(wrong string) φ(−n) = φ(n), φ(1) = 1, φ(0) = 0 || Mattias Andrée

From: <git_AT_suckless.org>
Date: Fri, 29 Jul 2016 00:30:19 +0200 (CEST)

commit 8092c767cb5f872b62a0cabbef793a08643497db
Author: Mattias Andrée <maandree_AT_kth.se>
AuthorDate: Thu Jul 28 22:55:43 2016 +0200
Commit: Mattias Andrée <maandree_AT_kth.se>
CommitDate: Thu Jul 28 22:55:43 2016 +0200

    φ(−n) = φ(n), φ(1) = 1, φ(0) = 0
    
    Signed-off-by: Mattias Andrée <maandree_AT_kth.se>

diff --git a/doc/exercises.tex b/doc/exercises.tex
index ebf8e91..14123d2 100644
--- a/doc/exercises.tex
+++ b/doc/exercises.tex
_AT_@ -262,10 +262,13 @@ which calculates the totient of $n$. Its
 formula is
 
 \( \displaystyle{
- \varphi(n) = n \prod_{p \in \textbf{P} : p | n}
+ \varphi(n) = |n| \prod_{p \in \textbf{P} : p | n}
     \left ( 1 - \frac{1}{p} \right ).
 }\)
 
+Note that, $\varphi(-n) = \varphi(n)$, $\varphi(0) = 0$,
+and $\varphi(1) = 1$.
+
 
 
 \end{enumerate}
_AT_@ -671,7 +674,8 @@ So, if we set $a = n$ and $b = 1$, then we iterate
 of all integers $p$, $2 \le p \le n$. For which $p$
 that is prime, we set $a \gets a \cdot (p - 1)$ and
 $b \gets b \cdot p$. After the iteration, $b | a$,
-and $\varphi(n) = \frac{a}{b}$.
+and $\varphi(n) = \frac{a}{b}$. However, if $n < 0$,
+then, $\varphi(n) = \varphi|n|$.
 
 
 
Received on Fri Jul 29 2016 - 00:30:19 CEST