Slight wording adjustments.
This commit is contained in:
@@ -153,12 +153,12 @@
|
|||||||
\abs{\int_{\gamma_j}\frac{f(z)}{z - z_0}dz - R_j(z)} < \frac{\eps}{n}
|
\abs{\int_{\gamma_j}\frac{f(z)}{z - z_0}dz - R_j(z)} < \frac{\eps}{n}
|
||||||
\]
|
\]
|
||||||
|
|
||||||
so
|
Thus
|
||||||
\[
|
\[
|
||||||
\abs{f(z_0) - \sum_{j = 1}^n R_j(z)} < \eps
|
\abs{f(z_0) - \sum_{j = 1}^n R_j(z)} < \eps
|
||||||
\]
|
\]
|
||||||
|
|
||||||
for all $z_0 \in K$. By the \hyperref[pole pushing lemma]{lemma:pole-pushing}, there exists $S \in \complex(z) \cap H(\complex_\infty \setminus P)$ such that $|S(z_0) - \sum_{j = 1}^n R_j(z_0)| < \eps$ for all $z_0 \in K$.
|
for all $z_0 \in K$. By the \hyperref[pole pushing lemma]{lemma:pole-pushing}, there exists $S \in \complex(z) \cap H(\complex_\infty \setminus P; \complex)$ such that $|S(z_0) - \sum_{j = 1}^n R_j(z_0)| < \eps$ for all $z_0 \in K$. Therefore $|S(z_0) - f(z_0)| < 2\eps$ for all $z_0 \in K$.
|
||||||
\end{proof}
|
\end{proof}
|
||||||
|
|
||||||
\begin{corollary}
|
\begin{corollary}
|
||||||
|
|||||||
Reference in New Issue
Block a user