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According to the Wikipedia entry and a few I've seen online, the remainder form with a $(n+1) \text{th}$ derivative can be used as long as $f: \mathbb R \to \mathbb R$, is $n+1$ times differentiable and $f^{(n)}$ is continuous. I am going to assume bounded intervals here since I think that is implicit in most cases. The remainder form is written like this: \begin{equation} f(x)=f(a)+\sum_{k=1}^{n}\frac{f^{(k)}(a)(x-a)^k}{k!}+ \frac{f^{(n+1)}(c)(x-a)^{n+1}}{(n+1)!} \end{equation} and so there is a $(n+1) \text{th}$ derivative involved, and my question is how do we know that the derivative is bounded? According to the comments on this answer: https://math.stackexchange.com/a/492165/463358, it needs to be asserted that the $(n+1) \text{th}$ derivative is bounded, but then someone comments that the assumptions above are enough to imply the boundedness.

I am currently thinking that the boundedness isn't implied by the conditions above, since $f^{(n)}$'s continuity seems to at best imply uniform continuity on bounded intervals, but not something stronger like absolute or Lipschitz. And since $f^{(n+1)}$ is not necessarily continuous we can't assert that it is bounded on a bounded interval.

I am not sure what I'm missing here and I've searched around a lot but can't find anything clear enough. Thanks for the help!

(There's a similar question here: Conditions of the Taylor Theorem, but it doesn't seem to address my particular question.)

Slade
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    If $;f;$ is $;n+1;$ times differentiable, then $;f^{(n)};$ is always continuous. – DonAntonio May 11 '19 at 23:22
  • Oh yeah, I was just copying the claim on Wikipedia. So I included what it said – Slade May 11 '19 at 23:23
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    Why do you need bounded (n+1)th derivative? Only its existence is required. Just think of $n=0$ and this becomes usual mean value theorem and it can be applied on $f(x) =\sqrt{x} $ on $[0,1]$. The derivative $f'$ exists in $(0,1)$ and is unbounded. – Paramanand Singh May 12 '19 at 01:58
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    The proof of Taylor's theorem using integrals assumes that $(n+1)$'th derivative is continuous (and therefore bounded) but this is a limitation of the proof. A proof without using integrals does not require the continuity of $f^{(n+1)}$. – Paramanand Singh May 12 '19 at 02:04
  • If you're trying to get an idea of the error of an approximation, if the derivative is infinite isn't that 'misleading' in a sense? For example if I use the above remainder theorem and say that the error is on the order of $(x-a)^{n+1}$, isn't this not true, if the derivative is infinite? – Slade May 12 '19 at 02:19
  • For example: https://en.wikipedia.org/wiki/Finite_difference#Higher-order_differences If the approximations are on certain orders due to Taylor's theorem then shouldn't it also assert that those derivatives are bounded? – Slade May 12 '19 at 02:25

2 Answers2

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As pointed out in comments, the derivative $f^{(n+1)}$ does not have to be bounded and the remainder form is still valid.

Consider the example where $n=0$ and $f(x) = x^2 \sin (x^{-2})$ for $0 < x \leqslant 1$ and $f(x) = 0$ for $x = 0$. Here, $f$ is continuous and differentiable at every point in $[0,1]$. In particular, $f(0) = f'(0) = 0$ and for $0 < x \leqslant 1$,

$$f'(x) = 2x \sin(x^{-2}) - 2x^{-1} \cos(x^{-2}) $$

where $f'$ is unbounded in a neighborhood of $x = 0$ due to the second term on the RHS.

However, for any $x \in (0,1]$ there exists $\xi$ such that

$$x^2 \sin(x^{-2}) = 2\xi \sin(\xi^{-2}) - 2\xi^{-1}\cos(\xi^{-2})$$

For example, taking $x=1$, check that $\xi \approx 0.0118822803500313$ works (and there are many more admissible values).

What this is telling you is that even if a low-order Taylor approximation is valid with this remainder, the relative error represented by that remainder can be extremely large-- as in this example of the constant approximation to a wildly oscillating function.

RRL
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  • If you read my last comments above, I guess then that is my real confusion. How can Taylor's theorem be applied to 'bound' errors to a certain order, if the derivatives are possible infinite? – Slade May 12 '19 at 02:35
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    If you look at applications like numerical integration, e.g., trapezoidal rule, the Taylor approximation with remainder is used when that last derivative is assumed bounded to get the $\mathcal{O}(h^2)$ accuracy. If the function is ill-behaved then that accuracy goes away for precisely this reason. – RRL May 12 '19 at 02:38
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    Look at how I got the classic error estimate for the mid-point integration rule here. I had to assume the derivative in the remainder was bounded. – RRL May 12 '19 at 02:42
  • Oh wow. Okay that clears everything up. Thanks a lot for your help! – Slade May 12 '19 at 02:45
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    @Slade: You're welcome. Another way to see it is that if the derivative in the remainder is unbounded the error of the approximation may not go to $0$ as $x \to a$ like $\mathcal{O}((x-a)^{n+1})$. It may reduce at a slower rate or even persist depending upon how fast the derivative grows. – RRL May 12 '19 at 02:59
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I came across an awesome PDF document (link at bottom) that shows the Taylor's Polynomial of the function sin(x) and how the remainder is generated. The PDF makes clear that the coefficient of the remainder is essentially just a number that bounds the error. Because the proof of based on Rolles theorem, changing the internal [a,b] even a bit could cause the c of f(c);to deviate strongly from whatever prior constant of the error term is assumed. So the example illustrates that K (proportional to the error) is a number that depends on the interval you choose, and that Roles theorem breaks the requirement for K to vary smoothly as the interval you consider changes.

Another fun fact. For any (n+1)th derivative, the remainder will be a finite number, so the existence of a finite number isn't enough to show that the function converge properly. You also have to show that Lim(nx>inf)R_(n+1)->0.

Link [Will Automatically Download or open PDF depending on browser]: https://www.google.com/url?sa=t&source=web&rct=j&url=https://math.dartmouth.edu/archive/m8w10/public_html/m8l02.pdf&ved=2ahUKEwiA0tbvlu7iAhVhw1kKHfiGDCoQFjAAegQIARAB&usg=AOvVaw0_Wr6Qd57AWe5hQwk7gZt5

I wrote this answer on my phone, hence the trash quality, but hopefully you will find it useful.