近代物理学|PHYS1002 Modern Physics代写 UWA代写

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We claim that this equation, for a positive operator $\omega_{1}$ and $d^{2}$ unitaries $U_{x}$, implies that $\omega_{1}=d^{-1} \mathbb{I}$. To see this, expand the operator $A=|\phi\rangle\left\langle e_{k}\right| \omega_{1}^{-1}$ in the basis $U_{x}$ according to the formula $A=\sum_{x} U_{x} \operatorname{tr}\left(U_{x}^{} A \omega_{1}\right)$ : $$\sum_{x}\left\langle e_{k}, U_{x}^{} \phi\right\rangle U_{x}=|\phi\rangle\left\langle e_{k}\right| \omega_{1}^{-1}$$
Taking the matrix element $\left\langle\phi|\cdot| e_{k}\right\rangle$ of this equation and summing over $k$, we find
$$\sum_{x, k}\left\langle e_{k}, U_{x}^{} \phi\right\rangle\left\langle\phi, U_{x} e_{k}\right\rangle=\sum_{x} \operatorname{tr}\left(U_{x}^{}|\phi\rangle\langle\phi| U_{x}\right)=d^{2}|\phi|^{2}=|\phi|^{2} \operatorname{tr}\left(\omega_{1}^{-1}\right)$$
Hence $\operatorname{tr}\left(\omega_{1}^{-1}\right)=d^{2}=\sum_{k} r_{k}^{-1}$, where $r_{k}$ are the eigenvalues of $\omega_{1}$. Using again the fact that the smallest value of this sum under the constraint $\sum_{k} r_{k}=1$ is attained only for constant $r_{k}$, we find $\omega_{1}=d^{-1}$ I. and $\Omega$ is indeed maximally entangled.

PHYS1002 COURSE NOTES ：

Prepare the ground state
$$\left|\varphi_{1}\right\rangle=|0 \ldots 0\rangle \otimes|0 \ldots 0\rangle$$
in both quantum registers.
Achieve equal amplitude distribution in the first register, for instance by an application of a Hadamard transformation to each qubit:
$$\left|\varphi_{2}\right\rangle=\frac{1}{\sqrt{2^{n}}} \sum_{x \in \mathbf{Z}{2}^{n}}|x\rangle \otimes|0 \ldots 0\rangle .$$ Apply $V{f}$ to compute $f$ in superposition. We obtain
$$\left|\varphi_{3}\right\rangle=\frac{1}{\sqrt{2^{n}}} \sum_{x \in \mathbf{Z}^{n}}|x\rangle|f(x)\rangle .$$

科学家和工程师的物理学|PHYS1001 Physics for Scientists and Engineers代写 UWA代写

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For computational purposes, let
$$\mathbf{Y}^{}=p\left(\mathbf{Y} \mid \mathscr{L}\left((\mathbf{A B}){11}, \ldots,(\mathbf{A B}){a b}\right)\right)=\sum_{i j} \bar{Y}{i j} \cdot(\mathbf{A B}){i j}$$
Then $\hat{\mathbf{Y}}{A B}=\mathbf{Y}^{}-\left(\hat{\mathbf{Y}}{0}+\hat{\mathbf{Y}}{A}+\hat{\mathbf{Y}}{B}\right)$ and by the Pythagorean Theorem,
$$\left|\hat{\mathbf{Y}}{A B}\right|^{2}=\left|\hat{\mathbf{Y}}^{}-\hat{\mathbf{Y}}{0}\right|^{2}-\left[\left|\hat{\mathbf{Y}}{A}\right|^{2}+\left|\hat{\mathbf{Y}}{B}\right|^{2}\right]$$
But $\mathbf{Y}^{}-\hat{\mathbf{Y}}{0}=\sum{i j}\left(\bar{Y}{i j} \ldots-\bar{Y}{\ldots} \ldots\right)(\mathbf{A B}){i j}$ so $$\left|\hat{\mathbf{Y}}^{}-\hat{\mathbf{Y}}{0}\right|^{2}=\sum_{i j}\left(\bar{Y}{i j} \ldots-\bar{Y}{\ldots} \ldots\right)^{2}(c m)=\left|\hat{\mathbf{Y}}^{}\right|^{2}-\left|\hat{\mathbf{Y}}{0}\right|^{2}=\sum{i j} \vec{Y}{i j}^{2} \ldots(c m)-\bar{Y}{\ldots}^{2} \ldots n$$

PHYS1001COURSE NOTES ：

Then
and
$$\mathrm{SSA}=\left|\hat{\mathbf{Y}}{A}\right|^{2}=\sum{1}^{l}\left[\left(a_{i}-\bar{a}\right)+\left(\bar{\varepsilon}{i} \cdot-\bar{\varepsilon} .\right)\right]^{2} J,$$ $$\mathrm{SSE}=|\mathbf{e}|^{2}=\sum{i j}\left(\varepsilon_{i j}-\bar{\varepsilon}{i}\right)^{2}$$ Let $W{i}=a_{i}+\vec{\varepsilon}{i, .}$ Then $W{i} \sim N\left(0, \sigma_{a}^{2}+\frac{\sigma^{2}}{J}\right)$ and the $W_{i}$ are independent. It follows that
$$\frac{\sum_{i}\left(W_{i}-\bar{W}\right)^{2}}{\sigma_{a}^{2}+\frac{\sigma^{2}}{J}}=\frac{\text { SSA }}{\sigma^{2}+J \sigma_{a}^{2}} \sim \chi_{I-1}^{2}$$
In addition, the $W_{i}$ are independent of the vector e, and therefore of SSE, which is the same as it is for the fixed effects model. Thus, SSE $/ \sigma^{2} \sim \chi_{(\Lambda-1)}^{2}$.

统计学习|STAT3064 Statistical Learning代写 UWA代写

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$c_{0} \beta_{0}+c_{1} \beta_{1}$, hence on $g(x)=\beta_{0}+\beta_{1} x$. Since $C=R_{2}, V_{1}=V=\mathscr{L}(\mathbf{J}, \mathbf{x})$. Thus, $100 \%$ simultaneous confidence intervals on $g(x)=\beta_{0}+\beta_{1} x$ are given by
$$\hat{g}(x) \pm K S\left(\hat{\eta}{c}\right) \quad \text { for } K=\sqrt{2 F{2, n-2, \gamma}}$$
where $\hat{g}(x)=\hat{\beta}{0}+\hat{\beta}{1} x$. Since $\operatorname{Var}(g(x))=h(x) \sigma^{2}$, where $h(x)=1 / n+(x-\bar{x})^{2} /$ $S_{x x}$, the simultaneous intervals are
$$\hat{g}(x) \pm\left[h(x) S^{2}\right]^{1 / 2} K$$
We earlier found that a $100 \% \%$ confidence interval on $g(x)$, holding for that $x$ only, is
$$g(x) \pm t h(x)^{1 / 2} S \quad \text { for } \quad t=t_{n-2,(t+n) 2}$$
Thus the ratio of the length of the simultaneous interval at $x$ to the individual interval is $(K / t)=\left(2 F, 2 t^{2}-3(1+w /)^{1 / 2}\right.$, which always exceeds one.

STAT3064 COURSE NOTES ：

\begin{aligned} \beta_{j}=\bar{\mu}{\cdot j}-\mu=\frac{1}{I} \sum{i} \mu_{i j}-\mu \ (\alpha \beta){i j} &=\mu{i j}-\left[\mu+\alpha_{i}+\beta_{j}\right] . \end{aligned}
Then $\mu_{i j}=\mu+\alpha_{i}+\beta_{j}+(\alpha \beta){i j}$. The full model then can be written as follows. $$\text { Full model: } \quad Y{i j k}=\mu+\alpha_{i}+\beta_{j}+(\alpha \beta){i j}+\varepsilon{i j k},$$
where
$$\sum_{1}^{l} \alpha_{i}=\sum_{1}^{J} \beta_{j}=\sum_{i}(\alpha \beta){i j}=\sum{j}(\alpha \beta){i j}=0, \quad \text { and } \quad \varepsilon{i j k} \sim N\left(0, \sigma^{2}\right)$$

空间统计和建模|STAT3063 Spatial Statistics and Modelling代写 UWA代写

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The distribution functions $D_{2}(r), D_{3}(r), \ldots$ of the distances to the $2 \mathrm{nd}, 3 \mathrm{rd}, \ldots$ nearest neighbours are
$$D_{k}(r)=1-\sum_{j=0}^{k-1} \exp \left(-\lambda \pi r^{2}\right) \frac{\left(\lambda \pi r^{2}\right)^{j}}{j !} \quad \text { for } r \geq 0$$
and the corresponding probability density functions are
$$d_{k}(r)=\frac{2\left(\lambda \pi r^{2}\right)^{k}}{r(k-1) !} \exp \left(-\lambda \pi r^{2}\right) \quad \text { for } r \geq 0$$The corresponding $j$ th moments are
$$m_{k, j}=\frac{\Gamma\left(k+\frac{1}{2} j\right)}{(k-1) !(\lambda \pi)^{j / 2}} \quad \text { for } j=1,2, \ldots,$$
and the position of the mode (maximum of density function) is
$$r_{k}=\sqrt{\frac{k-\frac{1}{2}}{\lambda \pi}}$$

STAT3063 COURSE NOTES ：

The so-called $L$-function is obtained by
$$L(r)=\left(\frac{K(r)}{b_{d}}\right)^{\frac{1}{2}}$$
as
$$L(r)=r \quad \text { for } r \geq 0,$$
and, similarly, the pair correlation function $g(r)$ is given by
$$g(r)=1 \quad \text { for } r \geq 0,$$
due to the general relation to $K(r)$,
$$g(r)=\frac{\mathrm{d} K(r)}{\mathrm{d} r} / d b_{d} .$$

统计科学|STAT3062 Statistical Science代写 UWA代写

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Now, $\hat{\mu}{m}$ is the value minimizing Taking the derivative of this equation, with $\xi$ given by and setting the result equal to zero, $\hat{\mu}{m}$ is determined by
$$2 \sum_{i=1}^{n} W\left(X_{i}-\mu_{m}\right)=0$$
where
$$\Psi(x)=\max [-K, \min (K, x)]$$
is Huber’s $\Psi$. (For a graph of Huber’s $\Psi$, see Chapter 2.) Of course, the constant 2 in is not relevant to solving for $\hat{\mu}{m}$, and typically is simplified to $$\sum{i=1}^{n} \Psi\left(X_{i}-\hat{\mu}_{m}\right)=0$$

STAT3062 COURSE NOTES ：

$$A(x)=\operatorname{sign}(|x-\theta|-\omega),$$
where $\theta$ is the population median and $\operatorname{sign}(x)$ equals $-1,0$, or 1 according to whether $x$ is less than, equal to, or greater than 0 . Let
$$B(x)=\operatorname{sign}(x-\theta),$$
and
$$C(x)=A(x)-\frac{B(x)}{f(\theta)}{f(\theta+\omega)-f(\theta-\omega)} .$$
The influence function of $\omega_{N}$ is
$$\mathrm{IF}{\omega{N}}(x)=\frac{C(x)}{2(.6745){f(\theta+\omega)+f(\theta-\omega)}} .$$

随机过程及其应用|STAT3061 Random Processes and their Applications代写 UWA代写

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For $p \geq 1 / 2$, we have to consider $f_{i}=\mathbb{P}{i}\left(T{i}<\infty\right)$. Writing $$\left.\mathbb{P}{0}\left(T{0}<\infty\right)=1-q+q \mathbb{P}{1} \text { (hit } 0\right),$$ we see that if $\mathbb{P}{1}($ hit 0$)<1$, the chain is transient. But $$\mathbb{P}{i}(\text { hit } i-1)=\frac{1-p}{p}, i \geq 1$$ see Section 1.5. Hence, for $p>1 / 2$ the chain is transient. It remains to check the case $p=1 / 2$. Here, $f{i}=1$, and the chain is recurrent. The invariance equations $$\pi_{i}=\frac{1}{2} \pi_{i-1}+\frac{1}{2} \pi_{i+1}, i>1,$$
have the general solution $\pi_{i}=A+B i, i \geq 1$. At $i=1,0$ they have the form
$$\pi_{1}=q \pi_{0}+\frac{1}{2} \pi_{2}, \pi_{0}=(1-q) \pi_{0}+\frac{1}{2} \pi_{1},$$
which yields $B=0$ and
$$\pi_{i} \equiv A, i \geq 1, \pi_{0}=\frac{1}{2 q} A$$

STAT3061 COURSE NOTES ：

these probabilities do not depend on the value of $i$ because of the homogeneous property of the chain. Conditioning on the first jump and using the strong Markov property we get
$$a=q+p b a^{2}, b=q+p b a$$
whence
$$b=\frac{q}{1-p a}, \text { and } a=q+\frac{p q a^{2}}{1-p a} .$$
Thus,
$$p(1+q) a^{2}-(p q+1) a+q=0$$
and the solutions are
$$a=1 \text { and } a=\frac{q}{1-q^{2}}$$
We are interested in the minimal solution
$$\frac{q}{1-q^{2}}<1 \text { if and only if } q<\frac{\sqrt{5}-1}{2} .$$
Therefore, the chain is recurrent if and only if $q \geq(\sqrt{5}-1) / 2$ and transient if and only if $q<(\sqrt{5}-1) / 2$.

几何学|MATH3033 Geometry代写 UWA代写

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Concretely parametrize the sphere in the usual way:
$$x_{1}=\sin \theta \sin \phi, \quad x_{2}=\sin \theta \cos \phi, \quad x_{3}=\cos \theta$$
then with the poles removed the range of values is $0<\theta<\pi, 0 \leq \phi<2 \pi$. The antipodal map is
$$\theta \mapsto \pi-\theta, \quad \phi \mapsto \phi+\pi .$$
We can therefore identify the space of lines in $\mathbf{R}^{2}$ as the pairs
$$(\theta, \phi) \in(0, \pi) \times[0, \pi]$$

MATH3033 COURSE NOTES ：

We can add symmetric bilinear forms: $(B+C)(v, w)=B(v, w)+C(v, w)$ and multiply by a scalar $(\lambda B)(v, w)=\lambda B(v, w)$ so they form a vector space isomorphic to the space of symmetric $n \times n$ matrices which has dimension $n(n+1) / 2$. If we take a different basis
$$w_{i}=\sum_{j} P_{j i} v_{j}$$
then
$$B\left(w_{i}, w_{j}\right)=B\left(\sum_{k} P_{k i} v_{k}, \sum_{\ell} P_{\ell j} v_{\ell}\right)=\sum_{k, \ell} P_{k i} B\left(v_{k}, v_{\ell}\right) P_{\ell j}$$
so that the matrix $\beta_{i j}=B\left(v_{i}, v_{j}\right)$ changes under a change of basis to
$$\beta^{\prime}=P^{T} \beta P .$$

拓扑结构和分析|MATH3032 Topology and Analysis代写 UWA代写

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By Fubini’s theorem, it follows that for $k \neq j$ that
$$\int_{\Omega_{n}}\left(X_{k}-x\right)\left(X_{j}-x\right) d \mu_{z}^{n}=\left[\int_{{0,1}}(\omega-x) d \mu_{z}(\omega)\right]^{2}=0$$
and
$$\int_{\Omega_{n}}\left(X_{k}-x\right)^{2} d \mu_{z}^{n}=\int_{{0,1}}(\omega-x)^{2} d \mu_{z}(\omega)=(1-x)^{2} x+x^{2}(1-x) \leq 2$$
Combining the last three displayed equations shows that
$$\mu_{z}^{n}\left(\left|S_{n}-x\right|>\epsilon\right) \leq \frac{1}{n^{2} \epsilon^{2}} 2 n=\frac{2}{n \epsilon^{2}}$$
which combined with $\mathrm{Eq}$. (5.4) implies that
$$\sup {z \in[0,1]}\left|f(x)-p{n}(x)\right| \leq \frac{4 M}{n \epsilon^{2}}+\delta_{c}$$
and therefore
$$\lim {n \rightarrow \infty} \sup \sup {z \in[0,1]}\left|f(x)-p_{n}(x)\right| \leq \delta_{c} \rightarrow 0 \text { as } \epsilon \rightarrow 0$$

MATH3032 COURSE NOTES ：

To prove the claim, let $V_{y}$ be an open neighborhood of $y$ such that $\left|f-g_{z y}\right|<\epsilon$ on $V_{y}$ so in particular $f<\epsilon+g_{z y}$ on $V_{y}$. By compactness, there exists $\Lambda \subset \subset X$ such that $X=\bigcup_{y \in \Lambda} V_{y}$. Set
$$g_{z}(z)=\max \left{g_{z y}(z): y \in \Lambda\right},$$
then for any $y \in \Lambda, f<\epsilon+g_{z y}<\epsilon+g_{z}$ on $V_{y}$ and therefore $f<\epsilon+g_{z}$ on $X$. Moreover, by construction $f(x)=g_{z}(x)$.

We now will finish the proof of the theorem. For each $x \in X$, let $U_{z}$ be a neighborhood of $x$ such that $\left|f-g_{z}\right|<\epsilon$ on $U_{z}$. Choose $\Gamma \subset \subset X$ such that $X=\bigcup_{z \in \Gamma} U_{z}$ and define $$g=\min \left{g_{z}: x \in \Gamma\right} \in \mathcal{A} .$$ Then $f0$ is arbitrary it follows that $f \in \overline{\mathcal{A}}=\mathcal{A}$.

代数结构和对称性|MATH3031 Algebraic Structures and Symmetry代写 UWA代写

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Let $A$ be a non-empty subset of a semihypergroup (H, )). We say that $A$ is a complete part of $(H$, $\circ)$ if, for every $n \in \mathbb{N}-{0}$ and $\left(x_{1}, x_{2}, \ldots, x_{n}\right) \in H^{n}$,
$$\left(x_{1} \circ \cdots \circ x_{n}\right) \cap A \neq \emptyset \Longrightarrow\left(x_{1} \circ \cdots \circ x_{n}\right) \subseteq A .$$
Clearly, the set $H$ is a complete part, and the intersection $\mathcal{C}(X)$ of all the complete parts containing a non-empty set $X$ is called the complete closure of $X$. If $X$ is a complete part of $(H, 0)$ then $\mathcal{C}(X)=X$. If $(H, 0)$ is a semihypergroup and $\varphi: H \rightarrow H / \beta^{}$ is the canonical projection, then, for every non-empty set $A \subseteq H$, we have $\mathcal{C}(A)=\varphi^{-1}(\varphi(A))$. Moreover, if $(H, 0)$ is a hypergroup, then $$\mathcal{C}(A)=\varphi^{-1}(\varphi(A))=A \circ \omega H=\omega_{H} \circ A$$ A hypergroup $(H, \circ)$ is said to be complete if $x \circ y=\mathcal{C}(x \circ y)$, for all $(x, y) \in H^{2}$. If $(H, \circ)$ is a complete hypergroup, then $$x \circ y=\mathcal{C}(a)=\beta^{}(a) .$$
for every $(x, y) \in H^{2}$ and $a \in x \circ y$

MATH3031 COURSE NOTES ：

1. The image of $A$ under $f$ is denoted by $f(A)$ or
$$C M_{f(A)}(y)= \begin{cases}\mathrm{V}{f(x)-y} C M{A}(x) & \text { if } f^{-1}(y) \neq \varnothing \ 0 & \text { otherwise. }\end{cases}$$
2. The inverse image of $B$ under $f$ is denoted by $f^{-1}(B)$ where $C M_{f-1}(B)(x)=C M_{B}(f(x))$.

复杂系统|MATH3024 Complex Systems代写 UWA代写

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Inserting for the $J_{i j}$ and cosidering $i \in I_{x}$ we get
$$m_{i}=\tanh \left[\beta \sum_{\boldsymbol{x}^{\prime}} Q\left(\boldsymbol{x}, \boldsymbol{x}^{\prime}\right) p_{\boldsymbol{x}^{\prime}} m_{\boldsymbol{x}^{\prime}}+\beta h_{i}\right]$$
where we have introduced sub-lattice (equilibrium) magnetizations $m_{x}$ via
$$m_{x}=\frac{1}{\left|I_{\boldsymbol{x}}\right|} \sum_{i \in I_{\boldsymbol{x}}} m_{i}$$
Inserting this into (5.13) we get
$$m_{\boldsymbol{x}}=\frac{1}{\left|I_{\boldsymbol{x}}\right|} \sum_{i \in I_{\boldsymbol{x}}} \tanh \left[\beta \sum_{\boldsymbol{x}^{\prime}} Q\left(\boldsymbol{x}, \boldsymbol{x}^{\prime}\right) p_{\boldsymbol{x}^{\prime}} m_{\boldsymbol{x}^{\prime}}+\beta h_{i}\right]=\left\langle\tanh \left[\beta \sum_{\boldsymbol{x}^{\prime}} Q\left(\boldsymbol{x}, \boldsymbol{x}^{\prime}\right) p_{\boldsymbol{x}^{\prime}} m_{\boldsymbol{x}^{\prime}}+\beta h\right]\right\rangle_{h},$$

MATH3024 COURSE NOTES ：

Introducing the overlap vector
$$m=\sum_{\xi} p_{\xi} \xi m_{\xi}=\left\langle\xi m_{\xi}\right\rangle_{\xi}$$
we see that the fpe’s can be written as
$$m_{\boldsymbol{\xi}}=\tanh [\beta \boldsymbol{\xi} \cdot \boldsymbol{m}]$$
which after multiplying by $\boldsymbol{\xi}$ and averaging over $\boldsymbol{\xi}$ gives
$$\boldsymbol{m}=\langle\boldsymbol{\xi} \tanh [\beta \boldsymbol{\xi} \cdot \boldsymbol{m}]\rangle_{\boldsymbol{\xi}},$$
or, in components
$$m_{\mu}=\left\langle\xi^{\mu} \tanh \left[\beta \sum_{\nu} \xi^{\nu} m_{\nu}\right]\right\rangle_{\xi},$$
Note that $m_{\mu}$ is nothing but the overlap of the equilibrium spin-configuration with the pattern $\xi_{i}^{\mu}$,
$$m_{\mu}=\frac{1}{N} \sum_{i} \xi_{i}^{\mu}\left\langle S_{i}\right\rangle=\sum_{\xi} p_{\xi} \xi^{\mu} m_{\xi}$$