User micah milinovich - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T23:55:31Z http://mathoverflow.net/feeds/user/3659 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/134029/on-the-critical-line-re-zetas-zetas-1-2-log-pi-1-2-re-psis-2/134047#134047 Answer by Micah Milinovich for On the critical line $\Re \zeta'(s)/\zeta(s) =? 1/2 \log(\pi) - 1/2 \Re \psi(s/2)$ ? Micah Milinovich 2013-06-18T13:03:53Z 2013-06-18T14:50:43Z <p>Your formula follows from Hadamard's product formula for $\zeta(s)$ and the corresponding partial fraction decomposition of $\zeta'(s)/\zeta(s)$. See, for instance, section 2 of Soundararajan's paper: <a href="http://arxiv.org/pdf/math/0612106v2.pdf" rel="nofollow">http://arxiv.org/pdf/math/0612106v2.pdf</a></p> http://mathoverflow.net/questions/131602/estimate-on-the-prime-counting-function-psix/131612#131612 Answer by Micah Milinovich for Estimate on the prime-counting function $\psi(x)$. Micah Milinovich 2013-05-23T16:12:47Z 2013-05-23T16:18:02Z <p>The reference for this (given in the 3rd edition of Davenport's <em>Multiplicative Number Theory</em>) is:</p> <p>Grosswald, Émile. "Sur l'ordre de grandeur des différences $\psi(x)-x$ et $\pi(x)-\ell i(x)$." (French) C. R. Acad. Sci. Paris <strong>260</strong> (1965), 3813–3816.</p> <p>It seems (according to the Math Review) that for $\alpha$ fixed, the error is $O(x^\alpha)$. So, two powers of the logarithm can be saved.</p> http://mathoverflow.net/questions/130247/closed-form-for-derivatives-zetan1-2/130294#130294 Answer by Micah Milinovich for Closed form for derivatives $\zeta^{(n)}(1/2)$ Micah Milinovich 2013-05-10T19:52:09Z 2013-05-10T20:27:17Z <p><strong>Edit</strong>: My original answer was incorrect.</p> <p>You can evaluate $\zeta'(\frac{1}{2})$ recursively in terms of $\zeta(\frac{1}{2})$ using the symmetric form of the functional equation:</p> <p>$$\zeta(s)\Gamma(\tfrac{s}{2}) \pi^{-s/2} = \zeta(1{-}s)\Gamma(\tfrac{1-s}{2}) \pi^{(s-1)/2}.$$</p> <p>Differentiating both sides sides of the equation, plugging in $s=\frac{1}{2}$, and then solving for $\zeta'(\frac{1}{2})$, I get the value listed on the MathWorld website.</p> <p>As Noam Elkies points out, taking higher derivatives, this process allows you to write $\zeta^{(2n+1)}(\frac{1}{2})$ in terms of the smaller even derivatives $\zeta(\frac{1}{2}),\zeta''(\frac{1}{2}), \zeta^{(4)}(\frac{1}{2}), \ldots, \zeta^{(2n)}(\frac{1}{2})$.</p> http://mathoverflow.net/questions/125647/riemann-z-function-bounds-on-number-of-non-trivial-zeros-along-horizontal-lines/125662#125662 Answer by Micah Milinovich for Riemann Z function, bounds on number of non-trivial zeros along horizontal lines, rather than vertical ones. Micah Milinovich 2013-03-26T20:46:37Z 2013-03-26T20:46:37Z <p>It $t$ is not an ordinate of a zero of $\zeta(s)$, define $$S(t) = \frac{1}{\pi} \arg \zeta(1/2+it) = -\frac{1}{\pi} \Im \int_{1/2}^\infty \frac{\zeta'}{\zeta}(\sigma+it) d\sigma$$ and define $$S(t)= \lim_{\delta\to 0} \frac{1}{2}\Big(S(t+\delta) + S(t-\delta)\Big)$$ otherwise. Then the number $N(T)$ of zeros of $\zeta(s)$ in the strip $0&lt;\Im s \le T$ is $$N(T) = \frac{T}{2\pi}\log \frac{T}{2\pi e} +\frac{7}{8}+S(T)+O(\frac{1}{T})$$ where the big-$O$ term is actually continuously differentiable. For a proof, look either in Titchmarsh's book on the zeta-function or in Montgomery &amp; Vaughan's "Multiplicative Number Theory, I."</p> <p>By continuity, the quantity you are looking for is precisely $$\lim_{\delta\to 0} \Big(S(t+\delta) - S(t-\delta)\Big).$$ Unconditionally, I think Tim Trudgian has the best results for this quantity showing that $$|S(t)| \le 0.111 \log t + 0.275 \log \log t + 2.450$$ for $t>e$ (so your quantity is bounded by essentially twice this amount). This can be sharpened if $t$ is allowed to tend to infinity.</p> <p>As is mentioned in previous comments/answers, assuming the Riemann hypothesis (RH) you are looking for bounds on the multiplicity of a zero. In this case, Goldston &amp; Gonek showed that $$\lim_{\delta\to 0} \Big(S(t+\delta) - S(t-\delta)\Big) \le \Big(\frac{1}{2}+o(1)\Big) \frac{\log t}{\log \log t}$$ as $t\to\infty$ using the Guinand-Weil explicit formula. </p> <p>References:</p> <p><a href="http://arxiv.org/pdf/1208.5846.pdf" rel="nofollow">http://arxiv.org/pdf/1208.5846.pdf</a></p> <p><a href="http://arxiv.org/pdf/math/0511092v1.pdf" rel="nofollow">http://arxiv.org/pdf/math/0511092v1.pdf</a></p> http://mathoverflow.net/questions/125250/recovering-sum-n-leq-x-an-from-sum-n-leq-x-ane-n-x/125548#125548 Answer by Micah Milinovich for Recovering $\sum_{n \leq x} a(n)$ from $\sum_{n \leq x} a(n)e^{-n/x}$ Micah Milinovich 2013-03-25T18:05:07Z 2013-03-25T18:11:15Z <p>In many applications, you can use the fact that $$e^{-n/x}=1 + O\Big(\frac{n}{x}\Big)$$ uniformly for $n\le x$ and therefore $$\sum_{n\le x} a_n e^{-n/x} = \sum_{n\le x} a_n+ O\left( \frac{1}{x} \sum_{n\le x}n \ a_n\right).$$</p> <p>This works well for sequences $a_n$ generated by $L$-functions near the $1$-line. For instance, if $$\sum_{n\le x} a_n \asymp (\log x)^k$$ for some $k$.</p> http://mathoverflow.net/questions/114888/bounds-of-chebyshevs-function-in-an-interval/114894#114894 Answer by Micah Milinovich for Bounds of Chebyshev's function in an interval. Micah Milinovich 2012-11-29T15:59:12Z 2012-11-29T16:15:52Z <p>Adding to Eric Naslund's comment: Let $\pi(x)$ denote the number of primes less than $x$, then Montgomery &amp; Vaughan proved that $$\pi(x+y)-\pi(x) \le 2 \pi(y)$$ for $x\ge 1$ and $y\ge 2.$ </p> http://mathoverflow.net/questions/24526/good-papers-books-essays-about-the-thought-process-behind-mathematical-research/107004#107004 Answer by Micah Milinovich for Good papers/books/essays about the thought process behind mathematical research Micah Milinovich 2012-09-12T12:28:19Z 2012-09-12T12:28:19Z <p>I know it is a bit dated, but I am fond of Littlewood's essay "The Mathematician's Art of Work" in his <em>Miscellany</em>.</p> http://mathoverflow.net/questions/106682/large-gaps-between-p2s/106686#106686 Answer by Micah Milinovich for Large gaps between P2s Micah Milinovich 2012-09-08T20:05:29Z 2012-09-08T20:05:29Z <p>Here is a result, due to Halberstam, Heath-Brown, and Richert that seems to be of the flavor you are looking for (<em>Almost-primes in short intervals</em>. Recent progress in analytic number theory, Vol. 1 (Durham, 1979), pp. 69–101, Academic Press, London-New York, 1981.)</p> <p><strong>Theorem.</strong> <em>For all sufficiently large $x$ the interval $(x−x^{0.455},x]$, contains at least $\frac{1}{121}\frac{x^{0.455}}{\log x}$ integers that are either primes or products of two primes.</em></p> <p>Reference chasing on MathSciNet, it seems that Iwaniec &amp; Laborde improved the exponent from $0.455$ to $0.45$ and Wenzhi Luo later improved this slightly. I am not sure if Luo's result is still the "best" exponent.</p> http://mathoverflow.net/questions/106503/sharpening-of-lindelof-hypothesis/106514#106514 Answer by Micah Milinovich for Sharpening of Lindelöf hypothesis Micah Milinovich 2012-09-06T14:24:29Z 2012-09-06T14:24:29Z <p>No. In fact it is known that for any $\varepsilon>0$ there are arbitrarily large values of $t$ with $$|\zeta(1/2+it)| \ge \exp\left( (1-\varepsilon) \sqrt{\frac{\log t}{\log \log t}} \ \right).$$ This is a recent result of Soundararajan, Math. Ann. (2008) 342:467–486. </p> <p>Ramachandra and Balasubramanian had previous proved an inequality of a similar form with the factor of $1-\varepsilon$ replaced by a smaller constant.</p> http://mathoverflow.net/questions/102536/axioms-for-riemann-zeta-function/102547#102547 Answer by Micah Milinovich for Axioms for Riemann $\zeta$ function Micah Milinovich 2012-07-18T14:48:49Z 2012-07-18T14:48:49Z <p>Perhaps you are looking for something like <em>Hamburger's Theorem</em>? </p> <p>It states, essentially, that the only Dirichlet series with a finite number of singularities satisfying the same functional equation as the zeta-function is the zeta-function. You can find the details in Titchmarsh's book. </p> <p>Googling I found the following link: <a href="http://www.mat.univie.ac.at/~esiprpr/Zetaproc/patterson.pdf" rel="nofollow">http://www.mat.univie.ac.at/~esiprpr/Zetaproc/patterson.pdf</a></p> http://mathoverflow.net/questions/98174/is-mertens-function-negatively-biased/98178#98178 Answer by Micah Milinovich for Is Mertens function negatively biased? Micah Milinovich 2012-05-28T07:56:00Z 2012-05-28T07:56:00Z <p>Assuming some well-known conjectures, Nathan Ng established the existence of a limiting distribution for $e^{-y/2}M(e^{y})$. This paper can also be found on the arXiv: <a href="http://arxiv.org/abs/math/0310381" rel="nofollow">http://arxiv.org/abs/math/0310381</a></p> <p>The paper on summatory function of the Louiville function, mentioned in kolik's answer, uses many of the same techniques.</p> http://mathoverflow.net/questions/97040/optimization-problem-arising-from-the-study-of-zeta-zeros Optimization problem arising from the study of zeta zeros Micah Milinovich 2012-05-15T19:57:06Z 2012-05-18T21:08:01Z <p><strong>Motivation</strong>: The following problem arose in <strong>[1]</strong> while studying the vertical distribution of the zeros of the Riemann zeta-function. At the time, my collaborators and I were unable to solve it and I have never been able to derive a "satisfying answer."</p> <p><strong>Set-up</strong>: Let $r\ge 1$ and let $f \in L^2[0,1]$ be a continuous real-valued function of bounded variation on $[0,1]$, normalized so that $$\int_0^1(1-u)^{r^2-1}f(u)^2 du = 1.$$ Further define $M(c)=M(c,f,r)$ as $$M(c):=c+\frac{2 r}{\pi}\int_0^1 (1-u)^{r^2-1}f(u) \int_0^u \frac{\sin(\pi c v)}{v} f(u-v) \ dv \ du.$$</p> <p><strong>Question</strong>: How does one choose $r$ and $f$ optimally so that $$M(c) >1$$ for $c$ as small as possible? </p> <p>An argument of Conrey, Ghosh, and Gonek <strong>[2]</strong> can be used to show that $M(c)&lt;1$ if $c&lt;\frac{1}{2}$ for any such $f$ and $r$. In <strong>[1]</strong>, choosing $f$ to be a polynomial of low degree ($\le 6$) and using Mathematica to numerically optimize the $r$ and the coefficients, we were able to find $f$ and $r$ such that $M(.5155)>1$.</p> <p>In the special case when $r=1$, Montgomery and Odlyzko <strong>[3]</strong> observed that this optimization problem had already been solved using prolate spheroidal wave functions (see comments below). Here is how they reduced the problem to one that was already solved. In this case, we have $$M(c) = c+ \int_0^1\int_0^1 f(u) f(v) \frac{\sin(\pi c(u-v))}{\pi(u-v)} \ dv \ du.$$ The double integral on the right-hand side is $$\int_{-c/2}^{c/2} \left| \int_0^1 f(v) e^{2\pi i t v} dv \right|^2 dt :=I(c),$$ say. They then observed that choosing $$f(x) = aR_{00}^{(1)}[\pi c/2,2x-1]$$ maximizes $I(c)$, where $R_{mn}^{(1)}[c,x]$ is the radial prolate spheroidal wave function of the first kind of order <em>m</em> and degree <em>n</em>, and <em>a</em> is a constant to be chosen according to our above normalization.</p> <p>These wave functions are very hard to study numerically, and Montgomery and Odlyzko approximated them using modified Bessel functions. In <strong>[1]</strong>, when $r=1$, we recovered their results to four decimal places using polynomials of degree four. So in this case it seems that polynomials of small degree work (almost) as well as more sophisticated techniques.</p> <p><em>References:</em></p> <p><strong>[1]</strong> H. M. Bui, M. B. Milinovich, and N. C. Ng, <em>A note on the gaps between consecutive zeros of the Riemann zeta-function</em>, Proc. Amer. Math. Soc. <strong>138</strong> (2010), no. 12, pp. 4167-4175.</p> <p><strong>[2]</strong> J. B. Conrey, A. Ghosh, and S. M. Gonek, <em>A note on gaps between zeros of the zeta function</em>, Bull. London Math. Soc. <strong>16</strong> (1984), 421–424.</p> <p><strong>[3]</strong> H. L. Montgomery and A. M. Odlyzko, <em>Gaps between zeros of the zeta function</em>, Colloq. Math. Soc. Janos Bolyai, 34. Topics in Classical Number Theory (Budapest, 1981), North-Holland, Amsterdam, 1984.</p> <p><strong>Edit/Additional Comments</strong>: The solution optimization problem when $r=1$ should probably be attributed to Slepian and Pollak and to Landau and Pollack in <em>Prolate spheroidal wave functions, Fourier analysis and uncertainty I</em> and <em>II,</em> Bell System Tech. J. (1961) <strong>40</strong>, pp. 43-61 (<em>I</em>) and pp. 65-84 (<em>II</em>). Among other things, they prove the following results.</p> <p><strong>Theorem:</strong> <em>Let $\alpha(c)$ be the least number such that</em> $$\int_{-c/2}^{c/2} \left|\int_0^1 f(x) e^{2\pi i t x} dx \right|^2 dt \le \alpha(c) \int_0^1 |f(x)|^2 dx$$ <em>for all $f\in L^2[0,1]$. Then $\alpha(c)$ is strictly increasing, $\alpha(c)\lt c$ for $c \gt 0$, $\alpha(c)\sim c$ as $c\to 0^+$, and $\alpha(c)\to 1$ as $c\to \infty$. Moreover, equality is achieved in the above inequality if</em> $f(x) = R_{00}^{(1)}[\pi c/2,2x-1]$.</p> <p>One of the possible hang-ups in solving the optimization problem when $r>1$ is that I have not been able to "complete" the double integral $$\int_0^1 (1-u)^{r^2-1}f(u) \int_0^u \frac{\sin(\pi c v)}{\pi v} f(u-v) \ dv \ du$$ into a double integral of the form $$\int_0^1 \int_0^1 [\text{nice integrand}] \ dv \ du,$$ which is the first step in Montgomery &amp; Odlyzko's argument.</p> http://mathoverflow.net/questions/97174/non-trivial-zeros-of-partial-zeta-functions/97179#97179 Answer by Micah Milinovich for non-trivial zeros of partial zeta functions Micah Milinovich 2012-05-17T02:04:50Z 2012-05-17T02:04:50Z <p>The answer to question 1 is classical: Any Dirichlet series which has a finite abscissa of absolute convergence has a zero-free half-plane.</p> <p>Suppose the Dirichlet series $$A(s)=\sum_{n=1}^\infty \frac{a_n}{n^s}$$ has an abscissa of absolute convergence $\sigma_a$. If $a_m$ is the first non-zero coefficient, then for $\Re(s)>\sigma_a$ $$m^s A(s) = a_m + a_{m+1} \Big(\frac{m}{m+1}\Big)^s + a_{m+2} \Big(\frac{m}{m+2}\Big)^s +\cdots \to a_m \ \text{ as } \ \Re(s) \to +\infty.$$ Hence, there exists an absolute constant $C$ such that for $\Re(s)>C$ we have $$\left|a_{m+1} \Big(\frac{m}{m+1}\Big)^s + a_{m+2} \Big(\frac{m}{m+2}\Big)^s +\cdots \right| \leq \frac{|a_m|}{2}.$$ Consequently, $m^s A(s)$ has no zeros in the half-plane $\Re(s)>C$.</p> http://mathoverflow.net/questions/94733/divergent-series-expansion-in-aperys-proof-of-the-irrationality-of-zeta2-an/94735#94735 Answer by Micah Milinovich for Divergent series expansion in Apéry's proof of the irrationality of $\zeta(2)$ and $\zeta(3)$ Micah Milinovich 2012-04-21T13:42:42Z 2012-04-21T13:42:42Z <p>I think that this is "amazing claim" number 1 in van der Poorten's paper "A proof that Euler missed ...."</p> <p><a href="http://www.ift.uni.wroc.pl/~mwolf/Poorten_MI_195_0.pdf" rel="nofollow">http://www.ift.uni.wroc.pl/~mwolf/Poorten_MI_195_0.pdf</a></p> http://mathoverflow.net/questions/93916/sum-of-sum-k1ndk2/93924#93924 Answer by Micah Milinovich for Sum of $\sum_{k=1}^nd(k^2)$ Micah Milinovich 2012-04-13T00:41:52Z 2012-04-17T15:20:55Z <p>In the case you are interested in there is a simple generating (Dirichlet) series: $$\sum_{n=1}^\infty \frac{d(n^2)}{n^s} = \frac{\zeta^3(s)}{\zeta(2s)}.$$ From this you can either use a convolution argument or a Perron formula type argument to get an asymptotic formula. In particular, I believe it follows that $$\sum_{n\leq x} d(n^2) = \frac{3}{\pi^2}x \log^2 x +O(x \log x).$$ With more work, you can get lower-order terms of size $\asymp x \log x$ and $\asymp x.$ </p> <p><strong>Edit:</strong> There seems to some disagreement on whether the coefficient of the leading order term is $\frac{3}{\pi^2}$ or $\frac{3}{2\pi^2}$. I believe that $\frac{3}{\pi^2}$ is correct. Here are three bits of reasoning: From the generating series, we have $$\sum_{n\leq x}d(n^2) = \sum_{n\leq x} \sum_{k\ell^2=n} d_3(k)\mu(\ell) = \sum_{\ell\leq \sqrt{x}} \mu(\ell) \sum_{k\leq x/\ell^2} d_3(k)$$ where $d_3(n)$ denotes the number of ways to write $n$ as a product of three positive divisors and $\mu(\ell)$ is the Moebius function. By a standard estimate $$\sum_{k\leq x/\ell^2} d_3(k) = \frac{x}{2\ell^2}\log^2(x/\ell^2) + O\left(\frac{x\log x}{\ell^2} \right)$$ from which it follows that $$\sum_{n\leq x}d(n^2) = \frac{x \log^2 x}{2} \sum_{\ell \leq \sqrt{x}} \frac{\mu(\ell)}{\ell^2} +O(x\log x) = \frac{3 x}{\pi^2}\log^2 x +O(x\log x).$$ Alternatively, a Perron formula (e.g. Prime Number Theorem) type argument can be used to show that $$\sum_{n\leq x}d(n^2) = \text{Res}_{s=1} \frac{\zeta^3(s)}{\zeta(2s)} \frac{x^s}{s} +o(x) = \frac{3x}{\pi^2}\log^2 x + O(x\log x).$$ Moreover, in Mathematica, you can use <em>DivisorSigma[0, n^2]</em> to calculate $d(n^2)$. For $x=1,000,000$ I get that $$\frac{\pi^2}{3x\log^2x}\sum_{n\leq x} d(n^2) \approx 1.27305392....$$ The slow convergence to 1 is from the influence of lower-order terms. Notice, however, that the value it is not anywhere near $1/2$. However, if I define $F(x)$ to be the residue of$\frac{\zeta^3(s)}{\zeta(2s)}\frac{x^s}{s}$ at $s=1$, I get that $$\frac{1}{F(x)}\sum_{n\leq x} d(n^2) \approx 1.0000073....$$</p> http://mathoverflow.net/questions/92599/distribution-of-primes-in-small-intervals/92642#92642 Answer by Micah Milinovich for Distribution of primes in small intervals Micah Milinovich 2012-03-30T03:41:47Z 2012-03-30T03:41:47Z <p>Assuming the Riemann Hypothesis, I believe the best known result is due to Cramer (I cannot figure out how to add the accent of the e) and it says the following:</p> <p>There is a constant $C > 0$ such that if if the Riemann Hypothesis is true, then for every $x \ge2$ the interval $(x, x + C \sqrt{x} \log x)$ contains at least $\sqrt{x}$ prime numbers.</p> <p>This is Theorem 13.3 in Montgomery and Vaughan's <em>Multiplicative Number Theory</em>.</p> <p>Translating things from $\pi(x)$ to $\psi(x)$, exercise 2, pp. 430-431 of the same book outlines a proof that the Riemann Hypothesis implies that</p> <p>$$\psi(x+y)-\psi(x)=y+O\left(\sqrt{x} \log x \log\left(\frac{2y}{\sqrt{x} \log x}\right) \right).$$</p> <p>Thus an asymptotic holds as soon as $\frac{y}{\sqrt{x} \log x} \to \infty$. This formula simultaneously implies both Cramer's result and von Koch's well-known result that $$\psi(x) = x + O(\sqrt{x}\log^2 x) \quad \text{equivalently } \quad \pi(x) = \int_2^x \frac{dt}{\log t} + O(\sqrt{x}\log x)$$ assuming the Riemann Hypothesis.</p> http://mathoverflow.net/questions/91280/is-this-sum-of-reciprocals-of-zeta-zeros-correct/91286#91286 Answer by Micah Milinovich for Is this sum of reciprocals of zeta zeros correct? Micah Milinovich 2012-03-15T14:03:11Z 2012-03-17T19:43:14Z <p><strong>Edit:</strong> I endorse Juan's answer to the original question. The sum $\displaystyle{\sum_{\rho} \tfrac{1}{|\rho|}}$, running over the non-trivial zeros $\rho$ of $\zeta(s)$, is known to diverge, so at best $\displaystyle{\sum_{\rho} \tfrac{1}{\rho}}$ is conditionally convergent so you cannot re-arrange the terms.</p> <p>In your second to last displayed equation, you removed the assumption that the sum runs over pairs of zeros $\rho$ and $1-\rho$. So it seems that Lagarias' result can be used to evaluate the sum $$\sum_{\rho} \frac{1}{\rho (1{-}\rho)}.$$</p> <p>As you observed, assuming the Riemann Hypothesis $1-\rho =\overline{\rho}$ for any non-trivial zero $\rho$ of $\zeta(s)$. This implies that</p> <p>$$\sum_{\rho} \frac{1}{\rho (1{-}\rho)}=\sum_{\rho} \frac{1}{|\rho|^2}.$$</p> http://mathoverflow.net/questions/91280/is-this-sum-of-reciprocals-of-zeta-zeros-correct/91299#91299 Answer by Micah Milinovich for Is this sum of reciprocals of zeta zeros correct? Micah Milinovich 2012-03-15T14:56:09Z 2012-03-15T14:56:09Z <p>To answer your modified question, according to Mathematica:</p> <p>$$\lim_{s\to 0} \left(\frac{\hat{\zeta}'}{\hat{\zeta}}(s)+\frac{1}{s}\right) = -\frac{\gamma}{2} + \tfrac{1}{2}\log(4\pi).$$</p> <p>This implies that</p> <p>$${\sum_\rho}'\frac{1}{\rho} = \sum_{\Im \rho >0} \frac{1}{\rho(1-\rho)}= 1 +\frac{\gamma}{2} - \tfrac{1}{2}\log(4\pi).$$</p> <p>Therefore $$\sum_{ \rho } \frac{1}{\rho(1-\rho)} = 2 \sum_{\Im \rho >0} \frac{1}{\rho(1-\rho)}= 2 +\gamma - \log(4\pi).$$</p> http://mathoverflow.net/questions/90079/prime-numbers-in-arithmetic-progressions-uniformity-with-respect-to-the-modulus/90121#90121 Answer by Micah Milinovich for Prime numbers in arithmetic progressions : uniformity with respect to the modulus Micah Milinovich 2012-03-03T15:25:08Z 2012-03-03T19:06:31Z <p>There is some very nice recent work of Dimitris Koukoulopoulos who uses "pretentious" methods to prove the Siegel-Walfisz Theorem. A preprint can be found here:</p> <p><a href="http://www.crm.umontreal.ca/~koukoulo/documents/publications/multfncs.pdf" rel="nofollow">http://www.crm.umontreal.ca/~koukoulo/documents/publications/multfncs.pdf</a></p> http://mathoverflow.net/questions/85351/possible-locations-for-non-trivial-zeroes-lying-off-the-critical-line/85370#85370 Answer by Micah Milinovich for Possible locations for non trivial zeroes lying off the critical line Micah Milinovich 2012-01-10T21:39:30Z 2012-01-10T21:47:08Z <p>Let $\chi(s)=2^s \pi^{s-1}\sin(\pi s/2) \Gamma(1-s)$ so that $\zeta(s)=\chi(s)\zeta(1-s)$. You are asking about the curve $|\chi(s)|=1$.</p> <p>As you have observed, $|\chi(1/2+it)|=1$ for real $t$. There is a partial converse to this statement, namely that there is a positive absolute constant $C_0$ such that if $|\chi(\sigma+it)| = 1$ with $0 \le \sigma \le 1$ and $|t| \ge C_0$, then $\sigma=1/2$. </p> <p>A simple proof can be found in Lemma 6.1 of S. M. Gonek "Finite Euler products and the Riemann hypothesis" <em>Trans. Amer. Math. Soc.</em> 364 (2012), 2157-2191. This paper is also on the arXiv. Gonek states that $C_0&lt;6.3$ so it seems that phenomena in your pictures stops shortly after the ranges you plotted.</p> http://mathoverflow.net/questions/84989/upper-bounds-on-the-difference-of-consecutive-zeta-zeros/85029#85029 Answer by Micah Milinovich for Upper bounds on the difference of consecutive zeta zeros Micah Milinovich 2012-01-06T04:53:39Z 2012-01-06T04:53:39Z <p>Littlewood was the first to prove that the gaps between the ordinates of successive zeros of $\zeta(s)$ tend to zero. This is proved, for instance, in Titchmarsh's book on the zeta-function (see Theorem 9.11).</p> <p>I believe the best known unconditional result states that $$\gamma_{n+1}-\gamma_n = O( 1/\log\log\log \gamma_n)$$ as $n\to \infty$. Assuming the Riemann Hypothesis, this can be improved to $O( 1/\log\log \gamma_n).$</p> http://mathoverflow.net/questions/84812/values-of-dirichlet-l-funcions-at-natural-numbers/84817#84817 Answer by Micah Milinovich for Values of Dirichlet L-funcions at natural numbers Micah Milinovich 2012-01-03T18:12:36Z 2012-01-03T18:12:36Z <p>Let $\chi$ be any Dirichlet character modulo $q$, and let $m$ be a positive integer. Then Theorem 4.2 of Washington's <em>Introduction to Cyclotomic Fields</em> states that</p> <p>$$L(1-m,\chi) = - \frac{q^{m-1}}{m} \sum_{a=1}^q \chi(a)B_{m}(\tfrac{a}{q}).$$</p> <p>Here $B_m(x)$ is the usual Bernoulli polynomial, defined by</p> <p>$$\frac{t e^{Xt}}{e^t-1} = \sum_{n=0}^\infty B_n(X) \frac{t^n}{n!}.$$</p> <p>As Stopple pointed out, you can use the functional equation for $L(s,\chi)$ to evaluate $L(m,\chi)$, for some values of $m$, if you know the value of $L(1-m,\chi).$</p> http://mathoverflow.net/questions/82635/explicit-formula-for-riemann-zeros-counting-function/82673#82673 Answer by Micah Milinovich for explicit formula for Riemann zeros counting function Micah Milinovich 2011-12-05T04:46:18Z 2011-12-05T04:46:18Z <p>Assuming the Riemann Hypothesis, you can use a smooth approximation to the characteristic function of an interval in the Guinand-Weil explicit formula to <em>approximately</em> count the number of zeros of the zeta-function in an interval on the critical line. This expresses the approximate number of such zeros in terms of an integral of your test function and a sum over primes, as you seek. In fact, this can be set-up in such a way that the sum over primes is finite. (This method can be used to give upper and lower bounds for the number of zeros, but not an exact formula.) </p> <p>The details are (essentially) contained in a paper by Goldston &amp; Gonek "A note on S(t) and the zeros of the Riemann zeta function" available on Dan Goldston's webpage: math.sjsu.edu/~goldston/publications.htm </p> http://mathoverflow.net/questions/71349/exponential-sums-related-to-cusp-forms Exponential sums related to cusp forms Micah Milinovich 2011-07-26T20:06:17Z 2011-07-26T20:06:17Z <p>Let $$f(z)=\sum_{n\geq 1} a_f(n) e^{2\pi n i z}$$ be a holomorphic newform on the upper half-plane of weight $k$ for $\Gamma_0(N)$ and of trivial character which is normalized so that $a_f(1)=1$. </p> <p>In Jutila's book "A method in the theory of exponential sums" he proves an estimate of the form:</p> <p>$$\sum_{n\leq x} a_f(n) e^{2\pi i n \frac{p}{q}} \ll q^{2/3}x^{k/2-1/6+\varepsilon}$$</p> <p>where $a_f(n)$ are the coefficients of a cusp form of weight $k$ for the full modular group ($N=1)$, and $p$ and $q$ are coprime integers. Conceivably, a similar estimate holds for coefficients of holomorphic cuspforms for congruence subgroups. Does anyone know a reference for an estimate of the form:</p> <p>$$\sum_{n\leq x} a_f(n) e^{2\pi i n\frac{p}{q}} \ll_{f,q} \ \ x^{k/2-\delta}$$</p> <p>where $a_f(n)$ are the coefficients of a holomorphic cusp form of weight $k$ for $\Gamma_0(N)$, $p$ and $q$ are coprime integers, and $\delta>0$? </p> http://mathoverflow.net/questions/71061/on-the-zeroes-of-hasse-weil-l-function/71062#71062 Answer by Micah Milinovich for on the Zeroes of Hasse -weil L-function Micah Milinovich 2011-07-23T12:13:36Z 2011-07-23T13:50:46Z <p>The $L$-function has about $\displaystyle{\frac{T}{\pi} \log T \ }$ zeros in the strip with $0 &lt; t &lt; T$. See section 5.3 of Iwaniec and Kowalski's "Analytic Number Theory," in particular Theorem 5.8.</p> <p>It should be possible, if it hasn't been done already, to show that a positive proportion of these zeros are on the critical line using Selberg's method. Hafner extended Selberg's method to various families of degree 2 $L$-functions in a series of papers in the 1980s.</p> http://mathoverflow.net/questions/69718/riemann-siegels-approximate-functional-equation-for-fixed-t-and-res1-2/69891#69891 Answer by Micah Milinovich for Riemann-Siegel's approximate functional equation for fixed t and Re(s)≠1/2 Micah Milinovich 2011-07-09T17:28:49Z 2011-07-09T18:07:39Z <p>The assumption $2\pi xy=t$ (or something similar) is certainly necessary to guarantee that you get an approximation to the zeta-function and not some other function. For instance, if we choose $x=y=t/2\pi$ then the approximate functional equation "approximates" twice the zeta-function. One can see this as follows:</p> <p>Inside the critical strip ($\frac{1}{4}\le\sigma\le \frac{3}{4}$, say), it is known that $$\zeta(s) = \sum_{n\leq \frac{t}{2\pi}} \frac{1}{n^s} +O(t^{-\sigma})$$ for sufficiently large $t.$ Hence, by the functional equation and Stirling's formula, we have $$\zeta(s) =\chi(s)\zeta(1-s) = \chi(s)\sum_{n\leq \frac{t}{2\pi}} \frac{1}{n^{1-s}} +O(t^{-1/2}),$$ as well. Therefore, $$2\zeta(s) = \sum_{n\leq \frac{t}{2\pi}} \frac{1}{n^s} + \chi(s)\sum_{n\leq \frac{t}{2\pi}} \frac{1}{n^{1-s}} +O(t^{-\sigma}) +O(t^{-1/2}),$$ as claimed.</p> <p>Edit: You can also see this choosing $x=\frac{t}{2\pi}$ and $y=1$ (and vice versa) in formula that you stated in your question.</p> http://mathoverflow.net/questions/67102/is-anyone-aware-of-a-good-exposition-of-the-gauss-kramer-model-of-integers/67145#67145 Answer by Micah Milinovich for Is anyone aware of a good exposition of the Gauss-Kramer model of Integers? Micah Milinovich 2011-06-07T15:25:55Z 2011-06-07T15:25:55Z <p>You might want to look at:</p> <p>1) "Harold Cramér and the distribution of prime numbers" by Andrew Granville <a href="http://www.dms.umontreal.ca/~andrew/PDF/cramer.pdf" rel="nofollow">http://www.dms.umontreal.ca/~andrew/PDF/cramer.pdf</a></p> <p>2) "The distribution of prime numbers" by K. Soundararajan <a href="http://arxiv.org/pdf/math/0606408v1" rel="nofollow">http://arxiv.org/pdf/math/0606408v1</a></p> http://mathoverflow.net/questions/64828/asymptotic-formula-for-a-mertens-style-sum/64862#64862 Answer by Micah Milinovich for Asymptotic Formula for a Mertens Style Sum Micah Milinovich 2011-05-13T03:16:13Z 2011-05-13T14:05:07Z <p>Here is an answer that is similar in spirit to Frank and Peter's answers, but possibly simpler.</p> <p>Summing by parts, we see that</p> <p>$$\sum_{p\le x} \frac{\log^k p}{p} = (\log x)^{k-1} \sum_{p\leq x} \frac{\log p}{p} -(k-1)\int_{2^-}^x (\log u)^{k-2}\sum_{p\le u} \frac{\log p}{p} \frac{du}{u}.$$</p> <p>Now use the formula $$\sum_{p\leq x} \frac{\log p}{p} = \log x + c_1 + O(\exp(-c_2\sqrt{\log x}))$$ and it is not hard to derive that $$\sum_{p\le x} \frac{\log^k p}{p} = \frac{\log^k x}{k} + c_3 + O(\exp(-c_4\sqrt{\log x})).$$</p> <p>This seems easier than dealing with $Li(x)$.</p> http://mathoverflow.net/questions/63078/zeros-of-primitive-functions-of-s/63081#63081 Answer by Micah Milinovich for Zeros of primitive functions of S Micah Milinovich 2011-04-26T22:02:52Z 2011-04-26T22:02:52Z <p>There is an example given in this paper:</p> <p>J. P. Buhler, B. H. Gross, and D. B. Zagier, "On the Conjecture of Birch and Swinnerton-Dyer for an Elliptic Curve of Rank 3," <em>Mathematics of Computation</em> <strong>44</strong> (1985), no. 170, pp. 473-481.</p> <p><a href="http://www.jstor.org/stable/2007967" rel="nofollow">http://www.jstor.org/stable/2007967</a></p> http://mathoverflow.net/questions/57037/are-there-any-rational-solutions-to-this-equation Are there any rational solutions to this equation? Micah Milinovich 2011-03-01T21:15:00Z 2011-03-07T19:38:13Z <p>I am not sure if this is an appropriate question, but I was asked this by a colleague today and do not know how to answer it.</p> <p>1) Are there any rational solutions to the following equation: $$x^3-8x^2+5x+1 = -7y^2(x-1)x$$</p> <p>2) Is it possible that this is an elliptic curve in disguise? I have noticed that after projectivizing, there are two points at infinity. Perhaps this is okay under some change of variables? (I plead ignorance on this.)</p> http://mathoverflow.net/questions/134141/how-is-large-defined-in-an-equality-for-the-modulus-of-riemann-zeta Comment by Micah Milinovich Micah Milinovich 2013-06-19T14:43:33Z 2013-06-19T14:43:33Z FYI, this inequality has been sharpened: <a href="http://blms.oxfordjournals.org/content/43/2/243.abstract" rel="nofollow">blms.oxfordjournals.org/content/43/2/243.abstract</a> http://mathoverflow.net/questions/134029/on-the-critical-line-re-zetas-zetas-1-2-log-pi-1-2-re-psis-2/134047#134047 Comment by Micah Milinovich Micah Milinovich 2013-06-18T14:51:03Z 2013-06-18T14:51:03Z Edited accordingly... http://mathoverflow.net/questions/81308/question-about-nyman-beurling-baez-duarte-equivalent-for-riemann-hypothesis/132061#132061 Comment by Micah Milinovich Micah Milinovich 2013-05-28T14:59:49Z 2013-05-28T14:59:49Z The optimal minimizing Dirichlet polynomial should be a mollifier, right? http://mathoverflow.net/questions/130247/closed-form-for-derivatives-zetan1-2/130294#130294 Comment by Micah Milinovich Micah Milinovich 2013-05-10T20:28:10Z 2013-05-10T20:28:10Z Thanks Noam. I'll edit accordingly. http://mathoverflow.net/questions/126296/how-i-can-prove-that-the-hasse-weil-l-function-vanishes-at-non-positive-integer Comment by Micah Milinovich Micah Milinovich 2013-04-02T17:29:26Z 2013-04-02T17:29:26Z The (analytic continuation) of the Dirichlet series does not satisfy that functional equation, the completed $L$-function does. The poles of the gamma-factors in the completed $L$-function imply the existence of the zeros you are interested in. http://mathoverflow.net/questions/125931/selbergs-orthonormality-conjecture-and-permutations Comment by Micah Milinovich Micah Milinovich 2013-03-29T19:52:58Z 2013-03-29T19:52:58Z I do not believe that the number of primitive elements of the Selberg class is countable. For instance, $F(s)=L(s+ia,\chi)$ is primitive when $\chi$ is a primitive Dirichlet character and $a \in \mathbb{R}$. http://mathoverflow.net/questions/120115/counting-square-free-numbers-smoothly Comment by Micah Milinovich Micah Milinovich 2013-01-28T15:25:36Z 2013-01-28T15:25:36Z Since the generating function for square-free numbers is $\zeta(s)/\zeta(2s)$, I believe you can get an error of $o(\sqrt{x})$ by using the classical zero-free region for the zeta-function (even without smoothing the sum). To get an error like $O(x^{1/2-\delta})$ for some $\delta&gt;0$ seems more or less equivalent to a quasi-Riemann hypothesis. http://mathoverflow.net/questions/114888/bounds-of-chebyshevs-function-in-an-interval/114894#114894 Comment by Micah Milinovich Micah Milinovich 2012-11-29T16:16:08Z 2012-11-29T16:16:08Z Thanks Eric ... corrected. http://mathoverflow.net/questions/110944/what-does-the-numerically-verified-part-of-the-riemann-hypothesis-tell-about-prim Comment by Micah Milinovich Micah Milinovich 2012-10-29T00:15:39Z 2012-10-29T00:15:39Z Knowing the location of first 2000 or so zeros of the zeta-function above the real axis to 75 digits of accuracy seems to have been essential in Odlyzko and te Riele's disproof of the Merten's conjecture. See: <a href="http://oai.cwi.nl/oai/asset/1823/1823A.pdf" rel="nofollow">oai.cwi.nl/oai/asset/1823/1823A.pdf</a> http://mathoverflow.net/questions/108942/uniqueness-of-factorization-in-the-selberg-class Comment by Micah Milinovich Micah Milinovich 2012-10-05T18:13:22Z 2012-10-05T18:13:22Z Selberg's conjectures imply the uniqueness of factorization. <a href="http://www.aimath.org/WWN/rh/articles/html/82a/" rel="nofollow">aimath.org/WWN/rh/articles/html/82a</a> http://mathoverflow.net/questions/106989/which-long-standing-open-problems-follow-immediately-from-the-proof-of-the-abc-co Comment by Micah Milinovich Micah Milinovich 2012-09-14T12:27:09Z 2012-09-14T12:27:09Z Here is a survey article by Andrew Granville and Tom Tucker: <a href="http://www.ams.org/notices/200210/fea-granville.pdf" rel="nofollow">ams.org/notices/200210/fea-granville.pdf</a> http://mathoverflow.net/questions/102536/axioms-for-riemann-zeta-function/102539#102539 Comment by Micah Milinovich Micah Milinovich 2012-07-19T19:20:15Z 2012-07-19T19:20:15Z Thanks for the clarification... http://mathoverflow.net/questions/102536/axioms-for-riemann-zeta-function/102539#102539 Comment by Micah Milinovich Micah Milinovich 2012-07-18T23:29:13Z 2012-07-18T23:29:13Z Robert: Do any (or all) of the proofs which classify degree one elements in the Selberg class in some way assume the existence of an Euler product? http://mathoverflow.net/questions/102430/number-theory-lectures Comment by Micah Milinovich Micah Milinovich 2012-07-17T13:13:47Z 2012-07-17T13:13:47Z <a href="http://www.numbertheory.org/ntw/lecture_notes.html" rel="nofollow">numbertheory.org/ntw/lecture_notes.html</a> http://mathoverflow.net/questions/102119/interplay-between-riemann-and-swinnerton-dyer Comment by Micah Milinovich Micah Milinovich 2012-07-14T16:52:30Z 2012-07-14T16:52:30Z There are precise formulae to count the number of zeros on L(E,S) in the critical strip, up to a height T say. See: <a href="http://mathoverflow.net/questions/71061/on-the-zeroes-of-hasse-weil-l-function/71062#71062" rel="nofollow" title="on the zeroes of hasse weil l function">mathoverflow.net/questions/71061/&hellip;</a> Moreover, it is generally believed that the nonreal zeros of L(E,s) are all simple.