Heuristic reason for Polya's conjecture - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T20:54:04Z http://mathoverflow.net/feeds/question/33246 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/33246/heuristic-reason-for-polyas-conjecture Heuristic reason for Polya's conjecture Peter Humphries 2010-07-25T02:49:31Z 2010-07-27T07:55:56Z <p>Let $\lambda(n)$ be Liouville's function, so that for each positive integer $n = p_1^{m_1}\cdots p_r^{m_r}$, we have that $\lambda(n) = (-1)^{\sum^{r}_{k=1}{m_k}}$. In 1919, Polya conjectured that $L(x) = \sum_{n \leq x}{\lambda(n)} \leq 0$ for all $x \geq 2$; his reasoning was based on some limited numerical evidence (up to $x = 1500$, I believe), its connection to the Riemann Hypothesis (it implies RH and the simplicity of the zeroes of $\zeta(s)$), and Polya showed that for $p \equiv 3 \pmod{4}$ with class number $h(-p) = 1$, $L(p) = 0$. Unfortunately, Polya's conjecture is false; it is known that the first counterexample occurs at $x = 906150257$ (so one can't really blame Polya for trying), and that there exist infinitely many positive integers $n$ such that $L(n) \geq 0.061867 \ldots$.</p> <p>Nevertheless, Polya's conjecture does seem to be <em>usually</em> true, in that $L(x) \leq 0$ "most" of the time. There are a couple of different arguments that give an indication of why one would expect $L(x)$ to often be negative. For example, standard methods show (under RH, of course) that $${\sum_{n \leq x}}'{\lambda(n)} = \frac{\sqrt{x}}{\zeta(1/2)} + \sum_{\rho}{\frac{\zeta(2\rho)}{\zeta'(\rho)}\frac{x^{\rho}}{\rho}} - 1 + O\left(\frac{1}{\sqrt{x}}\right),$$ and one expects the terms in the sum over the zeroes to generally be very small, whereas $1/\zeta(1/2) = -0.684765\ldots$, so it would be expected that $L(x)$ is "usually" negative. Another method is via Lambert series; I mentioned <a href="http://mathoverflow.net/questions/11074/partial-sums-of-multiplicative-functions/14233#14233" rel="nofollow">here</a> that one can show that $$\sum_{n=1}^{\infty}{\frac{\lambda(n)}{e^{n\pi/x}+1}} = \frac{1-\sqrt{2}}{2}\sqrt{x} + \frac{1}{2} + (\psi(x)-2\psi(x/2))\sqrt{x},$$ where $\psi(x) = \sum_{n=1}^{\infty}{e^{-\pi xn^2}} = O(e^{-\pi x})$; this Lambert series is in some sense a smoothed version of $L(x)$. Again, the leading term is negative, suggesting that $L(x) \leq 0$ often.</p> <p>My question is: what other methods (elementary, analytic, or probabilistic) can be used to show why we would expect $L(x)$ to usually be negative?</p> http://mathoverflow.net/questions/33246/heuristic-reason-for-polyas-conjecture/33484#33484 Answer by Greg Martin for Heuristic reason for Polya's conjecture Greg Martin 2010-07-27T07:55:56Z 2010-07-27T07:55:56Z <p>I'm not answering your actual question about other methods, but I can provide some information about the analytic approach.</p> <p>Suppose that the Riemann hypothesis is true and that all the imaginary parts of the zeros are linearly independent over the rational numbers. Then one can show, using the methods of Rubinstein and Sarnak ("Chebyshev's bias", Experiment. Math., 1994), that the function $L(x)/\sqrt x$ has the same limiting (logarithmic) distribution function as the random variable $$\frac1{\zeta(1/2)} + \sum_{\rho\colon \Im\rho>0} 2\bigg| \frac{\zeta(2\rho)}{\rho\zeta'(\rho)} \bigg| X_\rho,$$ where the $X_\rho$ are independent random variables each taking values in $[-1,1]$ according to the sine distribution (that is, each $X_\rho$ is the real part of a random variable uniformly distributed on the unit circle in the complex plane).</p> <p>The distribution of the sum is symmetric around 0, which explains why the distribution itself is predominantly negative (since $\zeta(1/2) \lt 0$).</p>