2
$\begingroup$

The functions in question are

$$L(s)=\sum_{k=1}^\infty \frac{\lambda(k)}{k^s}=\frac{\zeta(2s)}{\zeta(s)} \mbox{ and } L^*(s)=\frac{1}{2}\sum_{k=1}^\infty \frac{\lambda(k)+(-1)^{k+1}}{k^s}=\frac{L(s)+\eta(s)}{2},$$

where

An equivalent discussion could be based on the sister Moebius and Merterns function, but here our focus is on Liouville. The notations $L_n$ and $L_n^*$ are used to denote the first $n$ terms of the series representing respectively $L(s)$ and $L^*(s)$ when $s=0$. The asymptotic behavior of $L_n$ has well-known deep implications on the Riemann Hypothesis (RH), discussed later in this post.

Note: the series defining $L^*(s)$ might or might not converge if $\frac{1}{2}<s\leq 1$ (that's the purpose of my question), however the formula $L^*(s)=(L(s)+\eta(s))/2$ is valid only for $\Re(s)>1$ because the series defining $L(s)$ converges only for $\Re(s)>1$ (at least according to Wikipedia, see here).

My main question can be stated in simple words:

My question

Does the series representing $L^*(s)$ converge and is it an analytic function if $\Re(s)>\frac{1}{2}$? Probabilistic arguments in favor of a positive answer are discussed in the appendix. Assuming the answer is yes, then $L(s)=2L^*(s)-\eta(s)$ is an analytic continuation of the original $L(s)$, from $\Re(s)>1$ to $\Re(s)>\frac{1}{2}$. Consequences are discussed in the next section.

I computed $L^*(s)$ for various $s$ with $\frac{1}{2}<\Re(s)\leq 1$ using the first million terms of the series, and compared with $L^*(s)=\frac{1}{2}\Big(\eta(s)+\frac{\zeta(2s)}{\zeta(s)}\Big)$ computed by Mathematica: the first three digits are identical. In particular, $L^*(1)=\frac{\log 2}{2}$ and the analytic continuation gives $L(1)=0$.

Discussion

The sequence $\{\lambda(k)\}$ consists of $+1$ and $-1$ that seem rather randomly distributed, though not perfectly randomly, but supposedly randomly enough (see here) as to imply the Riemann Hypothesis (RH) - still a conjecture at this point. For instance, if you look at runs of $+1$ or $-1$ (subsequences of consecutive $+1$ or consecutive $-1$) the probability for a run to be of length $m>0$ is equal to $2^{-m}$, as in a sequence of i.i.i. Bernoulli trials. More about this is discussed in the appendix. Yet this is not enough to make the series for $L(s)$ converge if $\Re(s)\leq 1$. Also, it seems there is a little bias in the sequence $\{\lambda(k)\}$, which might favor $L_n$ to be negative more frequently than positive, unlike perfect random walks. For instance, if $1<n< 906150257$, then $L_n$ is always negative (see here), but not at $n=906380357$, thus disproving Polya's conjecture. That bias might disappear when $n$ becomes extremely large.

Now if you replace $\lambda(k)$ by $\lambda^*(k)=(\lambda(k)+(-1)^{k+1})/2$ corresponding to the terms in the series defining $L^*(s)$, the distribution of run lengths (if you ignore the terms $\lambda^*(k)$ equal to zero), is significantly altered. This would also be true, with the exact same impact, if you apply the same trick to i.i.d. Bernoulli trials or other sequences with similar distribution of $+1$ and $-1$. In the sequence $\{\lambda^*(k)\}$, assuming you omit the zero terms which do not contribute to the sums $L_n^*$ or $L^*(s)$, the probability for a run to be of length $m>0$ is now equal to $2\cdot 3^{-m}$, down from $2^{-m}$ for the original sequence $\{\lambda(k)\}$. This means that on average, runs are now much shorter, to the point that it makes the series $L^*(s)$ converge even if $\frac{1}{2}<\Re(s)\leq 1$. In essence, our trick is what gets us an analytic continuation from $\Re(s)>1$ to $\Re(s)>\frac{1}{2}$. Transforming $L(s)$ into $L^*(s)$ is the same trick as transforming $\zeta(s)$ into $\eta(s)$, though in the latter case, it extents analycity from $\Re(s)>1$ to $\Re(s)>0$.

An interesting question is whether you can infer some useful result about $\lim \sup_{n \rightarrow \infty} L_n$ from the smoother $L^*_n$, as the $\lim\sup$ in question is intimately connected to the Riemann Hypothesis (see my previous post here). Another curious facts is:

$$\sum_{k=1}^n \lambda(k)\Big\lfloor \frac{n}{k}\Big\rfloor =\lfloor \sqrt{n}\rfloor ,$$

where $\lfloor \cdot \rfloor$ denotes the integer part function. This formula is mentioned here, and it allows you to compute $\lambda(k)$ recursively without using a table of prime numbers. Finally,

$$\zeta(s)=\frac{\zeta(2s)}{2L^*(s)-\eta(s)} = \frac{1}{2L^*(s)-\eta(s)}\cdot\prod_{k=1}^\infty \frac{1}{L(2^k s)}$$

gives (by successive applications of the first equality) an infinite product for $\zeta(s)$ converging (and analytic) if $\Re(s)>\frac{1}{2}$. The roots of $\zeta(s)$ in the critical strip $\frac{1}{2}<\Re(s)<1$ (if any, RH says that there are none) are identical to the poles of $L^*(s)$, or in other words, to the roots of $1/L^*(s)$ in the same strip.

Appendix

Here we show what would happen if the numbers $\lambda(k)$ were replaced by independent random variables $X_k$ taking value $1$ with probability $\frac{1}{2}$, and $-1$ with the same probability. This is pretty much (but not exactly) the way the $\lambda(k)$'s are behaving. Let's define

$$Z=\sum_{k=1}^\infty \frac{X_k}{k^s}.$$

Here $Z$ is a complex random variable, and $s=\sigma +it$. Its expectation is zero, and its variance is given by

$$Var[\Re(Z)]=Var[X_1]\cdot\sum_{k=1}^\infty \frac{\cos^2(t\log k)}{k^{2\sigma}},\\ Var[\Im(Z)]=Var[X_1]\cdot\sum_{k=1}^\infty \frac{\sin^2(t\log k)}{k^{2\sigma}}.$$

Both series converge if $\sigma=\Re(s)>\frac{1}{2}$. In that case $Var[Z]=Var[\Re(Z)]+Var[\Im(Z)] =Var[X_1]\cdot \zeta(2\sigma)$. A formula related to the characteristic function, if $\tau$ is a real number, is the following:

$$E[\exp(i\tau Z)]=\prod_{k=1}^\infty \cos(\tau k^{-s}) .$$

$\endgroup$
11
  • 6
    $\begingroup$ Since the series for $\eta$ converges, the series for $L^*$ converges iff the one for $L$ converges. You've stumbled upon a statement equivalent to RH again. $\endgroup$ – Wojowu May 7 at 18:18
  • 2
    $\begingroup$ The function $L$ is equivalent to $\zeta$ in $\Re s >1$ in a well-defined sense (Bohr equivalence) and Turan prefered it in his RH research;for $C(x)=\sum_{n \le x}\frac{\lambda(n)}{n}$ there are famous results of Turan relating this function with RH (or weaker versions); it is known that if $C(x)$ would be nonnegative for large $x$ RH would follow but that result is false as $C$ takes infinitely many negative values; however estimates of how negative $C$ can be imply results about the zeroes of RZ (eg if for some $a,c>0$ we have $C(x) > -c\frac{\log^a x}{\sqrt x}, x \ge x_0$ then RH true) $\endgroup$ – Conrad May 7 at 18:59
  • 2
    $\begingroup$ Have you read Titchmarsh's book yet? (Probably not, because it takes about a year to read such a book.) $\endgroup$ – GH from MO May 7 at 21:18
  • 2
    $\begingroup$ If you have a series of the form $\sum a_k+b_k$ and you know $\sum a_k$ converges, then $\sum a_k+b_k$ converges iff $\sum b_k$ converges (and then $\sum a_k+b_k = \sum a_k+\sum b_k$). $\endgroup$ – Wojowu May 7 at 21:51
  • 4
    $\begingroup$ Questions are good - if they are of research level. Your questions are usually not of research level, and they tend to be lengthy, exactly because you are not familiar with the background theory. I don't want to discourage you to be curious and ask original questions. Those are very good things and part of being a scientist. It's just that it takes a lot of studying before one can come up with good questions. $\endgroup$ – GH from MO May 7 at 23:50
8
$\begingroup$

Let $s\in\mathbb{C}$ be any point with $\Re s>0$. The Dirichlet series of $\eta(s)$ converges, hence $L(s)$ converges if and only if $L^*(s)$ converges.

For $\sigma_0>1/2$, it is also known that $L(s)$ converges in the half-plane $\Re s>\sigma_0$ if and only if $\zeta(s)$ has no zero in that half-plane.

Combining the previous two statements, it follows that $L^*(s)$ converges in the half-plane $\Re s>1/2$ if and only if the RH is true.

The above results do not change when we restrict to real numbers $s$.

P.S. The Wikipedia page does not say that $L(s)$ only converges for $\Re s>1$.

$\endgroup$
2
  • $\begingroup$ I was talking about the third display in the section "Dirichlet Series" in the Wikipedia entry. It says that $|z|<2$ and $Re(s)>1$ (I assume for convergence) and here $z=-1$. But they could be wrong, they got the product wrong anyway I think; should be $1-z/p^s$, not $1+z/p^s$ in the product unless I am mistaken. I will redo my numerical computations anyway, because (unless some mistake), I could not get $L(0.9)$ to converge, but got $L^*(0.9)$ to converge to the correct value. $\endgroup$ – Vincent Granville May 7 at 22:32
  • 3
    $\begingroup$ @VincentGranville: The Wikipedia page defines the Dirichlet series for $\Re s>1$, because in that range we know convergence, and for $z=1$ you cannot replace $\Re s>1$ by $\Re s>0.999$, say. For $z=-1$ we conjecture that the Dirichlet series actually converges for $\Re s>1/2$: this is equivalent to the RH. $\endgroup$ – GH from MO May 7 at 23:08

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.