# Exponential of Sum/Real Numbers/Lemma

## 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

As $n \in \N_{> 0}$ we have that $n \ne 0$ and so the fractions in the expressions are defined.

 $\ds 1 + \frac {x + y} n + \frac {x y} {n^2}$ $=$ $\ds \frac {\paren {1 + \frac {x + y} n} \paren {1 + \frac {x + y} n + \frac {x y} {n^2} } } {1 + \frac {x + y} n}$ multiplying and dividing by $1 + \dfrac {x + y} n$ $\ds$ $=$ $\ds \paren {1 + \frac {x + y} n} \frac {\paren {1 + \frac {x + y} n + \frac {x y} {n^2} } } {\paren {1 + \frac {x + y} n} }$ extracting a factor $\ds$ $=$ $\ds \paren {1 + \frac {x + y} n} \paren {\frac {n^2 + n x + n y + x y} {n^2 + n x + n y} }$ multiplying top and bottom by $n^2$ $\ds$ $=$ $\ds \paren {1 + \frac {x + y} n} \paren {1 + \frac {x y} {n^2 + n x + n y} }$ Polynomial Long Division $\ds$ $=$ $\ds \paren {1 + \frac {x + y} n} \paren {1 + \frac {\paren {\frac {x y} {n + x + y} } } n}$ dividing top and bottom by $n + x + y$

That final step is justified, as we have that $n > -\paren {x + y}$ and so $n + x + y \ne 0$.

$\blacksquare$

## Also see

This lemma is used in:

Proof 2 of (Real) Exponential of Sum
Proof 3 of (Real) Exponential of Sum
$\map \exp {x + y} = \paren {\exp x} \paren {\exp y}$

for real $x$ and $y$.