Equivalence of Definitions of Real Exponential Function/Proof 2

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Theorem

The following definitions of the exponential function are equivalent.

Definition 1

The exponential function can be defined as a power series:

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

Definition 2

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

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

Definition 3

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 $x$.

Definition 4

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

From Logarithm is Strictly Increasing:

$\ln x$ is strictly increasing.

From Inverse of Strictly Monotone Function:

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:

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

Definition 5

The exponential function can be defined as the unique solution $y = f \left({x}\right)$ to the first order ODE:

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

satisfying the initial condition $f \left({0}\right) = 1$.


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 \exp \left({ 0 }\right) = 1$

on $\R$.

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


Thus they all are all equivalent.

$\blacksquare$