2
$\begingroup$

From Apostol - Introduction to analytic number theory (Theorem 3.3) we have $$ x\geq1, \sum_{n\leq x}d(n)=x\log x+(2\gamma-1)x+O(\sqrt{x}):=E(x), $$ I want to differentiate $E$ -- to get a rough estimate for $d(n)$ -- but I don't know how to deal with the big-o part (even by proceeding with its rigorous definition doesn't get me anywhere).

  • How to differentiate a function with a big-o?

Thanks

$\endgroup$
  • 2
    $\begingroup$ I don't think you can differentiate a function in a big-O notation: consider a function that is bounded by $\sqrt{x}$ (so that it is in $O(\sqrt{x})$) but with many oscillations $\endgroup$ – angryavian Sep 5 '14 at 17:35
  • $\begingroup$ ^ Exactly, take $\sin(x^2)$ for example. Belongs to $O(1)$ but its derivative $2x\cos(x^2)$ has no bound. $\endgroup$ – A.E Sep 5 '14 at 17:37
  • $\begingroup$ Understood. Thanks. $\endgroup$ – user173987 Sep 5 '14 at 17:45
1
$\begingroup$

As commenters said, a bound on a function does not give any bound in its derivative.

However, here we have partial sums of a nonnegative function of an integer argument. So at least something can be said (though it will be trivial):
$$\sum_{n\leq x}d(n)=x\log x+(2\gamma-1)x+O(\sqrt{x}) \tag{1}$$ implies $$d(x) \le x\log x+(2\gamma-1)x+O(\sqrt{x}) \tag2$$ Obviously, this is totally useless since by definition of $d$ we have $d(x) \le x$ anyway.

But if we didn't know anything about $d$ except (1) and $d\ge 0$, then (2) is the best bound one can have, since it is conceivable that $d$ is zero for $1,\dots,x-1$ and all of the partial sum in (1) is just $d(x)$.

| cite | improve this answer | |
$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.