The Unapologetic Mathematician

Taylor’s Theorem

I’ve decided I really do need one convergence result for Taylor series. In the form we’ll consider today, it’s an extension of the ideas in the Fundamental Theorem of Calculus.

Recall that if the function $f$ has a continuous derivative $f'$, then the Fundamental Theorem of Calculus states that

$\displaystyle f(x)-f(x_0)=\int\limits_{x_0}^xf'(t)dt$

Or, rearranging a bit

$\displaystyle f(x)=f(x_0)+\int\limits_{x_0}^xf'(t)dt$

That is, we start with the value at $x_0$, and we can integrate up the derivative to find how to adjust and find the value at the nearby point $x$. Now if $f'$ is itself continuously differentiable we can integrate by parts to find

$\displaystyle f(x)=f(x_0)+xf'(x)-x_0f'(x_0)-\int\limits_{x_0}^xtf''(t)dt$

Then we use the FToC to replace $f'(x)$

\displaystyle\begin{aligned}f(x)=f(x_0)+x\left(f'(x_0)+\int\limits_{x_0}^xf''(t)dt\right)-x_0f'(x_0)-\int\limits_{x_0}^xtf''(t)dt\\=f(x_0)+(x-x_0)f'(x_0)+\int\limits_{x_0}^x(x-t)f''(t)dt\end{aligned}

And if $f''$ is itself continuously differentiable we can proceed to find

$\displaystyle f(x)=f(x_0)+(x-x_0)f'(x_0)+\frac{1}{2}(x-x_0)^2f''(x_0)+\frac{1}{2}\int\limits_{x_0}^x(x-t)^2f'''(t)dt$

Is this starting to look familiar?

At the $n$th step we’ve got

$\displaystyle f(x)=\left(\sum\limits_{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k\right)+\int\limits_{x_0}^x\frac{f^{(n+1)}(t)}{n!}(x-t)^ndt$

and if $f^{(n+1)}$ is continuously differentiable we can integrate by parts and use the FToC to find

$\displaystyle f(x)=\left(\sum\limits_{k=0}^{n+1}\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k\right)+\int\limits_{x_0}^x\frac{f^{(n+2)}(t)}{(n+1)!}(x-t)^{n+1}dt$

The sum is the $n$th Taylor polynomial for $f$ — the beginning of the Taylor series of $f$ — at the point $x_0$, and the integral we call the “integral remainder term” $R_n(x)$. For infinitely-differentiable functions we can define $R_n$ for all $n$ and get a sequence. The function $f$ is then analytic if this sequence of errors converges to ${0}$ in a neighborhood of $x_0$.

September 30, 2008 - Posted by | Analysis, Calculus

1. […] 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 […]

Pingback by Taylor’s Theorem again « The Unapologetic Mathematician | October 1, 2008 | Reply

2. You assume that $f^{(n+1)}$ is continuously differentiable. But what if it is only differentiable? Is it still true?

Comment by Sune Kristian Jakobsen | October 6, 2008 | Reply

3. I think the general idea is that derivatives (not necessarily continuous) can behave pretty wildly, so it’s probably not true. An example I’d like to look into more carefully when I have a chance is f(x) = x^2 sin(1/x), which has a derivative everywhere but which I imagine is neither Riemann nor Lebesgue integrable over any interval containing the origin. If that’s the case, then much of John’s post would appear not to generalize in any straightforward way.

Comment by Todd Trimble | October 7, 2008 | Reply

4. I retract my previous example, but not the general idea đź™‚

Comment by Todd Trimble | October 7, 2008 | Reply

5. On the other hand, here is a slight modification: the function f(x) = |x|^{3/2} sin(1/x) has an everywhere defined derivative which I don’t think is Riemann integrable over any interval containing the origin.

Comment by Todd Trimble | October 7, 2008 | Reply

6. Thanks,
I thought every derivative was integrable. Do you know if it is sufficient to assume that $f^{(n+2)}$ is integrable?

Comment by Sune Kristian Jakobsen | October 7, 2008 | Reply

7. Sune, the fine details of the assumptions I’ll admit that I don’t know.

Here I make these assumptions so I can invoke integrations by parts. In the follow-up post, I note that the version there is true under more general assumptions, but that I’m explicitly not looking for the most general possible case.

Teasing out these sorts of pathologies is half of the reason I’m not an analyst.

Comment by John Armstrong | October 7, 2008 | Reply

8. […] which would give a nonzero value to . The second gives the result we’re really after, as Taylor’s theorem applied to shows us […]

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