Editing Exponential function (section)

=={{anchor|exp|expm1}}Computation==
The Taylor series definition above is generally efficient for computing (an approximation of) <math>e^x</math>. However, when computing near the argument <math>x=0</math>, the result will be close to 1, and computing the value of the difference <math>e^x-1</math> with [[floating-point arithmetic]] may lead to the loss of (possibly all) [[significant figures]], producing a large relative error, possibly even a meaningless result.

Following a proposal by [[William Kahan]], it may thus be useful to have a dedicated routine, often called <code>expm1</code>, which computes  {{math|''e<sup>x</sup>'' − 1}} directly, bypassing computation of {{math|''e''{{isup|''x''}}}}. For example,
one may use the Taylor series:
<math display="block">e^x-1=x+\frac {x^2}2 + \frac{x^3}6+\cdots +\frac{x^n}{n!}+\cdots.</math>

This was first implemented in 1979 in the [[Hewlett-Packard]] [[HP-41C]] calculator, and provided by several calculators,<ref name="HP48_AUR"/><ref name="HP50_AUR"/> [[operating system]]s (for example [[Berkeley UNIX 4.3BSD]]<ref name="Beebe_2017"/>), [[computer algebra system]]s, and programming languages (for example [[C99]]).<ref name="Beebe_2002"/>

In addition to base {{math|''e''}}, the [[IEEE 754-2008]] standard defines similar exponential functions near 0 for base 2 and 10: <math>2^x - 1</math> and <math>10^x - 1</math>.

A similar approach has been used for the logarithm; see [[log1p]].

An identity in terms of the [[hyperbolic tangent]],
<math display="block">\operatorname{expm1} (x) = e^x - 1 = \frac{2 \tanh(x/2)}{1 - \tanh(x/2)},</math>
gives a high-precision value for small values of {{math|''x''}} on systems that do not implement {{math|expm1(''x'')}}.

===Continued fractions===
The exponential function can also be computed with [[continued fraction]]s.

A continued fraction for {{math|''e''{{isup|''x''}}}} can be obtained via [[Euler's continued fraction formula|an identity of Euler]]:
<math display="block"> e^x = 1 + \cfrac{x}{1 - \cfrac{x}{x + 2 - \cfrac{2x}{x + 3 - \cfrac{3x}{x + 4 - \ddots}}}}</math>

The following [[generalized continued fraction]] for {{math|''e''{{isup|''z''}}}} converges more quickly:<ref name="Lorentzen_2008"/>
<math display="block"> e^z = 1 + \cfrac{2z}{2 - z + \cfrac{z^2}{6 + \cfrac{z^2}{10 + \cfrac{z^2}{14 + \ddots}}}}</math>

or, by applying the substitution {{math|1=''z'' = {{sfrac|''x''|''y''}}}}:
<math display="block">  e^\frac{x}{y} = 1 + \cfrac{2x}{2y - x + \cfrac{x^2} {6y + \cfrac{x^2} {10y + \cfrac{x^2} {14y + \ddots}}}}</math>
with a special case for {{math|1=''z'' = 2}}:
<math display="block">  e^2 = 1 + \cfrac{4}{0 + \cfrac{2^2}{6 + \cfrac{2^2}{10 + \cfrac{2^2}{14 + \ddots }}}} = 7 + \cfrac{2}{5 + \cfrac{1}{7 + \cfrac{1}{9 + \cfrac{1}{11 + \ddots }}}}</math>

This formula also converges, though more slowly, for {{math|''z'' > 2}}. For example:
<math display="block"> e^3 = 1 + \cfrac{6}{-1 + \cfrac{3^2}{6 + \cfrac{3^2}{10 + \cfrac{3^2}{14 + \ddots }}}} = 13 + \cfrac{54}{7 + \cfrac{9}{14 + \cfrac{9}{18 + \cfrac{9}{22 + \ddots }}}}</math>