# Real exponential function via exponential series/Introduction/Section

## Definition

For every ${}x\in \mathbb {R}$ , the series

$\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}$ is called the exponential series in ${}x$ .

So this is just the series

$1+x+{\frac {x^{2}}{2}}+{\frac {x^{3}}{6}}+{\frac {x^{4}}{24}}+{\frac {x^{5}}{120}}+\ldots .$ ## Theorem

For every ${}x\in \mathbb {R}$ , the exponential series

$\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}$ is

### Proof

For ${}x=0$ , the statement is clear. Else, we consider the fraction

${}\vert {\frac {\frac {x^{n+1}}{(n+1)!}}{\frac {x^{n}}{n!}}}\vert =\vert {\frac {x}{n+1}}\vert ={\frac {\vert {x}\vert }{n+1}}\,.$ This is, for ${}n\geq 2\vert {x}\vert$ , smaller than ${}1/2$ . By the ratio test, we get convergence.

$\Box$ Due to this property, we can define the real exponential function.

## Definition

The function

$\mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto \exp x:=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}},$ is called the (real)

exponential function.

The following statement is called the functional equation for the exponential function.

## Theorem

For real numbers ${}x,y\in \mathbb {R}$ , the equation

${}\exp {\left(x+y\right)}=\exp x\cdot \exp y\,$ holds.

### Proof

The Cauchy product of the two exponential series is

$\sum _{n=0}^{\infty }c_{n},$ where

${}c_{n}=\sum _{i=0}^{n}{\frac {x^{i}}{i!}}\cdot {\frac {y^{n-i}}{(n-i)!}}\,.$ This series is due to fact absolutely convergent and the limit is the product of the two limits. Furthermore, the ${}n$ -th summand of the exponential series of ${}x+y$ equals

${}{\frac {(x+y)^{n}}{n!}}={\frac {1}{n!}}\sum _{i=0}^{n}{\binom {n}{i}}x^{i}y^{n-i}=c_{n}\,,$ so that both sides coincide.

$\Box$ ## Corollary

$\mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto \exp x,$ fulfills the following properties.
1. ${}\exp 0=1$ .
2. For every ${}x\in \mathbb {R}$ , we have ${}\exp {\left(-x\right)}=(\exp x)^{-1}$ . In particular ${}\exp x\neq 0$ .
3. For integers ${}n\in \mathbb {Z}$ , the relation ${}\exp n=(\exp 1)^{n}$ holds.
4. For every ${}x$ , we have ${}\exp x\in \mathbb {R} _{+}$ .
5. For ${}x>0$ we have ${}\exp x>1$ , and for ${}x<0$ we have ${}\exp x<1$ .
6. The real exponential function is strictly increasing.

### Proof

(1) follows directly from the definition.
(2) follows from

${}\exp x\cdot \exp {\left(-x\right)}=\exp {\left(x-x\right)}=\exp 0=1\,$ using fact.
(3) follows for ${}n\in \mathbb {N}$ from fact by induction, and from that it follows with the help of (2) also for negative ${}n$ .
(4). Nonnegativity follows from

${}\exp x=\exp {\left({\frac {x}{2}}+{\frac {x}{2}}\right)}=\exp {\frac {x}{2}}\cdot \exp {\frac {x}{2}}={\left(\exp {\frac {x}{2}}\right)}^{2}\geq 0\,.$ (5). For real ${}x$ we have ${}\exp x\cdot \exp {\left(-x\right)}=1$ , so that because of (4), one factor must be ${}\geq 1$ and the other factor must be ${}\leq 1$ . For ${}x>0$ , we have

${}\exp x=\sum _{n=0}^{\infty }{\frac {1}{n!}}x^{n}=1+x+{\frac {1}{2}}x^{2}+\ldots >1\,$ as only positive numbers are added.
(6). For real ${}y>x$ , we have ${}y-x>0$ , and therefore, because of (5) ${}\exp {\left(y-x\right)}>1$ , hence

${}\exp y=\exp {\left(y-x+x\right)}=\exp {\left(y-x\right)}\cdot \exp x>\exp x\,.$ $\Box$ 