proof of Taylor’s Theorem

Let $f(x),\;a be a real-valued, $n$-times differentiable function, and let $a be a fixed base-point. We will show that for all $x\neq x_{0}$ 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 $x\neq x_{0}$ 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

 $F(\xi)=\sum_{k=0}^{n-1}f^{(k)}(\xi)\,\frac{(x-\xi)^{k}}{k!}+R\,\frac{(x-\xi)^{% n}}{n!},\quad a<\xi

We then have

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

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 ProofOfTaylorsTheorem 2013-03-22 12:33:59 2013-03-22 12:33:59 rmilson (146) rmilson (146) 8 rmilson (146) Proof msc 26A06