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?


  • 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

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 $d(x)$.

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