User micah milinovich - MathOverflow

Answer by Micah Milinovich for On the critical line $ \Re \zeta'(s)/\zeta(s) =? 1/2 \log(\pi) - 1/2 \Re \psi(s/2)$ ?

2013-06-18T13:03:53Z

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: http://arxiv.org/pdf/math/0612106v2.pdf

Answer by Micah Milinovich for Estimate on the prime-counting function $\psi(x)$.

2013-05-23T16:12:47Z

The reference for this (given in the 3rd edition of Davenport's Multiplicative Number Theory) is:

Grosswald, Émile. "Sur l'ordre de grandeur des différences $\psi(x)-x$ et $\pi(x)-\ell i(x)$." (French) C. R. Acad. Sci. Paris 260 (1965), 3813–3816.

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.

Answer by Micah Milinovich for Closed form for derivatives $\zeta^{(n)}(1/2)$

2013-05-10T19:52:09Z

Edit: My original answer was incorrect.

You can evaluate $\zeta'(\frac{1}{2})$ recursively in terms of $\zeta(\frac{1}{2})$ using the symmetric form of the functional equation:

$$ \zeta(s)\Gamma(\tfrac{s}{2}) \pi^{-s/2} = \zeta(1{-}s)\Gamma(\tfrac{1-s}{2}) \pi^{(s-1)/2}. $$

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.

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 $\zeta(\frac{1}{2}),\zeta''(\frac{1}{2}), \zeta^{(4)}(\frac{1}{2}), \ldots, \zeta^{(2n)}(\frac{1}{2})$.

Answer by Micah Milinovich for Riemann Z function, bounds on number of non-trivial zeros along horizontal lines, rather than vertical ones.

2013-03-26T20:46:37Z

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 $0<\Im s \le T$ 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 & Vaughan's "Multiplicative Number Theory, I."

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.

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 & 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.

References:

http://arxiv.org/pdf/1208.5846.pdf

http://arxiv.org/pdf/math/0511092v1.pdf

Answer by Micah Milinovich for Recovering $\sum_{n \leq x} a(n)$ from $\sum_{n \leq x} a(n)e^{-n/x}$

2013-03-25T18:05:07Z

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 $$ \sum_{n\le x} a_n e^{-n/x} = \sum_{n\le x} a_n+ O\left( \frac{1}{x} \sum_{n\le x}n \ a_n\right). $$

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 $k$.

Answer by Micah Milinovich for Bounds of Chebyshev's function in an interval.

2012-11-29T15:59:12Z

Adding to Eric Naslund's comment: Let $\pi(x)$ denote the number of primes less than $x$, then Montgomery & Vaughan proved that $$ \pi(x+y)-\pi(x) \le 2 \pi(y)$$ for $x\ge 1$ and $y\ge 2.$

Answer by Micah Milinovich for Good papers/books/essays about the thought process behind mathematical research

2012-09-12T12:28:19Z

I know it is a bit dated, but I am fond of Littlewood's essay "The Mathematician's Art of Work" in his Miscellany.

Answer by Micah Milinovich for Large gaps between P2s

2012-09-08T20:05:29Z

Here is a result, due to Halberstam, Heath-Brown, and Richert that seems to be of the flavor you are looking for (Almost-primes in short intervals. Recent progress in analytic number theory, Vol. 1 (Durham, 1979), pp. 69–101, Academic Press, London-New York, 1981.)

Theorem. 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.

Reference chasing on MathSciNet, it seems that Iwaniec & 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.

Answer by Micah Milinovich for Sharpening of Lindelöf hypothesis

2012-09-06T14:24:29Z

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.

Ramachandra and Balasubramanian had previous proved an inequality of a similar form with the factor of $1-\varepsilon$ replaced by a smaller constant.

Answer by Micah Milinovich for Axioms for Riemann $\zeta$ function

2012-07-18T14:48:49Z

Perhaps you are looking for something like Hamburger's Theorem?

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.

Googling I found the following link: http://www.mat.univie.ac.at/~esiprpr/Zetaproc/patterson.pdf

Answer by Micah Milinovich for Is Mertens function negatively biased?

2012-05-28T07:56:00Z

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: http://arxiv.org/abs/math/0310381

The paper on summatory function of the Louiville function, mentioned in kolik's answer, uses many of the same techniques.

Optimization problem arising from the study of zeta zeros

2012-05-15T19:57:06Z

Motivation: The following problem arose in [1] 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."

Set-up: 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 $$ M(c):=c+\frac{2 r}{\pi}\int_0^1 (1-u)^{r^2-1}f(u) \int_0^u \frac{\sin(\pi c v)}{v} f(u-v) \ dv \ du.$$

Question: How does one choose $r$ and $f$ optimally so that $$ M(c) >1$$ for $c$ as small as possible?

An argument of Conrey, Ghosh, and Gonek [2] can be used to show that $M(c)<1$ if $c<\frac{1}{2}$ for any such $f$ and $r$. In [1], 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 $M(.5155)>1$.

In the special case when $r=1$, Montgomery and Odlyzko [3] 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 m and degree n, and a is a constant to be chosen according to our above normalization.

These wave functions are very hard to study numerically, and Montgomery and Odlyzko approximated them using modified Bessel functions. In [1], 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.

References:

[1] H. M. Bui, M. B. Milinovich, and N. C. Ng, A note on the gaps between consecutive zeros of the Riemann zeta-function, Proc. Amer. Math. Soc. 138 (2010), no. 12, pp. 4167-4175.

[2] J. B. Conrey, A. Ghosh, and S. M. Gonek, A note on gaps between zeros of the zeta function, Bull. London Math. Soc. 16 (1984), 421–424.

[3] H. L. Montgomery and A. M. Odlyzko, Gaps between zeros of the zeta function, Colloq. Math. Soc. Janos Bolyai, 34. Topics in Classical Number Theory (Budapest, 1981), North-Holland, Amsterdam, 1984.

Edit/Additional Comments: The solution optimization problem when $r=1$ should probably be attributed to Slepian and Pollak and to Landau and Pollack in Prolate spheroidal wave functions, Fourier analysis and uncertainty I and II, Bell System Tech. J. (1961) 40, pp. 43-61 (I) and pp. 65-84 (II). Among other things, they prove the following results.

Theorem: Let $\alpha(c)$ be the least number such that $$ \int_{-c/2}^{c/2} \left|\int_0^1 f(x) e^{2\pi i t x} dx \right|^2 dt \le \alpha(c) \int_0^1 |f(x)|^2 dx $$ 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 $f(x) = R_{00}^{(1)}[\pi c/2,2x-1]$.

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 & Odlyzko's argument.

Answer by Micah Milinovich for non-trivial zeros of partial zeta functions

2012-05-17T02:04:50Z

The answer to question 1 is classical: Any Dirichlet series which has a finite abscissa of absolute convergence has a zero-free half-plane.

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 $\Re(s)>C$.

Answer by Micah Milinovich for Divergent series expansion in Apéry's proof of the irrationality of $\zeta(2)$ and $\zeta(3)$

2012-04-21T13:42:42Z

I think that this is "amazing claim" number 1 in van der Poorten's paper "A proof that Euler missed ...."

http://www.ift.uni.wroc.pl/~mwolf/Poorten_MI_195_0.pdf

Answer by Micah Milinovich for Sum of $\sum_{k=1}^nd(k^2)$

2012-04-13T00:41:52Z

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 $\asymp x.$

Edit: 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 DivisorSigma[0, n^2] 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 $$ \frac{1}{F(x)}\sum_{n\leq x} d(n^2) \approx 1.0000073....$$

Answer by Micah Milinovich for Distribution of primes in small intervals

2012-03-30T03:41:47Z

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:

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.

This is Theorem 13.3 in Montgomery and Vaughan's Multiplicative Number Theory.

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

$$\psi(x+y)-\psi(x)=y+O\left(\sqrt{x} \log x \log\left(\frac{2y}{\sqrt{x} \log x}\right) \right).$$

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.

Answer by Micah Milinovich for Is this sum of reciprocals of zeta zeros correct?

2012-03-15T14:03:11Z

Edit: 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.

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 $$ \sum_{\rho} \frac{1}{\rho (1{-}\rho)}.$$

As you observed, assuming the Riemann Hypothesis $1-\rho =\overline{\rho}$ for any non-trivial zero $\rho$ of $\zeta(s)$. This implies that

$$ \sum_{\rho} \frac{1}{\rho (1{-}\rho)}=\sum_{\rho} \frac{1}{|\rho|^2}.$$

Answer by Micah Milinovich for Is this sum of reciprocals of zeta zeros correct?

2012-03-15T14:56:09Z

To answer your modified question, according to Mathematica:

$$ \lim_{s\to 0} \left(\frac{\hat{\zeta}'}{\hat{\zeta}}(s)+\frac{1}{s}\right) = -\frac{\gamma}{2} + \tfrac{1}{2}\log(4\pi).$$

This implies that

$${\sum_\rho}'\frac{1}{\rho} = \sum_{\Im \rho >0} \frac{1}{\rho(1-\rho)}= 1 +\frac{\gamma}{2} - \tfrac{1}{2}\log(4\pi). $$

Therefore $$\sum_{ \rho } \frac{1}{\rho(1-\rho)} = 2 \sum_{\Im \rho >0} \frac{1}{\rho(1-\rho)}= 2 +\gamma - \log(4\pi).$$

Answer by Micah Milinovich for Prime numbers in arithmetic progressions : uniformity with respect to the modulus

2012-03-03T15:25:08Z

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:

http://www.crm.umontreal.ca/~koukoulo/documents/publications/multfncs.pdf

Answer by Micah Milinovich for Possible locations for non trivial zeroes lying off the critical line

2012-01-10T21:39:30Z

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 $|\chi(s)|=1$.

A simple proof can be found in Lemma 6.1 of S. M. Gonek "Finite Euler products and the Riemann hypothesis" Trans. Amer. Math. Soc. 364 (2012), 2157-2191. This paper is also on the arXiv. Gonek states that $C_0<6.3$ so it seems that phenomena in your pictures stops shortly after the ranges you plotted.

Answer by Micah Milinovich for Upper bounds on the difference of consecutive zeta zeros

2012-01-06T04:53:39Z

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).

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 $O( 1/\log\log \gamma_n).$

Answer by Micah Milinovich for Values of Dirichlet L-funcions at natural numbers

2012-01-03T18:12:36Z

Let $\chi$ be any Dirichlet character modulo $q$, and let $m$ be a positive integer. Then Theorem 4.2 of Washington's Introduction to Cyclotomic Fields states that

$$ L(1-m,\chi) = - \frac{q^{m-1}}{m} \sum_{a=1}^q \chi(a)B_{m}(\tfrac{a}{q}). $$

Here $B_m(x)$ is the usual Bernoulli polynomial, defined by

$$ \frac{t e^{Xt}}{e^t-1} = \sum_{n=0}^\infty B_n(X) \frac{t^n}{n!}.$$

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).$

Answer by Micah Milinovich for explicit formula for Riemann zeros counting function

2011-12-05T04:46:18Z

Assuming the Riemann Hypothesis, you can use a smooth approximation to the characteristic function of an interval in the Guinand-Weil explicit formula to approximately 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.)

The details are (essentially) contained in a paper by Goldston & 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

Exponential sums related to cusp forms

2011-07-26T20:06:17Z

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 $a_f(1)=1$.

In Jutila's book "A method in the theory of exponential sums" he proves an estimate of the form:

$$ \sum_{n\leq x} a_f(n) e^{2\pi i n \frac{p}{q}} \ll q^{2/3}x^{k/2-1/6+\varepsilon} $$

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:

$$ \sum_{n\leq x} a_f(n) e^{2\pi i n\frac{p}{q}} \ll_{f,q} \ \ x^{k/2-\delta} $$

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 $\delta>0$?

Answer by Micah Milinovich for on the Zeroes of Hasse -weil L-function

2011-07-23T12:13:36Z

The $L$-function has about $\displaystyle{\frac{T}{\pi} \log T \ }$ zeros in the strip with $0 < t < T$. See section 5.3 of Iwaniec and Kowalski's "Analytic Number Theory," in particular Theorem 5.8.

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.

Answer by Micah Milinovich for Riemann-Siegel's approximate functional equation for fixed t and Re(s)≠1/2

2011-07-09T17:28:49Z

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:

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.

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.

Answer by Micah Milinovich for Is anyone aware of a good exposition of the Gauss-Kramer model of Integers?

2011-06-07T15:25:55Z

You might want to look at:

1) "Harold Cramér and the distribution of prime numbers" by Andrew Granville http://www.dms.umontreal.ca/~andrew/PDF/cramer.pdf

2) "The distribution of prime numbers" by K. Soundararajan http://arxiv.org/pdf/math/0606408v1

Answer by Micah Milinovich for Asymptotic Formula for a Mertens Style Sum

2011-05-13T03:16:13Z

Here is an answer that is similar in spirit to Frank and Peter's answers, but possibly simpler.

Summing by parts, we see that

$$ \sum_{p\le x} \frac{\log^k p}{p} = (\log x)^{k-1} \sum_{p\leq x} \frac{\log p}{p} -(k-1)\int_{2^-}^x (\log u)^{k-2}\sum_{p\le u} \frac{\log p}{p} \frac{du}{u}.$$

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 $$ \sum_{p\le x} \frac{\log^k p}{p} = \frac{\log^k x}{k} + c_3 + O(\exp(-c_4\sqrt{\log x})).$$

This seems easier than dealing with $Li(x)$.

Answer by Micah Milinovich for Zeros of primitive functions of S

2011-04-26T22:02:52Z

There is an example given in this paper:

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," Mathematics of Computation 44 (1985), no. 170, pp. 473-481.

http://www.jstor.org/stable/2007967

Are there any rational solutions to this equation?

2011-03-01T21:15:00Z

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.

1) Are there any rational solutions to the following equation: $$x^3-8x^2+5x+1 = -7y^2(x-1)x$$

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.)

Comment by Micah Milinovich

2013-06-19T14:43:33Z

FYI, this inequality has been sharpened: blms.oxfordjournals.org/content/43/2/243.abstract

Comment by Micah Milinovich

2013-06-18T14:51:03Z

Edited accordingly...

Comment by Micah Milinovich

2013-05-28T14:59:49Z

The optimal minimizing Dirichlet polynomial should be a mollifier, right?

Comment by Micah Milinovich

2013-05-10T20:28:10Z

Thanks Noam. I'll edit accordingly.

Comment by Micah Milinovich

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.

Comment by Micah Milinovich

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 $a \in \mathbb{R}$.

Comment by Micah Milinovich

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 $\delta>0$ seems more or less equivalent to a quasi-Riemann hypothesis.

Comment by Micah Milinovich

2012-11-29T16:16:08Z

Thanks Eric ... corrected.

Comment by Micah Milinovich

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: oai.cwi.nl/oai/asset/1823/1823A.pdf

Comment by Micah Milinovich

2012-10-05T18:13:22Z

Selberg's conjectures imply the uniqueness of factorization. aimath.org/WWN/rh/articles/html/82a

Comment by Micah Milinovich

2012-09-14T12:27:09Z

Here is a survey article by Andrew Granville and Tom Tucker: ams.org/notices/200210/fea-granville.pdf

Comment by Micah Milinovich

2012-07-19T19:20:15Z

Thanks for the clarification...

Comment by Micah Milinovich

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?

Comment by Micah Milinovich

2012-07-17T13:13:47Z

numbertheory.org/ntw/lecture_notes.html

Comment by Micah Milinovich

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: mathoverflow.net/questions/71061/… Moreover, it is generally believed that the nonreal zeros of L(E,s) are all simple.