logarithm series

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

Integrating both from 0 to $x$ gives

 $\displaystyle\ln(1\!+\!x)\;=\;x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4% }+-\ldots\qquad\mbox{for}\;\;-1 (1)

which is valid on the whole open interval of convergence$-1  of this power series and in for  $x=1$, as one may prove.

Replacing $x$ with $-x$ in (1) yields the series

 $\displaystyle\ln(1\!-\!x)\;=\;-x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{% 4}-\ldots\qquad\mbox{for}\;\;-1 (2)

Subtracting (2) from (1) gives

 $\displaystyle\ln\frac{1\!+\!x}{1\!-\!x}\;=\;2\left(x+\frac{x^{3}}{3}+\frac{x^{% 5}}{5}+\frac{x^{7}}{7}+\ldots\right)$ (3)

which also is true for  $-1.  Here the inner function of the logarithm attains all positive real values when  $0 (its graph (http://planetmath.org/Graph2) is a hyperbola (http://planetmath.org/Hyperbola2) with asymptotes (http://planetmath.org/AsymptotesOfGraphOfRationalFunction)  $x=1$  and  $y=-1$).  Thus, in principle, the series (3) can be used for calculating any values of natural logarithm (http://planetmath.org/NaturalLogarithm2).  For this purpose, one could denote

 $\frac{1\!+\!x}{1\!-\!x}\;:=\;t,$

which implies

 $x\;=\;\frac{t\!-\!1}{t\!+\!1},$

and accordingly

 $\displaystyle\ln{t}\;=\;2\left[\frac{t\!-\!1}{t\!+\!1}+\frac{1}{3}\left(\frac{% t\!-\!1}{t\!+\!1}\right)^{3}+\frac{1}{5}\left(\frac{t\!-\!1}{t\!+\!1}\right)^{% 5}+\ldots\right]\!.$ (4)

For example,

 $\ln{3}\;=\;2\left(\frac{1}{2}+\frac{1}{3\cdot 2^{3}}+\frac{1}{5\cdot 2^{5}}+% \ldots\right)\!.$

The convergence of (4) is the slower the greater is $t$.

Title logarithm series LogarithmSeries 2013-03-22 18:56:13 2013-03-22 18:56:13 pahio (2872) pahio (2872) 8 pahio (2872) Topic msc 33B10 TaylorSeriesOfArcusSine TaylorSeriesOfArcusTangent