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