这是一份uwa西澳大学MATH3002的成功案例

网络科技|MATH3002 Network Science代写 UWA代写

If $A$ is the adjacency matrix of a graph $G, \alpha>0$, and $\lambda_{1}$ the largest eigenvalue of $A$, then
$$
\lambda_{1}<\frac{1}{\alpha} \Longleftrightarrow \sum_{k=1}^{\infty} \alpha^{k} A^{k} \text { converges. }
$$
For the proof see, e.g., [208].
Assuming convergence we find a closed form expression for the status index of Katz:
$$
c_{K}=\sum_{k=1}^{\infty} \alpha^{k}\left(A^{T}\right)^{k} \mathbf{1}{n}=\left(\left(I-\alpha A^{T}\right)^{-1}\right) \mathbf{1}{n}
$$
or, in another form
$$
\left(I-\alpha A^{T}\right) \boldsymbol{c}{K}=\mathbf{1}{n}
$$
an inhomogeneous system of linear equations emphasizing the feedback nature of the centrality: the value of $\boldsymbol{c}{K}(i)$ depends on the other centrality values $c{K}(j)$, $j \neq i$.

MATH3002 COURSE NOTES ：

Taking then $y(x)=e^{\frac{2}{3} x^{3 / 2}} h\left(\frac{2}{3} x^{3 / 2}\right)$ we get
$$
h^{\prime \prime}+\left(2+\frac{1}{3 s}\right) h^{\prime}+\frac{1}{3 s} h=0
$$
and with $H=\mathcal{L}^{-1}(h)$ we get
$$
p(p-2) H^{\prime}=\frac{5}{3}(1-p) H
$$
The solution is
$$
H=C p^{-5 / 6}(2-p)^{-5 / 6}
$$
and it can be easily checked that any integral of the form
$$
h=\int_{0}^{\infty e^{i \phi}} e^{-p s} H(p) d p
$$
for $\phi \neq 0$ and $\operatorname{Re}(p s)>0$, is a solution of yielding the expression