Fixed typo with 1/2pi i.

src/dg/complex/runge.tex

@@ -5,7 +5,7 @@
 \label{proposition:existence-curves}
     Let $K \subset \complex$ be compact and $U \in \cn_\complex(K)$, then there exists closed rectifiable curves $\seqf{\gamma_j}$ such that for any separated locally convex space $E$, $f \in H(U; E)$, and $z_0 \in K$, 
     \[
-    f(z) = \sum_{j = 1}^n \int_{\gamma_j} \frac{f(z)}{z - z_0}dz
+    f(z_0) = \frac{1}{2\pi i}\sum_{j = 1}^n \int_{\gamma_j} \frac{f(z)}{z - z_0}dz
     \]
 \end{proposition}
 \begin{proof}[Proof, {{\cite[Proposition VIII.1.1]{ConwayComplex}}}. ]
@@ -26,8 +26,8 @@
 
     and $\gamma_j = \gamma_{j, \downarrow} \cdot \gamma_{j, \leftarrow} \cdot \gamma_{j, \uparrow} \cdot \gamma_{j, \rightarrow}$ be their concatenation, then for each $z \in U \setminus \partial R_j$, 
     \[
-    \int_{\gamma_j} \frac{f(z)}{z - z_0}dz = \begin{cases}
-        f(z) &z \in R_j^o \\
+    \frac{1}{2\pi i}\int_{\gamma_j} \frac{f(z)}{z - z_0}dz = \begin{cases}
+        f(z_0) &z \in R_j^o \\
         0 &z \in U \setminus R_j
     \end{cases}
     \]
@@ -39,7 +39,7 @@
 
     then for each $z_0 \in \bigcup_{j = 1}^n R_j^o$ and $f \in H(U; E)$, 
     \[
-    f(z) = \sum_{j = 1}^N \int_{\mu_j}\frac{f(z)}{z - z_0}dz
+    f(z_0) = \frac{1}{2\pi i}\sum_{j = 1}^N \int_{\mu_j}\frac{f(z)}{z - z_0}dz
     \]
 
     From here, it is sufficient to eliminate line segments that intersect $K$ and ensure that the remaining segments form a collection of loops. Let $1 \le j \le k \le N$, then $(\mu_j, \mu_k)$ is \textit{redundant} if for each $t \in [0, 1]$, $\mu_j(t) = \mu_k(1 - t)$. If $(\mu_j, \mu_k)$ are redundant, then
@@ -52,7 +52,7 @@
 
     By relabeling, let $\bracs{\mu_j|1 \le j \le m}$ such that for each $1 \le j \le m$, there exists no $1 \le k \le n$ such that $(\mu_j, \mu_k)$ is redundant. By (5), for each $z_0 \in \bigcup_{j = 1}^n R_j^o$ and $f \in H(U; E)$, 
     \[
-    f(z) = \sum_{j = 1}^m \int_{\mu_j}\frac{f(z)}{z - z_0}dz
+    f(z_0) = \frac{1}{2\pi i}\sum_{j = 1}^m \int_{\mu_j}\frac{f(z)}{z - z_0}dz
     \]
 
     Let $1 \le j \le N$ such that $\mu_j([0, 1]) \cap K \ne \emptyset$. Since $V \in \cn_\complex(K)$, there exists $1 \le k \le N$ such that $(\mu_j, \mu_k)$ are redundant by (2) and (3). Therefore $\bigcup_{j = 1}^m \mu_j([0, 1]) \cap K = \emptyset$. Since $\bigcup_{j = 1}^n R_j^o$ is dense in $K$, the above also holds for all $z \in K$. 
@@ -145,12 +145,12 @@
 \begin{proof}
     Let $U \in \cn_\complex(K)$, $f \in H(U; \complex)$, and $\eps > 0$. By \autoref{proposition:existence-curves}, there exists closed rectifiable curves $\seqf{\gamma_j}$ in $U \setminus V$ such that for each $z_0 \in K$, 
     \[
-    f(z_0) = \sum_{j = 1}^n \int_{\gamma_j}\frac{f(z)}{z - z_0}dz
+    f(z_0) = \frac{1}{2\pi i}\sum_{j = 1}^n \int_{\gamma_j}\frac{f(z)}{z - z_0}dz
     \]
 
     Let $T$ be the union of the images of $\seqf{\gamma_j}$, then by \autoref{lemma:rational-curve-approximation}, there exists $\seqf{R_j} \subset \complex(z) \cap H(\complex \setminus T; \complex)$ such that for each $z_0 \in K$ and $1 \le j \le n$, 
     \[
-    \abs{\int_{\gamma_j}\frac{f(z)}{z - z_0}dz - R_j(z)} < \frac{\eps}{n}
+    \abs{\frac{1}{2\pi i}\int_{\gamma_j}\frac{f(z)}{z - z_0}dz - R_j(z)} < \frac{\eps}{n}
     \]
 
     Thus