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Mertens function has, by residues, an explicit formula of

$M(x)=\displaystyle\sum_{\rho}\frac{x^\rho}{\rho\zeta'(\rho)}-2+\sum_{n=1}^\infty\frac{(-1)^{2 n}(2\pi)^{2n}}{(2n)! n \zeta(2n+1)x^{2n}}$

where $\rho$ are the zeros of $\zeta(s)$, as usual.

Meanwhile, if we use this generalized identity for the number of divisors function, $d_z(n)=\displaystyle\prod_{p^\alpha | n}\frac{(z)(z+1)..(z+\alpha-1)}{\alpha!}$, it's not much work to see that the Moebius function $\mu(n)$ is equal to $d_{-1}(n)$, and with $D_z(n) = \sum_{j=1}^n d_z(j)$, that $M(n) = D_{-1}(n)$.

Is there an explicit formula, similar to that of $M(n)$, above, for the more general case of $D_z(n)$ that the formula for $M(n)$ is a specialization of?



Some more detail, in response to Eric N.:

I understand that we can't use residues to get an explicit formula, for the reasons mentioned. But does that lead naturally to the idea that there isn't / couldn't be explicit formulas for $D_k(n), k>0$ that use the Zeta Zeros?

I want to make a visual, intuitive argument here. Here's an identity for $D_z(n)$ for complex z.

$\displaystyle D_z(n) = \frac{z^0}{0!}1+\frac{z^1}{1!}\sum_{j=2}^n \kappa(j)+\frac{z^2}{2!}\sum_{j=2}^n \sum_{k=2}^{\lfloor \frac{n}{j} \rfloor} \kappa(j) \kappa(k)+\frac{z^3}{3!}\sum_{j=2}^n \sum_{k=2}^{\lfloor \frac{n}{j} \rfloor}\sum_{l=2}^{\lfloor \frac{n}{j k} \rfloor} \kappa(j) \kappa(k) \kappa(l)+\frac{z^4}{4!}...$

where $\kappa(n) = \frac{\Lambda(n)}{\log n}$. Define $\displaystyle P_k(n)=\sum_{j=2}^{n}\kappa(j) P_{k-1}(\lfloor \frac{n}{j} \rfloor)$ with $P_0(n)=1$, and restate that as

$\displaystyle D_z(n) = \frac{z^0}{0!}P_0(n)+\frac{z^1}{1!}P_1(n)+\frac{z^2}{2!}P_2(n)+\frac{z^3}{3!}P_3(n)+\frac{z^4}{4!}P_4(n)+...$

$P_k(n) = 0$ if $n < 2^k$, so only $\log_2 n$ terms are non-zero. This means, if you've computed those non-zero $P_k(n)$ terms, it's trivial to compute $D_z(n)$ for any z in $\log_2 n$ operations.

Now, use this identity to animate, in Mathematica, $\displaystyle\frac{(D_z(n)-1)}{z}$ over the range $z = 1$ to $z = -1$.

K[n_] := FullSimplify[MangoldtLambda[n]/Log[n]]
P[n_, k_] := P[n, k] = Sum[ K[j] P[Floor[n/j], k - 1], {j, 2, n}];P[n_, 0] := 1
DD[n_, k_] := Sum[ k^j/j! P[n, j], {j, 0, Log[2, n]}]
Animate[DiscretePlot[ (DD[n, z = Cos[k] ] - 1)/z, {n, 1, 100}], {k, 0, 2 Pi, .0001}]

What you'll see, if you watch this animation, is an animating line that starts as f(x)=(x-1), races down and at its fastest is the Riemann Prime Counting function right when z=0, and then finally comes to a halt at (1-Mertens Function), before it cycles back up - all in all, a nice gradual transformation between those three important functions.

I know it's only an appeal to visuals, but I feel like what's going on at $D_{-.2}(n)/-.2$ looks continuous with what's going on at $D_{.2}(n)/.2$. Here's a closer look at that.

Animate[DiscretePlot[(DD[n, z = Cos[k]*.2] - 1)/z, {n, 1, 400}], {k, 0, 2 Pi, .0001}]

I guess anything's possible, but it deeply offends my senses of symmetry to think the Zeta Zeros are accounting for the high frequency part of the line there, from -.2 to 0, and then in the blink of an eye, something else is accounting for what is almost the exact same high frequency information. Are my instincts wrong?

A Few More Notes About All This

That identity for $D_z(n)$ stems from Linnik's identity summed, inverted, and generalized a bit. There's a corresponding identity for $d_z(n)$ as well.

In the notation above, $P_1(n) = \Pi(n)$, the Riemann Prime Counting Function.

As was casually demonstrated above, $\displaystyle \lim_{z \to 0}\frac{D_z(n)-1}{z} = \Pi(n)$.

You can get this last result more easily by taking $d_z(n)=\displaystyle\prod_{p^\alpha | n}\frac{(z)(z+1)..(z+\alpha-1)}{\alpha!}$ and noting that $\displaystyle \lim_{z \to 0}\frac{d_z(n)}{z} = \frac{\Lambda(n)}{\log n}$ except at 1, where the limit is infinity.

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  • 1
    $\begingroup$ There are no such explicit formulas for $D_2(x)$ and higher. The zeros only contribute to $M(x)$ because they correspond to poles of the function $\frac{1}{\zeta(s)}$. If you are interested in what the left over term looks like, take a look at the Dirichlet Divisor problem. $\endgroup$ – Eric Naslund Aug 19 '13 at 16:43
  • $\begingroup$ Nathan - I think you make some interesting observations. Perhaps you should write it up in a paper and circulate it for comment. I guess what we can tell you is what kind of formulas can be easily proved using existing technology. For your more general observations and what theoretically might be possible to prove, maybe try cross posting to MathOverflow. The only thing I'll add is that meromorphic functions such as the zeta function are very "rigid", and trying to distinguish what part of their behavior comes from their zeros and what part comes from the "rest" is a difficult distinction. $\endgroup$ – John M Aug 20 '13 at 12:07
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I can give a partial answer to this. My apologies if I am telling you what you already know.

First, let's review how the explicit formula for the Merten's function is derived. We have that $$\frac{1}{\zeta(s)} = \sum_{n=1}^\infty \mu(n) n^{-s}.$$

We have $$M(x) = \frac{1}{2 \pi i}\int_{2-i\infty}^{2+i\infty} \frac{1}{\zeta(s)} \frac{x^s}{s} dx,$$ and the residue theorem gives you the explicit formula you mentioned above. Note that the zeros of $\zeta$ become poles of $\zeta^{-1}$, so the two summations in your formula correspond to the nontrivial zeros and the trivial zeros of Riemann zeta function respectively. Also the $1/s$ term gives us a pole at $s=0$, giving us a residue of $1/\zeta(0) = -2$.

More generally, we can define a function $d_k(n)$ by the Dirichlet series given by $\zeta^k$:

$$\zeta(s)^k = \sum_{n=1}^\infty d_k(n) n^{-s}.$$

Then, for $k\geq 1$, $d_k(n)$ gives the number of representations of $n$ as a product of $k$ positive integers, given explicitly by your product formula above. In particular, $d_2(n)$ gives the number of positive divisors of $n$. Also, as you mentioned, $d_{-1}(n) = \mu(n)$, the Möbius function.

(For $k<-1$, I am have not checked whether my definition using the Dirichlet series will coincide with your product formula.)

Now you can attempt to find an explicit formula by carrying out the contour integral $$D_k(x) = \sum_{n<x} d_k(n) = \frac{1}{2 \pi i}\int_{2-i\infty}^{2+i\infty} \zeta(s)^k \frac{x^s}{s} dx.$$

Now we can see that the explicit formula you will get will differ greatly depending on whether $k$ is positive or negative. If $k$ is positive, the main term will come from the residue of the pole at $s=1$, and this main term will be of the form $xP(\log x)$, where $P$ is a polynomial of degree $k-1$. If $k$ is negative, you'll pick up residues at all the zeros of $\zeta$, as in the case of $k=-1$.

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  • $\begingroup$ Thanks - I am generally familiar with the part of this that you covered, I'm most curious about the last step at the end, producing explicit formulas. Are there explicit formulas for D_k (say, for D_1 or D_2) that make use of zeta zeroes the way that Mertens function one does? And do you have idea about what happens when k is not an integer? Or if k is complex? $\endgroup$ – Nathan McKenzie Aug 19 '13 at 14:23
  • $\begingroup$ @NathanMcKenzie: For $k>0$, we get the main term mentioned above plus an error term, i.e. we only get an estimate, not an exact formula. See en.wikipedia.org/wiki/… for more information on this, and Ch 12 of Titchmarsh "The theory of the Riemann zeta function." I don't think these calcuations use zeta zeros directly. I don't know what could be done with non-integer $k$. Perhaps you could try to tackle the explicit formula for $D_{-2}$. I haven't seen that done before, but should be doable (but likely would be a complicated mess). $\endgroup$ – John M Aug 19 '13 at 14:59
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There are explicit formulas for the generalized divisor summatory function $D_z(n)$ defined in the question above for all $z\in\mathbb{Z}$, and there are also explicit formulas for the normal divisor summatory function $D_k(n)=\sum\limits_{n\le x}\sigma_k(n)$ for all $k\in \mathbb{Z}$. However, the theoretical convergence of these explicit formulas has not been established.


A specific example related to the question above is an explicit formula for $D_1(x)=\sum\limits_{n\le x}1=\lfloor x\rfloor$ which I illustrated in the following answer to one of my own questions.

Definition and Illustration of Explicit Formula for $D_1(x)=\lfloor x\rfloor$


The normal divisor summatory function $D_k(n)=\sum\limits_{n\le x}\sigma_k(n)$ is related to the Mertens function $M(x)=\sum\limits_{n\le x}\mu(n)$ and summatory square-free integer function $Q(x)=\sum\limits_{n\le x}|\mu(n)|$ as follows.

(1) $\quad D_k(n)=\sum\limits_{n=1}^x a(k,n)\,M\left(\frac{x}{n}\right),\quad a(k,n)=\sum\limits_{d|n}\sigma _k(d)$

(2) $\quad D_k(n)=\sum\limits_{n=1}^x b(k,n)\,Q\left(\frac{x}{n}\right),\quad b(k,n)=\sum\limits_{\substack{d|n\\\sqrt{d}\in\mathbb{Z}}}\left(\frac{n}{d}\right)^k$


The explicit formulas for $M(x)$ and $Q(x)$ are as follows.

(3) $\quad M_o(x)=\sum\limits_\rho\frac{x^{\rho}}{\rho\,\zeta'(\rho)}-2+\sum\limits_{n=1}^N\frac{x^{-2\,n}}{(-2\,n)\,\zeta'(-2\,n)}\,,\quad N\to\infty$

(4) $\quad Q_o(x)=\frac{6\,x}{\pi^2}+\sum\limits_\rho\frac{x^{\frac{\rho}{2}}\,\zeta\left(\frac{\rho}{2}\right)}{\rho\,\zeta'\rho)}+1+\sum\limits_{n=1}^N\frac{x^{-n}\,\zeta(-n)}{(-2\,n)\,\zeta'(-2\,n)}\,,\quad N\to\infty$


The relationships illustrated in (1) and (2) above suggest the explicit formulas defined in (3) and (4) above can be used to derive explicit formulas for $D_k(n)$ as follows.

(5) $\quad D_o(k,n)=\sum\limits_{n=1}^x a(k,n)\,M_o\left(\frac{x}{n}\right)$

(6) $\quad D_o(k,n)=\sum\limits_{n=1}^x b(k,n)\,Q_o\left(\frac{x}{n}\right)$


The formulas for $D_o(k,n)$ defined in formulas (5) and (6) above are illustrated below for $k\in\{-1,0,1\}$. The explicit formulas are illustrated in orange and the corresponding reference functions are illustrated in blue. The red discrete portions of the plots illustrate the evaluations of the explicit formulas at integer values of $x$. All plots are evaluated over the first 200 pairs of zeta zeros, and sums over $n$ in the corresponding underlying explicit formulas are also evaluated with the upper limit $N=200$.


The following three plots illustrate the explicit formula $D_o(k,n)$ derived from $M_o(x)$ as defined in (5) above for $k\in\{-1,0,1\}$.


Illustration of formula (5) for $D_o(-1,n)$

$\text{Figure (1): Illustration of formula (5) for $D_o(-1,n)$}$


Illustration of formula (5) for $D_o(0,n)$

$\text{Figure (2): Illustration of formula (5) for $D_o(0,n)$}$


Illustration of formula (5) for $D_o(1,n)$

$\text{Figure (3): Illustration of formula (5) for $D_o(1,n)$}$


The following three plots illustrate the explicit formula $D_o(k,n)$ derived from $Q_o(x)$ as defined in (6) above for $k\in\{-1,0,1\}$.


Illustration of formula (6) for $D_o(-1,n)$

$\text{Figure (4): Illustration of formula (6) for $D_o(-1,n)$}$


Illustration of formula (6) for $D_o(0,n)$

$\text{Figure (5): Illustration of formula (6) for $D_o(0,n)$}$


Illustration of formula (6) for $D_o(1,n)$

$\text{Figure (6): Illustration of formula (6) for $D_o(1,n)$}$


Note the explicit formula $D_o(k,n)$ derived from $M_o(x)$ illustrated in Figures (1), (2), and (3) above exhibits diverging negative spikes just prior to integer values of $x$, whereas the explicit formula $D_o(k,n)$ derived from $Q_o(x)$ illustrated in Figures (4), (5), and (6) above seem to converge much better. I've investigate several other explicit formulas analogous to the ones illustrated here, and the presence or absence and direction of these diverging spikes seems to be a function of the underlying explicit formula, the target function, and in some cases specific integer values of $x$. For example, see the illustrations I provided in an answer I posted at the following link.

Illustration of Explicit Formulas for $\pi(x)$ Derived from $M_o(x)$ and $Q_o(x)$

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The general divisor functions sum is given by

$$ \sum _{n \le x} \sigma _{k} (n)= \sum_{n=1}^{\infty}n^{k} [x/n].$$

No residue theorem is needed.

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  • $\begingroup$ This answer is not helpful. "Explicit formula" in analytic number theory means "formula in terms of zeroes of $L$-functions". $\endgroup$ – Peter Humphries Apr 15 '17 at 16:42

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