Tag Archives: 英国代写

量子物理和电磁学|PHYS2001 Quantum Physics and Electromagnetism代写 UWA代写

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这是一份uwa西澳大学PHYS2001的成功案例

量子物理和电磁学|PHYS2001 Quantum Physics and Electromagnetism代写 UWA代写


$$
c_{s}(t)=\frac{i}{\hbar}\left\langle s^{(0)}\left|\sum_{\alpha=1}^{N} \frac{q^{(\alpha)}}{m^{(\alpha)}} \Pi_{i}^{(\alpha)}\right| n^{(0)}\right\rangle \int_{0}^{t} A_{i}(0, t) e^{-i \omega_{n-s} t} d t
$$
The matrix element in is evaluated by using the commutator
$$
\left[H^{(0)}, p_{i}\right]=-i \hbar \sum_{\alpha=1}^{N} \frac{q^{(\alpha)}}{m^{(\alpha)}} \Pi_{i}^{(\alpha)}
$$
where $p_{i}$ is the electric dipole moment operator, and we have assumed a velocityindependent potential in $(2.3)$. Then
$$
\left\langle s^{(0)}\left|\sum_{\alpha=1}^{N} \frac{q^{(\alpha)}}{m^{(\alpha)}} \Pi_{i}^{(\alpha)}\right| n^{(0)}\right\rangle=i \omega_{s n}\left\langle p_{i}\right\rangle_{s n} .
$$
To proceed further we require an explicit form for the vector potential, which we take to be that for a harmonic plane wave
$$
\mathbf{A}(\mathbf{r}, t)=\mathbf{A}_{0} \cos (\mathbf{k} \cdot \mathbf{r}-\omega t) .
$$

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MATH2001 COURSE NOTES ：

$$
H^{(1)}=-\sum_{\alpha=1}^{N} \frac{q^{(\alpha)}}{2 m(\alpha)}\left(\mathbf{B} \times \mathbf{r}^{(\alpha)}\right) \cdot \mathbf{\Pi}^{(\alpha)}=-\mathbf{m} \cdot \mathbf{B}
$$
and
$$
H^{(2)}=-\sum_{\alpha=1}^{N} \frac{\left(q^{(\alpha)}\right)^{2}}{8 m^{(\alpha)}}\left[r_{i}^{(\alpha)} r_{j}^{(\alpha)}-\left(r^{(\alpha)}\right)^{2} \delta_{i j}\right] B_{i} B_{j}
$$
Here
$$
\mathbf{m}=\sum_{\alpha=1}^{N} \frac{q^{(\alpha)}}{2 m(\alpha)} \mathbf{r}^{(\alpha)} \times \boldsymbol{\Pi}^{(\alpha)}
$$













高级数学方法|MATH2501 Advanced Mathematical Methods代写 UWA代写

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这是一份uwa西澳大学MATH2501的成功案例

高级数学方法|MATH2501 Advanced Mathematical Methods代写 UWA代写


From the point $p \in S^{1}$ is mapped onto the point:
$$
\phi_{1}(p)=\frac{2 a}{\tan \frac{\theta}{2}} \in \mathbb{R}^{1} .
$$
This is defined for $\theta \neq 0$, that is, $p \in M_{1}$ where $M_{1}$ is the chart excluding $\theta=0$. For the opposite stereographic projection (upwards from the south pole), we will get
$$
\phi_{2}(p)=\frac{2 a}{\cot \frac{\theta}{2}}=2 a \tan \frac{\theta}{2}
$$
which is defined for $\theta \neq \pi$, hence for $p \in M_{2}$.
Now consider $\phi_{2} \circ \phi_{1}^{-1}: \mathbb{R} \rightarrow \mathbb{R}$. We have:
$$
\begin{aligned}
\phi_{1}^{-1}(0,2 \pi) &=2 \cot ^{-1} \frac{x}{2 a} \
\phi_{2} \circ \phi_{1}^{-1} &=2 a \tan \left(\cot ^{-1} \frac{x}{2 a}\right) \
&=\frac{4 a^{2}}{x}
\end{aligned}
$$
This is the inversion map: $\mathbb{R}-{0} \rightarrow \mathbb{R}-{0}$ which is infinitely differentiable. The map $\phi_{1} \circ \phi_{2}^{-1}$ works similarly.

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MATH2501 COURSE NOTES ：

There is an alternative way to look at this. For a given $a^{\prime} \in V^{*}$, we have a linear map
$$
\left\langle a^{\prime},\right\rangle: \quad V \rightarrow \mathbb{R}
$$
defined by
$$
\left\langle a^{\prime},\right\rangle: a \in V \rightarrow\left\langle a^{\prime}, a\right\rangle \in \mathbb{R}
$$
Thus the dual vector space to $V$ is the space of linear functionals on $V$.
Let us clarify what we mean by linear functionals. A functional is a map which takes an element of vector space to a number. In our case, clearly $\langle a,$, is a functional, and it has the additional property that:
$$
\left\langle a^{\prime},\right\rangle: \lambda a+\mu b \rightarrow \lambda\left\langle a^{\prime}, a\right\rangle+\mu\left\langle a^{\prime}, b\right\rangle
$$
which is just what we mean by linearity.
An important property of dual spaces is that if we have a map of vector spaces $V, W$ :
$$
f: V \rightarrow W
$$













物理学衔接单元|PHYS1030 Physics Bridging Unit代写 UWA代写

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这是一份uwa西澳大学PHYS1030的成功案例

物理学衔接单元|PHYS1030 Physics Bridging Unit代写 UWA代写


$$
p({\sigma})=\frac{e^{-\beta H({\sigma})}}{Z},
$$
where ${\sigma}$ labels a microstate (i.e. a configuration), $H({\sigma})$ is the system Hamiltonian, $\beta$ is the inverse temperature and $Z$ is the partition function. Suppose, now, that we sum over all microstates identifiable as belonging to a certain phase $\gamma$, then
$$
\begin{aligned}
p_{\gamma} &=\sum_{{\sigma} \in \gamma} p({\sigma}) \
&=\frac{1}{Z} \sum_{{\sigma} \in \gamma} e^{-\beta H({\sigma})} \
& \equiv \frac{Z_{\gamma}}{Z},
\end{aligned}
$$

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PHYS1030 COURSE NOTES ：

where the last step serves to define the configurational weight $Z_{\gamma}$ of the phase $\gamma$. The relative probabilities of two phases $A$ and $B$ is then
$$
\begin{aligned}
\mathcal{R}{A B} & \equiv \frac{p{A}}{p_{B}}=\frac{Z_{A}}{Z_{B}} \
& \equiv \frac{e^{-\beta F_{A}}}{e^{-\beta F_{B}}}
\end{aligned}
$$
where $F_{\gamma}$ denotes the free energy of phase $\gamma$. It follows that the free energy difference between two phases $A$ and $B$ is simply proportional to the logarithm of the ratio of their a-priori probabilities:
$$
F_{A}-F_{B}=-\frac{1}{\beta} \ln \mathcal{R}{A B} $$ Clearly this equation implies that precisely at coexistence $\left(F{A}=F_{B}\right)$ the system will be found with equal probability in each of the two phases.













数学基础|MATH1722 Mathematics Foundations: Specialist代写 UWA代写

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这是一份uwa西澳大学MATH1722的成功案例

数学基础|MATH1722 Mathematics Foundations: Specialist代写 UWA代写


$$
\hat{\mathbf{f}}(\mathbf{u})=\mathbf{f}(\mathbf{u})+h \mathbf{f}{1}(\mathbf{u})+h^{2} \mathbf{f}{2}(\mathbf{u})+\cdots
$$
Assume that at some initial time $\mathbf{u}=\hat{\mathbf{u}}$. Using a simple Taylor series we can expand the solution of the modified ordinary differential equation so that
$$
\hat{\mathbf{u}}(t+h)=\mathbf{u}+h\left(\mathbf{f}+h \mathbf{f}{2}+h^{2} \mathbf{f}{3}+\cdots\right)+\frac{h^{2}}{2}\left(\mathbf{f}^{\prime}+h \mathbf{f}^{\prime}+\cdots\right)(\mathbf{f}+\cdots)+\cdots
$$
We want to compare this solution with the action of the discrete operator $\Psi_{h}$. In general, for a consistent scheme, there will be functions $\mathbf{d}{k}(h)$ so that $$ \Psi{h} \mathbf{u}=\mathbf{u}+h \mathbf{f}(\mathbf{u})+h^{2} \mathbf{d}{2}(\mathbf{u})+h^{3} \mathbf{d}{3}(\mathbf{u})+\cdots
$$
We can now equate the above two expressions. Taking terms at orders $h^{2}$ and $h^{3}$ gives
$$
\mathbf{f}{2}=\mathbf{d}{2}-\frac{1}{2} \mathbf{f}^{\prime} \mathbf{f} \quad \text { and } \quad \mathbf{f}{3}=\mathbf{d}{3}-\frac{1}{6}\left(\mathbf{f}^{\prime}(\mathbf{f}, \mathbf{f})(\mathbf{u})+\mathbf{f}^{\prime} \mathbf{f}^{\prime} \mathbf{f}(\mathbf{u})\right)-\frac{1}{2}\left(\mathbf{f}^{\prime} \mathbf{f}{2}(\mathbf{u})+\mathbf{f}{2}^{\prime} \mathbf{f}(\mathbf{u})\right)
$$

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MATH1722 COURSE NOTES ：

$$
\frac{\mathrm{d}}{\mathrm{d} t} F\left(\mathbf{u}{i}(t)\right)=\left(D{i} F\right)\left(\mathbf{u}{i}(t)\right)=F^{\prime}\left(\mathbf{u}{i}\right) f_{i}\left(\mathbf{u}{i}\right), $$ where $\mathbf{u}{i}(t)$ is the solution of the associated differential equation. Then if $\mathrm{d} \mathbf{u}{i} / \mathrm{d} t=$ $\mathbf{f}{i}\left(\mathbf{u}{i}\right)$ we have $\mathrm{d} \mathbf{u}{i} / \mathrm{d} t=D_{i} \mathbf{u}{i}(t), \mathrm{d}^{2} \mathbf{u}{i} / \mathrm{d} t^{2}=D_{i}^{2} \mathbf{u}{i}(t)$, etc. As a consequence we may express the flow $\varphi{t}^{i}$ as an exponential of the form
$$
\varphi_{i}^{i} \mathbf{u}{0}=\exp \left(t D{i}\right) \mathbf{u}{0}=\sum{n} \frac{t^{n} D_{i}^{n}}{n !} \mathbf{u}{0} . $$ and compose two flows in the manner $$ \left(\varphi{t}^{1} \circ \varphi_{s}^{2}\right) u_{0}=\exp \left(t D_{1}\right) \exp \left(s D_{2}\right) \mathbf{u}_{0} .
$$
Thus we see how a flow can be thought of in terms of the exponential of an operator. Now suppose that $X$ and $Y$ are general operators with associated exponentials
$$
\exp (h X)=I+h X+\frac{h^{2}}{2} X^{2}+\cdots, \quad \exp (h Y)=I+h Y+\frac{h^{2}}{2} Y^{2}+\cdots,
$$













近代物理学|PHYS1002 Modern Physics代写 UWA代写

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这是一份uwa西澳大学PHYS1002的成功案例

近代物理学|PHYS1002 Modern Physics代写 UWA代写


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.

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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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这是一份uwa西澳大学PHYS1001的成功案例

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


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 $$ \left|\hat{\mathbf{Y}}^{}-\hat{\mathbf{Y}}{0}\right|^{2}=\sum_{i j}\left(\bar{Y}{i j} \ldots-\bar{Y}{\ldots} \ldots\right)^{2}(c m)=\left|\hat{\mathbf{Y}}^{}\right|^{2}-\left|\hat{\mathbf{Y}}{0}\right|^{2}=\sum{i j} \vec{Y}{i j}^{2} \ldots(c m)-\bar{Y}{\ldots}^{2} \ldots n
$$

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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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这是一份uwa西澳大学STAT3064的成功案例

统计学习|STAT3064 Statistical Learning代写 UWA代写

$c_{0} \beta_{0}+c_{1} \beta_{1}$, 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.


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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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这是一份uwa西澳大学STAT3063的成功案例

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

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
$$
r_{k}=\sqrt{\frac{k-\frac{1}{2}}{\lambda \pi}}
$$


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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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这是一份uwa西澳大学STAT3062的成功案例

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

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 $$ \sum{i=1}^{n} \Psi\left(X_{i}-\hat{\mu}_{m}\right)=0
$$


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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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这是一份uwa西澳大学STAT3061的成功案例

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

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
$$
\pi_{i} \equiv A, i \geq 1, \pi_{0}=\frac{1}{2 q} A
$$


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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 $q<(\sqrt{5}-1) / 2$.