The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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October 1, 2008 - Posted by | Analysis, Calculus

4 Comments »

  1. I like this series. It helps me rebuild my now-poor calculus skills.

    I got a small question though: in the second equation from the bottom: shouldn’t f in the right side should be in its (n+1) derivative?

    Comment by Eyal Ron | October 1, 2008 | Reply

  2. Good catch on the typo. And I’m glad you’re getting something out of this.

    Comment by John Armstrong | October 1, 2008 | Reply

  3. [...] does this series converge back to the exponential function? Taylor’s Theorem tells us [...]

    Pingback by The Taylor Series of the Exponential Function « The Unapologetic Mathematician | October 7, 2008 | Reply

  4. [...] for many of our purposes. However, there’s one thing it’s really good for: generalizing Taylor’s theorem. Specifically, the version of Taylor’s theorem that resembles the mean value theorem. And [...]

    Pingback by Taylor’s Theorem « The Unapologetic Mathematician | October 20, 2009 | Reply


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