Characterizations of the exponential function

Since

$${\displaystyle \lim _{n\to \infty }\left|{\frac {x^{n+1}/(n+1)!}{x^{n}/n!}}\right|=\lim _{n\to \infty }\left|{\frac {x}{n+1}}\right|=0<1.}$$

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

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

$${\displaystyle x\mapsto \int _{1}^{x}{\frac {dt}{t}}}$$

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

$${\displaystyle {\begin{aligned}\int _{1}^{\infty }{\frac {dt}{t}}&=\infty \\[8pt]\int _{1}^{0}{\frac {dt}{t}}&=-\infty \end{aligned}}}$$

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 following argument is adapted from a proof in Rudin, theorem 3.31, p. 63–65.

Let ${\displaystyle x\geq 0}$ be a fixed non-negative real number. Define

$${\displaystyle s_{n}=\sum _{k=0}^{n}{\frac {x^{k}}{k!}},\ t_{n}=\left(1+{\frac {x}{n}}\right)^{n}.}$$

By the binomial theorem,

$${\displaystyle {\begin{aligned}t_{n}&=\sum _{k=0}^{n}{n \choose k}{\frac {x^{k}}{n^{k}}}=1+x+\sum _{k=2}^{n}{\frac {n(n-1)(n-2)\cdots (n-(k-1))x^{k}}{k!\,n^{k}}}\\[8pt]&=1+x+{\frac {x^{2}}{2!}}\left(1-{\frac {1}{n}}\right)+{\frac {x^{3}}{3!}}\left(1-{\frac {1}{n}}\right)\left(1-{\frac {2}{n}}\right)+\cdots \\[8pt]&{}\qquad \cdots +{\frac {x^{n}}{n!}}\left(1-{\frac {1}{n}}\right)\cdots \left(1-{\frac {n-1}{n}}\right)\leq s_{n}\end{aligned}}}$$

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

$${\displaystyle \limsup _{n\to \infty }t_{n}\leq \limsup _{n\to \infty }s_{n}=e^{x}}$$

where e^x is in the sense of definition 2. Here, limsups must be used, because it is not known if t_n converges. For the other direction, by the above expression of t_n, if 2 ≤ m ≤ n,

$${\displaystyle 1+x+{\frac {x^{2}}{2!}}\left(1-{\frac {1}{n}}\right)+\cdots +{\frac {x^{m}}{m!}}\left(1-{\frac {1}{n}}\right)\left(1-{\frac {2}{n}}\right)\cdots \left(1-{\frac {m-1}{n}}\right)\leq t_{n}.}$$

Fix m, and let n approach infinity. Then

$${\displaystyle s_{m}=1+x+{\frac {x^{2}}{2!}}+\cdots +{\frac {x^{m}}{m!}}\leq \liminf _{n\to \infty }t_{n}}$$

(again, liminf's must be used because it is not known if t_n converges). Now, taking the above inequality, letting m approach infinity, and putting it together with the other inequality, this becomes

$${\displaystyle \limsup _{n\to \infty }t_{n}\leq e^{x}\leq \liminf _{n\to \infty }t_{n}}$$

so that

$${\displaystyle \lim _{n\to \infty }t_{n}=e^{x}.}$$

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

The error term of this limit-expression is described by

$${\displaystyle \left(1+{\frac {x}{n}}\right)^{n}=e^{x}\left(1-{\frac {x^{2}}{2n}}+{\frac {x^{3}(8+3x)}{24n^{2}}}+\cdots \right),}$$

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

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

$${\displaystyle {\frac {d}{dx}}\ln x={\frac {d}{dx}}\int _{1}^{x}{\frac {1}{t}}\,dt={\frac {1}{x}}.}$$

Besides, ${\textstyle \ln 1=\int _{1}^{1}{\frac {dt}{t}}=0}$

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

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

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

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

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

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

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}.)

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

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

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

$${\displaystyle \!\,=x.}$$

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 ${\displaystyle f(x)}$ satisfying ${\displaystyle f(x+y)=f(x)f(y)}$, and then one proves that continuity implies ${\displaystyle f(x)=e^{kx}}$ for some k, and finally ${\displaystyle f(1)=e}$ implies k = 1.

First, a few elementary properties from ${\displaystyle f(x)}$ satisfying ${\displaystyle f(x+y)=f(x)f(y)}$ are proven, and the assumption that ${\displaystyle f(x)}$ is not identically zero:

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

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

If ${\displaystyle f(x)}$ is a Lebesgue-integrable function, then

$${\displaystyle g(x)=\int _{0}^{x}f(x')\,dx'.}$$

It then follows that

$${\displaystyle g(x+y)-g(x)=\int _{x}^{x+y}f(x')\,dx'=\int _{0}^{y}f(x+x')\,dx'=f(x)g(y).}$$

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

$${\displaystyle {\begin{aligned}f(x+\delta )-f(x)&={\frac {[g(x+\delta +y)-g(x+\delta )]-[g(x+y)-g(x)]}{g(y)}}\\&={\frac {[g(x+y+\delta )-g(x+y)]-[g(x+\delta )-g(x)]}{g(y)}}\\&={\frac {f(x+y)g(\delta )-f(x)g(\delta )}{g(y)}}=g(\delta ){\frac {f(x+y)-f(x)}{g(y)}}.\end{aligned}}}$$

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

Now, ${\displaystyle f(q)=e^{kq}}$ can be proven, for some k, for all positive rational numbers q. Let q=n/m for positive integers n and m. Then

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

by elementary induction on n. Therefore, ${\displaystyle f(1/m)^{m}=f(1)}$ and thus

$${\displaystyle f\left({\frac {n}{m}}\right)=f(1)^{n/m}=e^{k(n/m)}.}$$

for ${\displaystyle k=\ln[f(1)]}$. If restricted to real-valued ${\displaystyle f(x)}$, then ${\displaystyle f(x)=f(x/2)^{2}}$ is everywhere positive and so k is real.

Finally, by continuity, since ${\displaystyle f(x)=e^{kx}}$ 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 ${\displaystyle f(1)=e}$ then k = 1. This is equivalent to characterization 1 (or 2, or 3), depending on which equivalent definition of e one uses.

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

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

Using Taylor series,

$${\displaystyle y=\sum _{n=0}^{\infty }{\frac {f^{(n)}(0)}{n!}}\,x^{n}=\sum _{n=0}^{\infty }{\frac {1}{n!}}\,x^{n}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}.}$$

This shows that definition 4 implies definition 2.

In the sense of definition 2,

$${\displaystyle {\begin{aligned}{\frac {d}{dx}}e^{x}&={\frac {d}{dx}}\left(1+\sum _{n=1}^{\infty }{\frac {x^{n}}{n!}}\right)=\sum _{n=1}^{\infty }{\frac {nx^{n-1}}{n!}}=\sum _{n=1}^{\infty }{\frac {x^{n-1}}{(n-1)!}}\\[6pt]&=\sum _{k=0}^{\infty }{\frac {x^{k}}{k!}},{\text{ where }}k=n-1\\[6pt]&=e^{x}\end{aligned}}}$$

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

In the sense of definition 2,[1]

$${\displaystyle {\begin{aligned}\lim _{h\to 0}{\frac {e^{h}-1}{h}}&=\lim _{h\to 0}{\frac {1}{h}}\left(\left(1+h+{\frac {h^{2}}{2!}}+{\frac {h^{3}}{3!}}+{\frac {h^{4}}{4!}}+\cdots \right)-1\right)\\&=\lim _{h\to 0}\left(1+{\frac {h}{2!}}+{\frac {h^{2}}{3!}}+{\frac {h^{3}}{4!}}+\cdots \right)\\&=1\end{aligned}}}$$

Characterisation 3 involves defining the natural logarithm before the exponential function is defined. First,

$${\displaystyle \log x:=\int _{1}^{x}{\frac {dt}{t}}}$$

This means that the natural logarithm of ${\displaystyle x}$ equals the (signed) area under the graph of ${\displaystyle 1/t}$ between ${\displaystyle t=1}$ and ${\displaystyle t=x}$. If ${\displaystyle x<1}$, then this area is taken to be negative. Then, ${\displaystyle \exp }$ is defined as the inverse of ${\displaystyle \log }$, meaning that

$${\displaystyle \exp(\log(x))=x{\text{ and }}\log(\exp(x))=x}$$

by the definition of an inverse function. If ${\displaystyle a}$ is a positive real number then ${\displaystyle a^{x}}$ is defined as ${\displaystyle \exp(x\log(a))}$. Finally, ${\displaystyle e}$ is defined as the number ${\displaystyle a}$ such that ${\displaystyle \log(a)=1}$. It can then be shown that ${\displaystyle e^{x}=\exp(x)}$:

$${\displaystyle e^{x}=\exp(x\log(e))=\exp(x)}$$

By the fundamental theorem of calculus, the derivative of ${\textstyle \log x={\frac {1}{x}}}$. We are now in a position to prove that ${\textstyle {\frac {d}{dx}}e^{x}=e^{x}}$, satisfying the first part of the initial value problem given in characterisation 4:

$${\displaystyle {\begin{aligned}{\text{Let }}y&=e^{x}=\exp(x)\\\log(y)&=\log(\exp(x))=x\\{\frac {1}{y}}{\frac {dy}{dx}}&=1\\{\frac {dy}{dx}}&=y=e^{x}\end{aligned}}}$$

Then, we merely have to note that ${\displaystyle e^{0}=\exp(0)=1}$, and we are done. Of course, it is much easier to show that characterisation 4 implies characterisation 3. If ${\displaystyle e^{x}}$ is the unique function ${\displaystyle f:\mathbb {R} \to \mathbb {R} }$ satisfying ${\displaystyle f'(x)=e^{x}}$, and ${\displaystyle f(0)=1}$, then ${\displaystyle \log }$ can be defined as its inverse. The derivative of ${\displaystyle \log }$ can be found in the following way:

$${\displaystyle y=\log x\implies x=e^{y}}$$

If we differentiate both sides with respect to ${\displaystyle y}$, we get

$${\displaystyle {\begin{aligned}{\frac {dx}{dy}}&=e^{y}\\{\frac {dy}{dx}}&={\frac {1}{e^{y}}}={\frac {1}{x}}\end{aligned}}}$$

Therefore,

$${\displaystyle \int _{1}^{x}{\frac {1}{t}}dt=\left[\log t\right]_{1}^{x}=\log x-\log 1=\log x-0=\log x}$$

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

$${\displaystyle f(0)=f(0+0)=f(0)f(0)}$$

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

$${\displaystyle {\begin{array}{rcccccc}f'(x)&=&\lim \limits _{h\to 0}{\frac {f(x+h)-f(x)}{h}}&=&\lim \limits _{h\to 0}{\frac {f(x)f(h)-f(x)}{h}}&=&\lim \limits _{h\to 0}f(x){\frac {f(h)-1}{h}}\\[1em]&=&f(x)\lim \limits _{h\to 0}{\frac {f(h)-1}{h}}&=&f(x)\lim \limits _{h\to 0}{\frac {f(0+h)-f(0)}{h}}&=&f(x)f'(0)=f(x).\end{array}}}$$

In the sense of definition 6,

$${\displaystyle {\frac {d}{dx}}e^{x}=\lim _{h\to 0}{\frac {e^{x+h}-e^{x}}{h}}=e^{x}\cdot \lim _{h\to 0}{\frac {e^{h}-1}{h}}=e^{x}.}$$

By the way ${\displaystyle e^{0}=1}$, therefore definition 6 implies definition 4.