Equivalence of Definitions of Real Exponential Function/Proof 2

Theorem

The following definitions of the exponential function are equivalent.

As a Power Series Expansion

The exponential function can be defined as a power series:

$\exp x := \ds \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$

As a Limit of a Sequence

The exponential function can be defined as the following limit of a sequence:

$\exp x := \displaystyle \lim_{n \mathop \to +\infty} \paren {1 + \frac x n}^n$

As an Extension of a Rational Exponential

Let $e$ denote Euler's number.

Let $f: \Q \to \R$ denote the real-valued function defined as:

$f \left({ x }\right) = e^x$

That is, let $f \left({ x }\right)$ denote $e$ to the power of $x$, for rational $x$.

Then $\exp : \R \to \R$ is defined to be the unique continuous extension of $f$ to $\R$.

$\exp \left({ x }\right)$ is called the exponential of $x$.

As the Inverse of the Natural Logarithm

Consider the natural logarithm $\ln x$, which is defined on the open interval $\openint 0 {+\infty}$.

$\ln x$ is strictly increasing.
the inverse of $\ln x$ always exists.

The inverse of the natural logarithm function is called the exponential function, which is denoted as $\exp$.

Thus for $x \in \R$, we have:

$y = \exp x \iff x = \ln y$

As the Solution of a Differential Equation

The exponential function can be defined as the unique solution $y = \map f x$ to the first order ODE:

$\dfrac {\d y} {\d x} = y$

satisfying the initial condition $\map f 0 = 1$.

Proof

From Derivative of Exponential Function and Exponential of Zero, each definition of $\exp$ satisfies the following:

$(1): \quad D_x \exp = \exp$
$(2): \quad \map \exp 0 = 1$

on $\R$.

From Exponential Function is Well-Defined, such a solution is unique.

Thus they all are all equivalent.

$\blacksquare$