这是一份uwa西澳大学MATH3021的成功案例

非线性动力学和混沌|MATH3021 Nonlinear Dynamics and Chaos代写 UWA代写

$$
\begin{aligned}
p\left(\mathbf{r} \mid \mathbf{s}{\mathbf{i}}\right) &=\prod{k=0}^{(N-1)} p\left(r_{k} \mid s_{i, k}\right) \
k &=0,1, \ldots,(N-1)
\end{aligned}
$$
leading to
$$
\begin{aligned}
p\left(\mathbf{r} \mid \mathbf{s}{\mathbf{i}}\right) &=\frac{1}{\sqrt{\pi N{0}}} \exp \left[-\frac{\sum_{k=0}^{(N-1)}\left(r_{k}-s_{i, k}\right)^{2}}{N_{0}}\right] \
i &=0,1, \ldots,(N-1)
\end{aligned}
$$
The term in the exponential is the distance between the received vector $\mathbf{r}$ and $\mathbf{s}{\mathrm{i}}$ where $$ \mathbf{r}=\left[r{0}, r_{1}, r_{2}, \ldots, r_{N-1}\right]
$$
and
$$
\mathbf{s}{\mathbf{i}}=\left[s{i, 0} s_{i, 1}, s_{i, 2}, \ldots s_{i,(N-1)}\right]
$$
where, for example,
$$
s_{3}=\left[0_{3,0} 0_{3,1}, s_{3,2}, 1_{3,3}, \ldots, 0_{3,(N-1)}\right]
$$

MATH3021 COURSE NOTES ：

$$
g_{n+1}-g_{n}=q_{n}=1-\left(\frac{1}{2}+n\right) \ln \left(1+\frac{1}{n}\right)=-\frac{1}{12 n^{2}}+\frac{1}{12 n^{3}}+\ldots
$$
First note that $\mathcal{L}^{-1} q=p^{-2} \mathcal{L}^{-1} q^{\prime \prime}$ which can be easily evaluated by residues since
$$
q^{\prime \prime}=\frac{1}{n}-\frac{1}{n+1}-\frac{1}{2}\left(\frac{1}{(n+1)^{2}}+\frac{1}{n^{2}}\right)
$$
Thus, with $\mathcal{L}^{-1} g_{n}:=G$ we get
$$
\begin{aligned}
&\left(e^{-p}-1\right) G(p)=\frac{1-\frac{p}{2}-\left(\frac{p}{2}+1\right) e^{-p}}{p^{2}} \
&g_{n}=\int_{0}^{\infty} \frac{1-\frac{p}{2}-\left(\frac{p}{2}+1\right) e^{-p}}{p^{2}\left(e^{-p}-1\right)} e^{-n p} d p
\end{aligned}
$$