Equivalence of Definitions of Real Exponential Function/Inverse of Natural Logarithm implies Limit of Sequence
Theorem
The following definition of the concept of the real exponential function:
As the Inverse of the Natural Logarithm
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:
- $y = \exp x \iff x = \ln y$
implies the following definition:
As the Limit of a Sequence
The exponential function can be defined as the following limit of a sequence:
- $\exp x := \ds \lim_{n \mathop \to +\infty} \paren {1 + \frac x n}^n$
Proof
Let $\exp x$ be the real function defined as the inverse of the natural logarithm:
- $y = \exp x \iff x = \ln y$
Let $\left \langle {x_n} \right \rangle$ be the sequence in $\R$ defined as:
- $x_n = \paren {1 + \dfrac x n}^n$
First it needs to be noted that $\sequence {x_n}$ does indeed converge to a limit.
From Equivalence of Definitions of Real Exponential Function: Limit of Sequence implies Power Series Expansion, we have:
- $\ds \lim_{n \mathop \to \infty} \paren {1 + \dfrac x n}^n = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$
From Series of Power over Factorial Converges, the right hand side is indeed shown to converge to a limit.
It will next be shown that:
- $\ds \map \ln {\lim_{n \mathop \to \infty} \sequence {x_n} } = x$
We have:
\(\ds \map \ln {\paren {1 + \frac x n}^n}\) | \(=\) | \(\ds n \, \map \ln {1 + x n^{-1} }\) | Logarithms of Powers | |||||||||||
\(\ds \) | \(=\) | \(\ds x \frac {\map \ln {1 + x n^{-1} } } {x n^{-1} }\) | multiplying by $1 = \dfrac {x n^{-1} } {x n^{-1} }$ |
From Limit of Sequence is Limit of Real Function, we can consider the differentiable analogue of the sequence.
From Derivative of Logarithm at One we have:
- $\ds \lim_{x \mathop \to 0} \frac {\map \ln {1 + x} } x = 1$
But $x n^{-1} \to 0$ as $n \to \infty$ from Sequence of Powers of Reciprocals is Null Sequence.
Thus:
- $\ds x \frac {\map \ln {1 + x n^{-1} } } {x n^{-1} } \to x$
as $n \to \infty$.
From Exponential Function is Continuous:
- $\paren {1 + \dfrac x n}^n = \exp \paren {n \map \ln {1 + \dfrac x n} } \to \exp x = e^x$
as $n \to \infty$.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 14.7 \ (3)$