1
$\begingroup$

Can the equality of these two be used as a second-order differentiability check?:

$$\lim_{h\downarrow 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}$$ And, $$\lim_{h\downarrow 0}\frac{f(x-2h)-2f(x-h)+f(x)}{h^2}$$

I think that just like when the limits $\lim\limits_{h\downarrow 0}\frac{f(x+h)-f(x)}{h}$ and $\lim\limits_{h\downarrow 0}\frac{f(x-h)-f(x)}{-h}$ are equal, then it means that the function is differentiable, the equality of these two limits should mean that the function is twice differentiable.

Please note that $h$ itself is assumed to be positive in the limits.

  • 2
    $$\dfrac{\dfrac{f(x+2h)-f(x+h)}h-\dfrac{f(x+h)-f(x)}h}h$$ – 2017-02-28
  • 0
    @labbhattacharjee What's that? – 2017-02-28
  • 0
    @labbhattacharjee You just gave a formula for evaluating the second derivative. I'm talking about differentiability. The limit you gave exists even when the second order derivative doesn't exist. – 2017-02-28
  • 0
    @Legoman Not at all. Have you read what I was asking? – 2017-02-28
  • 0
    The “two limits” are in fact the same, by substituting $-h$ for $h$. Hence, I agree with @Legoman on this one. (If you meat one-sided limit, you should have said so. But it won't change anything.) – 2017-02-28
  • 0
    @HaraldHanche-Olsen I think $\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$ and $\lim_{h\rightarrow 0}\frac{f(x-h)-f(x)}{-h}$ would also be the same by substituting $h=-h$ but they're used for differentiability check. – 2017-02-28
  • 0
    But you can't use the equality of a limit with itself as a check of *anything*!? – 2017-02-28
  • 0
    @HaraldHanche-Olsen What? Equality of the limits $\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$ and $\lim_{h\rightarrow 0}\frac{f(x-h)-f(x)}{-h}$ is used as a differntiability check. If those two limits are equal, then the function is said to be differentiable. But according to you, those two will be always equal by substituting $h=-h$. BTW, this question is at least not a duplicate. There's no mention of these two limits in the other post. My question is about these two limits. – 2017-02-28
  • 0
    I have no idea where you got that from. It's flat out wrong! It looks like a garbled version of the statement that a limit exists if and only if the corresponding *one sided* limits exists and are equal. If each limit had the form $\lim_{h\to0}…$, that would be better. As to the question of duplicate or not, it's not an exact duplicate. But if your question was repaired to make better sense, it *would* become a duplicate. – 2017-02-28
  • 0
    @HaraldHanche-Olsen I've edited my question. I've provided two other limits for a second-order differentiability check. Are these two also identical? – 2017-02-28
  • 0
    @HaraldHanche-Olsen Sorry. I had again made a mistake. I've edited it once again. – 2017-02-28
  • 0
    This time two limits you posted are exactly identical,you just multiplied by $(-1)$ both numerator and denominator. – 2017-02-28
  • 0
    @kingW3 Yeah, I just realized that this is a ridiculous question. I'm voting to close. – 2017-02-28
  • 0
    @kingW3 I again changed back to the previous limits. I think this question is fine now and not at all a duplicate. – 2017-02-28
  • 0
    Are you missing a $2$ in the first limit at $f(x+h)$ or is it supposed to be like that. – 2017-02-28
  • 0
    @kingW3 I've edited it. I know I've edited it too many times. I promise this one's my last edit and I'll not make more corrections. – 2017-02-28
  • 0
    @kingW3 Both these limits are different. Try both for the signum function. Both DON'T evaluate to the same thing always. – 2017-02-28
  • 0
    @HaraldHanche-Olsen Both these limits are different. Try both for the signum function. Both DON'T evaluate to the same thing always. – 2017-02-28
  • 0
    Both the limits in your question are always equal so it can not be used as a check for anything. Their equality follows from replacing $h$ by $-h$ and the fact that $h\to 0$. – 2017-02-28
  • 0
    @HaraldHanche-Olsen See this post of mine to see that these limits don't always evaluate to the same thing:http://math.stackexchange.com/questions/2165280/an-example-to-show-that-these-two-limits-can-be-used-as-a-second-order-different – 2017-02-28
  • 0
    @Dove: $\lim\limits_{h\rightarrow 0}g(h)$ and $\lim\limits_{h\rightarrow 0}g(-h)$ are essentially the same . Either both exist and are equal or none of them exists. But $\lim\limits_{h\downarrow 0}g(h)$ and $\lim\limits_{h\downarrow 0}g(-h)$ may be different. So maybe you are talking about the left and the right derivate $\lim\limits_{h \downarrow 0}\frac{f(x+h)-f(x)}{h}$ and $\lim\limits_{h \downarrow 0}\frac{f(x-h)-f(x)}{-h}(=\lim\limits_{h \downarrow 0}\frac{f(x)-f(x-h)}{h})$ – 2017-02-28
  • 0
    I see that I was right and you have added now the assumption that $h$ is positive – 2017-02-28

0 Answers 0