Characterizations of the exponential function#Characterizations

The six most common definitions of the exponential function $\backslash exp(x)=e^x$ for real values $x\backslash in\; \backslash mathbb\{R\}$ are as follows.

1. Product limit. Define $e^x$ by the limit:$$e^x\; =\; \backslash lim\_\{n\backslash to\backslash infty\}\; \backslash left(1+\backslash frac\; x\; n\; \backslash right)^n.$$
2. Power series. Define {{math|e^x}} as the value of the infinite series $$e^x\; =\; \backslash sum\_\{n=0\}^\backslash infty\; \{x^n\; \backslash over\; n!\}\; =\; 1\; +\; x\; +\; \backslash frac\{x^2\}\{2!\}\; +\; \backslash frac\{x^3\}\{3!\}\; +\; \backslash frac\{x^4\}\{4!\}\; +\; \backslash cdots$$ (Here {{math|n!}} denotes the factorial of {{mvar|n}}. One proof that e is irrational uses a special case of this formula.)
3. Inverse of logarithm integral. Define $e^x$ to be the unique number {{math|y > 0}} such that $$\backslash int\_1^y\; \backslash frac\{dt\}\{t\}\; =\; x.$$ That is, $e^x$ is the inverse of the natural logarithm function $x=\backslash ln(y)$, which is defined by this integral.
4. Differential equation. Define $y(x)=e^x$ to be the unique solution to the differential equation with initial value:$$y\text{'}\; =\; y,\backslash quad\; y(0)\; =\; 1,$$ where $y\text{'}=\backslash tfrac\{dy\}\{dx\}$ denotes the derivative of {{mvar|y}}.
5. Functional equation. The exponential function $e^x$ is the unique function {{math|f}} with the multiplicative property $f(x+y)=f(x)f(y)$ for all $x,y$ and $f\text{'}(0)=1$. The condition $f\text{'}(0)=1$ can be replaced with $f(1)=e$ together with any of the following regularity conditions:{{unordered list

| {{math|f }} is Lebesgue-measurable (Hewitt and Stromberg, 1965, exercise 18.46).

| {{math|f }} is continuous at any one point (Rudin, 1976, chapter 8, exercise 6).

| {{math|f }} is increasing on any interval. }} For the uniqueness, one must impose some regularity condition, since other functions satisfying $f(x+y)=f(x)f(y)$ can be constructed using a basis for the real numbers over the rationals, as described by Hewitt and Stromberg.

1. Elementary definition by powers. Define the exponential function with base $a>0$ to be the continuous function $a^x$ whose value on integers $x=n$ is given by repeated multiplication or division of $a$, and whose value on rational numbers $x=n/m$ is given by $a^\{n/m\}\; =\backslash \; \backslash \; \backslash sqrt[m]\{\backslash vphantom\{A^2\}a^n\}$. Then define $e^x$ to be the exponential function whose base $a=e$ is the unique positive real number satisfying: $$\backslash lim\_\{h\; \backslash to\; 0\}\; \backslash frac\{e^h\; -\; 1\}\{h\}\; =\; 1.$$

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 $e^x$ apply directly for $x$ 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 $2\backslash pi\; i$. As for definition (5), the additive property together with the complex derivative $f\text{'}(0)\; =\; 1$ are sufficient to guarantee $f(x)=e^x$. However, the initial value condition $f(1)=e$ together with the other regularity conditions are not sufficient. For example, for real x and y, the function$$f(x\; +\; iy)\; =\; e^x(\backslash cos(2y)\; +\; i\backslash sin(2y))\; =\; e^\{x\; +\; 2iy\}$$satisfies the three listed regularity conditions in (5) but is not equal to $\backslash exp(x+iy)$. A sufficient condition is that $f(1)=e$ and that $f$ is a conformal map at some point; or else the two initial values $f(1)=e$ and $f(i)\; =\; \backslash cos(1)\; +\; i\backslash sin(1)$ 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.

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.

The error of the product limit expression is described by:$$\backslash left(1+\backslash frac\; x\; n\; \backslash right)^n=e^x\; \backslash left(1-\backslash frac\{x^2\}\{2n\}+\backslash frac\{x^3(8+3x)\}\{24n^2\}+\backslash cdots\; \backslash right),$$

where the polynomial's degree (in x) in the term with denominator n^k is 2k.

Since

$$\backslash lim\_\{n\backslash to\backslash infty\}\; \backslash left|\backslash frac\{x^\{n+1\}/(n+1)!\}\{x^n/n!\}\backslash right|$$

= \lim_{n\to\infty} \left|\frac{x}{n+1}\right|

= 0 < 1.

it follows from the ratio test that $\backslash sum\_\{n=0\}^\backslash infty\; \backslash frac\{x^n\}\{n!\}$ converges for all x.

Since the integrand is an integrable function of {{mvar|t}}, the integral expression is well-defined. It must be shown that the function from $\backslash mathbb\{R\}^+$ to $\backslash mathbb\{R\}$ defined by

$$x\; \backslash mapsto\; \backslash int\_1^x\; \backslash frac\{dt\}\{t\}$$

is a bijection. Since {{math|1/t}} is positive for positive {{mvar|t}}, this function is strictly increasing, hence injective. If the two integrals

$$\backslash begin\{align\}$$

\int_1^\infty \frac{dt} t & = \infty \\[8pt]

\int_1^0 \frac{dt} t & = -\infty

\end{align}

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.

The definition depends on the unique positive real number $a=e$ satisfying: $$\backslash lim\_\{h\; \backslash to\; 0\}\; \backslash frac\{a^h\; -\; 1\}\{h\}\; =\; 1.$$This limit can be shown to exist for any $a$, and it defines a continuous increasing function $f(a)=\backslash ln(a)$ with $f(1)=0$ and $\backslash lim\_\{a\backslash to\backslash infty\}f(a)\; =\; \backslash infty$, so the Intermediate value theorem guarantees the existence of such a value $a=e$.

The following arguments demonstrate the equivalence of the above characterizations for the exponential function.

The following argument is adapted from Rudin, theorem 3.31, p. 63–65.

Let $x\; \backslash geq\; 0$ be a fixed non-negative real number. Define

$$t\_n=\backslash left(1+\backslash frac\; x\; n\; \backslash right)^n,\backslash qquad\; s\_n\; =\; \backslash sum\_\{k=0\}^n\backslash frac\{x^k\}\{k!\},\backslash qquad$$

e^x = \lim_{n\to\infty} s_n.

By the binomial theorem,

$$\backslash begin\{align\}$$

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}

(using x ≥ 0 to obtain the final inequality) so that:

$$\backslash limsup\_\{n\backslash to\backslash infty\}t\_n\; \backslash le\; \backslash limsup\_\{n\backslash to\backslash infty\}s\_n\; =\; e^x$$

One must use lim sup because it is not known if t_n converges.

For the other inequality, by the above expression for t_n, if 2 ≤ m ≤ n, we have:

$$1+x+\backslash frac\{x^2\}\{2!\}\backslash left(1-\backslash frac\{1\}\{n\}\backslash right)+\backslash cdots+\backslash frac\{x^m\}\{m!\}\backslash left(1-\backslash frac\{1\}\{n\}\backslash right)\backslash left(1-\backslash frac\{2\}\{n\}\backslash right)\backslash cdots\backslash left(1-\backslash frac\{m-1\}\{n\}\backslash right)\backslash le\; t\_n.$$

Fix m, and let n approach infinity. Then

$$s\_m\; =\; 1+x+\backslash frac\{x^2\}\{2!\}+\backslash cdots+\backslash frac\{x^m\}\{m!\}\; \backslash le\; \backslash liminf\_\{n\backslash to\backslash infty\}\backslash \; t\_n$$

(again, one must use lim inf because it is not known if t_n converges). Now, take the above inequality, let m approach infinity, and put it together with the other inequality to obtain:

$$\backslash limsup\_\{n\backslash to\backslash infty\}t\_n\; \backslash le\; e^x\; \backslash le\; \backslash liminf\_\{n\backslash to\backslash infty\}t\_n$$

so that

$$\backslash lim\_\{n\backslash to\backslash infty\}t\_n\; =\; e^x.$$

This equivalence can be extended to the negative real numbers by noting $\backslash left(1\; -\; \backslash frac\; r\; n\; \backslash right)^n\; \backslash left(1+\backslash frac\{r\}\{n\}\backslash right)^n\; =\; \backslash left(1-\backslash frac\{r^2\}\{n^2\}\backslash right)^n$ and taking the limit as n goes to infinity.

Here, the natural logarithm function is defined in terms of a definite integral as above. By the first part of fundamental theorem of calculus,

$$\backslash frac\; d\; \{dx\}\backslash ln\; x=\backslash frac\{d\}\{dx\}\; \backslash int\_1^x\; \backslash frac1\; t\; \backslash ,dt\; =\; \backslash frac\; 1\; x.$$

Besides, $\backslash ln\; 1\; =\; \backslash int\_1^1\; \backslash frac\{dt\}\{t\}\; =\; 0$

Now, let x be any fixed real number, and let

$$y=\backslash lim\_\{n\backslash to\backslash infty\}\backslash left(1+\backslash frac\{x\}\{n\}\backslash right)^n.$$

{{math|1=Ln(y) = x}}, which implies that {{math|1=y = e^x}}, where {{math|1=e^x}} is in the sense of definition 3. We have

$$\backslash ln\; y=\backslash ln\backslash lim\_\{n\backslash to\backslash infty\}\backslash left(1+\backslash frac\{x\}\{n\}\; \backslash right)^n\; =\; \backslash lim\_\{n\backslash to\backslash infty\}\; \backslash ln\backslash left(1+\backslash frac\{x\}\{n\}\backslash right)^n.$$

Here, the continuity of ln(y) is used, which follows from the continuity of 1/t:

$$\backslash ln\; y=\backslash lim\_\{n\backslash to\backslash infty\}n\backslash ln\; \backslash left(1+\backslash frac\{x\}\{n\}\; \backslash right)\; =\; \backslash lim\_\{n\backslash to\backslash infty\}\; \backslash frac\{x\backslash ln\backslash left(1+(x/n)\backslash right)\}\{(x/n)\}.$$

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 a^b can be defined for real b as e^{b lna}.)

$$=x\backslash cdot\backslash lim\_\{h\backslash to\; 0\}\backslash frac\{\backslash ln\backslash left(1+h\backslash right)\}\{h\}\; \backslash quad\; \backslash text\{\; where\; \}\; h\; =\; \backslash frac\{x\}\{n\}$$

$$=x\backslash cdot\backslash lim\_\{h\backslash to\; 0\}\backslash frac\{\backslash ln\backslash left(1+h\backslash right)-\backslash ln\; 1\}\{h\}$$

$$=x\backslash cdot\backslash frac\{d\}\{dt\}\; \backslash ln\; t\; \backslash Bigg|\_\{t=1\}$$

$$\backslash !\backslash ,\; =\; x.$$

Let $y(t)$ denote the solution to the initial value problem $y\text{'}\; =\; y,\backslash \; y(0)\; =\; 1$. Applying the simplest form of Euler's method with increment $\backslash Delta\; t\; =\; \backslash frac\{x\}\{n\}$ and sample points $t\; \backslash \; =\backslash \; 0,\backslash \; \backslash Delta\; t,\; \backslash \; 2\; \backslash Delta\; t,\; \backslash ldots,\; \backslash \; n\; \backslash Delta\; t$ gives the recursive formula:

$y(t+\backslash Delta\; t)\; \backslash \; \backslash approx\; \backslash \; y(t)\; +\; y\text{'}(t)\backslash Delta\; t\; \backslash \; =\backslash \; y(t)\; +\; y(t)\backslash Delta\; t\; \backslash \; =\backslash \; y(t)\backslash ,(1+\backslash Delta\; t).$

$y(x)\; =\; \backslash lim\_\{n\backslash to\backslash infty\}\backslash left(1+\backslash frac\{x\}\{n\}\backslash right)^n.$

Let n be a non-negative integer. In the sense of definition 4 and by induction, $\backslash frac\{d^ny\}\{dx^n\}=y$.

Therefore $\backslash frac\{d^ny\}\{dx^n\}\backslash Bigg|\_\{x=0\}=y(0)=1.$

Using Taylor series,

$$y=\; \backslash sum\_\{n=0\}^\backslash infty\; \backslash frac\; \{f^\{(n)\}(0)\}\{n!\}\; \backslash ,\; x^n\; =\; \backslash sum\_\{n=0\}^\backslash infty\; \backslash frac\; \{1\}\{n!\}\; \backslash ,\; x^n\; =\; \backslash sum\_\{n=0\}^\backslash infty\; \backslash frac\; \{x^n\}\{n!\}.$$ This shows that definition 4 implies definition 2.

In the sense of definition 2,

$$\backslash begin\{align\}$$

\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}

Besides, $e^0\; =\; 1\; +\; 0\; +\; \backslash frac\{0^2\}\{2!\}\; +\; \backslash frac\{0^3\}\{3!\}\; +\; \backslash cdots\; =\; 1.$ This shows that definition 2 implies definition 4.

In the sense of definition 2, the equation $\backslash exp(x+y)=\; \backslash exp(x)\backslash exp(y)$ 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:{{Cite web |title=Herman Yeung - Calculus - First Principle find d/Dx(e^x) 基本原理求 d/Dx(e^x) |url=https://www.youtube.com/watch?v=HHenriJHNMM&index=30&list=PLzDe9mOi1K8o2lveHTSM04WAhaGEZE7xB |website=YouTube}}

$$\backslash begin\{align\}$$

\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}

Characterisation 3 first defines the natural logarithm:$$\backslash log\; x\; \backslash \; \backslash \; \backslash stackrel\{\backslash text\{def\}\}\{=\}\backslash \; \backslash int\_\{1\}^\{x\}\backslash !\; \backslash frac\{dt\}\{t\},$$then $\backslash exp$ as the inverse function with $x=\backslash log(\backslash exp\; x)$. Then by the Chain rule:

$1=\backslash frac\{d\}\{dx\}[\; \backslash log(\backslash exp(x))\; ]\; =\; \backslash log\text{'}(\backslash exp(x))\backslash cdot\; \backslash exp\text{'}(x)\; =\; \backslash frac\{\backslash exp\text{'}(x)\}\{\backslash exp(x)\},$

Conversely, assume $y=\backslash exp(x)$ has $\backslash exp\text{'}(x)=\backslash exp(x)$ and $\backslash exp(0)=1$, and define $\backslash log(x)$ as its inverse function with $x\; =\; \backslash exp(\backslash log\; x)$ and $\backslash log(1)\; =\; 0$. Then:

$1=\backslash frac\{d\}\{dx\}[\; \backslash exp(\backslash log(x))\; ]\; =\; \backslash exp\text{'}(\backslash log(x))\backslash cdot\; \backslash log\text{'}(x)$

= \exp(\log(x))\cdot \log'(x) = x\cdot \log'(x),

The conditions {{math|1=f'(0) = 1}} and {{math|1=f(x + y) = f(x) f(y)}} imply both conditions in characterization 4. Indeed, one gets the initial condition {{math|1=f(0) = 1}} by dividing both sides of the equation

$$f(0)\; =\; f(0\; +\; 0)\; =\; f(0)\; f(0)$$

by {{math|f(0)}}, and the condition that {{math|1=f′(x) = f(x)}} follows from the condition that {{math|1=f′(0) = 1}} and the definition of the derivative as follows:

\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}

Assum characterization 5, the multiplicative property together with the initial condition $\backslash exp\text{'}(0)=\; 1$ imply that: $$\backslash begin\{array\}\{rcl\}$$

\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}

By inductively applying the multiplication rule, we get:

$$f\backslash left(\backslash frac\{n\}\{m\}\backslash right)^m=f\backslash left(\backslash frac\{n\}\{m\}+\backslash cdots+\backslash frac\{n\}\{m\}\; \backslash right)=f(n)=f(1)^n,$$

and thus

$$f\backslash left(\backslash frac\{n\}\{m\}\backslash right)=\backslash sqrt[m]\{f(1)^n\}\backslash \; \backslash stackrel\{\backslash text\{def\}\}=\backslash \; a^\{n/m\}$$

for $a=f(1)$. Then the condition $f\text{'}(0)=1$ means that $\backslash lim\_\{h\backslash to\; 0\}\backslash tfrac\{a^h-1\}\{h\}=1$, so $a=e$ by definition.

Also, any of the regularity conditions of definition 5 imply that $f(x)$ is continuous at all real $x$ (see below). The converse is similar.

Let $f(x)$ be a Lebesgue-integrable non-zero function satisfying the mulitiplicative property $f(x+y)=f(x)f(y)$ with $f(1)\; =\; e$. Following Hewitt and Stromberg, exercise 18.46, we will prove that Lebesgue-integrability implies continuity. This is sufficient to imply $f(x)\; =\; e^x$ according to characterization 6, arguing as above.

First, a few elementary properties:

1. If $f(x)$ is nonzero anywhere (say at $x=y$), then it is non-zero everywhere. Proof: $f(y)\; =\; f(x)\; f(y\; -\; x)\; \backslash neq\; 0$ implies $f(x)\; \backslash neq\; 0$.
2. $f(0)=1$. Proof: $f(x)=\; f(x+0)\; =\; f(x)\; f(0)$ and $f(x)$ is non-zero.
3. $f(-x)=1/f(x)$. Proof: $1\; =\; f(0)=\; f(x-x)\; =\; f(x)\; f(-x)$.
4. If $f(x)$ is continuous anywhere (say at $x=y$), then it is continuous everywhere. Proof: $f(x+\backslash delta)\; -\; f(x)\; =\; f(x-y)\; [\; f(y+\backslash delta)\; -\; f(y)]\; \backslash to\; 0$ as $\backslash delta\; \backslash to\; 0$ by continuity at $y$.

The second and third properties mean that it is sufficient to prove $f(x)=e^x$ for positive x.

Since$f(x)$ is a Lebesgue-integrable function, then we may define $g(x)\; =\; \backslash int\_0^x\; f(t)\backslash ,\; dt$. It then follows that

$$g(x+y)-g(x)\; =\; \backslash int\_x^\{x+y\}\; f(t)\backslash ,\; dt\; =\; \backslash int\_0^y\; f(x+t)\backslash ,\; dt\; =\; f(x)\; g(y).$$

Since $f(x)$ is nonzero, some {{mvar|y}} can be chosen such that $g(y)\; \backslash neq\; 0$ and solve for $f(x)$ in the above expression. Therefore:

$$\backslash begin\{align\}$$

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}

The final expression must go to zero as $\backslash delta\; \backslash to\; 0$ since $g(0)=0$ and $g(x)$ is continuous. It follows that $f(x)$ is continuous.

{{Reflist | refs =

{{cite book|author=Eli Maor|title=e: the Story of a Number|page=156}}

{{cite book |title=Real and complex analysis |author=Walter Rudin |date=1987 |publisher=McGraw-Hill |isbn=978-0-07-054234-1 |edition=3rd |location=New York |page=1 |url=https://archive.org/details/RudinW.RealAndComplexAnalysisMcGrawHillInternationalEdition3ndEd.19871Ed.1966Pp.428}}

}}

• Walter Rudin, Principles of Mathematical Analysis, 3rd edition (McGraw–Hill, 1976), chapter 8.
• Edwin Hewitt and Karl Stromberg, Real and Abstract Analysis (Springer, 1965).

Category:Mathematical analysis

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