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Let $f:D \to \mathbb{R}$ be a real function with $U \subset D$ an open interval. Suppose that $f$ is $k$ times differentiable at $a \in U$. Then

$$R_k(x) \in o(|x-a|^k) \ \text{as} \ x \to a$$

where $R_k(x)=f(x)-P_k(x)$ and $P_k$ is the $k$th order Taylor polynomial of $f$ evaluated at $a$.

Like many, I am somewhat averse to the usage of L'Hospital's rule. Is there a proof of the above without it? Note the minimal conditions on $f$.

Community
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MathematicsStudent1122
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  • If i remember, most just use an extension of the mean value theorem, rsther than lhopitals rulr – Andres Mejia Sep 03 '16 at 13:36
  • @AndresMejia Those are for the formulae for the remainder term. Is this a corollary of those? If so, it's not immediately apparent to me how. Both Spivak and Wiki use L'hopital for this (asymptotic) result. – MathematicsStudent1122 Sep 03 '16 at 13:37
  • Even though I have marked your question as duplicate of my own question, +1 for you because of your statement "like many, I am somewhat averse to the use of L'Hospital's Rule". – Paramanand Singh Sep 03 '16 at 21:45

1 Answers1

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Let's do it with $k=3$ and $a=0$ to illustrate. To simplify, let $g(x)=R_3(x).$ Then $0 = g(0)= g'(0)= g''(0)=g'''(0).$ By the MVT,

$$g(x)= g(x)-g(0) = g'(c_1)x = (g'(c_1)-g'(0))x$$ $$ = g''(c_2)c_1x = (g''(c_2)-g''(0))c_1x = \frac{g''(c_2)-g''(0)}{c_2}c_2c_1x.$$

As $x\to 0,$ $c_1,c_2$ are dragged along to $0$ with it. We have $c_2c_1x = O(x^3)$ and the fraction on the right $\to g'''(0) = 0.$ Thus $g(x) = o(x^3)$ as desired.

zhw.
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  • How do we know $g$ is differentiable at $c_1, c_2$ or any $x \neq 0$? The theorem assumes differentiability at a point, not a neighbourhood. – MathematicsStudent1122 Sep 08 '16 at 17:45
  • For $f'''(a)$ to exist, $f''$ must exist in a neighborhood of $a.$ – zhw. Sep 08 '16 at 17:54
  • Fine but I also have another issue. I don't think you can manipulate the mean value theorem derivative like that. As an example, consider simply $$h(x) := \frac{f(x) - f(a)}{x-a} = f'(c), c \in (a,x)$$

    where $f$ is differentiable. As $x \to a$, $h(x) \to f'(a)$. However, even though $c$ is being dragged to $a$, $f'(c)$ does not necessarily $\to$ $f'(a)$ (suppose $f \notin C^1$, for example).

     – MathematicsStudent1122 Sep 08 '16 at 18:52
  • Or, what I mean is, taking the limit with the mean value theorem quantity ("$c$") isn't justified (without regularity conditions on $f$). – MathematicsStudent1122 Sep 08 '16 at 19:06
  • I don't follow you. Please read the proof carefully. Indicate exactly where you spot the first difficulty. – zhw. Sep 08 '16 at 19:31
  • It's the term $\frac{g''(c_2) - g''(0)}{c_2}$. You are taking the limit as $x \to 0$ and arguing that $c_2$, the mean value theorem quantity, is being "dragged" to the limit $g'''(0) = 0$. The purpose of my comment is to suggest that this "dragging" may not be formally valid when taking limits. – MathematicsStudent1122 Sep 08 '16 at 21:13
  • It's not a question of "formally". And no, I'm not saying that. I am saying $c_2$ is being dragged to $0.$ Do you agree? (Recall $c_1$ is between $0$ and $x,$ and $c_2$ is between $0$ and $c_1.$) – zhw. Sep 08 '16 at 21:23
  • Yes, I do, if you read my above comment. I am saying that taking the limit with the $c_2$ quantity, however, as you do, is problematic. – MathematicsStudent1122 Sep 08 '16 at 21:26
  • No it's not. If $c_2 \to 0$ as both you and I agree, then what do you think $(g''(c_2) - g''(0))/c_2$ goes to? – zhw. Sep 08 '16 at 21:29
  • That's where my example comes in. If $h(x) = \frac{f(x) - f(a)}{x-a}$ and $f$ is differentiable, letting $x \to a$ is valid. $\lim_{x \to a} h(x) = f'(a)$. But $\frac{f(x) - f(a)}{x-a} = f'(c)$ for some $c \in (a,x)$. Now do you agree as $x \to a$, $c \to a$? You must also agree, then, that as $c \to a$, $f'(c) \to f'(a)$. But what if that limit does not exist? Say $f(x) = x^2\sin(\frac 1x)$. – MathematicsStudent1122 Sep 08 '16 at 21:35
  • No, you've misunderstood my argument. I'm looking at $g''(c_2)-g''(0))/c_2.$ I am not using the MVT on this. i'm just saying that, by definition, this $\to g'''(0).$ – zhw. Sep 08 '16 at 21:39