Exponential function: Difference between revisions

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&=\sum_{n=0}^\infty \frac{x^n}{n!},\end{align}</math>

[[Image:Exp series.gif|right|thumb|The exponential function (in blue), and the sum of the first {{math|''n'' + 1}} terms of its power series (in red)]]

where <math>n!</math> is the [[factorial]] of {{mvar|n}} (the product of the {{mvar|n}} first positive integers). This series is [[absolutely convergent]] for every <math>x</math> per the [[ratio test]]. So, the derivative of the sum can be computed by term-by-term derivationdifferentiation, and this shows that the sum of the series satisfies the above definition. This is a second existence proof, and shows, as a byproduct, that the exponential function is defined for every {{tmath|x}}, and is everywhere the sum of its [[Maclaurin series]].

===Functional equation===

''The exponential satisfies the [[functional equation]]:''

<math display=block>\exp(x+y)= \exp(x)\cdot \exp(y).</math>

This results from the uniqueness and the fact that the function

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===Limit of integer powers===

''The exponential function is the [[limit (mathematics)|limit]], as the integer {{mvar|n}} goes to infinity,<ref name="Maor"/><ref name=":0" />

<math display=block>\exp(x)=\lim_{n \to +\infty} \left(1+\frac xn\right)^n,.</math>

where <math>n</math> takes only integer values (otherwise, the exponentiation would require the exponential function to be defined). By continuity of the logarithm, this can be proved by taking logarithms and proving

<math display=block>x=\lim_{n\to\infty}\ln \left(1+\frac xn\right)^n= \lim_{n\to\infty}n\ln \left(1+\frac xn\right),</math>

for example with [[Taylor's theorem]].

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The two first characterizations are equivalent, since, if {{tmath|1=b=e^k}} and {{tmath|1= k=\ln b}}, one has

s.<math display=block>e^{kx}= (e^k)^x= b^x.</math>

The basic properties of the exponential function (derivative and functional equation) implies immediately the third and thsthe last condititoncondition.

Suppose that the third condition is verified, and let {{tmath|k}} be the constant value of <math>f'(x)/f(x).</math> Since <math display = inline>\frac {\partial e^{kx}}{\partial x}=ke^{kx},</math> the [[quotient rule]] for derivation

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==Complex exponential==

{{anchor|On the complex plane|Complex plane}}

[[File:The exponential function e^z plotted in the complex plane from -2-2i to 2+2i.svg|alt=The exponential function {{math|''e^''{{isup|''z''}}}} plotted in the complex plane from -{{math|−2 − 2-2i''i''}} to {{math|2 +2i 2''i''}}|thumb|The exponential function {{math|''e^''{{isup|''z''}}}} plotted in the complex plane from -{{math|−2 − 2-2i''i''}} to {{math|2 +2i 2''i''}}]]

[[Image:Exp-complex-cplot.svg|thumb|right|A [[Domain coloring|complex plot]] of <math>z\mapsto\exp z</math>, with the [[Argument (complex analysis)|argument]] <math>\operatorname{Arg}\exp z</math> represented by varying hue. The transition from dark to light colors shows that <math>\left|\exp z\right|</math> is increasing only to the right. The periodic horizontal bands corresponding to the same hue indicate that <math>z\mapsto\exp z</math> is [[periodic function|periodic]] in the [[imaginary part]] of <math>z</math>.]]

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<math display="block">e^z = \lim_{n\to\infty}\left(1+\frac{z}{n}\right)^n</math>

As with the real exponential function (see {{slink||Functional equation}} above), the complex exponential satisfies the functional equation

The functional equation

<math display="block">e^{w+\exp(z}+w)=e^we^ \exp(z)\cdot \exp(w).</math>

Among complex functions, it is the unique solution which is [[holomorphic]] at the point {{tmath|1= z = 0}} and takes the derivative {{tmath|1}} there.<ref>{{cite book |last=Hille |first=Einar |year=1959 |title=Analytic Function Theory |volume=1 |place=Waltham, MA |publisher=Blaisdell |chapter=The exponential function |at=§ 6.1, {{pgs|138–143}} }}</ref>

holds for every complex numbers {{tmath|w}} and {{tmath|z}}. The complex exponential is the unique [[continuous function]] that satisfies this functional equation and has the value {{tmath|1}} for {{tmath|1=z=0}}.

The [[complex logarithm]] is a [[left inverse function|right-inverse function ]] of the complex exponential:

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<math display="block">\overline{e^z}=e^{\overline z}.</math>

Its modulus is

<math display="block">|e^z|= e^{|\Re (z)|},</math>

where {{tmath|\Re(z)}} denotes the real part of {{tmath|z}}.

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<math display="block">e^{it} =\cos(t)+i\sin(t). </math>

This formula provides the decomposition of complex exponentialexponentials into [[real and imaginary parts]]:

<math display="block">e^{x+iy} = e^{x}e^{iy} = e^x\,\cos y + i e^x\,\sin y.</math>

The trigonometric functions can be expressed in terms of complex exponentials:

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<gallery caption="3D plots of real part, imaginary part, and modulus of the exponential function" class="center" mode="packed" style="text-align:left" heights="150px">

Image:ExponentialAbs_real_SVG.svg| {{math|1=''z'' = Re(''e''^{{{isup|''x'' + ''iy''}}})}}

Image:ExponentialAbs_image_SVG.svg| {{math|1=''z'' = Im(''e''^{{{isup|''x'' + ''iy''}}})}}

Image:ExponentialAbs_SVG.svg| {{math|1=''z'' = {{abs|''e''^{{{isup|''x'' + ''iy''}}}}}}}

</gallery>

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==Matrices and Banach algebras==

The power series definition of the exponential function makes sense for square [[matrix (mathematics)|matrices]] (for which the function is called the [[matrix exponential]]) and more generally in any unital [[Banach algebra]] {{math|''B''}}. In this setting, {{math|1=''e''^{{isup|0}} = 1}}, and {{math|''e''^{{isup|''x''}}}} is invertible with inverse {{math|''e''^{{{isup|−''x''}}}}} for any {{math|''x''}} in {{math|''B''}}. If {{math|1=''xy'' = ''yx''}}, then {{math|1=''e''^{{{isup|''x'' + ''y''}}} = ''e''^{{isup|''x''}}''e''^{{isup|''y''}}}}, but this identity can fail for noncommuting {{math|''x''}} and {{math|''y''}}.

Some alternative definitions lead to the same function. For instance, {{math|''e''^{{isup|''x''}}}} can be defined as

<math display="block">\lim_{n \to \infty} \left(1 + \frac{x}{n} \right)^n .</math>

Or {{math|''e''^{{isup|''x''}}}} can be defined as {{math|''f''_''x''(1)}}, where {{math|''f''_''x'' : '''R''' → ''B''}} is the solution to the differential equation {{math|1={{sfrac|''df''_''x''''|''dt''}}(''t'') = ''x{{space|hair}}f''_''x''(''t'')}}, with initial condition {{math|1=''f''_''x''(0) = 1}}; it follows that {{math|1=''f''_''x''(''t'') = ''e''^{{{isup|''tx''}}}}} for every {{mvar|t}} in {{math|'''R'''}}.

==Lie algebras==

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==Transcendency==

The function {{math|''e''^{{isup|''z''}}}} is a [[transcendental function]], which means that it is not a [[polynomial root|root]] of a polynomial over the [[ring (mathematics)|ring]] of the [[rational fraction]]s <math>\C(z).</math>

If {{math|''a''₁, ..., ''a''_''n''<nowiki/>}} are distinct complex numbers, then {{math|''e''^{''a''₁''z''}, ..., ''e''^{''a''_''n''''z''}<nowiki/>}} are linearly independent over <math>\C(z)</math>, and hence {{math|''e''^{{isup|''z''}}}} is [[transcendental function|transcendental]] over <math>\C(z)</math>.

=={{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^x'' − 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>

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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>