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

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 …
Lucia's user avatar
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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 $\mod q$, …
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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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 …
Lucia's user avatar
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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 …
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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 …
Lucia's user avatar
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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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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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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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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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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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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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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 …
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