From 6053d367c8d626d81885fba6cd439c10a7f9901f Mon Sep 17 00:00:00 2001 From: Richard Kreckel Date: Fri, 21 Jan 2000 13:55:04 +0000 Subject: [PATCH] - Little addition about something unmentioned but ovious: x^ax^b=x^(a+b) --- doc/powerlaws.tex | 40 +++++++++++++++++++++------------------- 1 file changed, 21 insertions(+), 19 deletions(-) diff --git a/doc/powerlaws.tex b/doc/powerlaws.tex index 75ce8c92..1943957a 100644 --- a/doc/powerlaws.tex +++ b/doc/powerlaws.tex @@ -7,7 +7,7 @@ \subsection{Definitions} Definitions for power and log: - +\label{powerdef} x^a \equiv e^{a \ln x} @@ -19,18 +19,20 @@ x^a \equiv e^{a \ln x} e^x e^y = e^{x+y} -for arbitrary complex $x$ and $y$ +for arbitrary complex $$x$$ and $$y$$ (with~(\ref{powerdef}) we obtain +the rule $$x^ax^b=x^{a+b}$$ since $$x^ax^b\equiv e^{a\ln x}e^{b\ln x} = +e^{(a+b)\ln x}\equiv x^{a+b}$$ for arbitrary complex $$a,b,x$$) $$x^{-a} = \frac{1}{x^a}$$ -for arbitrary complex $x$ and $a$ +for arbitrary complex $$x$$ and $$a$$ -\subsection{$(ax)^b=a^b x^b$} +\subsection{$$(ax)^b=a^b x^b$$} -\subsubsection{$b$ integer, $x$ and $a$ arbitrary complex} +\subsubsection{$$b$$ integer, $$x$$ and $$a$$ arbitrary complex} -assume $b>0$ +assume $$b>0$$ \begin{eqnarray} (ax)^b & = & \underbrace{(ax) \cdots (ax)}_{b \times} @@ -41,7 +43,7 @@ assume $b>0$ & = & a^b x^b \mbox{ q.e.d.} \end{eqnarray} -if $b<0$ (so $b=-|b|$) +if $$b<0$$ (so $$b=-|b|$$) \begin{eqnarray} (ax)^b & = & \frac{1}{(ax)^{|b|}} \nonumber\\ @@ -52,7 +54,7 @@ if $b<0$ (so $b=-|b|$) & = & a^b x^b \end{eqnarray} -\subsubsection{$a>0$, $x$ and $b$ arbitrary complex} +\subsubsection{$$a>0$$, $$x$$ and $$b$$ arbitrary complex} \begin{eqnarray} (ax)^b & = & e^{b \ln(ax)} @@ -60,7 +62,7 @@ if $b<0$ (so $b=-|b|$) & = & e^{b (\ln |ax| + i \arg(ax))} \end{eqnarray} -if $a$ is real and positive: +if $$a$$ is real and positive: $$\ln |ax| = \ln |a| + \ln |x| = \ln a + \ln |x|$$ @@ -81,11 +83,11 @@ e^{b (\ln a + \ln |x| + i \arg(x))} & = & a^b x^b \mbox{ q.e.d.} \end{eqnarray} -\subsection{$(x^a)^b = x^{ab}$} +\subsection{$$(x^a)^b = x^{ab}$$} -\subsubsection{$b$ integer, $x$ and $a$ arbitrary complex} +\subsubsection{$$b$$ integer, $$x$$ and $$a$$ arbitrary complex} -assume $b>0$ +assume $$b>0$$ \begin{eqnarray} (x^a)^b & = & \underbrace{(x^a) \cdots (x^a)}_{b \times} @@ -99,7 +101,7 @@ assume $b>0$ & = & x^{ab} \mbox{ q.e.d.} \end{eqnarray} -if $b<0$ (so $b=-|b|$) +if $$b<0$$ (so $$b=-|b|$$) \begin{eqnarray} (x^a)^b & = & \frac{1}{(x^a)^{|b|}} \nonumber\\ @@ -110,13 +112,13 @@ if $b<0$ (so $b=-|b|$) & = & x^{ab} \end{eqnarray} -\subsubsection{$-1 < a \le 1$, $x$ and $b$ arbitrary complex} +\subsubsection{$$-1 < a \le 1$$, $$x$$ and $$b$$ arbitrary complex} We have $$x^a=e^{a \ln|x| + ia\arg(x)}$$ -if $a$ is real +if $$a$$ is real $$|x^a|=e^{a\ln|x|}$$ @@ -124,19 +126,19 @@ and $$\arg(x^a)-a\arg(x)=2k\pi$$ -now if $-1 < a \le 1$, then $-\pi < a\arg(x) \le \pi$, -and so $k=0$, i.e. +now if $$-1 < a \le 1$$, then $$-\pi < a\arg(x) \le \pi$$, +and so $$k=0$$, i.e. $$\arg(x^a)=a\arg(x)$$ -(Note that for $a=-1$ this may not be true, as $-1 \arg(x)$ may be equal to $-\pi$.) +(Note that for $$a=-1$$ this may not be true, as $$-1 \arg(x)$$ may be equal to $$-\pi$$.) So \begin{eqnarray} \ln(x^a) & = & \ln|x^a| + i\arg(x^a) \nonumber\\ & = & \ln (e^{a\ln|x|})+ia\arg(x) \nonumber\\ -& = & a \ln |x| + ia\arg(x) \mbox{ (because $a\ln|x|$ is real)} +& = & a \ln |x| + ia\arg(x) \mbox{ (because $$a\ln|x|$$ is real)} \nonumber\\ & = & a\ln x \end{eqnarray} -- 2.35.3