1

Consider the mean-value form of the Taylor Theorem:

Let $k \geq 1$ be an integer and let the function $f : \mathbb{R} → \mathbb{R}$ be $k+1$ times differentiable on the open interval with $f^{(k)}$ continuous on the closed interval between $a$ and $x$.

$ f(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k + \frac{f^{(k+1)}(\xi)}{k!}(x-a)^{n+1}$ for some $\xi \in [a,x]$

Is it necessarily true that

$\lim_{x\rightarrow a}f^{(k+1)}(\xi) = f^{(k+1)}(a)$ ?

I think that this is true because we look at the interval shrinking to $[a,a]$ first. My friend on the other end feels that the limit does not exist since $f^{(k+1)}$ is not continuous. In other words, the function that you are taking limits with respect to should be considered first. However, we are both not sure what is the correct take to this $\xi$ in the remainder term of the Taylor theorem. If someone knows the correct answer to this, do let me know in the comments below!

** Edit **

With regards to the discussion below, I should provide an example to explain why I think this claim is true. The standard counterexample that you guys have provided is $f(x) = x^2 \sin(\frac{1}{x})$. Using Taylor's theorem, we have

$f(x) = f(0) + f'(\xi)x$ for $\xi \in [0,x]$

i.e

$x^2 \sin(\frac{1}{x}) = f'(\xi)x \rightarrow f'(\xi) = x \sin(\frac{1}{x})$ for some $\xi \in [0,x]$. The argument provided is that $f'(\xi(x))$ is not continuous with respect to $x$ at $x=0$. However, we can see that $\lim_{x \rightarrow 0} f'(\xi) = 0$ even if you were to argue from the perspective of "continuous $f'$" or shrinking interval.

HK Tan
  • 175
  • 7
  • Take the same example $x^2 \sin(1/x)$ in which its first derivative is not continuous. We first do the taylor expansion: $f(x) = f(0) + f'(\xi)(x)$ for $\xi \in (0,x)$ Next, $f'(\xi) = f(x)/x$

    Clearly, we now take limit as $x \rightarrow 0$ on both sides. $f(x)/x \rightarrow 0$ means that the limit of $f'(\xi)$ exists and tends to 0 right?

     – HK Tan Sep 24 '19 at 02:19

2 Answers2

1

Your friend is correct that $f^{(k+1)}$ is not necessarily continuous, so we can't say that $f^{(k+1)}(\xi) \to f^{(k+1)}(a)$.

In general, you will want to prove some upper bound on $f^{(k+1)}$ on the interval $[a,x]$ in order to bound the error of the Taylor approximation.

kccu
  • 20,808
  • 1
  • 22
  • 41
  • However, all the standard way to think about continuity is that $f$ is valid for a fixed interval right? For instance, the counterexample on using $x^2 sin(1/x)$ is whereby the derivative is defined everywhere on $x \neq 0$, so our only concern would be the derivative function value converging itself. However, in this situation, it is compounded by the fact that the interval itself shrinks, which would force us to pick $f^{(k+1)}$ at the point $a$, even though the derivative function might not be continuous? – HK Tan Sep 24 '19 at 02:03
  • I'm sorry, I'm not sure I understand what you mean by "$f$ is valid for a fixed interval." – kccu Sep 24 '19 at 02:49
  • Ok to put it this way, most of the time if $f$ is continuous at say $0$, we have $\lim_{x\rightarrow0}f(x) = f(0)$ right? This works because since $f$ is continuous, we can interchange the limit and $f$ to obtain $f(\lim_{x\rightarrow0}x) = f(0)$. This might not work if $f$ is not continuous.

    However, this scenario is slightly different in the sense that I am looking at $\lim_{x\rightarrow0}f(\xi)$ where $\xi \in (0,x)$ depends on $x$. Even though I cannot interchange the limit, I know that since $\xi$ depends on $x$, $\xi $ will go to $0$ under this limit.

     – HK Tan Sep 24 '19 at 03:17
  • @HKTan It sounds like you're talking about a squeeze theorem sort of argument. But it won't work because you're not looking at $f(\xi)$, you're looking at $f^{(k+1)}(\xi)$, and $f^{(k+1)}$ is not necessarily continuous. – kccu Sep 24 '19 at 12:32
  • Yes over here, I didn't use the fact that $f$ is continuous to interchange the limit. I argue on the fact that the interval shrinks towards $(0,0)$ without even considering its expression.

    If what you say is true, we consider the Mean Value Theorem. $f'(c) = \frac{f(x) - f(a)}{x-a}$ for $c \in (a,x)$ and $f$ is differentiable but $f'$ is not continuous. Taking limits on both side, we realise that $\lim_{x\rightarrow a}\frac{f(x) - f(a)}{x-a} = f'(a)$, which is the definition of the derivative.

     – HK Tan Sep 24 '19 at 13:41
  • The limit exists because clearly, $f'$ is differentiable at $a$ and that is the definition of the derivative at $a$. Hence, since the limit of $f'(c)$ exists and is equals to $f'(a)$, can't we conclude based on the "shrinking interval" argument? Of course unless whatever I have argued here in this example on MVT is incorrect. – HK Tan Sep 24 '19 at 13:42
  • 1
    That would imply that every derivative is continuous, which is not the case. Your argument only tells you that there exists a sequence of points $c_n$ such that $c_n \to a$ and $f'(c_n) \to f'(a)$. However, it does not tell you that every sequence of points $c_n \to a$ satisfies $f'(c_n) \to f'(a)$. There could be all sorts of points $c \in [a,x]$ for which $f'(c)$ is nowhere close to $f'(a)$, and that could continue to be the case as you take $x \to a$. – kccu Sep 24 '19 at 16:52
  • Sorry about that, I'm still not convinced. If the limit exists and is equals to a value, shouldn't this means that for all sequences $c_n \rightarrow a$, we have $f'(c_n) \rightarrow f'(a)$? Of course, I agree with you that the logic should never be the continuity of $f'$, but in the way I have worked it out there is no assumption on the continuity of $f'$, but instead, am looking at the shrinkage in the interval. – HK Tan Sep 24 '19 at 17:25
  • Furthermore, you argument only works if the interval is fixed, ie say $c \in [a,b]$ rather than $[a,x]$. The standard argument in a normal analysis/calculus class wouldn't immediately apply since the interval itself depends on $x$. – HK Tan Sep 24 '19 at 17:27
  • The mean value theorem only guarantees that there exists $c \in [a,x]$ such that $f'(c)=\frac{f(x)-f(a)}{x-a}$, but it doesn't tell you anything about all $c \in [a,x]$. Take a look at this function. $f'(0)=0$, and certainly for any $x>0$ there exists $c \in [0,x]$ such that $f'(c)=0$, but it is not the case that $\lim_{c \to 0} f'(c)=0$. – kccu Sep 24 '19 at 17:38
  • My argument does not rely on a fixed interval. It relies on a sequence of shrinking intervals, say $[a,a+1/n]$. – kccu Sep 24 '19 at 17:39
  • Well, the correct expression should be $\lim_{x\rightarrow 0}f'(c)$ (with subscript $x$) . Here, indeed, for any $x > 0$, such a $c$ exists and $\lim_{c\rightarrow 0}f'(c)$ does not exists. But what about $\lim_{x\rightarrow 0}f'(c)$?

    Furthermore, yes it does not tell me anything about all $c\in [a,x]$. But as long as I know that such a $c$ exists and $x \rightarrow a$, then this $c$ must take the value of $a$ even though I have no information about $c$ except for the fact that it is in that interval of choice.

     – HK Tan Sep 24 '19 at 18:40
  • @HKTan: it is best to write $\lim_{x\to 0}f'(c_x)$ and your issue is that you are not convinced that $\lim_{x\to 0}f'(x)$ and $\lim_{x\to 0}f'(c_x)$ are different things. The former may not exist and the latter may. But if the former exists it must equal $f'(0)$ (via the use of mean value theorem as mentioned in your comments). This is a key property of derivatives that they don't have jump derivatives. – Paramanand Singh Sep 25 '19 at 01:57
  • @HKTan: also regarding your last comment $c_x$ does not necessarily equal $a$ but rather as $x\to a$ we have $c_x\to a$. – Paramanand Singh Sep 25 '19 at 02:00
  • Hmm I think I am referring them to different things right? It is precisely the fact that $\lim_{x\rightarrow 0} f'(x)$ might not exists but $\lim_{x\rightarrow0}f'(c_x)$ may which allows me to conclude $f'(c_n) \rightarrow f'(0)$? Because I am not applying nor using the fact that the derivative is continuous. I am just looking at the fact that the interval itself is shrinking, so you will end up at evaluating the expression at the point anyway? – HK Tan Sep 25 '19 at 06:57
  • @HKTan You absolutely can conclude that $\lim_{x \to 0} f'(c_x)$ exists and equals $f'(0)$ for all the reasons you stated. But this does not imply that $\lim_{x \to 0} f'(x)$ exists, since the values $c_x$ might "skip over" points $x$ for which $f'(x)$ is nowhere near $f'(0)$. I.e., if you define a function $x \mapsto c_x$ which outputs a value $c_x$ for which $\frac{f(x)-f(0)}{x-0}=f'(c_x)$, this function need not be continuous. – kccu Sep 25 '19 at 18:06
1

The claim $\lim_{x\to a} f^{(k+1)}(\xi)=f^{(k+1)}(a)$ is true provided that $f^{(k+1)}(a)$ exists (this goes by the name of Taylor's theorem with Peano's form of remainder). Note that existence of $f^{(k+1)}$ at $a$ is not part of your hypothesis.

Paramanand Singh
  • 87,309
  • Yup actually this was mentioned in the question when it specifies that $f$ is $k+1$ times differentiable (ie $f^{(k+1)}$ exists). My bad for not specifying the interval in which this is valid...

    Hence, from your answer, what you are saying is that even though $f^{(k+1)}$ is not continuous on $[a,x]$ but exists, I can still conclude $\lim_{x\rightarrow a} f^{(k+1)}(\xi) = f^{(k+1)}(a)$ right?

     – HK Tan Sep 25 '19 at 06:59
  • @HKTan: continuity of $f^{(k+1)}$ at $a$ is needed to ensure $\lim_{x\to a} f^{(k+1)}(x)=f^{(k+1)}(a)$ (well that's actually the definition of continuity). Your question however deals with $f^{(k+1)}(\xi)$ which by definition is $$f^{(k+1)}(\xi)=\frac{(k+1)!}{(x-a)^{k+1}} \left(f(x)-\sum_{n=0}^{k}f^{(n)} (a) \frac{(x-a) ^n} {n!} \right)$$ and it can be proved (see link in my answer) that the right hand side of above equation tends to $f^{(k+1)}(a)$ as $x\to a$ provided the derivative $f^{(k+1)}(a)$ exists. – Paramanand Singh Sep 25 '19 at 08:51
  • Alright, thanks a lot! The equation itself that you have included makes a lot of sense now. – HK Tan Sep 25 '19 at 09:02