>From 69881c2d7792c59d8dfbed3a799186a95ef835fb Mon Sep 17 00:00:00 2001
From: Akira Yokosawa <akiyks@xxxxxxxxx>
Date: Sat, 30 Sep 2017 17:37:45 +0900
Subject: [PATCH 02/10] debugging: Use upright font for Euler's number
Also use \ln for natural logarithm.
Signed-off-by: Akira Yokosawa <akiyks@xxxxxxxxx>
---
debugging/debugging.tex | 24 ++++++++++++------------
perfbook.tex | 2 ++
2 files changed, 14 insertions(+), 12 deletions(-)
diff --git a/debugging/debugging.tex b/debugging/debugging.tex
index 5747656..7a5f71d 100644
--- a/debugging/debugging.tex
+++ b/debugging/debugging.tex
@@ -1116,7 +1116,7 @@ is reassuring.
\QuickQuiz{}
In Equation~\ref{eq:debugging:Binomial Number of Tests Required},
- are the logarithms base-10, base-2, or base-$e$?
+ are the logarithms base-10, base-2, or base-$\euler$?
\QuickQuizAnswer{
It does not matter.
You will get the same answer no matter what base of logarithms
@@ -1201,7 +1201,7 @@ The fundamental formula for failure probabilities is the Poisson
distribution:
\begin{equation}
- F_m = \frac{\lambda^m}{m!} e^{-\lambda}
+ F_m = \frac{\lambda^m}{m!} \euler^{-\lambda}
\label{eq:debugging:Poisson Probability}
\end{equation}
@@ -1225,14 +1225,14 @@ In this case, $\lambda$ is zero, so that
Equation~\ref{eq:debugging:Poisson Probability} reduces to:
\begin{equation}
- F_0 = e^{-\lambda}
+ F_0 = \euler^{-\lambda}
\end{equation}
Solving this requires setting $F_0$
to 0.01 and solving for $\lambda$, resulting in:
\begin{equation}
- \lambda = - \log 0.01 = 4.6
+ \lambda = - \ln 0.01 = 4.6
\end{equation}
Because we get $0.3$ failures per hour, the number of hours required
@@ -1246,11 +1246,11 @@ is a good and sufficient substitute for the Poisson distribution in
a great many situations.
More generally, if we have $n$ failures per unit time, and we want to
-be P\,\% certain that a fix reduced the failure rate, we can use the
+be $P$\,\% certain that a fix reduced the failure rate, we can use the
following formula:
\begin{equation}
- T = - \frac{1}{n} \log \frac{100 - P}{100}
+ T = - \frac{1}{n} \ln \frac{100 - P}{100}
\label{eq:debugging:Error-Free Test Duration}
\end{equation}
@@ -1287,14 +1287,14 @@ Equation~\ref{eq:debugging:Poisson Probability} as follows:
\begin{equation}
F_0 + F_1 + \dots + F_{m - 1} + F_m =
- \sum_{i=0}^m \frac{\lambda^i}{i!} e^{-\lambda}
+ \sum_{i=0}^m \frac{\lambda^i}{i!} \euler^{-\lambda}
\end{equation}
This is the Poisson cumulative distribution function, which can be
written more compactly as:
\begin{equation}
- F_{i \le m} = \sum_{i=0}^m \frac{\lambda^i}{i!} e^{-\lambda}
+ F_{i \le m} = \sum_{i=0}^m \frac{\lambda^i}{i!} \euler^{-\lambda}
\label{eq:debugging:Possion CDF}
\end{equation}
@@ -1341,18 +1341,18 @@ that the fix actually had some relationship to the bug.\footnote{
Indeed it should.
And it does.
- To see this, note that $e^{-\lambda}$ does not depend on $i$,
+ To see this, note that $\euler^{-\lambda}$ does not depend on $i$,
which means that it can be pulled out of the summation as follows:
\begin{equation}
- e^{-\lambda} \sum_{i=0}^\infty \frac{\lambda^i}{i!}
+ \euler^{-\lambda} \sum_{i=0}^\infty \frac{\lambda^i}{i!}
\end{equation}
The remaining summation is exactly the Taylor series for
- $e^\lambda$, yielding:
+ $\euler^\lambda$, yielding:
\begin{equation}
- e^{-\lambda} e^\lambda
+ \euler^{-\lambda} \euler^\lambda
\end{equation}
The two exponentials are reciprocals, and therefore cancel,
diff --git a/perfbook.tex b/perfbook.tex
index 84e48eb..da9cfa8 100644
--- a/perfbook.tex
+++ b/perfbook.tex
@@ -137,6 +137,8 @@
\newcommand{\nf}[1]{\textnormal{#1}} % to return to normal font
\newcommand{\qop}[1]{{\sffamily #1}} % QC operator such as H, T, S, etc.
+\DeclareRobustCommand{\euler}{\ensuremath{\mathrm{e}}}
+
\newcommand{\Epigraph}[2]{\epigraphhead[65]{\rmfamily\epigraph{#1}{#2}}}
\input{ushyphex} % Hyphenation exceptions for US English from hyphenex package
--
2.7.4
--
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