Английская Википедия:Characterizations of the exponential function
In mathematics, the exponential function can be characterized in many ways. The following characterizations (definitions) are most common. This article discusses why each characterization makes sense, and why they are all equivalent to each other. As a special case of these considerations, it will be demonstrated that the three most common definitions for the mathematical constant e are equivalent to each other.
Characterizations
The six most common definitions of the exponential function <math>\exp(x)=e^x</math> for real values <math>x\in \mathbb{R}</math> are as follows.
- Product limit. Define <math>e^x</math> by the limit:<math display="block">e^x = \lim_{n\to\infty} \left(1+\frac x n \right)^n.</math>
- Power series. Define Шаблон:Math as the value of the infinite series <math display="block">e^x = \sum_{n=0}^\infty {x^n \over n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots</math> (Here Шаблон:Math denotes the factorial of Шаблон:Mvar. One [[proof that e is irrational|proof that Шаблон:Math is irrational]] uses a special case of this formula.)
- Inverse of logarithm integral. Define <math>e^x</math> to be the unique number Шаблон:Math such that <math display="block">\int_1^y \frac{dt}{t} = x.</math> That is, <math>e^x</math> is the inverse of the natural logarithm function <math>x=\ln(y)</math>, which is defined by this integral.
- Differential equation. Define <math>y(x)=e^x</math> to be the unique solution to the differential equation with initial value:<math display="block">y' = y,\quad y(0) = 1,</math> where <math>y'=\tfrac{dy}{dx}</math> denotes the derivative of Шаблон:Mvar.
- Functional equation. The exponential function <math>e^x</math> is the unique function Шаблон:Math with <math>f(x+y)=f(x)f(y)</math> for all <math>x,y</math> and <math>f'(0)=1</math>. The condition <math>f'(0)=1</math> can be replaced with <math>f(1)=e</math> together with any of the following regularity conditions:Шаблон:Unordered list For the uniqueness, one must impose some regularity condition, since other functions satisfying <math>f(x+y)=f(x)f(y)</math> can be constructed using a basis for the real numbers over the rationals, as described by Hewitt and Stromberg.
- Elementary definition by powers. Define the exponential function with base <math>a>0</math> to be the continuous function <math>a^x</math> whose value on integers <math>x=n</math> is given by repeated multipication or division of <math>a</math>, and whose value on rational numbers <math>x=n/m</math> is given by <math>a^{n/m} =\ \ \sqrt[m]{\vphantom{A^2}a^n}</math>. Then define <math>e^x</math> to be the exponential function whose base <math>a=e</math> is the unique positive real number satisfying: <math display="block">\lim_{h \to 0} \frac{e^h - 1}{h} = 1.</math>
Larger domains
One way of defining the exponential function over the complex numbers is to first define it for the domain of real numbers using one of the above characterizations, and then extend it as an analytic function, which is characterized by its values on any infinite domain set.
Also, characterisations (1), (2), and (4) for <math>e^x</math> apply directly for <math>x</math> a complex number. Definition (3) presents a problem because there are non-equivalent paths along which one could integrate; but the equation of (3) should hold for any such path modulo <math>2\pi</math>. As for definition (5), the additive property together with the complex derivative <math>f'(0) = 1</math> are sufficient to guarantee <math>f(x)=e^x</math>. However, the initial value condition <math>f(1)=e</math> together with the other regularity conditions are not sufficient. For example, for real x and y, the function<math display="block"> f(x + iy) = e^x(\cos(2y) + i\sin(2y)) = e^{x + 2iy} </math>satisfies the three listed regularity conditions in (5) but is not equal to <math>\exp(x+iy)</math>. A sufficient condition is that <math>f(1)=e</math> and that f is a conformal map at some point; or else the two initial values <math>f(1)=e</math> and <math display="inline"> f(i) = \cos(1) + i\sin(1) </math> together with the other regularity conditions.
One may also define the exponential on other domains, such as matrices and other algebras. Definitions (1), (2), and (4) all make sense for arbitrary Banach algebras.
Proof that each characterization makes sense
Some of these definitions require justification to demonstrate that they are well-defined. For example, when the value of the function is defined as the result of a limiting process (i.e. an infinite sequence or series), it must be demonstrated that such a limit always exists.
Characterization 1
The error of the product limit expression is described by:<math display="block">\left(1+\frac x n \right)^n=e^x \left(1-\frac{x^2}{2n}+\frac{x^3(8+3x)}{24n^2}+\cdots \right),</math> where the polynomial's degree (in x) in the term with denominator nk is 2k.
Characterization 2
Since <math display="block">\lim_{n\to\infty} \left|\frac{x^{n+1}/(n+1)!}{x^n/n!}\right|
= \lim_{n\to\infty} \left|\frac{x}{n+1}\right|
= 0 < 1.</math>
it follows from the ratio test that <math display="inline">\sum_{n=0}^\infty \frac{x^n}{n!}</math> converges for all x.
Characterization 3
Since the integrand is an integrable function of Шаблон:Mvar, the integral expression is well-defined. It must be shown that the function from <math>\mathbb{R}^+</math> to <math>\mathbb{R}</math> defined by <math display="block">x \mapsto \int_1^x \frac{dt}{t}</math> is a bijection. Since Шаблон:Math is positive for positive Шаблон:Mvar, this function is strictly increasing, hence injective. If the two integrals <math display="block">\begin{align} \int_1^\infty \frac{dt} t & = \infty \\[8pt] \int_1^0 \frac{dt} t & = -\infty \end{align}</math> hold, then it is surjective as well. Indeed, these integrals do hold; they follow from the integral test and the divergence of the harmonic series.
Characterization 6
The defnition depends on the unique positive real number <math>a=e</math> satisfying: <math display="block">\lim_{h \to 0} \frac{a^h - 1}{h} = 1.</math>This limit can be shown to exist for any <math>a</math>, and it defines a continuous increasing function <math>f(a)=\ln(a) </math> with <math>f(1)=0</math> and <math>\lim_{a\to\infty}f(a) = \infty </math>, so the Intermediate value theorem guarantees the existence of such a value <math>a=e</math>.
Equivalence of the characterizations
The following arguments demonstrate the equivalence of the above characterizations for the exponential function.
Characterization 1 ⇔ characterization 2
The following argument is adapted from Rudin, theorem 3.31, p. 63–65.
Let <math>x \geq 0</math> be a fixed non-negative real number. Define <math display="block">t_n=\left(1+\frac x n \right)^n,\qquad s_n = \sum_{k=0}^n\frac{x^k}{k!},\qquad e^x = \lim_{n\to\infty} s_n.</math>
By the binomial theorem, <math display="block">\begin{align} t_n & =\sum_{k=0}^n{n \choose k}\frac{x^k}{n^k}=1+x+\sum_{k=2}^n\frac{n(n-1)(n-2)\cdots(n-(k-1))x^k}{k!\,n^k} \\[8pt] & = 1+x+\frac{x^2}{2!}\left(1-\frac{1}{n}\right)+\frac{x^3}{3!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)+\cdots \\[8pt] & {}\qquad \cdots +\frac{x^n}{n!}\left(1-\frac{1}{n}\right)\cdots\left(1-\frac{n-1}{n}\right)\le s_n \end{align}</math> (using x ≥ 0 to obtain the final inequality) so that: <math display="block">\limsup_{n\to\infty}t_n \le \limsup_{n\to\infty}s_n = e^x</math> On must use lim sup because it is not known if tn converges.
For the other inequality, by the above expression for tn, if 2 ≤ m ≤ n, we have: <math display="block">1+x+\frac{x^2}{2!}\left(1-\frac{1}{n}\right)+\cdots+\frac{x^m}{m!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{m-1}{n}\right)\le t_n.</math>
Fix m, and let n approach infinity. Then <math display="block">s_m = 1+x+\frac{x^2}{2!}+\cdots+\frac{x^m}{m!} \le \liminf_{n\to\infty}\ t_n</math> (again, one must use lim inf because it is not known if tn converges). Now, take the above inequality, let m approach infinity, and put it together with the other inequality to obtain: <math display="block">\limsup_{n\to\infty}t_n \le e^x \le \liminf_{n\to\infty}t_n </math> so that <math display="block">\lim_{n\to\infty}t_n = e^x. </math>
This equivalence can be extended to the negative real numbers by noting <math display="inline">\left(1 - \frac r n \right)^n \left(1+\frac{r}{n}\right)^n = \left(1-\frac{r^2}{n^2}\right)^n </math> and taking the limit as n goes to infinity.
Characterization 1 ⇔ characterization 3
Here, the natural logarithm function is defined in terms of a definite integral as above. By the first part of fundamental theorem of calculus, <math display="block">\frac d {dx}\ln x=\frac{d}{dx} \int_1^x \frac1 t \,dt = \frac 1 x.</math>
Besides, <math display="inline">\ln 1 = \int_1^1 \frac{dt}{t} = 0</math>
Now, let x be any fixed real number, and let <math display="block">y=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n.</math>
Шаблон:Math, which implies that Шаблон:Math, where Шаблон:Math is in the sense of definition 3. We have <math display="block">\ln y=\ln\lim_{n\to\infty}\left(1+\frac{x}{n} \right)^n = \lim_{n\to\infty} \ln\left(1+\frac{x}{n}\right)^n.</math>
Here, the continuity of ln(y) is used, which follows from the continuity of 1/t: <math display="block">\ln y=\lim_{n\to\infty}n\ln \left(1+\frac{x}{n} \right) = \lim_{n\to\infty} \frac{x\ln\left(1+(x/n)\right)}{(x/n)}.</math>
Here, the result lnan = nlna has been used. This result can be established for n a natural number by induction, or using integration by substitution. (The extension to real powers must wait until ln and exp have been established as inverses of each other, so that ab can be defined for real b as eb lna.) <math display="block">=x\cdot\lim_{h\to 0}\frac{\ln\left(1+h\right)}{h} \quad \text{ where } h = \frac{x}{n}</math> <math display="block">=x\cdot\lim_{h\to 0}\frac{\ln\left(1+h\right)-\ln 1}{h}</math> <math display="block">=x\cdot\frac{d}{dt} \ln t \Bigg|_{t=1}</math> <math display="block">\!\, = x.</math>
Characterization 1 ⇔ characterization 4
Let <math>y(t) </math> denote the solution to the initial value problem <math>y' = y,\ y(0) = 1</math>. Applying the simplest form of Euler's method with increment <math>\Delta t = \frac{x}{n}</math> and sample points <math>t \ =\ 0,\ \Delta t, \ 2 \Delta t, \ldots, \ n \Delta t </math> gives the recursive formula:
<math>y(t+\Delta t) \ \approx \ y(t) + y'(t)\Delta t \ =\ y(t) + y(t)\Delta t \ =\ y(t)\,(1+\Delta t).</math>
This recursion is immediately solved to give the approximate value <math>y(x) = y(n\Delta t) \approx (1+\Delta t)^n</math>, and since Euler's Method is known to converge to the exact solution, we have:
<math>y(x) = \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n. </math>
Characterization 1 ⇔ characterization 5
The following proof is a simplified version of the one in Hewitt and Stromberg, exercise 18.46. First, one proves that measurability (or here, Lebesgue-integrability) implies continuity for a non-zero function <math>f(x)</math> satisfying <math>f(x+y)=f(x)f(y)</math>, and then one proves that continuity implies <math>f(x) = e^{kx}</math> for some k, and finally <math>f(1) = e</math> implies Шаблон:Math.
First, a few elementary properties from <math>f(x)</math> satisfying <math>f(x+y)=f(x)f(y)</math> are proven, and the assumption that <math>f(x)</math> is not identically zero:
- If <math>f(x)</math> is nonzero anywhere (say at x=y), then it is non-zero everywhere. Proof: <math>f(y) = f(x) f(y - x) \neq 0</math> implies <math>f(x) \neq 0</math>.
- <math>f(0)=1</math>. Proof: <math>f(x)= f(x+0) = f(x) f(0)</math> and <math>f(x)</math> is non-zero.
- <math>f(-x)=1/f(x)</math>. Proof: <math>1 = f(0)= f(x-x) = f(x) f(-x)</math>.
- If <math>f(x)</math> is continuous anywhere (say at x = y), then it is continuous everywhere. Proof: <math>f(x+\delta) - f(x) = f(x-y) [ f(y+\delta) - f(y)] \to 0</math> as <math>\delta \to 0</math> by continuity at y.
The second and third properties mean that it is sufficient to prove <math>f(x)=e^x</math> for positive x.
If <math>f(x)</math> is a Lebesgue-integrable function, then <math display="block">g(x) = \int_0^x f(x')\, dx'.</math>
It then follows that <math display="block">g(x+y)-g(x) = \int_x^{x+y} f(x')\, dx' = \int_0^y f(x+x')\, dx' = f(x) g(y). </math>
Since <math>f(x)</math> is nonzero, some Шаблон:Mvar can be chosen such that <math>g(y) \neq 0</math> and solve for <math>f(x)</math> in the above expression. Therefore: <math display="block">\begin{align} f(x+\delta)-f(x) & = \frac{[g(x+\delta+y)-g(x+\delta)]-[g(x+y)-g(x)]}{g(y)} \\ & =\frac{[g(x+y+\delta)-g(x+y)]-[g(x+\delta)-g(x)]}{g(y)} \\ & =\frac{f(x+y)g(\delta)-f(x)g(\delta)}{g(y)}=g(\delta)\frac{f(x+y)-f(x)}{g(y)}. \end{align}</math>
The final expression must go to zero as <math>\delta \to 0</math> since <math>g(0)=0</math> and <math>g(x)</math> is continuous. It follows that <math>f(x)</math> is continuous.
Now, <math>f(q) = e^{kq}</math> can be proven, for some k, for all positive rational numbers q. Let q=n/m for positive integers n and m. Then <math display="block">f\left(\frac{n}{m}\right)=f\left(\frac{1}{m}+\cdots+\frac{1}{m} \right)=f\left(\frac{1}{m}\right)^n</math> by elementary induction on n. Therefore, <math>f(1/m)^m = f(1)</math> and thus <math display="block">f\left(\frac{n}{m}\right)=f(1)^{n/m}=e^{k(n/m)}.</math> for <math>k = \ln [f(1)]</math>. If restricted to real-valued <math>f(x)</math>, then <math>f(x) = f(x/2)^2</math> is everywhere positive and so k is real.
Finally, by continuity, since <math>f(x) = e^{kx}</math> for all rational x, it must be true for all real x since the closure of the rationals is the reals (that is, any real x can be written as the limit of a sequence of rationals). If <math>f(1) = e</math> then k = 1. This is equivalent to characterization 1 (or 2, or 3), depending on which equivalent definition of e one uses.
Characterization 2 ⇔ characterization 4
Let n be a non-negative integer. In the sense of definition 4 and by induction, <math>\frac{d^ny}{dx^n}=y</math>.
Therefore <math>\frac{d^ny}{dx^n}\Bigg|_{x=0}=y(0)=1.</math>
Using Taylor series, <math display="block">y= \sum_{n=0}^\infty \frac {f^{(n)}(0)}{n!} \, x^n = \sum_{n=0}^\infty \frac {1}{n!} \, x^n = \sum_{n=0}^\infty \frac {x^n}{n!}.</math> This shows that definition 4 implies definition 2.
In the sense of definition 2, <math display="block">\begin{align} \frac{d}{dx}e^x & = \frac{d}{dx} \left(1+\sum_{n=1}^\infty \frac {x^n}{n!} \right) = \sum_{n=1}^\infty \frac {nx^{n-1}}{n!} =\sum_{n=1}^\infty \frac {x^{n-1}}{(n-1)!} \\[6pt] & =\sum_{k=0}^\infty \frac {x^k}{k!}, \text{ where } k=n-1 \\[6pt] & =e^x \end{align}</math>
Besides, <math display="inline">e^0 = 1 + 0 + \frac{0^2}{2!} + \frac{0^3}{3!} + \cdots = 1.</math> This shows that definition 2 implies definition 4.
Characterization 2 ⇒ characterization 5
In the sense of definition 2, the equation <math>\exp(x+y)= \exp(x)\exp(y)</math> follows from the term-by-term manipulation of power series justified by uniform convergence, and the resulting equality of coefficients is just the Binomial theorem. Furthermore:[1] <math display="block">\begin{align} \exp'(0) & = \lim_{h\to 0} \frac{e^h-1}{h} \\
& =\lim_{h\to 0} \frac{1}{h} \left (\left (1+h+ \frac{h^2}{2!}+\frac{h^3}{3!}+\frac{h^4}{4!}+\cdots \right) -1 \right) \\
& =\lim_{h\to 0} \left(1+ \frac{h}{2!}+\frac{h^2}{3!}+\frac{h^3}{4!}+\cdots \right) \ =\ 1.\\
\end{align}</math>
Characterization 3 ⇔ characterization 4
Characterisation 3 involves defining the natural logarithm before the exponential function is defined. First, <math display="block">\log x := \int_{1}^{x} \frac{dt}{t}</math> This means that the natural logarithm of <math>x</math> equals the (signed) area under the graph of <math>1/t</math> between <math>t = 1</math> and <math>t=x</math>. If <math>x<1</math>, then this area is taken to be negative. Then, <math>\exp</math> is defined as the inverse of <math>\log</math>, meaning that <math display="block">\exp(\log(x))=x \text{ and } \log(\exp(x))=x</math> by the definition of an inverse function. If <math>a</math> is a positive real number then <math>a^x</math> is defined as <math>\exp(x\log(a))</math>. Finally, <math>e</math> is defined as the number <math>a</math> such that <math>\log(a)=1</math>. It can then be shown that <math>e^x=\exp(x)</math>: <math display="block">e^x=\exp(x\log(e))=\exp(x)</math> By the fundamental theorem of calculus, the derivative of <math display="inline">\log x = \frac{1}{x}</math>. We are now in a position to prove that <math display="inline">\frac{d}{dx} e^x=e^x</math>, satisfying the first part of the initial value problem given in characterisation 4: <math display="block">\begin{align} \text{Let }y&=e^x=\exp(x) \\ \log(y)&=\log(\exp(x))=x \\ \frac{1}{y}\frac{dy}{dx}&=1 \\ \frac{dy}{dx}&=y=e^x \end{align}</math> Then, we merely have to note that <math>e^0=\exp(0)=1</math>, and we are done. Of course, it is much easier to show that characterisation 4 implies characterisation 3. If <math>e^x</math> is the unique function <math>f:\mathbb{R}\to\mathbb{R}</math> satisfying <math>f'(x)=e^x</math>, and <math>f(0)=1</math>, then <math>\log</math> can be defined as its inverse. The derivative of <math>\log</math> can be found in the following way: <math display="block">y = \log x \implies x=e^y</math> If we differentiate both sides with respect to <math>y</math>, we get <math display="block">\begin{align} \frac{dx}{dy} &= e^y \\ \frac{dy}{dx} &= \frac{1}{e^y} = \frac{1}{x} \end{align}</math> Therefore, <math display="block">\int_{1}^{x}\frac{1}{t}dt=\left[\log t\right]_{1}^{x} = \log x - \log 1 = \log x - 0 = \log x</math>
Characterization 5 ⇒ characterization 4
The conditions Шаблон:Math and Шаблон:Math imply both conditions in characterization 4. Indeed, one gets the initial condition Шаблон:Math by dividing both sides of the equation <math display="block">f(0) = f(0 + 0) = f(0) f(0)</math> by Шаблон:Math, and the condition that Шаблон:Math follows from the condition that Шаблон:Math and the definition of the derivative as follows: <math display="block"> \begin{array}{rcccccc} f'(x) & = & \lim\limits_{h\to 0}\frac{f(x+h)-f(x)} h
& = & \lim\limits_{h\to 0}\frac{f(x)f(h)-f(x)} h
& = & \lim\limits_{h\to 0}f(x)\frac{f(h)-1} h
\\[1em]
& = & f(x)\lim\limits_{h\to 0}\frac{f(h)-1} h
& = & f(x)\lim\limits_{h\to 0}\frac{f(0+h)-f(0)} h
& = & f(x)f'(0) = f(x).
\end{array} </math>
Characterization 5 ⇒ characterization 4
In the sense of definition 5, the multiplicative property together with the initial condition <math>\exp'(0)= 1 </math> imply that: <math display="block">\begin{array}{rcl} \frac{d}{dx}\exp(x) &=& \lim_{h \to 0} \frac{\exp(x{+}h)-\exp(x)}{h}\\ & = & \exp(x) \cdot \lim_{h \to 0}\frac{\exp(h)-1}{h}\\ & = & \exp(x) \exp'(0) =\exp(x) . \end{array}</math>
Characterization 5 ⇔ characterization 6
The multiplicative property <math>f(x+y)=f(x)f(y)</math> of definition 5 implies that <math>f(0)=1</math>, and that <math>f(x)=a^x</math> according to the multiplication/division and root definition of exponentiation for rational <math>x=n/m</math> in definition 6, where <math>a=f(1)</math>. Then the condition <math>f'(0)=1</math> means that <math>\lim_{h\to 0}\tfrac{a^h-1}{h}=1</math>. Also any of the conditions of definition 5 imply that <math>f(x)</math> is continuous at all real <math>x</math>. The converse is similar.
References
- Walter Rudin, Principles of Mathematical Analysis, 3rd edition (McGraw–Hill, 1976), chapter 8.
- Edwin Hewitt and Karl Stromberg, Real and Abstract Analysis (Springer, 1965).
