Exponential of Sum/Real Numbers

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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:

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


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:

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


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


First we introduce a lemma:


By definition:

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

$\blacksquare$


Proof 3

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

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


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


By definition:

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

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:

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

Now, as $n \to +\infty$, we use the Combination Theorem for Limits of Real 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:

$\ds \sum_{k \mathop = 1}^n \, \size {\binom n k n^{-k} \paren {\frac {x y} {n + x + y} }^k}$


From $\dbinom n k$ is not greater than $n^k$:

\(\ds \binom n k\) \(\le\) \(\ds n^k\) for all $n, k$ here considered
\(\ds \leadsto \ \ \) \(\ds \binom n k n^{-k}\) \(\le\) \(\ds 1\) divide both sides by $n^k$

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

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


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

\(\ds \frac {\dfrac {x y} {n + x + y} } {1 - \dfrac {x y} {n + x + y} }\) \(=\) \(\ds \frac {x y} {n + x + y - x y}\)
\(\ds \) \(\to\) \(\ds 0\) 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:

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

From Exponential of Real Number is Strictly Positive, $f$ is well-defined.


So:

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


Thus $f$ satisfies $(1)$.


Further:

\(\ds \map f 0\) \(=\) \(\ds \frac {\map \exp {0 + y} } {\map \exp y}\)
\(\ds \) \(=\) \(\ds \frac {\map \exp y} {\map \exp y}\)
\(\ds \) \(=\) \(\ds 1\)


So $f$ satisfies $(2)$.


From Exponential Function is Well-Defined:

$f = \exp$

That is:

\(\ds \frac {\map \exp {x + y} } {\map \exp y}\) \(=\) \(\ds \map \exp x\)
\(\ds \leadsto \ \ \) \(\ds \map \exp {x + y}\) \(=\) \(\ds \map \exp x \, \map \exp y\)

$\blacksquare$


Proof 5

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


Then:

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

$\blacksquare$


Proof 6

Fix $a \in \R$ and define the function $f_a : \R \to \R$ by:

$\map {f_a} x = \map \exp {a - x} \exp x$

for all $x \in \R$.

We aim to establish that:

$\map {f_a} x = \map \exp {a - x} \exp x = \exp a$

for all $a, x \in \R$.

Then, we can fix $x, y \in \R$ and set $a = x + y$ to obtain:

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

which is the claim.

Note that $f_a$ is differentiable and we have:

\(\ds \map {f_a'} x\) \(=\) \(\ds \map {\frac \d {\d x} } {\map \exp {a - x} } \exp x + \map {\frac \d {\d x} } {\exp x} \map \exp {a - x}\) Product Rule for Derivatives
\(\ds \) \(=\) \(\ds -\map \exp {a - x} \exp x + \exp x \map \exp {a - x}\) Chain Rule for Derivatives, Derivative of Exponential Function
\(\ds \) \(=\) \(\ds 0\)

From Zero Derivative implies Constant Function, $f_a$ is constant.

That is, there exists $C \in \R$ such that:

$\map {f_a} x = C$

for all $x \in \R$.

We have:

\(\ds \map {f_a} a\) \(=\) \(\ds \map \exp {a - a} \exp a\)
\(\ds \) \(=\) \(\ds \exp 0 \exp a\)
\(\ds \) \(=\) \(\ds \exp a\) Exponential of Zero

So:

$\map {f_a} x = \exp a$

for all $a, x \in \R$.

That is:

$\map \exp {a - x} \map \exp x = \exp a$

for all $a, x \in \R$.

Hence the result.

$\blacksquare$

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