Equivalence of Definitions of Euler's Number/Proof 1

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The following definitions of the concept of Euler's Number are equivalent:

Limit of Series

The series $\displaystyle \sum_{n \mathop = 0}^\infty \frac 1 {n!}$ 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

The 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

The number $e$ can be defined as the number satisfied by:

$e := \exp 1 = e^1$

where $\exp 1$ denotes the exponential function of $1$.


See Equivalence of Definitions of Real Exponential Function: Inverse of Natural Logarithm implies Limit of Sequence for how $\displaystyle \lim_{n \mathop \to \infty} \left({1 + \frac 1 n}\right)^n = e$ 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 $\displaystyle e = \sum_{n \mathop = 0}^\infty \frac 1 {n!}$ follows from $\displaystyle \lim_{n \mathop \to \infty} \left({1 + \frac 1 n}\right)^n = e$.

Now suppose $e$ is defined as $\displaystyle e = \sum_{n \mathop = 0}^\infty \frac 1 {n!}$.

Let us consider the series $\displaystyle f \left({x}\right) = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$.

From Series of Power over Factorial Converges, this is convergent for all $x$.

We differentiate $f \left({x}\right)$ with respect to $x$ term by term (justified by Power Series is Differentiable on Interval of Convergence), and get:

\(\displaystyle D_x \left({f \left({x}\right)}\right)\) \(=\) \(\displaystyle D_x \left({1}\right) + D_x \left({\frac {x} {1!} }\right) + D_x \left({\frac {x^2} {2!} }\right) + D_x \left({\frac {x^3} {3!} }\right) + \cdots + D_x \left({\frac {x^n} {n!} }\right) + D_x \left({\frac {x^{n+1} } {\left({n+1}\right)!} }\right) + \cdots\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 0 + 1 + \frac {2 x} {2!} + \frac {3 x^2} {3!} + \cdots + \frac {n x^{n-1} } {n!} + \frac {\left({n+1}\right) x^n} {\left({n+1}\right)!} + \cdots\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 1 + \frac x {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n-1} } {\left({n-1}\right)!} + \frac {x^n} {n!} \cdots\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle f \left({x}\right)\) $\quad$ $\quad$

Thus we have:

$D_x \left({f \left({x}\right)}\right) = f \left({x}\right)$

From Derivative of Exponential Function:

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

From Derivative of Inverse Function:

$D_x \left({f^{-1} \left({x}\right)}\right) = \dfrac 1 {f^{-1} \left({x}\right)}$

Hence from Derivative of Natural Logarithm Function:

$f^{-1} \left({x}\right) = \ln x$

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.