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Suppose I have a multiplicative function $f(n),$ and I want to understand the behavior of $$ \sum_{n<x} f(n^k), $$ for some integer $k.$ This seems like it should be easy (since the Dirichlet series seems to be just evaluated at $ks$), but it can't be too easy, since, for example, for $f(n) = \mu(n),$ $f(n^k)$ is identically $0$ for every $k>1.$

EDIT I have no idea why this question engendered such a negative reaction. The perfect answer was given by socalledfriendDon in the comments to the one answer (this is now the accepted answer): under certain conditions, you have an asymptotic formula for average value of multiplicative functions, as given by Wirsing's theorem. Certain conditions are: positivity, having a mean on primes, and having slow growth on prime powers.

For the function $f(n) = \tau(n^k)$ this gives an average value of $\log^k(x).$ Many of the pooh-poohers gave completely incorrect references. One gave a reference to Erdos' paper from 1952, where Uncle Paul analyzes the behavior of $\tau(P(n))$ where $P$ is an irreducible polynomial (AND, while Erdos' result is correct, his proof is buggy). A better reference (which no-one gave) is a 1939 paper by van der Corput, but that gives bounds, and not asymptotics.

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    $\begingroup$ It's not as easy as you think - you don't just evaluate the Dirichlet series at $ks$. In the case that $f(n)$ is the coefficient of a modular form, Iwaniec says (at the very end of Topics in Classical Automorphic Forms), that the Dirichlet series does not admit an analytic continuation past $s = 0$ if $k$ is large enough. $\endgroup$
    – Jeremy Rouse
    Commented Oct 23, 2015 at 21:04
  • $\begingroup$ @JeremyRouse Well, as I said, I wasn't sure if it should be easy or hard, being unwise in this sort of thing (though the Moebius function really makes one wonder). By the way, the specific $f$ which brought this to mind was $f(n) = \tau(n)$ (the number of divisors). There might be a trick which works there... $\endgroup$
    – Igor Rivin
    Commented Oct 23, 2015 at 21:16
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    $\begingroup$ I would have thought that if $f(n^k)$ is identically zero, then the behavior of your sum would be exceptionally easy to understand. $\endgroup$
    – Gerry Myerson
    Commented Oct 23, 2015 at 23:43
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    $\begingroup$ It's not identically 0. $f(1)=1$. (Friday p.m. pedantry) $\endgroup$
    – Anthony Quas
    Commented Oct 24, 2015 at 0:51
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    $\begingroup$ For $\tau(n)$ the count of divisors, $F(s) := \sum \frac{\tau(n^k)}{n^s} = \prod_p \Big(1+ \frac{\tau(p^k)}{p^s} + \frac{\tau(p^{2k})}{p^{2s}}+\cdots\Big)$. But $\tau(p^{k\ell}) = k\ell+1$, so with a little work (if I make no mistake with the algebra) one may check that $F(s) = \zeta(s)^{k+1}\prod_p (1+ (k-1)p^{-s})(1-p^{-s})^{k-1}$, where the latter Euler product converges absolutely for $\Re s > 1/2$. I think this should be enough to get an asymptotic formula by residue calculus for instance. $\endgroup$
    – Brad Rodgers
    Commented Oct 24, 2015 at 23:46

2 Answers 2

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For each fixed $k$, the function $n\mapsto \tau(n^k)$ is a nonnegative-valued multiplicative function. There are quite general results in the literature about mean values of such functions. One classic paper in this area is

Wirsing, Eduard. Das asymptotische Verhalten von Summen über multiplikative Funktionen. (German) Math. Ann. 143 1961 75–102

His main theorem applies to $\tau(n^k)$ and implies that the partial sums in this case are asymptotic to $c x(\log{x})^k$ for an explicit nonzero constant $c$.

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  • $\begingroup$ By the way, the hypotheses to Wirsing's theorem seem tailor-made for this function (which works for me...) but one wonders how widely this is applicable... $\endgroup$
    – Igor Rivin
    Commented Oct 25, 2015 at 20:22
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    $\begingroup$ There's a beautiful expository paper by Moree and Cazaran that touches on this; see cl.ly/0V012j0O1a1B $\endgroup$
    – so-called friend Don
    Commented Oct 25, 2015 at 20:26
  • $\begingroup$ Ah, thanks, that's extremely interesting! $\endgroup$
    – Igor Rivin
    Commented Oct 25, 2015 at 20:33
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Actually it can be arbitrarily hard or arbitrarily easy. To elaborate: For a typical multiplicative function $f$ what determines the behavior of $$ \sum_{n < x} f(n) $$ is the behavior of $f(p)$ (while the $f(p^2)$ don't have much effect unless the function is extremely large at prime squares). Similarly what determines the behavior of $$ \sum_{n \leq x} f(n^2) $$ is mostly the behavior of $f(p^2)$. Therefore if you want the first sum to be easy and the second to be hard take a multiplicative function with say $f(p) = 1$ and $f(p^2) = -1$ (admitedly the Mobius function is hard!). If you want the first sum to be hard and the second to be easy then take $f(p) = -1$ and $f(p^2) = 1$. Further adjustments can be made if you consider dealing with the Mobius function to be "easy".

As an example which occurs in real life take $f$ to be the coefficient of a high symmetric power of a Maass form (say the 8-th symmetric power). Then the behavior of $\sum_{n < x} f(n)$ is somewhat understood (but still poorly as we know continuation only up to the 1-line) but that of $\sum_{n < x} f(n^2)$, which (I think) roughly corresponds to a 16-th symmetric power, is a complete mystery...

EDIT: On the other hand if you have something more specific in mind then please give us the details!

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  • $\begingroup$ Thanks. The multiplicative function I had in mind when asking the question was $\tau$ (the number of divisors). There are many estimates for sums of $\tau(P(n)),$ but no asymptotics that I can find (with the possible exception of $P(n) = n^2,$ though even that is not clear), which might just mean that the question is hard. $\endgroup$
    – Igor Rivin
    Commented Oct 24, 2015 at 22:19
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    $\begingroup$ P. Erdos. "On the sum $\sum^{x} _{k=1}d( f (k))$. J. London Math. Soc. 27 (1952), 7–15. $\endgroup$
    – Alexey Ustinov
    Commented Oct 25, 2015 at 1:29
  • $\begingroup$ @AlexeyUstinov Yes, I am aware of this reference. $f$ is an irreducible polynomial, and the result is a bound, not asymptotic. $\endgroup$
    – Igor Rivin
    Commented Oct 25, 2015 at 10:06
  • $\begingroup$ Perhaps I don't understand the question. But $\tau(n^k)$ is a nonnegative multiplicative function. And there are very general theorems giving asymptotics for partial sums of nonnegative multiplicative functions; e.g., there is a beautiful theorem of Wirsing that immediately applies to this question. See Wirsing, Eduard Das asymptotische Verhalten von Summen über multiplikative Funktionen. (German) Math. Ann. 143 1961 75–102. (But I think Wirsing is "overkill" here, in the sense that Brad Rodgers's suggestion above also works and is simpler.) $\endgroup$
    – so-called friend Don
    Commented Oct 25, 2015 at 18:01
  • $\begingroup$ @so-calledfriendDon That was precisely the sort of thing I was asking about! I will check out the Wirsing reference. $\endgroup$
    – Igor Rivin
    Commented Oct 25, 2015 at 18:18

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