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<record version="5" id="11791">
 <title>logarithm series</title>
 <name>LogarithmSeries</name>
 <created>2009-05-17 18:35:22</created>
 <modified>2009-09-30 20:44:04</modified>
 <type>Topic</type>
<parent id="2666">natural logarithm</parent>
 <creator id="2872" name="pahio"/>
 <author id="2872" name="pahio"/>
 <classification>
	<category scheme="msc" code="33B10"/>
 </classification>
 <related>
	<object name="TaylorSeriesOfArcusSine"/>
	<object name="TaylorSeriesOfArcusTangent"/>
 </related>
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 <content>The derivative of\, $\ln(1\!+\!x)$\, is $\displaystyle\frac{1}{1\!+\!x}$, which can be represented as the sum of geometric series:
$$\frac{1}{1\!+\!x} \;=\; 1-x+x^2-x^3+-\ldots \qquad\mbox{for}\;\; -1 &lt; x &lt; 1.$$
Integrating both \PMlinkescapetext{sides} from 0 to $x$ gives
\begin{align}
\ln(1\!+\!x) \;=\; x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+-\ldots \qquad\mbox{for}\;\; -1 &lt; x &lt; 1.
\end{align}
which is valid on the whole open interval of convergence \,$-1 &lt; x &lt; 1$\, of this power series and in 
\PMlinkescapetext{addition} for\, $x = 1$, as one may prove.

Replacing $x$ with $-x$ in (1) yields the series
\begin{align}
\ln(1\!-\!x) \;=\; -x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\ldots \qquad\mbox{for}\;\; -1 &lt; x &lt; 1.
\end{align}
Subtracting (2) from (1) gives
\begin{align}
\ln\frac{1\!+\!x}{1\!-\!x} \;=\; 2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\ldots\right)
\end{align}
which also is true for\, $-1 &lt; x &lt; 1$.\, Here the inner function of the logarithm attains all positive real values when\, $0 &lt; x &lt; 1$ (its \PMlinkname{graph}{Graph2} is a \PMlinkname{hyperbola}{Hyperbola2} with \PMlinkname{asymptotes}{AsymptotesOfGraphOfRationalFunction} \,$x = 1$\, and\, $y = -1$).\, Thus, in principle, the series (3) can be used for calculating any values of \PMlinkname{natural logarithm}{NaturalLogarithm2}.\, For this purpose, one could denote
$$\frac{1\!+\!x}{1\!-\!x} \;:=\; t,$$
which implies
$$x \;=\;\frac{t\!-\!1}{t\!+\!1},$$
and accordingly
\begin{align}
\ln{t} \;=\; 2\left[\frac{t\!-\!1}{t\!+\!1}
            +\frac{1}{3}\left(\frac{t\!-\!1}{t\!+\!1}\right)^3
            +\frac{1}{5}\left(\frac{t\!-\!1}{t\!+\!1}\right)^5+\ldots\right]\!.
\end{align}
For example, 
$$\ln{3} \;=\; 2\left(\frac{1}{2}+\frac{1}{3\cdot2^3}+\frac{1}{5\cdot2^5}+\ldots\right)\!.$$
The convergence of (4) is the slower the greater is $t$.


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