# Exponential of Sum/Real Numbers/Proof 3

## 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

### Lemma

Let $x, y \in \R$.

Let $n \in \N_{> 0}$ such that $n > -\paren {x + y}$.

Then:

$1 + \dfrac {x + y} n + \dfrac {x y} {n^2} = \paren {1 + \dfrac {x + y} n} \paren {1 + \dfrac {\paren {\frac {x y} {n + x + y} } } n}$

$\Box$

This proof assumes the definition of $\exp$ as defined by a limit of a sequence:

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

From Powers of Group Elements we can presuppose the Exponent Combination Laws for natural number indices.

By definition:

 $\displaystyle \paren {\exp x} \paren {\exp y}$ $=$ $\displaystyle \lim_{n \mathop \to +\infty} \paren {1 + \frac x n}^n \lim_{n \mathop \to +\infty} \paren {1 + \frac y n}^n$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to +\infty} \paren {\paren {1 + \frac x n} \paren {1 + \frac y n} }^n$ Combination Theorem for Limits of Functions $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to +\infty} \paren {1 + \frac{x + y} n + \frac{x y} {n^2} }^n$

Intuitively, the $\paren {1 + \dfrac {x + y} n}$ term is the most influential of the terms involved in the limit, and:

$\displaystyle \paren {1 + \frac {x + y} n + \frac {x y} {n^2} }^n \to \paren {1 + \frac {x + y} n}^n$ as $n \to +\infty$

To formalize this claim:

$\map \exp {x + y} = \exp x \cdot \exp y \iff \dfrac {\exp x \cdot \exp y} {\map \exp {x + y} } = 1$
 $\displaystyle \frac {\paren {1 + \frac {x + y} n + \frac {x y} {n^2} }^n} {\paren {1 + \frac{x + y} n}^n}$ $=$ $\displaystyle \paren {1 + \frac{x y} {n^2 + n x + n y} }^n$ Lemma $\displaystyle$ $=$ $\displaystyle \sum_{k \mathop = 0}^n \binom n k \paren {\frac {x y} {n^2 + n x + n y} }^k$ Binomial Theorem $\displaystyle$ $=$ $\displaystyle 1 + \sum_{k \mathop = 1}^n \binom n k n^{-k} \paren {\frac {x y} {n + x + y} }^k$

Now, as $n \to +\infty$, we use the Combination Theorem for Limits of Functions to investigate the behavior of this sequence, term by term.

As $1$ trivially converges to $1$, consider now the other terms of the sequence.

We invoke the Squeeze Theorem for Absolutely Convergent Series.

Hence it will suffice to investigate the limit behaviour of:

$\displaystyle \sum_{k \mathop = 1}^n \, \size {\binom n k n^{-k} \paren {\frac {x y} {n + x + y} }^k}$
 $\displaystyle \binom n k$ $\le$ $\displaystyle n^k$ for all $n,k$ here considered $\displaystyle \leadsto \ \$ $\displaystyle \binom n k n^{-k}$ $\le$ $\displaystyle 1$ divide both sides by $n^k$

Therefore, we may conclude, using Absolute Value is Bounded Below by Zero:

$\displaystyle 0 \le \sum_{k \mathop = 1}^n \, \size { {n \choose k} n^{-k} \paren {\frac {x y} {n + x + y} }^k} \le \sum_{k \mathop = 1}^n \size {\frac {x y} {n + x + y} }^k$

From Sum of Infinite Geometric Progression, the right hand term converges to:

 $\displaystyle \frac {\dfrac {x y} {n + x + y} } {1 - \dfrac {x y} {n + x + y} }$ $=$ $\displaystyle \frac {x y} {n + x + y - x y}$ $\displaystyle$ $\to$ $\displaystyle 0$ as $n \to +\infty$
$0 \to 0$ as $n \to +\infty$, trivially.

This means:

$\dfrac {\paren {1 + \dfrac{x + y} n + \dfrac {x y} {n^2} }^n} {\paren {1 + \dfrac {x + y} n}^n} \to 1$ as $n \to +\infty$

which is equivalent to our hypothesis:

$\paren {1 + \dfrac{x + y} n + \dfrac {x y} {n^2} }^n \to \paren {1 + \dfrac {x + y} n}^n$ as $n \to +\infty$

$\blacksquare$