# Ordinary Generating Function for Mobius

Is there any information known for the Ordinary Generating Function for Mobius? $$\sum_{n=1}^{\infty} {\mu(n)}x^n$$ I know that

1. It has radius of convergence 1.

2. Does not have limit as $x\rightarrow 1$.

My question is

1. Does it have limit as $x\rightarrow \textrm{exp}(i\theta)\neq 1$?

2. Is there a similar result for $\textrm{exp}(i\theta)\neq 1$ with $$\sum_{n\leq x}\mu(n)=o(x)$$ i.e. is there a result of this type? $$\sum_{n\leq x}\mu(n)\textrm{exp}(in\theta) = o(x)$$

EDIT1: Aside from my questions, there are some known results for the OGF for Mobius.

EDIT2: Found a reference for (III)

Reference: JASON P. BELL, NILS BRUIN, AND MICHAEL COONS "TRANSCENDENCE OF GENERATING FUNCTIONS WHOSE COEFFICIENTS ARE MULTIPLICATIVE" available at arXiv:1003.2221v2

P. Borwein, T. Erd´elyi, and F. Littman, Polynomials with coefficients from a finite set available at http://www.cecm.sfu.ca/personal/pborwein/PAPERS/P195.pdf

(I) This function has the unit circle as natural boundary.

(II) This function is a transcendental function.

(III) It is not bounded on any open sector($\{z:|z|<1, z=re^{i\theta}, \alpha<\theta<\beta\}$)of unit disc.

This result (III) together with Baire Category answers my question 1 with a dense set of $\theta$'s, but as Prof Tao mentioned, there can be a possibility that it is bounded for some values of $\theta$.

-

Mobius randomness heuristics suggest that $\sum_n \mu(n) (r e^{i\theta})^n$ does not converge to a limit as $r \to 1^-$ for any (or at least almost any) $\theta$. If it did converge for some $\theta$, then we would have

$$\sum_n \mu(n) e^{in\theta} \psi_k(n) \to 0$$ as $k \to \infty$, where

$$\psi_k(n) := (1-2^{-k-1})^n - (1-2^{-k})^n.$$

The cutoff function $\psi_k$ is basically a smoothed out version of the indicator function $1_{[2^k,2^{k+1}]}$. The Mobius randomness heuristic (discussed for instance in this article of Sarnak) then suggests that the sum $\sum_n \mu(n) e^{in\theta} \psi_k(n)$ should have a typical size of $2^{k/2}$, and so should not decay to zero as $k \to \infty$.

Note from Plancherel's theorem that the $L^2_\theta$ mean of $\sum_n \mu(n) e^{in\theta} \psi_k(n)$ is indeed comparable to $2^{k/2}$. This does not directly preclude the (very unlikely) scenario that this exponential sum is very small for many $\theta$ and only large for a small portion of the $\theta$, but if one optimistically applies various versions of the Mobius randomness heuristic (with square root gains in exponential sums) to guess higher moments of $\sum_n \mu(n) e^{in\theta} \psi_k(n)$ in $\theta$, one is led to conjecture a central limit theorem type behaviour for the distribution of this quantity (as $\theta$ ranges uniformly from $0$ to $2\pi$), i.e. it should behave like a complex gaussian with mean zero and variance $\sim 2^k$, and in particular it should only be $O(1)$ about $O(2^{-k})$ of the time, and Borel-Cantelli then suggests divergence for almost every $\theta$ at least. Unfortunately this is all very heuristic, and it seems difficult with current technology to unconditionally rule out a strange conspiracy that makes $\sum_n \mu_n (re^{i\theta})^n$ bounded as $r \to 1$ for some specific value of $\theta$ (say $\theta = \sqrt{2} \pi$), though this looks incredibly unlikely to me. (One can use bilinear sums methods, e.g. Vaughan identity, to get some nontrivial pointwise upper bounds on these exponential sums when $\theta$ is highly irrational, but I see no way to get pointwise lower bounds, since $L$-function methods will not be available in this setting, and there may well be some occasional values of $\theta$ and $k$ for which these sums are actually small. But it is likely at least that the $\theta=0$ theory can be extended to rational values of $\theta$.)

-

For (2), the answer is yes - see for example this paper of Baker and Harman.

-
Thank you for this. Actually their paper is conditional. Indeed, unconditional result by H. Davenport is qjmath.oxfordjournals.org/content/os-8/1/313.full.pdf. Anyway, that answers 2. –  i707107 Apr 3 '13 at 21:01
Fair enough, although on the first page it cites unconditional results as well. –  Greg Martin Apr 4 '13 at 5:57
Yeah. They cited Davenport's result. I enjoyed reading it. –  i707107 Apr 4 '13 at 18:27