0
$\begingroup$

I was investigating variations on defining the second derivative of a function $f$ with respect to another function $g$, using the quotient rule. But the quotient rule doesn't hold if the function in the denominator is not continuous. I'm not sure how this affects second derivatives. The two limits below should be generalizations of the second derivative of $f$ with respect to $g$ and with independent variable $x$, but I'm not sure how discontinuities in $g$ could affect the second limit.

Do there exist two functions $f$ and $g$ such that, for some value of $x$, at least one of the two limits $$\lim_{h \to 0}\frac{\frac{f\left(x+2h\right)-2f\left(x+h\right)+f\left(x\right)}{g\left(x+h\right)-g\left(x\right)}\left(g\left(x+h\right)-g\left(x\right)\right)-\frac{g\left(x+2h\right)-2g\left(x+h\right)+g\left(x\right)}{g\left(x+h\right)-g\left(x\right)}\left(f\left(x+h\right)-f\left(x\right)\right)}{\left(g\left(x+h\right)-g\left(x\right)\right)\left(\left(g\left(x+h\right)-g\left(x\right)\right)+\left(g\left(x+2h\right)-2g\left(x+h\right)+g\left(x\right)\right)\right)}$$ or $$\lim_{h \to 0}\frac{\frac{f\left(x+2h\right)-2f\left(x+h\right)+f\left(x\right)}{g\left(x+h\right)-g\left(x\right)}\left(g\left(x+h\right)-g\left(x\right)\right)-\frac{g\left(x+2h\right)-2g\left(x+h\right)+g\left(x\right)}{g\left(x+h\right)-g\left(x\right)}\left(f\left(x+h\right)-f\left(x\right)\right)}{\left(g\left(x+h\right)-g\left(x\right)\right)^2}$$ is defined, but the two limits are not equal? In other words, are these two limits always equivalent?

$\endgroup$
  • $\begingroup$ Is this better? $\endgroup$ – Georgelemental Apr 25 at 16:35
  • $\begingroup$ Already better, yes. Your edit has placed the question in the Review Queue, so that users can see what you've added and vote to reopen or leave a comment for feedback if they see fit (I just did). Alternatively some people might see the question on the front page due to the activity and do the same. You may have to wait a little bit before the question is reopened. $\endgroup$ – Arnaud D. Apr 25 at 17:10
  • $\begingroup$ I've just realized you've been a member for some time on another SE site (the association bonus should have been a hint for me...), so sorry if you knew all this. $\endgroup$ – Arnaud D. Apr 25 at 17:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.