Editing Natural logarithm (section)

== Series ==

[[File:LogTay.svg|290px|thumb|right|The Taylor polynomials for {{math|ln(1 + ''x'')}} only provide accurate approximations in the range {{math|−1 < ''x'' ≤ 1}}. Beyond some {{math|''x'' > 1}}, the Taylor polynomials of higher degree are increasingly ''worse'' approximations.]]

Since the natural logarithm is undefined at 0, <math>\ln(x)</math> itself does not have a [[Maclaurin series]], unlike many other elementary functions. Instead, one looks for Taylor expansions around other points. For example, if <math>\vert x - 1 \vert \leq 1 \text{ and } x \neq 0, </math> then<ref>{{cite web| url = http://www.math2.org/math/expansion/log.htm| title = "Logarithmic Expansions" at Math2.org}}</ref>
<math display="block">\begin{align}
   \ln x &= \int_1^x \frac{1}{t} \, dt = \int_0^{x - 1} \frac{1}{1 + u} \, du \\
   &= \int_0^{x - 1} (1 - u + u^2 - u^3 + \cdots) \, du \\
   &= (x - 1) - \frac{(x - 1)^2}{2} + \frac{(x - 1)^3}{3} - \frac{(x - 1)^4}{4} + \cdots \\
   &= \sum_{k=1}^\infty \frac{(-1)^{k-1} (x-1)^k}{k}.
 \end{align}</math>

This is the [[Taylor series]] for <math>\ln x</math> around 1. A change of variables yields the [[Mercator series]]:
<math display="block">\ln(1+x)=\sum_{k=1}^\infty \frac{(-1)^{k-1}}{k} x^k = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots,</math>
valid for <math>|x| \leq 1</math> and <math>x\ne -1.</math>

[[Leonhard Euler]],<ref>[[Leonhard Euler]], Introductio in Analysin Infinitorum. Tomus Primus. Bousquet, Lausanne 1748. Exemplum 1, p. 228; quoque in: Opera Omnia, Series Prima, Opera Mathematica, Volumen Octavum, Teubner 1922</ref> disregarding <math>x\ne -1</math>, nevertheless applied this series to <math>x=-1</math> to show that the [[harmonic series (mathematics)|harmonic series]] equals the natural logarithm of <math>\frac{1}{1-1}</math>; that is, the logarithm of infinity. Nowadays, more formally, one can prove that the harmonic series truncated at {{mvar|N}} is close to the logarithm of {{mvar|N}}, when {{mvar|N}} is large, with the difference converging to the [[Euler–Mascheroni constant]].

The figure is a [[Graph of a function|graph]] of {{math|ln(1 + ''x'')}} and some of its [[Taylor polynomial]]s around 0. These approximations converge to the function only in the region {{math|−1 < ''x'' ≤ 1}}; outside this region, the higher-degree Taylor polynomials devolve to ''worse'' approximations for the function.

A useful special case for positive integers {{mvar|n}}, taking <math>x = \tfrac{1}{n}</math>, is:
<math display="block"> \ln \left(\frac{n + 1}{n}\right) = \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k n^k}  = \frac{1}{n} - \frac{1}{2 n^2} + \frac{1}{3 n^3} - \frac{1}{4 n^4} + \cdots</math>

If <math>\operatorname{Re}(x) \ge 1/2,</math> then
<math display="block">\begin{align}
   \ln (x) &= - \ln \left(\frac{1}{x}\right) = - \sum_{k=1}^\infty \frac{(-1)^{k-1} (\frac{1}{x} - 1)^k}{k} = \sum_{k=1}^\infty \frac{(x - 1)^k}{k x^k} \\
   &= \frac{x - 1}{x} + \frac{(x - 1)^2}{2 x^2} +  \frac{(x - 1)^3}{3 x^3} + \frac{(x - 1)^4}{4 x^4} + \cdots
 \end{align}</math>

Now, taking  <math>x=\tfrac{n+1}{n}</math> for positive integers {{mvar|n}}, we get:
<math display="block"> \ln \left(\frac{n + 1}{n}\right) = \sum_{k=1}^\infty \frac{1}{k (n + 1)^k} = \frac{1}{n + 1} + \frac{1}{2 (n + 1)^2} + \frac{1}{3 (n + 1)^3} + \frac{1}{4 (n + 1)^4} + \cdots</math>

If <math>\operatorname{Re}(x) \ge 0 \text{ and } x \neq 0,</math> then
<math display="block"> \ln (x) = \ln \left(\frac{2x}{2}\right) = \ln\left(\frac{1 + \frac{x - 1}{x + 1}}{1 - \frac{x - 1}{x + 1}}\right) = \ln \left(1 + \frac{x - 1}{x + 1}\right) - \ln \left(1 - \frac{x - 1}{x + 1}\right). </math>
Since
<math display="block">\begin{align}
\ln(1+y) - \ln(1-y)&= \sum^\infty_{i=1}\frac{1}{i}\left((-1)^{i-1}y^i - (-1)^{i-1}(-y)^i\right) = \sum^\infty_{i=1}\frac{y^i}{i}\left((-1)^{i-1} +1\right)  \\
&= y\sum^\infty_{i=1}\frac{y^{i-1}}{i}\left((-1)^{i-1} +1\right)\overset{i-1\to 2k}{=}\; 2y\sum^\infty_{k=0}\frac{y^{2k}}{2k+1},
\end{align}</math> 
we arrive at
<math display="block">\begin{align}
   \ln (x) &= \frac{2(x - 1)}{x + 1} \sum_{k = 0}^\infty \frac{1}{2k + 1} {\left(\frac{(x - 1)^2}{(x + 1)^2}\right)}^k \\
&= \frac{2(x - 1)}{x + 1} \left( \frac{1}{1} + \frac{1}{3} \frac{(x - 1)^2}{(x + 1)^2} + \frac{1}{5} {\left(\frac{(x - 1)^2}{(x + 1)^2}\right)}^2 + \cdots \right) .
 \end{align}</math>
Using the substitution <math>x=\tfrac{n+1}{n}</math> again for positive integers {{mvar|n}}, we get: 
<math display="block">\begin{align}
   \ln \left(\frac{n + 1}{n}\right) &= \frac{2}{2n + 1} \sum_{k=0}^\infty \frac{1}{(2k + 1) ((2n + 1)^2)^k}\\
&= 2 \left(\frac{1}{2n + 1} + \frac{1}{3 (2n + 1)^3} + \frac{1}{5 (2n + 1)^5} + \cdots \right).
\end{align}</math>

This is, by far, the fastest converging of the series described here.

The natural logarithm can also be expressed as an infinite product:<ref>{{cite web |last1=RUFFA |first1=Anthony |title=A PROCEDURE FOR GENERATING INFINITE SERIES IDENTITIES |url=https://www.emis.de/journals/HOA/IJMMS/2004/65-683653.pdf |website=International Journal of Mathematics and Mathematical Sciences |access-date=27 February 2022}} (Page 3654, equation 2.6)</ref>
<math display="block">\ln(x)=(x-1) \prod_{k=1}^\infty \left ( \frac{2}{1+\sqrt[2^k]{x}} \right )</math>

Two examples might be:
<math display="block">\ln(2)=\left ( \frac{2}{1+\sqrt{2}} \right )\left ( \frac{2}{1+\sqrt[4]{2}} \right )\left ( \frac{2}{1+\sqrt[8]{2}} \right )\left ( \frac{2}{1+\sqrt[16]{2}} \right )...</math>
<math display="block">\pi=(2i+2)\left ( \frac{2}{1+\sqrt{i}} \right )\left ( \frac{2}{1+\sqrt[4]{i}} \right )\left ( \frac{2}{1+\sqrt[8]{i}} \right )\left ( \frac{2}{1+\sqrt[16]{i}} \right )...</math>

From this identity, we can easily get that:
<math display="block">\frac{1}{\ln(x)}=\frac{x}{x-1}-\sum_{k=1}^\infty\frac{2^{-k}x^{2^{-k}}}{1+x^{2^{-k}}}</math>

For example:
<math display="block">\frac{1}{\ln(2)} = 2-\frac{\sqrt{2}}{2+2\sqrt{2}}-\frac{\sqrt[4]{2}}{4+4\sqrt[4]{2}}-\frac{\sqrt[8]{2}}{8+8\sqrt[8]{2}}  \cdots</math>