# Taylor series/R/One variable/Introduction/Section

## Definition

Let ${\displaystyle {}I\subseteq \mathbb {R} }$ denote an interval,

${\displaystyle f\colon I\longrightarrow \mathbb {R} }$

an infinitely often differentiable function, and ${\displaystyle {}a\in I}$. Then

${\displaystyle \sum _{k=0}^{\infty }{\frac {f^{(k)}(a)}{k!}}(x-a)^{k}}$
is called the Taylor series of ${\displaystyle {}f}$ in the point ${\displaystyle {}a}$.

## Theorem

Let ${\displaystyle {}\sum _{n=0}^{\infty }c_{n}x^{n}}$ denote a power series which converges on the interval ${\displaystyle {}]-r,r[}$, and let

${\displaystyle f\colon ]-r,r[\longrightarrow \mathbb {R} }$

denote the function defined via fact. Then ${\displaystyle {}f}$ is infinitely often differentiable, and the Taylor series of ${\displaystyle {}f}$ in ${\displaystyle {}0}$ coincides with the given power series.

### Proof

That ${\displaystyle {}f}$ is infinitely often differentiable, follows directly from fact by induction. Therefore, the Taylor series exists in particular in the point ${\displaystyle {}0}$. Hence, we only have to show that the ${\displaystyle {}n}$-th derivative has ${\displaystyle {}c_{n}n!}$ as its value. But this follows also from fact.

${\displaystyle \Box }$

## Example

We consider the function

${\displaystyle f\colon \mathbb {R} \longrightarrow \mathbb {R} ,x\longmapsto f(x),}$

given by

${\displaystyle {}f(x):={\begin{cases}0,\,{\text{ if }}x\leq 0\,,\\e^{-{\frac {1}{x}}},\,{\text{ if }}x>0\,.\end{cases}}\,}$

We claim that this function is infinitely often differentiable, which is only in ${\displaystyle {}0}$ not directly clear. We first show, by induction, that all derivatives of ${\displaystyle {}e^{-{\frac {1}{x}}}}$ have the form ${\displaystyle {}p{\left({\frac {1}{x}}\right)}e^{-{\frac {1}{x}}}}$ with certain polynomials ${\displaystyle {}p\in \mathbb {R} [Z]}$, and that therefore the limit for ${\displaystyle {}x\rightarrow 0,\,x>0}$ equals ${\displaystyle {}0}$ (see exercise and exercise). Therefore, the limit exists for all derivatives and is ${\displaystyle {}0}$. So all derivatives in ${\displaystyle {}0}$ have value ${\displaystyle {}0}$, and therefore the Taylor series in ${\displaystyle {}0}$ is just the zero series. However, the Function ${\displaystyle {}f}$ is in no neighborhood of ${\displaystyle {}0}$ the zero function, since ${\displaystyle {}e^{-{\frac {1}{x}}}>0}$.