# Exponential of Sum/Real Numbers

## Theorem

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

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

Then:

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

### Corollary

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

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

Then:

$\map \exp {x - y} = \dfrac {\exp x} {\exp y}$

## Lemma

Let $x, y \in \R$.

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

Then:

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

$\Box$

## Proof 1

This proof assumes the definition of $\exp$ as:

$\exp x = y \iff \ln y = x$

where:

$\ln y = \displaystyle \int_1^y \dfrac 1 t \rd t$

Let $X = \exp x$ and $Y = \exp y$.

From Sum of Logarithms, we have:

$\ln X Y = \ln X + \ln Y = x + y$

From the Exponential of Natural Logarithm:

$\map \exp {\ln x} = x$

Thus:

$\map \exp {x + y} = \map \exp {\ln X Y} = X Y = \paren {\exp x} \paren {\exp y}$

$\blacksquare$

Alternatively, this may be proved directly by investigating:

$\map D {\map \exp {x + y} / \exp x}$

## Proof 2

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.

First we introduce a lemma:

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 Sequences $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to +\infty} \paren {1 + \frac {x + y} n + \frac {x y} {n^2} }^n$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to +\infty} \paren {\paren {1 + \frac {x + y} n} \paren {1 + \frac {\paren {\frac {x y} {n + x + y} } } n} }^n$ Lemma: Without loss of generality let $n > - x - y$: therefore $n + x + y > 0$ $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to +\infty} \paren {1 + \frac {x + y} n}^n \lim_{n \mathop \to +\infty} \paren {1 + \frac {\paren {\frac {x y} {n + x + y} } } n}^n$ Combination Theorem for Sequences $\displaystyle$ $=$ $\displaystyle \lim_{n \mathop \to +\infty} \paren {1 + \frac {x + y} n}^n$ Null Sequence in Exponential Sequence $\displaystyle$ $=$ $\displaystyle \map \exp {x + y}$

$\blacksquare$

## Proof 3

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:

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

$\blacksquare$

## Proof 4

This proof assumes the definition of $\exp$ as defined by an initial value problem.

That is, suppose $\exp$ satisfies:

$(1): \quad D_x \exp x = \exp x$
$(2): \quad \exp 0 = 1$

on $\R$.

Consider the real function $f: \R \to \R$ defined by:

$\map f x := \dfrac {\map \exp {x + y} } {\map \exp y}$

So:

 $\displaystyle D_x \, \map f x$ $=$ $\displaystyle D_x \frac {\map \exp {x + y} } {\map \exp y}$ $\displaystyle$ $=$ $\displaystyle \frac 1 {\map \exp y} D_x \, \map \exp {x + y}$ Derivative of Constant Multiple $\displaystyle$ $=$ $\displaystyle \frac {\map \exp {x + y} } {\map \exp y}$ Chain Rule for Derivatives $\displaystyle$ $=$ $\displaystyle \map f x$

Thus $f$ satisfies $(1)$.

Further:

 $\displaystyle \map f 0$ $=$ $\displaystyle \frac {\map \exp {0 + y} } {\map \exp y}$ $\displaystyle$ $=$ $\displaystyle \frac {\map \exp y} {\map \exp y}$ $\displaystyle$ $=$ $\displaystyle 1$

So $f$ satisfies $(2)$.

$f = \exp$

That is:

 $\displaystyle \frac {\map \exp {x + y} } {\map \exp y}$ $=$ $\displaystyle \map \exp x$ $\displaystyle \leadsto \ \$ $\displaystyle \map \exp {x + y}$ $=$ $\displaystyle \map \exp x \, \map \exp y$

$\blacksquare$

## Proof 5

This proof assumes the definition of $\exp$ as a series.

Then:

 $\displaystyle \map \exp {x + y}$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \frac 1 {n!} \paren {x + y}^n$ $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \frac 1 {n!} \sum_{k \mathop = 0}^n \frac {n!} {k! \paren {n - k}!} x^k y^{n - k}$ Binomial Theorem $\displaystyle$ $=$ $\displaystyle \sum_{n \mathop = 0}^\infty \sum_{k \mathop = 0}^n \paren {\frac 1 {k!} x^k} \paren {\frac 1 {\paren {n - k}!} y^{n - k} }$ $\displaystyle$ $=$ $\displaystyle \paren {\sum_{n \mathop = 0}^\infty \frac {x^n} {n!} } \paren {\sum_{n \mathop = 0}^\infty \frac {y^n} {n!} }$ Definition of Cauchy Product $\displaystyle$ $=$ $\displaystyle \map \exp x \, \map \exp y$

$\blacksquare$