Equivalence of Definitions of Real Exponential Function/Limit of Sequence implies Extension of Rational Exponential

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The following definition of the concept of the real exponential function:

As the Limit of a Sequence

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

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

implies the following definition:

As an Extension of the 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 $x$.


Let the restriction of the exponential function to the rationals be defined as:

$\displaystyle \exp \restriction_\Q: x \mapsto \lim_{n \mathop \to +\infty}\left ({1 + \frac x n}\right)^n$

Thus, let $e$ be Euler's Number defined as:

$e = \displaystyle \lim_{n \mathop \to +\infty}\left ({1 + \frac 1 n}\right)^n$

For $x = 0$:

\(\displaystyle \exp \restriction_\Q \paren 0 \ \ \) \(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to +\infty} \paren {1 + \frac 0 n}^n\)
\(\displaystyle \) \(=\) \(\displaystyle 1\)
\(\displaystyle \) \(=\) \(\displaystyle e^0\)

For $x \ne 0$:

\(\displaystyle \exp \restriction_\Q \paren x \ \ \) \(\displaystyle \) \(=\) \(\displaystyle \lim_{n \mathop \to +\infty}\paren {1 + \frac x n}^n\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{\paren {n/x} \mathop \to +\infty}\paren {\paren {1 + \frac 1 {\paren {n/x} } }^{\paren {n/x} } }^x\) Exponent Combination Laws
\(\displaystyle \) \(=\) \(\displaystyle e^x\)

For $x \in \R \setminus \Q$, we invoke Power Function to Rational Power permits Unique Continuous Extension.