Exponential function: Difference between revisions

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In [[mathematics]], the '''exponential function''' is the unique [[real function]] which maps [[0|zero]] to [[1|one]] and has a [[derivative (mathematics)|derivative]] everywhere equal to its value. The exponential of a variable {{tmath|x}} is denoted {{tmath|\exp x}} or {{tmath|e^x}}, with the two notations used interchangeably. It is called ''exponential'' because its argument can be seen as an [[exponent (mathematics)|exponent]] to which a constant [[e (mathematical constant)|number {{math|''e'' ≈ 2.718}}]], the base, is raised. There are several other definitions of the exponential function, which are all equivalent although being of very different nature.

The exponential function converts sums to products: it maps the [[additive identity]] {{math|0}} to the [[multiplicative identity]] {{math|1}}, and the exponential of a sum is equal to the product of separate exponentials, {{tmath|1=\exp(x + y) = \exp x \cdot \exp y }}. Its [[inverse function]], the [[natural logarithm]], {{tmath|\ln}} or {{tmath|\log}}, converts products to sums: {{tmath|1= \ln(x\cdot y) = \ln x + \ln y}}.

The exponential function is occasionally called the '''natural exponential function''', matching the name ''natural logarithm'', for distinguishing it from some other functions that are also commonly called ''exponential functions''. These functions include the functions of the form {{tmath|1=f(x) = b^x}}, which is [[exponentiation]] with a fixed base {{tmath|b}}. More generally, and especially in applications, functions of the general form {{tmath|1=f(x) = ab^x}} are also called exponential functions. They [[exponential growth|grow]] or [[exponential decay|decay]] exponentially in that the amountrate that {{tmath|f(x)}} changes when {{tmath|x}} is increased is ''proportional'' to the current value of {{tmath|f(x)}}.

The exponential function can be generalized to accept [[complex number]]s as arguments. This reveals relations between multiplication of complex numbers, rotations in the [[complex plane]], and [[trigonometry]]. [[Euler's formula]] {{tmath|1= \exp i\theta = \cos\theta + i\sin\theta}} expresses and summarizes these relations.

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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>