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