Taylor series/R/One variable/Introduction/Section

Definition

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

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

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

$\sum _{k=0}^{\infty }{\frac {f^{(k)}(a)}{k!}}(x-a)^{k}$

is called the Taylor series of ${}f$ in the point ${}a$.

Theorem

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

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

denote the function defined via fact. Then ${}f$ is infinitely often differentiable, and the Taylor series

of ${}f$ in ${}0$ coincides with the given power series.

Proof

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

$\Box$

Example

We consider the function

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

given by

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

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