# The Unapologetic Mathematician

## Fall Break

WKU is off for a few days, so I’ll take a break too. I should be back Monday.

## Taylor’s Theorem again

What you say? Taylor’s Theorem again? Well, yes. But not quite. Now instead of showing it as an extension of the Fundamental Theorem of Calculus we’ll see it’s an extension of the Differential Mean Value Theorem. And this despite the most strenuous objections of my resident gadfly that the DMVT and FToC have absolutely nothing whatsoever to do with each other, never have, and never will.

Remember that Taylor’s theorem tells us that if $f$ is $n+1$ times continuously differentiable we can write

$\displaystyle f(x)=\left(\sum\limits_{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k\right)+R_n(x)$

where the remainder term $R_n(x)$ is given in the integral form

$\displaystyle R_n(x)=\int\limits_{x_0}^x\frac{f^{(n+1)}(t)}{n!}(x-t)^ndt$

What I want to say today is that it can also be given in a different form. Specifically, there is some $\xi$ between $x_0$ and $x$ so that

$\displaystyle R_n(x)=\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}$

Now, we could actually weaken the assumptions on $f$ slightly and work up a whole new induction from the ground up using a souped-up version of the DMVT. In particular, notice that the $n=0$ case says there is a $\xi$ between $x_0$ and $x$ so that

$\displaystyle f(x)=f(x_0)+f'(\xi)(x-x_0)$

which is just a different way of writing the DMVT itself.

Instead of doing all that hard work, I’ll prove that this form of the remainder term holds under the same conditions as the integral version. And I’ll do this by using a form of the Integral Mean Value Theorem for Riemann-Stieltjes integration.

Specifically, we’ve got an integral to evaluate

$\displaystyle\int\limits_{x_0}^xf^{(n+1)}(t)\frac{(x-t)^n}{n!}dt$

We set

$\displaystyle d\alpha=\frac{(x-t)^n}{n!}dt$

and antidifferentiate to find

$\displaystyle\alpha=-\frac{(x-t)^{n+1}}{(n+1)!}$

Now the IMVT tells us that there is a $\xi$ between $x_0$ and $x$ so that

\displaystyle\begin{aligned}f^{n+1}(\xi)=\frac{1}{-\frac{(x-x)^{n+1}}{(n+1)!}+\frac{(x-x_0)^{n+1}}{(n+1)!}}\int\limits_{x_0}^xf^{(n+1)}(t)\frac{(x-t)^n}{n!}dt\\\frac{(n+1)!}{(x-x_0)^{n+1}}\int\limits_{x_0}^xf^{(n+1)}(t)\frac{(x-t)^n}{n!}dt\end{aligned}

That is

$\displaystyle\frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1}=\int\limits_{x_0}^xf^{(n+1)}(t)\frac{(x-t)^n}{n!}dt=R_n(x)$

Again, if we can make $R_n(x)$ converge to ${0}$ in a neighborhood of $x_0$, then the infinitely-differentiable function $f$ is actually analytic. Here the limit gets fidgety because we don’t know where between $x_0$ and $x$ we might pick $\xi$, and the $\xi$ jumps around as we use larger and larger $n$. But if we can keep good control on the size of all of the derivatives of $f$ near $x_0$, this can be a useful strategy.

October 1, 2008 Posted by | Analysis, Calculus | 4 Comments