Exponential of Sum/Real Numbers/Proof 7

Theorem

Let $x, y \in \R$ be real numbers.

Let $\exp x$ be the exponential of $x$.


Then:

$\map \exp {x + y} = \paren {\exp x} \paren {\exp y}$


Proof

Let $\exp$ be defined as the limit of a sequence:

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


Then:

\(\ds \paren {\exp x} \paren {\exp y}\) \(=\) \(\ds \paren {\lim_{n \mathop \to \infty} \paren {1 + \frac x n}^n} \paren {\lim_{n \mathop \to \infty} \paren {1 + \frac y n}^n}\) Definition of Real Exponential Function
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \paren {1 + \frac x n}^n \paren {1 + \frac y n}^n\) Product Rule for Real Sequences
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \paren {1 + \frac {x + y} n + \frac {xy} {n^2} }^n\)

Now, let:

$\nu_n = n$
$\xi_n = \dfrac {x + y} n + \dfrac {xy} {n^2}$

for all $n \in \N$

Since:

\(\ds \lim_{n \mathop \to \infty} \nu_n \xi_n\) \(=\) \(\ds \lim_{n \mathop \to \infty} x + y + \dfrac {xy} n\)
\(\ds \) \(=\) \(\ds x + y\)

it follows from Generalized Exponential Limit that:

$\ds\lim_{n \mathop \to \infty} \paren {1 + \frac {x + y} n + \frac {xy} {n^2} }^n = \map \exp {x + y}$

with $\exp$ defined as a power series.


Hence, by Equivalence of Definitions of Real Exponential Function:

$\paren {\exp x} \paren {\exp y} = \map \exp {x + y}$

$\blacksquare$


Sources

  • 1926: T.J.I'A. Bromwich: An Introduction to the Theory of Infinite Series (2nd ed.): Art. 58