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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 …
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 …
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 …
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 …
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 …
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 $\mod q$, …
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 …
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 …
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 …
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
$$
\le \binom{2n}{n} |xy|^n \sum_{j=0}^{n} |y/x|^j \le
\binom{2n}{n} |x …
10
votes
Accepted
more on "sinc-ing" integrals and sums
Results like this hold for small values of $j$, but are eventually false. Put $f(x)$ to be the indicator function of the interval $[-1/2,1/2]$, and put $f_j(x)$ to be the convolution of $f$ with itse …
8
votes
Accepted
Why would the roots of the generating functions of the number of k-almost primes less than x...
In any bounded region, for large $x$, the polynomial can only take on zeros very near the negative real axis (and indeed near the non-positive integers). This follows from the work of Selberg (Note on …
7
votes
Accepted
A conjecture regarding the integral of the square of an entire function
It's false. Take for example
$$
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 …
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 …