Exponential of Sum/Real Numbers/Proof 6

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

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.

$\blacksquare$