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Complex analysis, holomorphic functions, automorphic group actions and forms, pseudoconvexity, complex geometry, analytic spaces, analytic sheaves.

48 votes
Accepted

Is this equivalent to RH - Riemann hypothesis?

Yes, this is equivalent to RH (but not in any significant way). Recall the completed Riemann $\xi$-function $$ \xi(s) = s(s-1) \pi^{-s/2} \Gamma(s/2) \zeta(s), $$ which, by Hadamard's factorizati …
Lucia's user avatar
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43 votes

Why does the Riemann zeta function have non-trivial zeros?

Rather than the problem of why the zeta function has non-trivial zeros, let me address Gowers's question of why the error term in the prime number theorem needs to be large. The short answer that …
Lucia's user avatar
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23 votes

Dirichlet series with a single zero

That such a Dirichlet series exists was a conjecture of Balazard, which was recently resolved by Hilberdink and Saias. If the Riemann Hypothesis is true, then $1/\zeta(s)$ would provide such an exam …
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17 votes
Accepted

Reference for Lindelöf Hypothesis implying finitely many zeros off critical line?

The Lindelof hypothesis (LH) does not seem to give such precise information about the zeros of $\zeta(s)$. Here are three known implications of Lindelof on the zeros, but they will be seen to fall fa …
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16 votes

Zeros of MacLaurin polynomials for the exponential function

Indeed the polynomial $\exp_n(z)$ has no purely imaginary zeros. Write $$ \exp_n(ix) = C_n(x) + i S_n(x) $$ in the obvious notation, with $C_n$ for truncations of cosine, and $S_n$ for truncation …
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14 votes
Accepted

Hadamard factorization of L-functions

I believe you are correct and $b$ is zero, although I find it inexplicable why this is not better known (certainly I didn't know it before). Let's stick to a primitive Dirichlet character $\mod q$, …
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12 votes

Objections to and arguments for the simplicity of all Riemann zeros

Quite a lot is known in the literature about zeros of $\zeta(s)$ and $\zeta^{\prime}(s)$, and in particular one can show that zeros of $\zeta^{\prime}(s)$ do get close to zeros of $\zeta(s)$. However …
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11 votes
Accepted

Upper bound for maximum modulus of polynomial on unit circle in term of the distribution of ...

Let me first start with the other side: does the maximum being small guarantee that the roots are equidistributed? This is indeed the case, and is a beautiful theorem of Erdos and Turan. For a recen …
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11 votes
Accepted

$\pm1$-polynomials with a maximal non-real root

A closely related problem was considered by Odlyzko and Poonen who looked at the class of polynomials with $0$ or $1$ coefficients. On page 330, they discuss the question of the smallest (in size) no …
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10 votes
Accepted

Bound on sum of complex summands involving binomial coefficients

Assuming that $|x+y|<1$ and $4|xy| \le 1$, here's a proof of the decay. First suppose that $|x|> |y|$. The desired sum is $$ \le \binom{2n}{n} |xy|^n \sum_{j=0}^{n} |y/x|^j \le \binom{2n}{n} |x …
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10 votes
Accepted

more on "sinc-ing" integrals and sums

Results like this hold for small values of $j$, but are eventually false. Put $f(x)$ to be the indicator function of the interval $[-1/2,1/2]$, and put $f_j(x)$ to be the convolution of $f$ with itse …
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8 votes
Accepted

Why would the roots of the generating functions of the number of k-almost primes less than x...

In any bounded region, for large $x$, the polynomial can only take on zeros very near the negative real axis (and indeed near the non-positive integers). This follows from the work of Selberg (Note on …
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7 votes
Accepted

A conjecture regarding the integral of the square of an entire function

It's false. Take for example $$ f(x) =\sum_{n\in {\Bbb Z}} e^{-n(x-n)^2}. $$ Clearly $f(n) \ge 1$ for all integers $n$. Since in intervals of length $1$ the function $f$ is large only in a small …
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2 votes

Asymptotic value of a multivariate integral

I'll consider the integrals extended from $-\infty$ to $\infty$ and show that the integral of $G$ is exponentially small compared to the integral of $|G|$ -- precisely, the oscillating integral is sm …
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