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29 March 2024
 
  » arxiv » 1706.8868

 Article overview


On the zeros of Riemann $Xi(z)$ function
Yaoming Shi ;
Date 18 Jun 2017
AbstractThe Riemann $Xi(z)$ function (even in $z$) admits a Fourier transform of an even kernel $Phi(t)=4e^{9t/2} heta’’(e^{2t})+6e^{5t/2} heta’(e^{2t})$. Here $ heta(x):= heta_3(0,ix)$ and $ heta_3(0,z)$ is a Jacobi theta function, a modular form of weight $frac{1}{2}$. (A) We discover a family of functions ${Phi_n(t)}_{ngeqslant 2}$ whose Fourier transform on compact support $(-frac{1}{2}log n, frac{1}{2}log n)$, ${F(n,z)}_{ngeqslant2}$, converges to $Xi(z)$ uniformly in the critical strip $S_{1/2}:={|Im(z)|< frac{1}{2}}$. (B) Based on this we then construct another family of functions ${H(14,n,z)}_{ngeqslant 2}$ and show that it uniformly converges to $Xi(z)$ in the critical strip $S_{1/2}$. (C) Based on this we construct another family of functions ${W(n,z)}_{ngeqslant 8}:={H(14,n,2z/log n)}_{ngeqslant 8}$ and show that if all the zeros of ${W(n,z)}_{ngeqslant 8}$ in the critical strip $S_{1/2}$ are real, then all the zeros of ${H(14,n,z)}_{ngeqslant 8}$ in the critical strip $S_{1/2}$ are real. (D) We then show that $W(n,z)=U(n,z)-V(n,z)$ and $U(n,z^{1/2})$ and $V(n,z^{1/2})$ have only real, positive and simple zeros. And there exists a positive integer $Ngeqslant 8$ such that for all $ngeqslant N$, the zeros of $U(n,x^{1/2})$ are strictly left-interlacing with those of $V(n,x^{1/2})$. Using an entire function equivalent to Hermite-Kakeya Theorem for polynomials we show that $W(ngeqslant N,z^{1/2})$ has only real, positive and simple zeros. Thus $W(ngeqslant N,z)$ have only real and imple zeros. (E) Using a corollary of Hurwitz’s theorem in complex analysis we prove that $Xi(z)$ has no zeros in $S_{1/2}setminusmathbb{R}$, i.e., $S_{1/2}setminus mathbb{R}$ is a zero-free region for $Xi(z)$. Since all the zeros of $Xi(z)$ are in $S_{1/2}$, all the zeros of $Xi(z)$ are in $mathbb{R}$, i.e., all the zeros of $Xi(z)$ are real.
Source arXiv, 1706.8868
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