Exponential of Sum/Real Numbers

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

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

Then:
 * $\exp \left({x + y}\right) = \left({\exp x}\right) \left({\exp y}\right)$

Proof 1
This proof assumes the definition of $\exp$ as:


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


 * $\ln y = \displaystyle \int_1^y \frac 1 t \ \mathrm dt$

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

From Sum of Logarithms, we have:
 * $\ln XY = \ln X + \ln Y = x + y$

From the Exponential of Natural Logarithm:
 * $\exp \left({\ln x}\right) = x$

Thus:
 * $\exp \left({x + y}\right) = \exp \left({\ln XY}\right) = XY = \left({\exp x}\right) \left({\exp y}\right)$

Alternatively, this may be proved directly by investigating:
 * $D \left({\exp \left({x + y}\right) / \exp x}\right)$

Proof 2
This proof assumes the definition of $\exp$ as defined by a limit:


 * $\exp x = \displaystyle \lim_{n \to +\infty} \left({1 + \frac x n}\right)^n$

Note that from Powers of Group Elements we can presuppose the exponent combination laws for natural number indices.

By definition:

Intuitively, the $\left({1 + \frac{x + y}{n}}\right)$ term is the most influential of the terms involved in the limit, and:

$\displaystyle \left({1 + \frac{x + y}{n}} + \frac {xy}{n^2}\right)^n \to \left({1 + \frac{x + y}{n}}\right)^n$ as $n \to +\infty$

To formalize this claim:


 * $\exp \left({x + y}\right) = \exp x \cdot \exp y \iff \dfrac {\exp x \cdot \exp y} {\exp \left({x + y}\right)} = 1$

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.

From Negative of Absolute Value and Absolute Value Bounded Below by Zero:


 * $\displaystyle 0 \le -\left\vert{\sum_{k=1}^n {n \choose k}n^{-k}\left({\frac{xy}{n + x + y} }\right)^k}\right\vert \le \sum_{k=1}^n {n \choose k}n^{-k}\left({\frac{xy}{n + x + y} }\right)^k \le +\left\vert{\sum_{k=1}^n {n \choose k}n^{-k}\left({\frac{xy}{n + x + y} }\right)^k}\right\vert$

From $n$ choose $k$ is less than $n^k$:


 * $\implies \displaystyle 0 \le \sum_{k=1}^n \left\vert{ {n \choose k}n^{-k}\left({\frac{xy}{n + x + y} }\right)^k}\right\vert \le\sum_{k=1}^n \left\vert{\frac{xy}{n + x + y} }\right\vert^k$

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


 * $0 \to 0$ as $n \to +\infty$, trivially.

Which means:


 * $\displaystyle \frac{\left({1 + \frac{x + y}{n} } + \frac {xy}{n^2}\right)^n}{\left({1 + \frac{x + y}{n} }\right)^n} \to 1$ as $n \to +\infty$

Which is equivalent to our hypothesis:


 * $\displaystyle \left({1 + \frac{x + y}{n}} + \frac {xy}{n^2}\right)^n \to \left({1 + \frac{x + y}{n}}\right)^n$ as $n \to +\infty$