Equivalence of Definitions of Euler's Number
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Theorem
The following definitions of the concept of Euler's Number are equivalent:
Limit of Series
The series $\ds \sum_{n \mathop = 0}^\infty \frac 1 {n!}$ converges to a limit.
This limit is Euler's number $e$.
Limit of Sequence
The sequence $\sequence {x_n}$ defined as $x_n = \paren {1 + \dfrac 1 n}^n$ converges to a limit as $n$ increases without bound.
That limit is called Euler's number and is denoted $e$.
Base of Logarithm
Euler's number $e$ can be defined as the number satisfied by:
- $\ln e = 1$
where $\ln e$ denotes the natural logarithm of $e$.
That $e$ is unique follows from Logarithm is Strictly Increasing.
Exponential Function
Euler's number $e$ can be defined as the number satisfied by:
- $e := \exp 1 = e^1$
where $\exp 1$ denotes the exponential function of $1$.
Proof 1
See Equivalence of Definitions of Real Exponential Function: Inverse of Natural Logarithm implies Limit of Sequence for how $\ds \lim_{n \mathop \to \infty} \paren {1 + \frac 1 n}^n = e$ follows from the definition of $e$ as the number satisfied by $\ln e = 1$.
See Euler's Number: Limit of Sequence implies Limit of Series for how $\ds e = \sum_{n \mathop = 0}^\infty \frac 1 {n!}$ follows from $\ds \lim_{n \mathop \to \infty} \paren {1 + \frac 1 n}^n = e$.
Now suppose $e$ is defined as $\ds e = \sum_{n \mathop = 0}^\infty \frac 1 {n!}$.
Let us consider the series $\ds \map f x = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$.
From Series of Power over Factorial Converges, this is convergent for all $x$.
We differentiate $\map f x$ with respect to $x$ term by term (justified by Power Series is Differentiable on Interval of Convergence), and get:
| \(\ds \map {D_x} {\map f x}\) | \(=\) | \(\ds \map {D_x} 1 + \map {D_x} {\frac x {1!} } + \map {D_x} {\frac {x^2} {2!} } + \map {D_x} {\frac {x^3} {3!} } + \cdots + \map {D_x} {\frac {x^n} {n!} } + \map {D_x} {\frac {x^{n + 1} } {\paren {n + 1}!} } + \cdots\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 0 + 1 + \frac {2 x} {2!} + \frac {3 x^2} {3!} + \cdots + \frac {n x^{n - 1} } {n!} + \frac {\paren {n + 1} x^n} {\paren {n + 1}!} + \cdots\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1 + \frac x {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n - 1} } {\paren {n - 1}!} + \frac {x^n} {n!} \cdots\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map f x\) |
Thus we have:
- $\map {D_x} {\map f x} = \map f x$
From Derivative of Exponential Function:
- $\map f x = e^x$
From Derivative of Inverse Function:
- $\map {D_x} {\map {f^{-1} } x} = \dfrac 1 {\map {f^{-1} } x}$
Hence from Derivative of Natural Logarithm Function:
- $\map {f^{-1} } x = \ln x$
It follows that $e$ can be defined as that number such that $\ln e = 1$.
Hence all the definitions of $e$ as given here are equivalent.
$\blacksquare$
Proof 2
1 implies 2
This is proved in Euler's Number: Limit of Sequence implies Limit of Series.
$\Box$
2 implies 3
This is proved in Euler's Number: Limit of Sequence implies Base of Logarithm.
$\Box$
3 implies 4
Let $e$ be the unique solution to the equation $\ln x = 1$.
We want to show that $\exp 1 = e$, where $\exp$ is the exponential function.
| \(\ds \exp 1 = 0\) | \(\leadstoandfrom\) | \(\ds \map \ln {\exp 1} = \ln e\) | Logarithm is Strictly Increasing and Strictly Concave: Corollary | |||||||||||
| \(\ds \) | \(\leadstoandfrom\) | \(\ds 1 = \ln e\) | Exponential is Inverse of Logarithm and Inverse of Inverse |
where the final equation holds by hypothesis.
Hence the result.
$\Box$
4 implies 1
Let $e = \exp 1$, where $\exp$ denotes the exponential function.
We want to show that:
- $\ds \sum_{n \mathop = 0}^\infty \frac 1 {n!} = e$
- $\ds \sum_{n \mathop = 0}^\infty \frac 1 {n!} = \exp 1$
We also have that $\exp 1 = e$ by hypothesis.
Hence the result.
$\blacksquare$