Editing Logarithm (section)

===Power series===
====Taylor series====

[[File:Taylor approximation of natural logarithm.gif|right|thumb|The Taylor series of {{math|ln(''z'')}} centered at {{math|''z'' {{=}} 1}}. The animation shows the first&nbsp;10 approximations along with the 99th and 100th. The approximations do not converge beyond a distance of 1 from the center.|alt=An animation showing increasingly good approximations of the logarithm graph.]]
For any real number {{mvar|z}} that satisfies {{math|0 < ''z'' ≤ 2}}, the following formula holds:{{refn|The same series holds for the principal value of the complex logarithm for complex numbers {{mvar|z}} satisfying {{math|{{!}}''z'' − 1{{!}} < 1}}.|group=nb}}<ref name=AbramowitzStegunp.68>{{Harvard citations|editor1-last=Abramowitz|editor2-last=Stegun|year=1972 |nb=yes|loc=p. 68}}</ref>

<math display="block">
\begin{align}\ln (z)  &= \frac{(z-1)^1}{1} - \frac{(z-1)^2}{2} + \frac{(z-1)^3}{3} - \frac{(z-1)^4}{4} + \cdots \\
&= \sum_{k=1}^\infty (-1)^{k+1}\frac{(z-1)^k}{k}.
\end{align}
</math>

Equating the function {{math|ln(''z'')}} to this infinite sum ([[series (mathematics)|series]]) is shorthand for saying that the function can be approximated to a more and more accurate value by the following expressions (known as [[partial sum]]s):

<math display=block>
(z-1),\ \ (z-1) - \frac{(z-1)^2}{2},\ \ (z-1) - \frac{(z-1)^2}{2} + \frac{(z-1)^3}{3},\ \ldots
</math>

For example, with {{math|''z'' {{=}} 1.5}} the third approximation yields {{math|0.4167}}, which is about {{math|0.011}} greater than {{math|ln(1.5) {{=}} 0.405465}}, and the ninth approximation yields {{math|0.40553}}, which is only about {{math|0.0001}} greater. The {{mvar|n}}th partial sum can approximate {{math|ln(''z'')}} with arbitrary precision, provided the number of summands {{mvar|n}} is large enough.

In elementary calculus, the series is said to [[convergent series|converge]] to the function {{math|ln(''z'')}}, and the function is the [[limit (mathematics)|limit]] of the series. It is the [[Taylor series]] of the [[natural logarithm]] at {{math|1=''z'' = 1}}. The Taylor series of {{math|ln(''z'')}} provides a particularly useful approximation to {{math|ln(1 + ''z'')}} when {{mvar|z}} is small, {{math|{{!}}''z''{{!}} < 1}}, since then
<math display="block">
\ln (1+z) = z - \frac{z^2}{2}  +\frac{z^3}{3} -\cdots \approx z.
</math>

For example, with {{math|1=''z'' = 0.1}} the first-order approximation gives {{math|ln(1.1) ≈ 0.1}}, which is less than {{math|5%}} off the correct value {{math|0.0953}}.

====Inverse hyperbolic tangent====
Another series is based on the [[area hyperbolic tangent#Inverse hyperbolic tangent|inverse hyperbolic tangent]] function:
<math display="block">
\ln (z) = 2\cdot\operatorname{artanh}\,\frac{z-1}{z+1} = 2 \left ( \frac{z-1}{z+1} + \frac{1}{3}{\left(\frac{z-1}{z+1}\right)}^3 + \frac{1}{5}{\left(\frac{z-1}{z+1}\right)}^5 + \cdots \right ),
</math>
for any real number {{math|''z'' > 0}}.{{refn|The same series holds for the principal value of the complex logarithm for complex numbers {{mvar|z}} with positive real part.|group=nb}}<ref name=AbramowitzStegunp.68 /> Using [[sigma notation]], this is also written as
<math display="block">\ln (z) = 2\sum_{k=0}^\infty\frac{1}{2k+1}\left(\frac{z-1}{z+1}\right)^{2k+1}.</math>
This series can be derived from the above Taylor series. It converges quicker than the Taylor series, especially if {{mvar|z}} is close to 1. For example, for {{math|1=''z'' = 1.5}}, the first three terms of the second series approximate {{math|ln(1.5)}} with an error of about {{val|3|e=-6}}. The quick convergence for {{mvar|z}} close to 1 can be taken advantage of in the following way: given a low-accuracy approximation {{math|''y'' ≈ ln(''z'')}} and putting
<math display="block">A = \frac z{\exp(y)},</math>
the logarithm of {{mvar|z}} is:
<math display="block">\ln (z)=y+\ln (A).</math>
The better the initial approximation {{mvar|y}} is, the closer {{mvar|A}} is to 1, so its logarithm can be calculated efficiently. {{mvar|A}} can be calculated using the [[exponential function|exponential series]], which converges quickly provided {{mvar|y}} is not too large. Calculating the logarithm of larger {{mvar|z}} can be reduced to smaller values of {{mvar|z}} by writing {{math|''z'' {{=}} ''a'' · 10<sup>''b''</sup>}}, so that {{math|ln(''z'') {{=}} ln(''a'') + {{mvar|b}} · ln(10)}}.

A closely related method can be used to compute the logarithm of integers. Putting <math>\textstyle z=\frac{n+1}{n}</math> in the above series, it follows that:
<math display="block">\ln (n+1) = \ln(n) + 2\sum_{k=0}^\infty\frac{1}{2k+1}\left(\frac{1}{2 n+1}\right)^{2k+1}.</math>
If the logarithm of a large integer&nbsp;{{mvar|n}} is known, then this series yields a fast converging series for {{math|log(''n''+1)}}, with a [[rate of convergence]] of <math display="inline">\left(\frac{1}{2 n+1}\right)^{2}</math>.