From 7f48bcbef53515f0028bb8b1412e3460b091af7e Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?V=C3=ADctor?= <victor.ballester.ribo@gmail.com>
Date: Sat, 24 Jun 2023 17:12:30 +0200
Subject: [PATCH] corrected mathematical analysis

---
 Mathematics/2nd/Mathematical_analysis/Mathematical_analysis.tex | 2 +-
 1 file changed, 1 insertion(+), 1 deletion(-)

diff --git a/Mathematics/2nd/Mathematical_analysis/Mathematical_analysis.tex b/Mathematics/2nd/Mathematical_analysis/Mathematical_analysis.tex
index 167e1eb..049f65c 100644
--- a/Mathematics/2nd/Mathematical_analysis/Mathematical_analysis.tex
+++ b/Mathematics/2nd/Mathematical_analysis/Mathematical_analysis.tex
@@ -407,7 +407,7 @@
     Let $f:\RR \rightarrow\CC $ be a $T$-periodic function. Then: $$\int_a^{a+T}f(x)\dd{x}=\int_0^Tf(x)\dd{x}$$ where $a\in\RR $. In particular, $$\int_a^{a+kT}f(x)\dd{x}=k\int_0^Tf(x)\dd{x}$$
   \end{proposition}
   \begin{lemma}\label{MA:periodicbounded}
-    Let $f:\RR \rightarrow\CC $ be a $T$-periodic continuous function. Then, $|f|$ is bounded.
+    Let $f:\RR \rightarrow\CC $ be a $T$-periodic continuous function. Then, $\abs{f}$ is bounded.
   \end{lemma}
   \begin{sproof}
     Use \mnameref{RVF:weierstrass} on the interval $[0,T]$ and the periodicity of $f$.