这是一份nottingham诺丁汉大学MATH2012作业代写的成功案例

We can assume that the local coordinates are chosen so that the separatrix $L$ is a coordinate axis, $L={y=0}$. Then $\omega=\lambda y(1+O(1)) d x+(x+$ $O(2)) d y$, where $O(1), O(2)$ denote terms of order $\geqslant 1$ and $\geqslant 2$ respectively and $\lambda$ is the negative ratio of eigenvalues.

A curve $\gamma$ tangent to ${y=0}$ with multiplicity $k \geqslant 0$, is defined by the
$$
\tau_{0}(\mathcal{F}, \gamma)=\operatorname{ord}_{x=0}\left[\lambda b(x)(1+O(1))-b^{\prime}(x)(x+O(2))\right]=k+1
$$
if $\lambda \neq k+1$, i.e., if the singular point is not a saddle with the ratio of eigenvalues $-1:(k+1)$.

Using the tangency order, one can combine into a single identity valid for both $n>1$ and $n=1$. Assume that the origin is a dicritical singularity of a holomorphic foliation $\mathcal{F}$. Denote by $\Sigma$ the singular locus of its blow-up $\tilde{\mathcal{F}}$ and by $T$ the collection of the tangency points between $\tilde{\mathcal{F}}$ and the exceptional divisor.

MATH2012 COURSE NOTES ：

$$
f(x, y)=a(x) y+O\left(y^{2}\right), \quad g(x, y)=b(x)+O(y)
$$
with the holomorphic function $b(x)$ having an isolated root at $x=0$. Linearization of the Pfaffian equation ${\omega=0}$ on $S$ (i.e., keeping only the first order terms in powers of $y$ and $d y$ ) yields the equation
$$
y a(x) d x+b(x) d y=0,
$$
which corresponds to the linear ordinary (nonautonomous) equation
$$
\frac{d y}{d x}=r(x) y, \quad r(x)=-\frac{a(x)}{b(x)}
$$
The point $x=0$ is a pole of the meromorphic function $r(x)$. The meromorphic 1 -form $\theta \in \Lambda^{1}(S, 0)$ on the curve $S$, defined as
$$
\theta=-\frac{a(x)}{b(x)} d x
$$
is called the linearization form of $\omega$ along $S$; note, that this form depends on the choice of the local coordinates $(x, y)$ used in the construction.