User peter humphries - MathOverflow

Answer by Peter Humphries for Is there a stronger (but widely believed) version of the Chowla conjecture?

2013-06-16T22:53:12Z

I imagine the "correct" conjecture is the Möbius $s$-tuples conjecture in the form that if $s \in \mathbb{N}$, $\alpha_1, \ldots, \alpha_s \in \mathbb{N}$ with at least one $\alpha_i$ odd, and $d_1, \ldots, d_s \in \mathbb{Z}$ distinct, then for all $\varepsilon > 0$ we have that $$\sum_{n \leq x}{\mu(n + d_1)^{\alpha_1} \mu(n + d_2)^{\alpha_2} \cdots \mu(n + d_s)^{\alpha_s}} \ll_{\varepsilon} x^{1/2 + \varepsilon}$$ uniformly for all $|d_i| \leq x$. However, I don't imagine there is any good evidence for this other than the fact that the Riemann hypothesis implies the case $s = 1$. I don't think this problem has really been studied enough for there to be a widely-believed generalisation of Chowla's conjecture.

A form of this conjecture is assumed by Ng in this paper on the distribution of the Möbius function in short intervals; he assumes that this conjecture holds with "for all $\varepsilon > 0$" replaced by "for some $0 < \beta_0 < 1/2$".

Modular Forms of Weight One

2011-04-11T05:10:26Z

Consider the space of newforms $S^{\mathrm{new}}_k(\Gamma_1(q))$ of weight $k$ and level $q$ for the congruence subgroup $\Gamma_1(q)$ of $\mathrm{SL}_2(\mathbb{Z})$; for simplicity's sake, let's assume that $q$ is prime. Then for $k \geq 2$, it is known via Riemann-Roch that $$\dim S^{\mathrm{new}}_k(\Gamma_1(q)) = \frac{k - 1}{24} (q^2 - 1) + E(q,k)$$ for an error term $E(q,k)$. This error term can be calculated explicitly (though not particularly neatly): see Theorem 13 of http://www.math.ubc.ca/~gerg/papers/downloads/DSCFN.pdf. So for $k \geq 2$, it is certainly possible to determine $\dim S^{\mathrm{new}}_k(\Gamma_1(q))$ precisely.

For $k = 1$, on the other hand, no such precise equations seem to exist, as the method used to prove the $k \geq 2$ case breaks down. Instead, it is conjectured (see Conjecture 2.1 of http://arxiv.org/pdf/0906.4579v1) that $$\dim S^{\mathrm{new}}_1(\Gamma_1(q)) = \frac{q - 2}{2} h(K_q) + O_{\varepsilon}(q^{\varepsilon}),$$ for any $\varepsilon > 0$ with the error term is uniform in $q$, and where $h(K_q)$ is the class number of $\mathbb{Q}(\sqrt{-q})$; here the leading term comes from the dihedral modular forms, while the error term is due to the others (icosahedral etc.).

Now note that the leading term in the formula for $S^{\mathrm{new}}_k(\Gamma_1(q))$ for $k \geq 2$ vanishes when $k = 1$, so if that formula where to be valid for $k = 1$, we would be left with the error term $E(q,k)$, which we can explicitly compute.

Question: Is there a reason why we should not expect $\dim S^{\mathrm{new}}_1(\Gamma_1(q)) = E(q,1)$? Obviously a quick check on Magma or Sage should prove that this is not the case, but unfortunately I don't have either installed.

If not, is there any chance that we will one day find a closed form for $\dim S^{\mathrm{new}}_1(\Gamma_1(q))$?

Answer by Peter Humphries for On the oscillation of the summatory totient about its average

2013-04-08T20:52:43Z

I am not sure if this result is explicitly mentioned in the literature, but it certainly is classical.

Let $$R(x) = \sum_{n \leq x}{\varphi(n)} - \frac{3x^2}{\pi^2}, \qquad H(x) = \sum_{n \leq x}{\frac{\varphi(n)}{n}} - \frac{6x}{\pi^2}.$$

Then by partial summation, $$\int^{x}_{0}{\frac{R(t)}{t^2} \: dt} = H(x) - \frac{R(x)}{x}.$$ A classical result of Chowla states that $$H(x) - \frac{R(x)}{x} = O\left((\log x)^{-4}\right).$$ See Lemma 13 of S. Chowla, "Contributions to the analytic theory of numbers", Mathematische Zeitschrift 35:1 (1932), 279-299. (If you have access to Springer Link then it is available here.)

From a cursory glance of Chowla's proof, the negative powers of a logarithm stem from the prime number theorem applied to the summatory function of the Möbius function, so it is likely that this bound could be improved with more modern estimates for this.

For what it's worth, I answered a question closely related to this here.

Answer by Peter Humphries for Prime race in 2 dimensions

2013-02-04T18:30:51Z

This is not an answer, but rather an explanation of why this question is so difficult.

For positive coprime integers $a,q$, let $$\pi(x;q,a) = \# \{p \leq x : p \equiv a \pmod{q}\}.$$ For $k \in \mathbb{Z}$, let $$A_k = \{n \in \mathbb{N} : \pi(n;8,1) - \pi(n;8,5) = k\},$$ and let $$B_k = \{\pi(n;8,3) - \pi(n;8,7) \in \mathbb{Z} : n \in A_k\}.$$ Then your conjecture that the function $$f(n) = \sum_{p \leq n}{e^{\pi i(p - 1)/4}}$$ is surjective on $\mathbb{Z}[i]$ is equivalent to the conjecture that $B_k = \mathbb{Z}$ for each $k \in \mathbb{Z}$.

For this to happen, the set $A_k$ must be countably infinite; that is, the equality $\pi(n;8,1) = \pi(n;8,5)$ must occur infinitely often. This is a difficult result, but it is in fact known unconditionally: it is covered by Theorem 5.1 of "Comparative prime-number theory. II" by S. Knapowski, and P. Turán. Apparently, it has now been proven unconditionally by Jason Sneed that $\pi(x;q,a) - \pi(x;q,b)$ changes sign infinitely often for all $q \leq 100$, but this is yet to appear in print (see this paper for a discussion).

If one assumes two strong conjectures, the Grand Riemann hypothesis, and the Linear Independence hypothesis (namely that the imaginary parts of the nontrivial zeroes of all Dirichlet $L$-functions are linearly independent over the rationals), then one can say a lot more. Rubinstein and Sarnak's paper on Chebyshev's bias shows that not only are there infinitely many sign changes, but the function $$\left(\frac{\log x}{\sqrt{x}} \left(\pi(x;q,a_1) - \mathrm{Li}(x)\right), \ldots, \frac{\log x}{\sqrt{x}} \left(\pi(x;q,a_r) - \mathrm{Li}(x)\right)\right)$$ has a limiting logarithmic distribution. In particular, they can say roughly how likely $(\log x / \sqrt{x}) \pi(x;8,1)$ and $(\log x / \sqrt{x}) \pi(x;8,5)$ are to be in particular regions; unfortunately, this doesn't really tell you anything about the set $A_k$ for each integer $k$.

Once you have that $A_k$ is countably infinite, you still need to ensure that there is no "conspiracy" happening, in that the other prime number race $\pi(x;8,3) - \pi(x;8,7)$ could avoid certain configurations whenever $x$ is a zero of the prime number race $\pi(x;8,1) - \pi(x;8,5)$. This seems extremely difficult, and I don't know how one might attempt to analyse this. That being said, questions peripherally related to this were studied by Knapowski and Turán, so it is possible that there might be something in the literature that can deal with this type of problem.

As an aside, one interesting modification of this conjecture is the following. Let $\chi$ be a Dirichlet character modulo $q$, so that $\chi$ is generated by some root of unity $\zeta_Q$. Is the function $$f_{\chi}(n) = \sum_{p \leq n}{\chi(p)}$$ surjective on $\mathbb{Z}[\zeta_Q]$?

Answer by Peter Humphries for Density of the "multiplicative odd numbers"

2012-12-13T22:59:45Z

Gerry has the right idea here: you are asking about the limiting behaviour of the sum $$\frac{1}{x} \sum_{n \leq x}{\frac{1 - (-1)^{\Omega(n)}}{2}}.$$ The arithmetic function $\lambda(n) = (-1)^{\Omega(n)}$ is known as Liouville's function. It is well-known (and equivalent to the prime number theorem!) that the summatory function of the Liouville function, $$L(x) = \sum_{n \leq x}{\lambda(n)},$$ satisfies the asymptotic $$L(x) = o(x)$$ as $x$ tends to infinity. (In fact, one can probably improve this slightly in the usual way to get better error terms in the prime number theorem.) So it is indeed true that $$d(A) = \lim_{x \to \infty} \frac{1}{x} \sum_{n \leq x}{\frac{1 - (-1)^{\Omega(n)}}{2}} = \frac{1}{2},$$ and with a little work you could actually say something slightly stronger about the rate at which this converges.

From Zeta Functions to Curves

2011-07-18T09:21:33Z

Let $C$ be a nonsingular projective curve of genus $g \geq 0$ over a finite field $\mathbb{F}_q$ with $q$ elements. From this curve, we define the zeta function $$Z_{C/{\mathbb{F}}_q}(u) = \exp\left(\sum^{\infty}_{n = 1}{\frac{\# C(\mathbb{F}_{q^n})}{n} u^n}\right),$$ valid for all $|u| < q^{-1}$. This zeta function extends meromophicially to $\mathbb{C}$ via the equation $$Z_{C / \mathbb{F}_q}(u) = \frac{P_{C / \mathbb{F}_q}(u)}{(1 - u) (1 - qu)}$$ for some polynomial with coefficients in $\mathbb{Z}$ that factorises as $$P_{C/\mathbb{F}_q}(u) = \prod^{2g}_{j = 1}{(1 - \gamma_j u)}$$ with $|\gamma_j| = \sqrt{q}$ and $\gamma_{j + g} = \overline{\gamma_j}$ for all $1 \leq j \leq g$. This last point tells us that $Z_{C / \mathbb{F}_q}(u)$ has a functional equation and satisfies a version of the Riemann hypothesis.

What happens if we run this construction in reverse? What if we start with a set of numbers $\gamma_j$, $1 \leq j \leq 2g$, such that $|\gamma_j| = \sqrt{q}$, $\gamma_{j + g} = \overline{\gamma_j}$ for all $1 \leq j \leq g$, and such that the polynomial $$P(u) = \prod^{2g}_{j = 1}{(1 - \gamma_j u)}$$ has coefficients in $\mathbb{Z}$? Is there a way of telling whether the function $$\frac{P(u)}{(1 - u) (1 - qu)}$$ is the zeta function of some curve $C$? Furthermore, what is this curve exactly?

A simple case of this is if we look at the function $$\frac{1 - au + qu^2}{(1 - u) (1 - qu)}$$ for some $a \in \mathbb{Z}$ with $|a| \leq 2 \sqrt{q}$. How do we determine whether this function is the zeta function $Z_{C / \mathbb{F}_q}(u)$ of an elliptic curve $C$ over $\mathbb{F}_q$ ? If it is indeed equal to $Z_{C / \mathbb{F}_q}(u)$ , what is the Weierstrass equation for $C$ (assuming $\mathrm{char}(q) \geq 5$ )?

Answer by Peter Humphries for Is Mertens function negatively biased?

2012-10-12T21:13:14Z

Somehow I missed this question when it was originally asked. I'm not entirely sure what you mean by a negative bias; the bits you've highlighted in the graph of $M(x)$ aren't when $M(x)$ is negative, but rather where is is, in some sense, "decreasing on average", and I don't know how to formalise that notion. If you're actually interested in the set of $x$ for which $M(x)$ is negative, then it seems clear from the graph that this happens roughly half the time. Here's how to make that notion formal.

In Nathan Ng's paper linked in Micah Milinovich's answer, Ng (very conditionally!) proves the existence of a limiting logarithmic distribution of $M(x)/\sqrt{x}$, so that there exists a measure $\nu$ satisfying $$\lim_{X \to \infty} \frac{1}{\log X} \int_{1}^{X}{f\left(\frac{M(x)}{\sqrt{x}}\right) \ \frac{dx}{x}} = \int_{\mathbb{R}}{f(x) \ d\nu(x)}$$ for every continuous bounded $f : \mathbb{R} \to \mathbb{R}$; equivalently, for every Borel $B \subset \mathbb{R}$ whose boundary has $\nu$-measure zero, $$\lim_{X \to \infty} \frac{1}{\log X} \int\limits_{\{x \in [1,X] : M(x)/\sqrt{x} \in B\}}{ \ \frac{dx}{x}} = \nu(B).$$ Furthermore, he calculates the Fourier transform of $\nu$ explicitly and shows, among other things, that $\widehat{\nu}$ is even about the origin. This implies that $$\lim_{X \to \infty} \frac{1}{\log X} \int\limits_{\{x \in [1,X] : M(x) < 0\}}{ \ \frac{dx}{x}} = \nu((-\infty,0)) = \nu((0,\infty)) = \lim_{X \to \infty} \frac{1}{\log X} \int\limits_{\{x \in [1,X] : M(x) > 0\}}{ \ \frac{dx}{x}} = \frac{1}{2}.$$ To show this, we need to know that the set $\{0\}$ has $\nu$-measure zero, but this follows as Ng's explicit formula for $\widehat{\nu}$ is in $L^1(\mathbb{R})$, and hence that $\nu$ is absolutely continuous with respect to the Lebesgue measure on $\mathbb{R}$. (I don't think Ng actually includes this argument, but it's in my paper that kolik linked in his answer.)

So this shows (conditionally) that the median on $\nu$ is $0$, and hence that $M(x)$ is unbiased, in the sense that the logarithmic density of the set of points where $M(x)$ is negative is the same as that of the set of points where $M(x)$ is positive.

Interestingly, if you consider instead the weighted sum $$M_{1/2}(x) = \sum_{n \leq x}{\frac{\mu(n)}{\sqrt{n}}},$$ then there is a negative bias; this follows from the same methods in my article that kolik linked to, together with the explicit expression $$M_{1/2}(x) = \frac{1}{\zeta(1/2)} + \sum_{\rho}{\frac{x^{\rho - 1/2}}{(\rho - 1/2) \zeta'(\rho)}} + R(x),$$ where the sum is over the nontrivial zeroes of the Riemann zeta function, and $R(x)$ is some small error term. Other than this, I don't know much about other biases, though it should be the same in number fields, and I recently wrote a paper about this kind of thing in function fields, based on previous work of Byungchul Cha (in which he does not explicitly state the analogous result of there being no bias, though it is clear from the results).

Edit: I forgot to mention Brent and van de Lune's recent paper, where they look at a form of the Lambert series generated by $\mu(n)$ and show that it is negative as $x$ tends to $1$ from below (as juan mentioned in a comment above). But this really isn't telling you anything other than that the Riemann zeta function is negative at $s = 0$, which is a much weaker statement than, say, a pole whose residue is negative (as is the case with the summatory function of the Liouville function).

Answer by Peter Humphries for equidistribution on the unit circle of particular sequences of finite subsets

2012-07-04T14:51:10Z

Gerry's reference turned out to be quite useful. Theorem 2.7 of Uniform Distribution of Sequences by Kuipers and Niederreiter states that if $a$ and $b$ are integers with $a < b$, and if $f$ is twice differentiable on $[a,b]$ with $|f''(x)| \geq \rho > 0$ on $[a,b]$, then $$\left|\sum_{n = a}^{b}{e^{2\pi i f(n)}}\right| \leq \left(\left|f'(b) - f'(a)\right| + 2\right)\left(\frac{4}{\sqrt{\rho}} + 3\right).$$ So if we assume that $g : [0,1] \to \mathbb{R}$ is a continuous twice-differentiable function with $\lambda = \inf_{x \in [0,1]} g''(x) > 0$, then by taking $a = 1$, $b = N$, $f(x) = m N g(x/N)$, we find that $$\left|\sum_{n = 1}^{N}{e^{2\pi i m N g(n/N)}}\right| \leq \left(m \left|g'(1) - g'(1/N)\right| + 2\right)\left(\frac{4}{\sqrt{m N \lambda}} + 3\right).$$

So now let $\mu_N$ be the probability measure on $[0,1]$ given by $$\mu_N(B) = \frac{1}{N} \# \left\{1 \leq n \leq N : N g(n/N) \in B \pmod{1}\right\}$$ for each Borel set $B \subset [0,1]$, and let $\mu$ denote the Lebesgue measure on $[0,1]$. Then the Erdős–Turán inequality states that for any positive integer $M$, the discrepency $$D(N) = \sup_{B \in [0,1]} \left|\mu_N(B) - \mu(B)\right|$$ satisfies $$D(N) \leq C \left(\frac{1}{M} + \frac{1}{N} \sum_{m = 1}^{M}{\left| \sum_{n = 1}^{N}{e^{2\pi i m N g(n/N)}} \right|}\right)$$ for some absolute constant $C > 0$ (independent of $N$ and $M$). Taking $M = \lfloor N^{1/3}\rfloor$ and using the earlier bound on $\sum_{n = 1}^{N}{e^{2\pi i m N g(n/N)}}$ shows that $$D(N) = O\left(N^{-1/3}\right)$$ and hence that $\mu_N$ converges weakly to $\mu$ as $N$ tends to infinity.

It may be possible to relax some of these conditions on $g$ by modifying the proof of this theorem in the book of Kuipers and Niederreiter, but I haven't checked too closely yet.

Answer by Peter Humphries for Error to sum of Euler phi-functions

2012-05-04T11:31:09Z

This is an interesting question. I don't think anyone has worked out what the distribution of the error term $$\frac{E(x)}{x} = \frac{1}{x}\left(\sum_{n \leq x}{\phi(n)} - \frac{3x^2}{\pi^2}\right)$$ actually looks like in any useful sense; from what I can make out, it seems to essentially be the same as the distribution of $$\sum_{n = 1}^{\infty}{\frac{\mu(n)}{n} \left\{\frac{x}{n}\right\}}$$ where $\mu(n)$ is the Möbius function and $\{x\}$ is the fractional part of $x$, but this doesn't really seem to tell you anything particularly useful.

Nevertheless, it certainly is known that $E(x)/x$ has a distribution function; this is proved on p.13 of "On the Existence of Limiting Distributions of Some Number-Theoretic Error Terms" by Yuk-Kam Lau. This also follows quite easily from the fact that $$\frac{E(x)}{x} = H(x) + O\left((\log x)^{-4}\right)$$ where $$H(x) = \sum_{n \leq x}{\frac{\phi(n)}{n}} - \frac{6 x}{\pi^2}$$ combined with the main result of the paper "The Existence of a Distribution Function for an Error Term Related to the Euler Function" by Erdős and Shapiro. What is better known is the average behaviour of $E(x)$, in the form of the asymptotics $$\sum_{n \leq x}{E(x)} \sim \frac{3 x^2}{2 \pi^2}$$ and $$\int^{x}_{0}{E(t)^2 dt} \sim \frac{x^3}{6 \pi^2}.$$

Probably the best reference for what is currently known about this error term is the paper "Oscillations of the remainder term related to the Euler totient function" by Kaczorowski and Wiertelak (but unfortunately this paper isn't available for free online).

Answer by Peter Humphries for A curious definite integral.

2012-03-14T12:49:36Z

This really isn't particularly remarkable. By definition, $$\zeta(s) = \sum_{n = 1}^{\infty}{\frac{1}{n^s}} = s \int_{1}^{\infty}{\frac{\lfloor x \rfloor}{x^s} \frac{dx}{x}}$$ for $\Re(s) > 1$, where the second inequality follows by a partial summation argument, and $\lfloor x \rfloor$ is the floor function. We can write $\lfloor x \rfloor = x - \{x\}$, where $\{x\}$ is the fractional part of $x$, and split up the integral above in order to find that $$\zeta(s) = \frac{1}{s - 1} + 1 - s \int_{1}^{\infty}{\frac{\{ x \}}{x^s} \frac{dx}{x}}.$$ This is now defines a meromorphic function on $\Re(s) > 0$ with just a simple pole at $s = 1$ with residue $1$. In any case, this is a well-known integral representation of $\zeta(s)$.

This should explain your result for $0 < n < 1$, by making the change of variables $n = 1/s$, and then making the change of variables $y = x^{-1/n}$ in the integral. Also, the $n = 1$ case is definitional, as the Euler--Mascheroni constant is defined to be $$1 - \int_{1}^{\infty}{\frac{\{ x \}}{x^2} dx}$$ A partial summation argument should show why this is the same as $\lim_{x \to \infty} \left( \sum_{n \leq x}{\frac{1}{n}} - \log x\right)$.

You can also see why $$\zeta(\sigma) < \frac{\sigma}{\sigma - 1}$$ for all $\sigma > 0$; this is simply because $$\sigma \int_{1}^{\infty}{\frac{\{ x \}}{x^{\sigma}} \frac{dx}{x}} > 0.$$

Note that none of this really uses complex analysis, apart from the fact that $\zeta(s)$ isn't implicitly defined for $0 < \Re(s) < 1$ initially, so you need to take a leap of faith to believe that the integral representation of $\zeta(s)$ is actually valid in this strip.

Answer by Peter Humphries for Dedekind Zeta function: behaviour at 1

2012-02-08T08:42:19Z

This is called the (generalised) Euler constant of the number field $K$, denoted $\gamma_K$, as for $K = \mathbb{Q}$ we have $\gamma_K = \gamma_0$, the Euler--Mascheroni constant. There are many estimates known for $\gamma_K$. For example, page 61 of this paper has an upper bound for $|\gamma_K|$, which basically states that $$|\gamma_K| \leq 2 \mathrm{Res}_{s = 1} \zeta_K(s).$$ Some other useful references are here and here.

As for a function field, I am not so sure, but I'm guessing things should be similar but slightly easier (as there are only finitely many zeroes for $\zeta_{C/\mathbb{F}_q}(s)$).

EDIT: This paper deals with bounds for $\gamma_K$ for function fields.

Answer by Peter Humphries for Error term of the Prime Number Theorem and the Riemann Hypothesis

2011-07-19T07:37:52Z

It is not hard to show that $$\mathrm{Li}(x) = \frac{x}{\log x} \sum_{k=0}^{m - 1}{\frac{k!}{(\log x)^k}} + O\left(\frac{x}{(\log x)^{m + 1}}\right)$$ for any $m \geq 0$ (just use the definition of $\mathrm{Li}(x)$ and repeated integration by parts). Thus $$\pi(x) = \frac{x}{\log x} \sum_{k=0}^{m - 1}{\frac{k!}{(\log x)^k}} + O\left(\frac{x}{(\log x)^{m + 1}}\right).$$ It is not possible to improve on this (this is true unconditionally; you don't even need the Riemann hypothesis). So $\mathrm{Li}(x)$ really is the "better" approximation to $\pi(x)$ compared to $x/\log x$.

Answer by Peter Humphries for Learning roadmap for harmonic analysis

2011-06-03T09:37:08Z

It depends very much on what areas of harmonic analysis you're interested in, of course. Grafakos' books are excellent and really quite advanced, and if you wish to continue in that style of harmonic analysis, then there's not much else you can do other than start reading many of the articles that he cites. On the other hand, there are interesting areas in harmonic analysis not covered by Grafakos. I'd recommend a couple of textbooks by Stein: Singular Integrals and Differentiability Properties of Functions and Harmonic Analysis: Real-Variable Methods, Orthogonality, and Oscillatory Integrals. There are probably some other interesting textbooks on singular integral operators that might be useful (though I can't think of any off the top of my head). One other interesting (and very modern) area is wavelets: Mayer's book Wavelets and Operators is probably the place to start there. Other useful resources are lecture notes or survey articles about harmonic analysis available online. For example, Pascal Auscher taught a course at ANU on harmonic analysis using real-variable methods last year, and one of the students in the class typed up notes, which are available here. Similarly, Terry Tao taught a course a few years ago, and he has lecture notes here and here. Finally, if you want to learn about harmonic analysis with an operator-theoretic bent, there are useful lecture notes here and here.

Answer by Peter Humphries for Asymptotic Formula for a Mertens Style Sum

2011-05-13T02:45:40Z

Expanding on Frank's answer: by partial summation, we have that $$\sum_{p \leq x} \frac{(\log p)^k}{p} = \frac{(\log x)^k}{x} \pi(x) - \int_{2}^{x} \pi(t) \left(\frac{kt (\log t)^{k-1} - (\log t)^k}{t^2}\right) dt.$$ Using the fact that $\pi(x) = \mathrm{Li}(x) + E(x)$, where $E(x) = O(e^{-c\sqrt{\log x}})$, we have that the first term is equal to $$\frac{(\log x)^k}{x} \mathrm{Li}(x) + O((\log x)^k e^{-c\sqrt{\log x}}) = \frac{(\log x)^k}{x} \mathrm{Li}(x) + O(e^{-c_k \sqrt{\log x}}).$$ For the second term, $$- \int_{2}^{x} \pi(t) \left(\frac{kt (\log t)^{k-1} - (\log t)^k}{t^2}\right) dt = R_1(x) + R_2 + R_3(x),$$ where $$R_1(x) = - \int_{2}^{x} \mathrm{Li}(t) \left(\frac{kt (\log t)^{k-1} - (\log t)^k}{t^2}\right) dt,$$ $$R_2 = - \int_{2}^{\infty} E(t) \left(\frac{kt (\log t)^{k-1} - (\log t)^k}{t^2}\right) dt,$$ $$R_3(x) = \int_{x}^{\infty} E(t) \left(\frac{kt (\log t)^{k-1} - (\log t)^k}{t^2}\right) dt.$$ Let's look at $R_1(x)$ first. Using the fact that $\mathrm{Li}(t) = \int_{2}^{s} \frac{ds}{\log s}$ and interchanging the order of integration, then evaluating the integral with respect to $t$, we obtain $$R_1(x) = - \frac{(\log x)^k}{x} \mathrm{Li}(x) + \frac{(\log x)^k}{k} - \frac{(\log 2)^k}{k}.$$ For $R_3(x)$, a simple calculation shows that $$R_3(x) \ll - \int_{x}^{\infty}{(\log t)^{k-1} e^{-c\sqrt{\log t}} \left(\frac{kt - \log t}{t^2}\right) dt} = - \int_{\log x}^{\infty}{u^{k-1} e^{-c\sqrt{u}} \left(\frac{ke^{u} - u}{e^u}\right) du}.$$ We can rewrite this integral as $$- \int_{\log x}^{\infty}{\left(ku^{k-1} e^{-c\sqrt{u}} - \frac{c}{2} u^{k - 1/2} e^{-c\sqrt{u}}\right) + \left(\frac{c}{2} u^{k - 1/2} e^{-c\sqrt{u}} - u^k e^{-c\sqrt{u} - u}\right) du}$$ and from this it is clear that $$R_3(x) \ll_k - \int_{\log x}^{\infty}{\left(ku^{k-1} e^{-c\sqrt{u}} - \frac{c}{2} u^{k - 1/2} e^{-c\sqrt{u}}\right) du} = (\log x)^k e^{-c \sqrt{\log x}} \ll e^{-c_k \sqrt{\log x}}.$$ Finally, it's clear from the estimate for $R_3(x)$ that the integral defining $R_2$ converges, and so we obtain $$\sum_{p \leq x} \frac{(\log p)^k}{p} = \frac{(\log x)^k}{k} - \frac{(\log 2)^k}{k} + R_2 + O_k(e^{-c\sqrt{\log x}}).$$

Note that it's possible to make this valid for much more than just nonnegative integers $k$ - you could also have $k$ complex.

Answer by Peter Humphries for Given: $\phi(s) = \sum_p \frac{\log p}{p^s}$ Why is: $\lim_{e \rightarrow 0} e \phi(1+e)$ = 1 ?

2011-04-17T02:26:25Z

I think you're approaching the question in the wrong way. The whole point is that you can show that for $\Re(s) > 1$, $$\Phi(s) = \sum_{p}{\frac{\log p}{p^s}} = \frac{1}{s - 1} + E(s),$$ where the function $E(s)$ is meromorphic on the open half-plane $\Re(s) > 1/2$ with poles possibly at the zeroes of $\zeta(s)$; in fact, Zagier shows that $$E(s) = - \frac{\zeta'(s)}{\zeta(s)} - \frac{1}{s - 1} - \sum_{p}{\frac{\log p}{p^s (p^s - 1)}}.$$

Basically, this is saying that $\Phi(s)$ is meromorphic on an open neighbourhood of $\Re(s) \geq 1$ with a simple pole at $s = 1$, and the expansion above shows that the residue of $\Phi(s)$ at $s = 1$ is equal to $1$. This is precisely the same as saying that $$\lim_{\varepsilon \to 0} \varepsilon \Phi(1 + \varepsilon) = 1.$$ Indeed, we have that $$\lim_{\varepsilon \to 0} \varepsilon \Phi(1 + \varepsilon) = \lim_{\varepsilon \to 0} \frac{\varepsilon}{1 + \varepsilon - 1} + \lim_{\varepsilon \to 0} \varepsilon E(1 + \varepsilon),$$ and the first limit tends to $1$ (obviously) while the second limit tends to zero (as $E(1 + \varepsilon)$ tends to something finite).

If you don't understand this method at all (i.e. all about meromorphic extensions of functions, poles, and residues), then this is probably due to a lack of background in complex analysis. Seeing as this proof of the prime number theorem is all about complex analysis, I'd recommend reading up on all these basics beforehand.

Answer by Peter Humphries for Determining the asymptotic behavior of a series

2011-04-12T09:20:17Z

Using partial summation, we have that $$\sum_{k = 0}^{K}{\mu^k(1 - \mu^k t)^n} = \frac{(1 - \mu^{K + 1})(1 - \mu^K t)^n}{1 - \mu} - \frac{nt \log \mu^{-1}}{1 - \mu} \int_{0}^{K}{\mu^x (1 - \mu^{\lfloor x \rfloor + 1}) (1 - \mu^x t)^{n-1} \: dx},$$ where $\lfloor x \rfloor$ is the integer part of $x$. By taking the limit as $K$ tends to infinity, $$\sum_{k = 0}^{\infty}{\mu^k(1 - \mu^k t)^n} = \frac{1}{1 - \mu} - \frac{nt \log \mu^{-1}}{1 - \mu} \int_{0}^{\infty}{\mu^x (1 - \mu^{\lfloor x \rfloor + 1}) (1 - \mu^x t)^{n-1} \: dx}.$$ Now a simple calculation shows that $$\frac{1}{1 - \mu} - \frac{nt \log \mu^{-1}}{1 - \mu} \int_{0}^{\infty}{\mu^x (1 - \mu^x t)^{n-1} \: dx} = \frac{(1 - t)^n}{1 - \mu}$$ by making the substitution $u = 1 - \mu^x t$. So the tricky part is the other bit of the integral, which is $$E = \frac{nt \mu \log \mu^{-1}}{1 - \mu} \int_{0}^{\infty}{\mu^{x + \lfloor x \rfloor} (1 - \mu^x t)^{n-1} \: dx}.$$ Note that as $x - 1 < \lfloor x \rfloor \leq x$, we have the bounds $$A \leq E \leq \frac{1}{\mu} A$$ with $$A = \frac{nt \mu \log \mu^{-1}}{1 - \mu} \int_{0}^{\infty}{\mu^{2x} (1 - \mu^x t)^{n-1} \: dx}.$$ Once again, this isn't tricky to calculate: making the same substitution as earlier, we find that $$A = \frac{\mu}{t (1 - \mu)} \frac{1 - (nt + 1) (1 - t)^n}{n + 1}.$$

So piecing everything together, we obtain $$\sum_{k = 0}^{\infty}{\mu^k(1 - \mu^k t)^n} = \frac{(1 - t)^n}{1 - \mu} + E$$ with $$E \asymp_{\mu} \frac{1 - (nt + 1) (1 - t)^n}{t(n + 1)}.$$ This doesn't yield a closed form for $\lim_{n \to \infty} n f_n(t)$, unfortunately, but it does show that $$\frac{\mu}{t(1 - \mu)} \leq \liminf_{n \to \infty} n f_n(t) \leq \limsup_{n \to \infty} n f_n(t) \leq \frac{1}{t (1 - \mu)}.$$

Answer by Peter Humphries for Discrete Fourier Transform of the Möbius Function

2011-03-04T08:42:15Z

Expanding on Matt's answer, it is possible to show without too much difficulty (see here, Exercise 3 of section 18.2.1) that if $(a,q) = 1$, then $$\sum_{n \leq x}{\mu(n) e^{2\pi i an/q}} = \sum_{d \mid q} \frac{\mu(d)}{\varphi(q/d)} \sum_{\chi \pmod{q/d}} \tau(\overline{\chi}) \chi(a) M\left(\frac{x}{d}; \chi \chi_{0(d)}\right),$$ where the inner sum is over all Dirichlet characters modulo $q/d$, $\tau(\overline{\chi})$ is the Gauss sum of $\overline{\chi}$, $\chi_{0(d)}$ is the principal character modulo $d$, and $$M(x ; \chi) = \sum_{n \leq x}{\mu(n) \chi(n)}.$$ A pretty simple calculation (using Euler products and the fact that $\mu(n) \chi(n)$ is multiplicative) shows that for $\Re(s) > 1$, $$\frac{1}{L(s,\chi)} = s \int_{1}^{\infty} \frac{M(x ; \chi)}{x^s} \: \frac{dx}{x}.$$ So one can then use Perron's formula to invert this relationship, and then apply the classic method of pushing the integral to the left of the line $\Re(s) > 1$ and use estimates on $1/L(s,\chi)$ in the zero-free region to obtain the desired bound on $$\sum_{n \leq x}\mu(n) e^{2\pi i an/q}.$$ Though this doesn't seem to give a uniform bound, Montgomery and Vaughan's notes outline (Q1-Q5 of the same section) how to obtain uniform bounds in $\alpha$.

Note also that it is quite easy to show that for every $\varepsilon > 0$, $$\sum_{n \leq x}{\mu(n) e^{2\pi i n \alpha}} = O(x^{1/2 + \varepsilon})$$ for almost every $\alpha \in \mathbb{R}$; this is just a simple application of Carleson's theorem on the pointwise almost-everywhere convergence of an $L^2$-periodic function to its Fourier series. Unfortunately, this is not quantitative; we cannot say that this applies to rational $\alpha$ (otherwise we would be able to prove GRH).

Answer by Peter Humphries for Most memorable titles

2010-11-02T11:17:37Z

I'm quite surprised that no one has mentioned The Joy of Sets.

Answer by Peter Humphries for Jokes in the sense of Littlewood: examples?

2010-09-17T04:13:55Z

In the same vein as the "Freshman's dream" $$(a + b)^p = a^p + b^p,$$ which is true in characteristic $p$, there is also the "Sophomore's dream", which is the identity $$\int_{0}^{1}{x^{-x} \: dx} = \sum_{n = 1}^{\infty}{n^{-n}}.$$ Surprisingly enough, this identity is actually correct.

Heuristic reason for Polya's conjecture

2010-07-25T02:49:31Z

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 $L(n) \geq 0.061867 \ldots$.

Nevertheless, Polya's conjecture does seem to be usually 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 here 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.

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?

Answer by Peter Humphries for What are the analytic properties of Dirichlet Euler products restricted to arithmetic progressions?

2010-06-13T08:03:14Z

It's best to split this up into two cases.

Case 1: $\chi(a) = 1$. Then for $\Re(s) > 1$, $$\prod_{p \equiv a \pmod{q}} \left(1 - \frac{\chi(p)}{p^s}\right)^{-1} = \prod_{p \equiv a \pmod{q}} \left(1 - \frac{1}{p^s}\right)^{-1} = \sum_{n \in \left\langle \mathcal{P} \right\rangle} \frac{1}{n^s},$$ where $\left\langle \mathcal{P} \right\rangle$ is the multiplicative semigroup generated by the set of primes $\mathcal{P}$ consisting of all $p \equiv a \pmod{q}$. So this is just the Burgess zeta function $\zeta_{\mathcal{P}}(s)$. Now there exist Burgess zeta functions that cannot be holomorphically extended to $1 + it$ for any $t \in \mathbb{R}$ (this is mentioned for example in Terry Tao's paper "A Remark on Partial Sums Involving the Mobius Functions", which I'm pretty sure is available somewhere on the arxiv). In this case, however, I have no idea; perhaps some of the relevant literature discusses it.

Case 2: $\chi(a) = -1$. Then $$\prod_{p \equiv a \pmod{q}} \left(1 - \frac{\chi(p)}{p^s}\right)^{-1} = \prod_{p \equiv a \pmod{q}} \left(1 + \frac{1}{p^s}\right)^{-1} = \sum_{n \in \left\langle \mathcal{P} \right\rangle} \frac{\lambda(n)}{n^s},$$ where $\lambda(n)$ is the Liouville function. Equivalently, $$\prod_{p \equiv a \pmod{q}} \left(1 - \frac{\chi(p)}{p^s}\right)^{-1} = \frac{\zeta_{\mathcal{P}}(2s)}{\zeta_{\mathcal{P}}(s)},$$ so it comes down to the same thing; determining whether $\zeta_{\mathcal{P}}(s)$ extends holomorphically to the line $\Re(s) = 1$ and beyond.

EDIT: Recall that the prime number theorem for arithmetic progressions says that $$\pi(x;q,a) = \frac{1}{\varphi(q)} \mathrm{li}(x) + O_A(x \exp(-c_1 (\log x)^{1/2})$$ for fixed $A > 0$ with $q \leq (\log x)^A$. An application of a result of Diamond (cf. Asymptotic Distribution of Beurling's Generalized Integers) then implies that $$N_{\mathcal{P}}(x) = \sum_{n \in \left\langle \mathcal{P} \right\rangle, \; n \leq x}{1} = a x + O_A(x \exp(-c_2 (\log x \log \log x)^{1/3})$$ for some particular $a > 0$. By partial summation, we have that for $\Re(s) > 1$, $$\zeta_{\mathcal{P}}(s) = \frac{as}{s-1} + s \int_{1}^{\infty} \frac{N_{\mathcal{P}}(x) - ax}{x^{s+1}} \: dx .$$ Diamond's result implies that this integral is uniformly convergent for $\Re(s) \geq 1$, and so it is continuous in this half-plane. Thus $\zeta_{\mathcal{P}}(s) = c/(s-1) + r_0(s)$ with $r_0(s)$ continuous for $\Re(s) \geq 1$, and so $\zeta_{\mathcal{P}}(s)$ extends to $\Re(s) \geq 1$ with a singularity at $s = 1$. Moreover, it is not difficult to show that $\zeta_{\mathcal{P}}(1+it) \neq 0$ for all $t \in \mathbb{R}$; a version of this is shown in Montgomery and Vaughan's Multiplicative Number Theory I: Classical Theory section 8.4.

Note also that assuming the generalised Riemann Hypothesis, it is possible to strengthen this meromorphic extension of $\zeta_{\mathcal{P}}(s)$ to $\Re(s) > 1/2$ with $\zeta_{\mathcal{P}}(s)$ nonvanishing in this open half-plane; see Titus W. Hilberdink and Michel L. Lapidus, Beurling Zeta Functions, Generalised Primes and Fractal Membranes.

Answer by Peter Humphries for probability in number theory

2010-06-11T04:50:47Z

Tenenbaum's book is indeed one of the best on the subject; it's well-motivated and quite accessible. If you go a bit further back, there are also the Probabilistic Number Theory books by P. D. T. A. Elliot; volume I is on Mean Value Theorems, while volume II is on Central Limit Theorems. These are both a bit more specialised and slightly outdated. Even older still is Probabilistic Methods In the Theory of Numbers by J. Kubilius.

Answer by Peter Humphries for Shortest/Most elegant proof for $L(1,\chi)\neq 0$

2010-05-27T03:11:22Z

Everyone seems to have their own favourite here. Mine is slightly similar to a couple of those mentioned earlier; it involves studying the properties of $\zeta(s) L(s,\chi)$ where $\chi$ is a real character (as mentioned earlier, it's quite simple to prove the case where $\chi$ is complex) --- it is the approach taken in Montgomery and Vaughan's Multiplicative Number Theory I. Classical Theory, for example.

It is in some sense quite natural to look at the product $\zeta(s) L(s,\chi)$, because $\zeta(s)$ is the "simplest" case of an $L$-series having a simple pole at $s=1$, while we are assuming that $L(s,\chi)$ has a zero at $s = 1$, and so these will cancel out and hence $\zeta(s) L(s,\chi)$ will extend holomorphically to $\Re(s) > 0$. A simple calculation shows that the coefficients $c(n) = \sum_{d \mid n}{\chi(d)}$ of the Dirichlet series for $\zeta(s) L(s,\chi)$ are positive whenever $c(n)$ is a perfect square. This Dirichlet series is valid only when $\zeta(s), L(s,\chi)$ are absolutely convergent (i.e. for $\Re(s) > 1$), but as the assumption $L(1,\chi) = 0$ implies that $\zeta(s) L(s,\chi)$ extends holomorphically to $\Re(s) > 0$, a theorem of Landau implies that this Dirichlet series is conditionally convergent for $\Re(s) > 0$. But $\sum^{\infty}_{n=1}{c(n)n^{-1/2}}$ diverges as $c(n^2) > 0$, which yields the result.

Of course, the idea behind this proof does not differ markedly from the proof in Serre's book, and even Monsky's proof uses some of the same ideas.

Answer by Peter Humphries for Residues of $1/\zeta$

2010-05-27T02:36:10Z

This is actually a very difficult problem, and currently most results are highly conjectural. It essentially comes down to finding useful bounds on discrete moments of the Riemann zeta function of the form $$J_k(T) = \sum_{0 < \Im(\rho) < T}{|\zeta'(\rho)|^{2k}},$$ as one can then choose the correct value of $k$ and apply partial summation.

For positive $k$, recent results of Milinovich and Milinovich and Ng show under the Riemann Hypothesis that $$T(\log T)^{(k+1)^2} \ll J_k(T) \ll T(\log T)^{(k+1)^2 + O(1/\log \log \log T)};$$ and slightly stronger results are known for $k = 0,1,2$ (with the latter two being under the Riemann Hypothesis).

For negative $k$, the situation is much more difficult. A conjecture of Gonek-Hejhal suggests that for all $k > -3/2$, $$J_k(T) \asymp T(\log T)^{(k+1)^2}$$ and this has been refined significantly by Hughes, Keating, and O'Connell. But it seems out of reach to prove anything significantly useful in this area; the best result so far has been by Gonek, who proved that $J_{-1}(T) \gg T$ assuming the Riemann Hypothesis and the simplicity of the zeroes of $\zeta(s)$. But this isn't useful for most applications, where an upper bound is needed. I believe it is possible to show $J_{-1} \ll T^{2+\varepsilon}$ under the Riemann Hypothesis and the simplicity of the zeroes of $\zeta(s)$, though I don't have a reference for this. Also it is quite possible that this result also holds if we replace $|\zeta'(\rho)|^{-2}$ by $\left|\mathrm{Res}_{s = \rho} \zeta(\rho)^{-1}\right|^2$, though again I don't know of a reference for this.

Answer by Peter Humphries for Shortest Paths on fractals

2010-05-26T05:27:06Z

For the Sierpinski carpet, using the metric induced by its embedding in $\mathbb{R}^2$ is quite difficult computationally. There are alternative metrics: for fractals arising from hyperbolic iterated function systems (of which the Sierpinski carpet is one), one can identify points via its tops code space, which involves determining which subcopies of the Sierpinski carpet a point lies in. It's not difficult to place a metric on the tops code space of such a fractal, and this in some sense tells us how "close" two points are, in terms of how many iterations of the function system have to occur before the two points are sent to different subcopies of the fractal. It's a bit difficult to make this more precise without going into a fair bit of detail; I'm pretty sure this is all covered in Michael Barnsley's textbook Fractals Everywhere though. Note also of course that this only works for fractals arising from hyperbolic iterated function systems, which includes other fractals like the Koch curve and the Menger sponge.

Answer by Peter Humphries for Partial sums of multiplicative functions

2010-02-05T05:08:01Z

This paper shows that $L(n) > .061867\sqrt{n}$ for infinitely many $n$.

As for somewhat elementary methods (in the sense of avoiding the Riemann zeta function) to show that $L(n)$ is "usually" of order $\sqrt{n}$, one can use the Lambert series $$\sum_{n=1}^{\infty}{\frac{\lambda(n)q^n}{1-q^n}} = \sum_{n=1}^{\infty}{q^{n^2}}.$$

As $$\frac{q^n}{1+q^n} = \frac{q^n}{1-q^n} - 2\frac{q^{2n}}{1-q^{2n}},$$ we have $$\sum_{n=1}^{\infty}{\frac{\lambda(n)}{q^{-n}+1}} = \sum_{n=1}^{\infty}{q^{n^2}} - 2\sum_{n=1}^{\infty}{q^{2n^2}}$$

or equivalently, letting $q = e^{-\pi/x}$ and $\psi(x) = \sum_{n=1}^{\infty}{e^{-\pi xn^2}}$, where $x$ is large, $$\sum_{n=1}^{\infty}{\frac{\lambda(n)}{e^{n\pi/x}+1}} = \psi(1/x) - 2\psi(2/x)$$

Now $\psi(x)$ satisfies the functional equation $$\frac{1+2\psi(x)}{1+2\psi(1/x)} = \frac{1}{\sqrt{x}},$$

and so we can rewrite this as $$\sum_{n=1}^{\infty}{\frac{\lambda(n)}{e^{n\pi/x}+1}} = \frac{1-\sqrt{2}}{2}\sqrt{x} + \frac{1}{2} + (\psi(x)-2\psi(x/2))\sqrt{x}.$$

For large $x$, the left-hand side "looks like" $L(x)$, whereas the right-hand side is dominated by the term $\frac{1-\sqrt{2}}{2}\sqrt{x}$. This also explains why $L(n)$ is predominantly negative, as $\frac{1-\sqrt{2}}{2}$ is negative.

Comment by Peter Humphries

2013-04-08T21:09:00Z

Yep, thanks. All fixed now.

Comment by Peter Humphries

2013-02-04T21:34:55Z

@Greg: You are right about the first part; this was actually proved by Knapowski and Turán in 1962. I'll edit my answer accordingly.

Comment by Peter Humphries

2013-01-24T17:46:23Z

It's also pretty common for number theory conferences to be promoted on the number theory mailing list: listserv.nodak.edu/cgi-bin/wa.exe?A0=NMBRTHRY

Comment by Peter Humphries

2012-12-14T00:00:10Z

Also, you may be interested in the contents of this paper: staff.science.uu.nl/~dahme104/…

Comment by Peter Humphries

2012-12-13T23:59:44Z

I don't know of a direct reference of such a proof, but this theorem is folklore and it's pretty easy to see why it's true: one can show (say, via comparing Euler products) that $\sum_{n=1}^{\infty}\lambda(n)n^{-s}=\zeta(2s)/\zeta(s)$ for $\Re(s)>1$, then use the fact that this extends meromorphically to the entire complex plane and has no poles in the region $\Re(s)>1-c/\log(|\Im(s)|+2)$. So basically the usual way of proving the prime number theorem, just with a different Dirichlet series.

Comment by Peter Humphries

2012-11-08T19:04:56Z

For what it's worth, the reference for David's answer (and Noam's generalisation) is William C. Waterhouse, "Abelian Varieties over Finite Fields" (numdam.org/item?id=ASENS_1969_4_2_4_521_0). There is also a recent paper classifying the possible polynomials $P(u)$ for hyperelliptic curves: the paper is Everett W. Howe, Enric Nart, and Christophe Ritzenthaler, "Jacobians in Isogeny Classes of Abelian Surfaces over Finite Fields" (dx.doi.org/10.5802/aif.2430).

Comment by Peter Humphries

2012-07-04T08:50:47Z

By the Erdos-Turan inequality, this would follow if you could show that $\sum_{n = 1}^{N}{e^{2\pi i m N g(n/N)}} = o(N)$ uniformly in $m$. I'm not sure how one would go about showing this though.

Comment by Peter Humphries

2012-07-03T14:16:09Z

I imagine this has been studied before. The probability ought to be $1/\zeta_K(2)$, where $\zeta_K(s)$ is the Dedekind zeta function of the algebraic number field $K$.

Comment by Peter Humphries

2012-06-20T10:44:33Z

This is a proof of a different result: the original question is showing how big ($\Omega(x \log \log x)$) the error term must be, not how small ($O(x \log x)$) it must be.

Comment by Peter Humphries

2012-05-17T03:38:05Z

I think you have a couple of typos; shouldn't it be $e^{2\pi i t}$ in place of $e^{2\pi i \nu}$ (and also in place of $e^{2\pi i x}$ later on)?

Comment by Peter Humphries

2012-05-15T10:36:15Z

To the best of my knowledge, the furthest $L(x)$ has been calculated up to is $x = 2 \cdot 10^{14}$, as reported in this paper: davidson.edu/math/mossinghoff/…. Out of curiosity, may I ask why you are interested in this summatory function?

Comment by Peter Humphries

2012-05-13T02:47:35Z

I'm not sure what this post is actually trying to say. It is certainly well known that the Riemann Hypothesis is equivalent to the bound $M(x) = O(x^{1/2 + \varepsilon})$, but no one has managed to get even close to proving such a bound unconditionally. Furthermore, the square root bound $M(x) = O(\sqrt{x})$ is almost certainly false, as it would contradict the Linear Independence hypothesis.

Comment by Peter Humphries

2012-04-26T04:22:01Z

Another relevant question: mathoverflow.net/questions/28000/…

Comment by Peter Humphries

2012-02-10T10:31:56Z

And of course the functional equation implies that $P(q^{-1}) = q^{-g} P(1) = q^{-g} h_F$, where $g$ is the genus of the underlying curve and $h_F$ is the class number of $F$.

Comment by Peter Humphries

2012-02-08T08:57:07Z

Hopefully the link should be fixed now.