb{R}$, and because $\mathrm{R}$ is complete,
$$L=\lim {n \rightarrow \infty} d\left(p{n}, q_{n}\right)$$
exists. Let $\left(p_{n}^{\prime}\right)$ and $\left(q_{n}^{\prime}\right)$ be sequences that are co-Cauchy with $\left(p_{n}\right)$ and $\left(q_{n}\right)$, and let
$$L^{\prime}=\lim {n \rightarrow \infty} d\left(p{n}^{\prime}, q_{n}^{\prime}\right) .$$

Then
$$\left|L-L^{\prime}\right| \leq\left|L-d\left(p_{n}, q_{n}\right)\right|+\left|d\left(p_{n}, q_{n}\right)-d\left(p_{n}^{\prime}, q_{n}^{\prime}\right)\right|+\left|d\left(p_{n}^{\prime}, q_{n}^{\prime}\right)-L^{\prime}\right| .$$
As $n \rightarrow \infty$, the first and third terms tend to 0 . the middle term is
$$\left|d\left(p_{n}, q_{n}\right)-d\left(p_{n}^{\prime}, q_{n}^{\prime}\right)\right| \leq d\left(p_{n}, p_{n}^{\prime}\right)+d\left(q_{n}, q_{n}^{\prime}\right) .$$

## MATH41220-WE01/MATH1051-WE01/MATH3011-WE01COURSE NOTES ：

$$|n|{p}=\frac{1}{p^{k}}$$ where $p^{k}$ is the largest power of $p$ that divides $n$. (The norm of 0 is by definition 0 .) The more factors of $p$, the smaller the $p$-norm. Similarly, if $x=a / b$ is a fraction, we factor $x$ as $$x=p^{k} \cdot \frac{r}{s}$$ where $p$ divides neither $r$ nor $s$, and we set $$|x|{p}=\frac{1}{p^{k}} .$$
The $p$-adic metric on $Q$ is
$$d_{p}(x, y)=|x-y|_{p} .$$

# 量化方法|MA10214 Quantitative methods代写

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The laws of Probability, Bayes’ Theorem, Decision Trees. Binomial, Poisson and Normal distributions and their applications; the relationship between these distributions. Different types of sample; sampling distributions of means, standard deviations and proportions. Confidence intervals and hypothesis testing; types of error, significance levels and P values. Power. Quality control: Acceptance sampling and Shewhart charts.

Let the exercise price be $E$ and the barrier be at $B$ where $B<E$.
Since the Black-Scholes partial differential equation governs the price of the option we can, as before, look for solutions of the form:
$$c(S, E){d o}=A{1} S^{m_{1}}+A_{2} S^{m_{2}}$$
subject to the boundary conditions: (i) $c_{d b}(B, E)=0$ and (ii) $c(\infty, E){d o}=S$, see the previous section. From (i) we have: $$c{d b}(B, E)=A_{1} B^{m_{1}}+A_{2} B^{m_{2}}=0, \text { so } \quad A_{1}=-A_{2} B^{m_{2}-m_{n}}$$
Therefore
$$c_{d m}(S, E)=-A_{2} B^{m_{2}-m_{n}} S^{m_{1}}+A_{2} S^{m_{2}}$$
From (ii), as $S \rightarrow \infty$ :
$$c_{d m}(S, E)=-A_{2} B^{m_{2}-m_{n}} S^{m_{1}}+A_{2} S^{m_{2}}=S$$
However, since $m_{2}<0$, we have $A_{2} S^{m_{2}} \rightarrow 0$, as $S \rightarrow \infty$, giving
$$c_{d o}(S, E)=-A_{2} B^{m_{2}-m_{1}} S^{m_{n}}=S$$

## MA10214 COURSE NOTES ：

The Black approximation, $C_{B L}$, can be expressed more concisely in terms of our previously defined notation as:
$$C_{B L}(S, E, \tau)=\max \left(v_{1}, v_{2}\right)$$
where $v_{1}$ and $v_{2}$ are the following European calls
$$v_{1}=c\left(S_{D}, E, \tau\right) \quad \text { and } \quad v_{2}=c\left(S_{D}^{+}, E, \tau_{1}\right), \quad \tau=T-t \quad \tau_{1}=T-t_{n}$$
and
$$S_{D}=S-\sum_{i=1}^{n} D_{i} \quad \text { and } \quad S_{D}^{+}=S-\sum_{i=1}^{n-1} D_{i}$$

# 数学分析|MA20219 Analysis 2B代写

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If we adapt the ideas seen in previous analysis units, and in particular the idea of a derivative, to complex variables, then we find a remarkably beautiful theory. In particular, a lot of the complications coming from limited differentiability or from various notions of convergence fall away. Of course, this comes at a price, which is that complex differentiability is much more restrictive than the real counterpart. So we will have a very powerful theory, but it will apply only to a relatively small set of functions (which still includes rational, exponential, and trigonometric functions and much more).

Let $U=\mathbb{C} \backslash{1-i,-1-i, \log 2+\pi i}$. Consider the function $f: U \rightarrow \mathbb{C}$ given by
$$f(z)=\frac{e^{z}+2}{(z-\log 2-\pi i)\left(z^{2}+2 i z-2\right)}, \quad z \in U .$$
Let $\gamma:[-\pi / 2, \pi / 2+20] \rightarrow \mathbb{C}$ such that
$$\gamma(t)= \begin{cases}10 e^{i t} & \text { if }-\pi / 2 \leq t \leq \pi / 2 \ (10+\pi / 2-t) i & \text { if } \pi / 2<t \leq \pi / 2+20\end{cases}$$
Set $\Gamma=\gamma([-\pi / 2, \pi / 2+20])$.
You may use any result from the lectures and exercises to answer the following questions. If you need to determine any winding numbers, you may use geometric intuition.
(a) Show that $f$ has poles of order 1 at $1-i$ and $-1-i$.
(b) Sketch the curve $\Gamma$, including its position relative to the points $1-i$ and $-1-i$. Indicate the orientation of $\gamma$.
(c) The singularity of $f$ at $\log 2+\pi i$ is removable. Using this information (for which no proof is required), or otherwise, find
$$\int_{\gamma} f(z) d z .$$

## MA20219 COURSE NOTES ：

i. The singularities of the given function are 0 and 2 . The distance from the centre to the nearest singularity (apart from the centre itself) is therefore 2. So by a result from the lecture notes, the punctured disc of convergence has radius 2 and therefore is $B_{2}(0) \backslash{0}$.
ii. We have (using the geometric series) for $z \in B_{2}(0)$
$$\frac{1}{z-2}=\frac{-1}{2} \frac{1}{1-\frac{z}{2}}=\frac{-1}{2} \sum_{k=0}^{\infty}\left(\frac{z}{2}\right)^{k}=\sum_{k=0}^{\infty} \frac{-1}{2^{k+1}} z^{k}$$
so that for $z \in B_{2}(0) \backslash{0}$
$$\frac{1}{z(z-2)}=\sum_{k=0}^{\infty} \frac{-1}{2^{k+1}} z^{k-1}=\sum_{n=-1}^{\infty} \frac{-1}{2^{n+2}} z^{n}$$

# 代数 Algebra 1B MA10210

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An equivalent system is, therefore,
\begin{aligned} &x+y-z=0 \ &-4 y+2 z=0 \end{aligned}

This has infinitely many solutions: $2 y=z, x=z-y=2 y-y=y$. Hence, $\operatorname{ker} T={(y, y, 2 y): y \in \mathbb{R}}$
$$={y(1,1,2): y \in \mathbb{R}}=\langle(1,1,2)\rangle,$$
and nullity $T=1$. Now $T((1,0,0))=(1,3,5), T((0,1,0))=(1,-1,1)$, $T((0,0,1))=(-1,-1,-3)$. But $(1,0,0),(0,1,0),(0,0,1)$ is a basis for $\mathbb{R}^{3}$, so $(1,3,5),(1,-1,1),(-1,-1,-3)$ is a spanning sequence for im $T$, by Theorem 6.4.3. These vectors must be linearly dependent, since, by the dimension theorem,c

## MA10210 COURSE NOTES ：

to above is $T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ given by
$$T((x, y, z))=(x-y+2 z, 2 x+y-z, x-4 y+7 z)$$
Now $T((1,0,0))=(1,2,1), \quad T((0,1,0))=(-1,1,-4), \quad T((0,0,1))=(2,-1,7)$ span im $T$, by Theorem $6.4 .3$. However,
$$(1,2,1)=5(-1,1,-4)+3(2,-1,7),$$
so these three vectors are linearly dependent. Since $(1,2,1),(-1,1,-4)$ are linearly independent, these two vectors form a basis for im $T$. Hence $\operatorname{rank} A=\operatorname{rank} T=2$.

# 数论 5E: Number Theory/4H: Number Theory MATHS5074_1/MATHS4108_1

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If $f:[a, b] \rightarrow \mathbb{R}$ is continuous, then the function $(f \circ \phi) \phi^{\prime}:[\alpha, \beta] \rightarrow \mathbb{R}$ is integrable and
$$\int_{\phi(\alpha)}^{\phi(\beta)} f(x) d x=\int_{\alpha}^{\beta} f(\phi(t)) \phi^{\prime}(t) d t$$

If $f:[a, b] \rightarrow \mathbb{R}$ is integrable and $\phi^{\prime}(t) \neq 0$ for every $t \in(\alpha, \beta)$, then the function $(f \circ \phi)\left|\phi^{\prime}\right|:[\alpha, \beta] \rightarrow \mathbb{R}$ is integrable and
$$\int^{b} f(x) d x=\int^{\beta} f(\phi(t))\left|\phi^{\prime}(t)\right| d t$$

## MATHS5074_1/MATHS4108_1 COURSE NOTES ：

Now take $\lambda \in \operatorname{Sp}(T)$. Put
$$\lambda \neq z \Rightarrow g(z):=\frac{f(\lambda)-f(z)}{\lambda-z} ; \quad g(\lambda):=f^{\prime}(\lambda) .$$
Clearly, $g$ is a holomorphic function (the singularity is “removed”). obtain
$$g(T)(\lambda-T)=(\lambda-T) g(T)=f(\lambda)-f(T) .$$
Consequently, if $f(\lambda) \in \operatorname{res}(f(T))$ then the operator $R(f(T), f(\lambda)) g(T)$ is inverse to $\lambda-T$. In other words, $\lambda \in \operatorname{res}(T)$, which is a contradiction. Thus,
$$f(\lambda) \in \mathbb{C} \backslash \operatorname{res}(f(T))=\operatorname{Sp}(f(T))$$
i.e., $f(\operatorname{Sp}(T)) \subset \operatorname{Sp}(f(T)) . \triangleright$

# 数学分析 Analysis 1 MA10207

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Let $X$ be a (nonzero) complex Banach space and let $T$ be a bounded endomorphism of $X$; i.e., $T \in B(X)$. For $h \in \mathscr{H}(\mathrm{Sp}(T))$, the contour integral with kernel the resolvent $R(T, \cdot)$ of $T$ is denoted by
$$\mathscr{R}_{T} h:=\frac{1}{2 \pi i} \oint h(z) R(T, z) d z$$

and called the Riesz-Dunford integral (of the germ $h$ ). If $f$ is a tunction holomorphic in a neighborhood about $\operatorname{Sp}(T)$ then put $f(T):=\mathscr{R}{T} f:=\mathscr{R}{T} \bar{f}$. We also use more suggestive designations like
$$f(T)=\frac{1}{2 \pi i} \oint \frac{f(z)}{z-T} d z$$

## MA10207 COURSE NOTES ：

Now take $\lambda \in \operatorname{Sp}(T)$. Put
$$\lambda \neq z \Rightarrow g(z):=\frac{f(\lambda)-f(z)}{\lambda-z} ; \quad g(\lambda):=f^{\prime}(\lambda) .$$
Clearly, $g$ is a holomorphic function (the singularity is “removed”). obtain
$$g(T)(\lambda-T)=(\lambda-T) g(T)=f(\lambda)-f(T) .$$
Consequently, if $f(\lambda) \in \operatorname{res}(f(T))$ then the operator $R(f(T), f(\lambda)) g(T)$ is inverse to $\lambda-T$. In other words, $\lambda \in \operatorname{res}(T)$, which is a contradiction. Thus,
$$f(\lambda) \in \mathbb{C} \backslash \operatorname{res}(f(T))=\operatorname{Sp}(f(T))$$
i.e., $f(\operatorname{Sp}(T)) \subset \operatorname{Sp}(f(T)) . \triangleright$

# 数学分析|Mathematical Analysis代写 MATH0048

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According to $5.6 .9$, for $|\lambda|>|T|$ there is an operator $(1-T / \lambda)^{-1}$ presenting the sum of the Neumann series; i.e.,
$$R(T, \lambda)=\frac{1}{\lambda}\left(1-{ }^{T} / \lambda\right)^{-1}=\frac{1}{\lambda} \sum_{k=0}^{\infty} \frac{T^{k}}{\lambda^{k}}$$

It is clear that
$$|R(T, \lambda)| \leq \frac{1}{|\lambda|} \cdot \frac{1}{1-|T | /| \lambda \mid} . \triangleright$$

\begin{aligned} (y, x)=&(y, x){\mathbb{R}}-i(i y, x){\mathbb{R}}=(x, y){\mathbb{R}}-i(x, i y){\mathbb{R}} \ &=(x, y){\mathbb{R}}+i(i x, y){\mathbb{R}}=(x, y)^{*} \end{aligned}
$(x, i y){\mathbf{R}}=1 / 4\left(|x+i y|^{2}-|x-i y|^{2}\right)$ ${ }^{1}{ }^{1} / 4\left(|i||y-i x|^{2}-|-i||i x+y|^{2}\right)=-(i x, y){\mathbb{R}}$.