# proof of Taylor’s Theorem

Let $$ be a real-valued, $n$-times differentiable function, and let $$ be a fixed base-point. We will show that for all $x\ne {x}_{0}$ in the domain of the function, there exists a $\xi $, strictly between ${x}_{0}$ and $x$ such that

$$f(x)=\sum _{k=0}^{n-1}{f}^{(k)}({x}_{0})\frac{{(x-{x}_{0})}^{k}}{k!}+{f}^{(n)}(\xi )\frac{{(x-{x}_{0})}^{n}}{n!}.$$ |

Fix $x\ne {x}_{0}$ and let $R$ be the remainder defined by

$$f(x)=\sum _{k=0}^{n-1}{f}^{(k)}({x}_{0})\frac{{(x-{x}_{0})}^{k}}{k!}+R\frac{{(x-{x}_{0})}^{n}}{n!}.$$ |

Next, define

$$ |

We then have

${F}^{\prime}(\xi )$ | $={f}^{\prime}(\xi )+{\displaystyle \sum _{k=1}^{n-1}}\left({f}^{(k+1)}(\xi ){\displaystyle \frac{{(x-\xi )}^{k}}{k!}}-{f}^{(k)}(\xi ){\displaystyle \frac{{(x-\xi )}^{k-1}}{(k-1)!}}\right)-R{\displaystyle \frac{{(x-\xi )}^{n-1}}{(n-1)!}}$ | ||

$={f}^{(n)}(\xi ){\displaystyle \frac{{(x-\xi )}^{n-1}}{(n-1)!}}-R{\displaystyle \frac{{(x-\xi )}^{n-1}}{(n-1)!}}$ | |||

$={\displaystyle \frac{{(x-\xi )}^{n-1}}{(n-1)!}}({f}^{(n)}(\xi )-R),$ |

because the sum telescopes. Since, $F(\xi )$ is a differentiable function, and since $F({x}_{0})=F(x)=f(x)$, Rolle’s Theorem imples that there exists a $\xi $ lying strictly between ${x}_{0}$ and $x$ such that ${F}^{\prime}(\xi )=0$. It follows that $R={f}^{(n)}(\xi )$, as was to be shown.

Title | proof of Taylor’s Theorem |
---|---|

Canonical name | ProofOfTaylorsTheorem |

Date of creation | 2013-03-22 12:33:59 |

Last modified on | 2013-03-22 12:33:59 |

Owner | rmilson (146) |

Last modified by | rmilson (146) |

Numerical id | 8 |

Author | rmilson (146) |

Entry type | Proof |

Classification | msc 26A06 |