Editing Exponential function (section)

==General exponential functions==

A function is commonly called ''an exponential function''{{mdash}}with an indefinite article{{mdash}}if it has the form {{tmath|x \mapsto b^x}}, that is, if it is obtained from [[exponentiation]] by fixing the base and letting the ''exponent'' vary.

More generally and especially in applied contexts, the term ''exponential function'' is commonly used for functions of the form {{tmath|1=f(x) = ab^x}}. This may be motivated by the fact that, if the values of the function represent [[quantities]], a change of [[measurement unit]] changes the value of {{tmath|a}}, and so, it is nonsensical to impose {{tmath|1=a=1}}.

These most general exponential functions are the [[differentiable function]]s that satisfy the following equivalent characterizations.
* {{tmath|1=f(x) = ab^x}} for every {{tmath|x}} and some constants {{tmath|a}} and {{tmath|b>0}}.
* {{tmath|1=f(x)=ae^{kx} }} for every {{tmath|x}} and some constants {{tmath|a}} and {{tmath|k}}.
* The value of <math>f'(x)/f(x)</math>  is independent of <math>x</math>.
* For every <math>d,</math> the value of <math>f(x+d)/f(x)</math> is independent of <math>x;</math> that is, <math display=block>\frac{f(x+d)}{f(x)}= \frac{f(y+d)}{f(y)}</math> for every {{mvar|x}}, {{mvar|y}}.<ref>G. Harnett, ''Calculus 1'', 1998, Functions continued:
"General exponential functions have the property that the ratio of two outputs depends only on the difference of inputs. The ratio of outputs for a unit change in input is the base."</ref>

[[Image:Exponenciala priklad.png|thumb|200px|right|Exponential functions with bases 2 and 1/2]]
The ''base'' of an exponential function  is the ''base'' of the [[exponentiation]] that appears in it when written as {{tmath|x\to ab^x}}, namely {{tmath|b}}.<ref>G. Harnett, ''Calculus 1'', 1998; Functions continued / Exponentials & logarithms: "The ratio of outputs for a unit change in input is the ''base'' of a general exponential function."</ref> The base is {{tmath|e^k}} in the second characterization, <math display=inline>\exp \frac{f'(x)}{f(x)}</math> in the third one, and <math display=inline>\left(\frac{f(x+d)}{f(x)}\right)^{1/d}</math> in the last one.

===In applications===
The last characterization is important in [[empirical science]]s, as allowing a direct [[experimental]] test whether a function is an exponential function. 

Exponential [[exponential growth|growth]] or [[exponential decay]]{{mdash}}where the variable change is [[proportionality (mathematics)|proportional]] to the variable value{{mdash}}are thus modeled with exponential functions. Examples are unlimited population growth leading to [[Malthusian catastrophe]], [[compound interest#Continuous compounding|continuously compounded interest]], and [[radioactive decay]].

If the modeling function has the form {{tmath|x\mapsto ae^{kx},}} or, equivalently, is a solution of the differential equation {{tmath|1=y'=ky}}, the constant {{tmath|k}} is called, depending on the context, the ''decay constant'', ''disintegration constant'',<ref name="Serway-Moses-Moyer_1989" /> ''rate constant'',<ref name="Simmons_1972" /> or ''transformation constant''.<ref name="McGrawHill_2007" />

===Equivalence proof===
For proving the equivalence of the above properties, one can proceed as follows.

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 ths last condititon

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
implies that
<math display=block>\frac \partial{\partial x}\,\frac{f(x)}{e^{kx}}=0,</math> and thus that there is a constant {{tmath|a}} such that <math>f(x)=ae^{kx}.</math> 

If the last condition is verified, let <math display=inline>\varphi(d)=f(x+d)/f(x),</math> which is independent of {{tmath|x}}. Using {{tmath|1=\varphi (0)=1}}, one gets
<math display=block>\frac{f(x+d)-f(x)}{d} = f(x)\,\frac{\varphi(d)-\varphi(0)}{d}. </math>
Taking the limit when {{tmath|d}} tends to zero, one gets that the third condition is verified with {{tmath|1=k=\varphi'(0)}}. It follows therefore that {{tmath|1=f(x)= ae^{kx} }} for some {{tmath|a,}} and {{tmath|1=\varphi(d)= e^{kd}.}} As a byproduct, one gets that 
<math display=block>\left(\frac{f(x+d)}{f(x)}\right)^{1/d}=e^k</math>
is independent of both {{tmath|x}} and {{tmath|d}}.