Exponent of Sum

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:

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

$\blacksquare$

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({\exp x}\right) \left({\exp y}\right)\)	\(=\)	\(\displaystyle \lim_{n \to +\infty} \left({1 + \frac x n}\right)^n \lim_{n \to +\infty} \left({1 + \frac y n}\right)^n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lim_{n \to +\infty} \left({\left({1 + \frac x n}\right)\left({1 + \frac y n}\right)}\right)^n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Combination Theorem for Sequences
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lim_{n \to +\infty} \left({1 + \frac{x + y}{n} + \frac{xy}{n^2} }\right)^n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lim_{n \to +\infty} \left({\left({1 + \frac{x + y} n}\right) \left({1 + \frac{\left({\frac{xy}{n+x+y} }\right)} n} \right)}\right)^n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Lemma: Let WLOG $n > -x-y$: therefore $n+x+y > 0$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lim_{n \to +\infty} \left(1 + \frac{x + y} n \right)^n \lim_{n \to +\infty} \left(1 + \frac{\left(\frac{xy}{n+x+y}\right)} n \right)^n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Combination Theorem for Sequences
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lim_{n \to +\infty} \left(1 + \frac{x + y} n \right)^n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Null Sequence in Exponential Sequence
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \exp(x + y)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Proof 3

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({\exp x}\right) \left({\exp y}\right)\)	\(=\)	\(\displaystyle \lim_{n \to +\infty} \left({1 + \frac x n}\right)^n \lim_{n \to +\infty} \left({1 + \frac y n}\right)^n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lim_{n \to +\infty} \left({\left({1 + \frac x n}\right)\left({1 + \frac y n}\right)}\right)^n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Combination Theorem for Limits of Functions
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \lim_{n \to +\infty} \left({1 + \frac{x + y} n + \frac{xy}{n^2} }\right)^n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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:

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac{\left({1 + \frac{x + y} n} + \frac {xy}{n^2}\right)^n}{\left({1 + \frac{x + y} n}\right)^n}\)	\(=\)	\(\displaystyle \left({1 + \frac{xy}{n^2 + nx + ny} }\right)^n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Exponent of Sum/Lemma
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{k=0}^n {n \choose k}\left({\frac{xy}{n^2 + nx + ny} }\right)^k\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Binomial Theorem
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1 + \sum_{k=1}^n {n \choose k}n^{-k} \left({\frac{xy}{n + x + y} }\right)^k\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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 $n^k$:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle {n \choose k}\)	\(\le\)	\(\displaystyle n^k\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	for all $n,k$ here considered
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle {n \choose k}n^{-k}\)	\(\le\)	\(\displaystyle 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	divide both sides by $n^k$

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

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac{\frac{xy}{n + x + y} }{1 - \frac{xy}{n + x + y} }\)	\(=\)	\(\displaystyle \frac {xy}{n + x + y -xy}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\to\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	as $n \to +\infty$

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

This 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 $n \to +\infty$

$\blacksquare$