Johan Andersson
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Registered User
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Analytic number theorist. I work at Stockholm university as a temporary associate professor. Interested in zeta-functions, universality, prime numbers, power sums, automorphic forms, moment estimates. I am especially interested in the interplay beween analytic number theory and harmonic analysis.
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May 14 |
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polynomial zero within a square This does not seem to work since we do not longer know that $|f(c(1+i))|>|f(0)|$ after scaling with the constant $c$. By using the lagrange interpolation formula in a differnt way (as in my new answer above) to be say $1$ for $1,i,1+i$ etc and $0$ for $z=0$ and adding a sufficiently large positive constant $k$, we are sure that $0<p(0)<p(z)$ (and of course the same inequality holds when taking absolute values) when $z=1,i,1+i$, etc and also that the polynomial is zero-free. |
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May 14 |
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polynomial zero within a square Actually if we consider the generalized problem of $n$ points $z_k$ and want to find a polynomial that is zero-free on a disc D such that $|f(0)|<|f(z_k)|$ a similar construction $p(z)=z^N+k$ works. First use Kroneckers approximation theorem to find $N$ such that $N \arg(z_k)$ mod $2 \pi$ lies in the interval $(-\pi/2,\pi/2)$. Then find $k$ sufficiently large such that the polynomial is zero free. |
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May 14 |
awarded | ● Citizen Patrol |
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May 12 |
revised |
polynomial zero within a square added 754 characters in body; edited body |
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May 12 |
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polynomial zero within a square unit disc was a mistake that I just corrected. Thanks, I meant the unit square. The same result holds however for the unit disc, for example for any closed Jordan domain with the inequality holding for finitely many boundary points. |
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May 12 |
revised |
polynomial zero within a square added 2 characters in body |
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May 12 |
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polynomial zero within a square Gerald's answer is much better and simpler than my. I was considering deleting my answer, but decided to leave it. |
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May 12 |
answered | polynomial zero within a square |
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May 9 |
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Asymptotics of a function and by Stirling's formula this gives us $f(n)=\sqrt{2 \pi n} (\frac{n}{4e \ln n})^n (1+O(n^{-1/2}))$. |
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Apr 14 |
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Are there refuted analogues of the Riemann hypothesis? See also the question and my answer mathoverflow.net/questions/45912/… |
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Apr 14 |
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Hejhal’s algorithm and computational methods for non-classical Maass wave forms The second reference should be Booker, Strömbergsson and Venkatesh, not just Venkatesh |
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Feb 22 |
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The Paley-Wiener theorem and exponential decay. Koosis, The logarithmic integral part I has a good treatment of this theory (part II treats the somewhat more complicated Beurling-Malliavin theorem). |
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Feb 10 |
revised |
Pair correlation for the Riemann zeros and $(\zeta^\prime(s)/\zeta(s))^\prime$ edited body |
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Feb 10 |
answered | Pair correlation for the Riemann zeros and $(\zeta^\prime(s)/\zeta(s))^\prime$ |
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Feb 1 |
awarded | ● Necromancer |
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Feb 1 |
revised |
Saying things rapidly about integer factorisations added 88 characters in body |
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Feb 1 |
answered | Saying things rapidly about integer factorisations |
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Jan 14 |
accepted | On the location of zeros of L functions from modular forms |
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Jan 14 |
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On the location of zeros of L functions from modular forms In the first section I state that. In the second section I mention the analogue of the Davenport-Heilbronn theorem that there are zeroes for Re(s)>1. It should follow in a similar way as the case of the Hurwitz zeta-function for rational parameter. |
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Jan 14 |
answered | On the location of zeros of L functions from modular forms |
