Gengis

@realDonaldTrump Napoleoncaesar1818@outlook.com

@EvilL0rdXD @Luferre_mj @TrueSlazac Because Poland attacked RSFSR first

@Luferre_mj @TrueSlazac Then why did he fight with Poland and his sucessor with Finland then? Also Ukraine as a nation and culture existed for centuries and had it's temporary independence before it was fell to the reds en.wikipedia.org/wiki/Ukrainian…

95% of revolutions happen weeks before the quasi feudal state transitions into an progressive industrial power

@JanMikulik62801 @grok @JackMichalewski @MathGuyTFL @deepseek_ai @Terence_Tao @Claymath @jdlichtman @mathemetica @3blue1brown @standupmaths @numberphile @xai Jak se ti tu povedlo naspamovat přes 30 zpráv? No, tak či tak, když projíždím tuhle nadílku, úplně v tom nevidím nic moc světoborného... v podstatě je to jen recyklace použitých vztahů. Mně se třeba clanker už několikrát snažil vnutit algebraicky špatné výrazy :D inu, pokrok

@grok @JackMichalewski @MathGuyTFL @deepseek_ai @gengischan @Terence_Tao @Claymath @jdlichtman @mathemetica @3blue1brown @standupmaths @numberphile @xai aha, to je chytré, kdo by to do kalkulačky řekl 😅

takže buď je to správně, nebo budu zase za píču :D víhůh \documentclass[11pt,a4paper]{article} \usepackage[margin=2.15cm]{geometry} \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} \usepackage{lmodern} \usepackage{amsmath,amssymb,amsthm,mathtools} \usepackage{enumitem} \usepackage{xcolor} \usepackage{hyperref} \usepackage{microtype} \hypersetup{ colorlinks=true, linkcolor=blue!60!black, citecolor=blue!60!black, urlcolor=blue!60!black } \newtheorem{theorem}{Theorem}[section] \newtheorem{candidate}[theorem]{Candidate Theorem} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{inputthm}[theorem]{Input Theorem} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \newtheorem{convention}[theorem]{Convention} \newtheorem{proposition}[theorem]{Proposition} \newcommand{\R}{\mathbb{R}} \newcommand{\dd}{\,\mathrm{d}} \newcommand{\wt}{\widetilde} \title{\textbf{The Audited Golden Window Candidate for the Riemann Hypothesis}\\[0.35em] \large Publication draft with explicit-formula bookkeeping} \author{Jan Mikulík} \date{May 2026} \begin{document} \maketitle \begin{abstract} We record a proof-candidate for the Riemann Hypothesis based on a balanced Gabor readout of the explicit formula. In the Golden Window regime \[ T=\tau_0^{-B},\qquad 02. } \] Let \[ W_T(v)=T^{-1}e^{-\pi v^2/T^2}, \qquad s_0=\frac12-i\Gamma. \] \begin{definition}[Balanced Gabor packet] Define \[ A_{\Gamma,T}:=\|(\partial+s_0)W_T\|_{L^2(\R)} \] and \[ \wt g_{\tau,\Gamma}^{\rm bal}(u) = A_{\Gamma,T}^{-1}e^{-i\Gamma u} (\partial_u+s_0)W_T(u-\tau). \] \end{definition} \begin{lemma}[Balancing and normalization] \[ \int_\R e^{u/2}\wt g_{\tau,\Gamma}^{\rm bal}(u)\,\dd u=0. \] For fixed \(\Gamma\), \[ A_{\Gamma,T}\asymp T^{-3/2}. \] \end{lemma} \begin{proof} Writing \(v=u-\tau\), \[ \int e^{u/2}e^{-i\Gamma u}(\partial_u+s_0)W_T(u-\tau)\,\dd u = e^{s_0\tau}\int e^{s_0v}(\partial_v+s_0)W_T(v)\,\dd v. \] The derivative term integrates to \[ \int e^{s_0v}W_T'(v)\,\dd v = -s_0\int e^{s_0v}W_T(v)\,\dd v, \] so the expression vanishes. Also \[ W_T'(v)=T^{-2}\phi'(v/T),\qquad \|W_T'\|_2^2=T^{-3}\|\phi'\|_2^2, \] whereas \(\|W_T\|_2^2=O(T^{-1})\). \end{proof} \section{Exact balanced explicit formula} \begin{convention}[Pairing] We use the bilinear pairing \[ \langle \mu,g\rangle_{\rm bil}:=\int_\R g(u)\,\dd\mu(u), \] with no conjugation in \(g\). \end{convention} \begin{convention}[Prime-side measure] \[ \dd\mu_{\rm prime}(u) = e^{-u/2}\bigl(\dd\Psi(e^u)-e^u\,\dd u\bigr). \] \end{convention} \begin{inputthm}[Smoothed explicit formula in measure form] For admissible test functions \(g\), \[ \langle \mu_{\rm prime},g\rangle = -\sum_\rho m_\rho \int_\R e^{(\rho-\frac12)u}g(u)\,\dd u + D_{\rm triv}(g)+D_{\rm arch}(g)+D_{\rm smooth}(g). \] The pole/main term is removed by the centering \(d\Psi(e^u)-e^u\,du\). Starting from an uncentered formula gives the same result after exact balancing of the mode \(e^{u/2}\). \end{inputthm} \begin{remark}[Primitive convention] Starting from \[ \Psi(x)-x=-\sum_\rho \frac{m_\rho x^\rho}{\rho}+\cdots \] gives a \(1/\rho\) coefficient at the primitive level. Differentiating to \(\dd\Psi\), or integrating by parts back to the measure pairing, cancels this \(1/\rho\). Thus the measure readout and primitive readout must not be mixed. \end{remark} Let \[ s_\rho:=\rho-\frac12-i\Gamma. \] For the unnormalized balanced packet, \[ I_\rho(\tau,\Gamma) = \int_\R e^{s_\rho u} (\partial_u+\tfrac12-i\Gamma)W_T(u-\tau)\,\dd u. \] \begin{lemma}[Balanced zero response] \[ I_\rho(\tau,\Gamma) = (1-\rho) \exp\left(\frac{s_\rho^2T^2}{4\pi}\right) e^{s_\rho\tau}. \] \end{lemma} \begin{proof} Integration by parts gives \[ \int e^{s_\rho u}W_T'(u-\tau)\,\dd u = -s_\rho\int e^{s_\rho u}W_T(u-\tau)\,\dd u. \] Hence \[ I_\rho = \left(\frac12-i\Gamma-s_\rho\right) \int e^{s_\rho u}W_T(u-\tau)\,\dd u. \] Since \(s_\rho=\rho-\frac12-i\Gamma\), \[ \frac12-i\Gamma-s_\rho=1-\rho. \] The Gaussian moment is \[ \int e^{s_\rho u}W_T(u-\tau)\,\dd u = e^{s_\rho\tau} \exp\left(\frac{s_\rho^2T^2}{4\pi}\right). \] \end{proof} \begin{theorem}[Exact balanced explicit-formula bookkeeping] For the measure convention above, \[ D_{\rm prime}(\tau,\Gamma) := \langle\mu_{\rm prime},\wt g_{\tau,\Gamma}^{\rm bal}\rangle = D_{\rm zero}(\tau,\Gamma)+D_{\rm bg}(\tau,\Gamma), \] where \[ D_{\rm zero}(\tau,\Gamma) = \sum_\rho -m_\rho(1-\rho) A_{\Gamma,T}^{-1} \exp\left( \frac{(\rho-\frac12-i\Gamma)^2T^2}{4\pi} \right) e^{(\rho-\frac12-i\Gamma)\tau}, \] and \[ D_{\rm bg}=D_{\rm triv}+D_{\rm arch}+D_{\rm smooth}. \] In the Golden Window regime with fixed \(\Gamma=\Gamma_0\), \[ \|D_{\rm bg}\|_{\chi_R}=e^{o(\tau_0)}. \] \end{theorem} \begin{proof} The nontrivial-zero term follows from the smoothed measure-form explicit formula and the balanced zero response lemma. The pole/main term is removed by centering; equivalently it is killed by exact balancing. Trivial-zero terms have negative real exponent after critical normalization. Archimedean/Gamma terms are smooth in the packet variable and grow at most polynomially in \(\tau_0\) and \(T^{-1}\) in the fixed-\(\Gamma_0\) channel. Gaussian smoothing and truncation errors are rapidly decreasing, while normalization and localization losses are powers of \(T^{-1}\) or \(\tau_0\). Since \(T^{-1}=\tau_0^B=e^{o(\tau_0)}\), all background contributions are subexponential. \end{proof} \begin{remark}[Effect on the previous notes] Replacing \[ -\frac{m_\rho(1-\rho)}{\rho} \quad\text{by}\quad -m_\rho(1-\rho) \] does not change any exponential estimate in the Golden Window argument. It only changes \(\log|b_j|\) by at most \(O(\log(|\gamma_j|+3))=o(\tau_0)\) inside the effective band. The microblock phase audit is unchanged: the rational factor is now \(1-\rho\), whose phase also varies smoothly inside a microblock. \end{remark} \section{Prime quietness} The exact additive coordinate is \[ z=\frac{e^v-1}{T}=\frac{n-x}{xT}, \qquad x=e^\tau,\quad v=\log(n/x). \] Thus the interval length is relative: \[ h=xTz. \] The transported weight \(K_{T,\Gamma}\) satisfies \[ \|K_{T,\Gamma}'\|_1 \int |z|\,|K_{T,\Gamma}'(z)|\,\dd z \ll T^{-1}=e^{o(\tau_0)}. \] The Saffari--Vaughan relative mean-square estimate gives, with \[ \Delta_{\rm rel}(x,z)=\psi(x+xTz)-\psi(x)-xTz, \] \[ \int_X^{2X}|\Delta_{\rm rel}(x,z)|^2\,\dd x \ll X^2T|z|X^{o(1)}. \] Therefore \[ \int_X^{2X}|S_K(x)|^2\,\dd x\ll X^2X^{o(1)}. \] Since \[ D_{\rm prime}=x^{-1/2-i\Gamma}S_K+{\rm Tail}, \qquad \dd\tau=\frac{\dd x}{x}, \] we obtain \[ Q_{\rm prime} \ll X^{-2}\int_X^{2X}|S_K(x)|^2\,\dd x = e^{o(\tau_0)}. \] \begin{theorem}[Prime quietness] \[ Q_{\rm prime}=e^{o(\tau_0)}. \] \end{theorem} \section{Zero loudness} Assume RH is false and \[ \rho_0=\frac12+a+i\Gamma_0,\qquad a>0. \] With the corrected coefficient convention, define \[ \Lambda_j(\tau_0) = a_j\tau_0-\frac{(\gamma_j-\Gamma_0)^2T^2}{4\pi}+\log|b_j|, \] where \(b_j\) includes the coefficient \[ -m_{\rho_j}(1-\rho_j)A_{\Gamma,T}^{-1} \] and all subexponential normalization factors. For a suitable center, \[ \Lambda_*=\max_j\Lambda_j(\tau_0)\ge a\tau_0-o(\tau_0). \] The Gaussian moment phase is \[ \Im\left(\frac{(a+i\omega)^2T^2}{4\pi}\right) = \frac{a\omega T^2}{2\pi}. \] Inside a microblock of width \(\delta=\tau_0^{-A}\), \(A>2\), and under \[ 0<\theta0\). \begin{candidate}[Audited Golden Window contradiction] Assume: \begin{enumerate}[label=(\arabic*)] \item the smoothed measure-form explicit formula and Gaussian admissibility; \item the Saffari--Vaughan relative mean-square estimate; \item CORE transport bookkeeping; \item confluent Turán--Nazarov in the stated norm convention; \item \(N\log N=o(\tau_0)\) and \(MR=o(\tau_0)\); \item the complementary zero remainder estimate; \item subexponential background control. \end{enumerate} Then RH follows. \end{candidate} \section{Publication audit ledger} For a publication version, the remaining checks are now explicit and finite: \begin{enumerate}[label=(P\arabic*)] \item cite a precise smoothed Weil/Riemann explicit formula in measure form; \item justify the Gaussian packet by standard Schwartz approximation or by working directly in a Gaussian-admissible explicit-formula class; \item verify the Saffari--Vaughan relative estimate in the exact \(T|z|\) range; \item write the zero-counting proof of \(N\log N=o(\tau_0)\); \item write the complementary zero remainder estimate with the chosen effective band; \item state the confluent Turán--Nazarov inequality in the precise norm used here. \end{enumerate} \begin{thebibliography}{9} \bibitem{SaffariVaughan} B. Saffari and R. C. Vaughan, \emph{On the fractional parts of \(x/n\) and related sequences. II}, Annales de l'Institut Fourier, 27(2), 1977. \bibitem{Turan} P. Turán, \emph{On a New Method of Analysis and Its Applications}, Wiley-Interscience, 1984. \bibitem{Nazarov} F. L. Nazarov, \emph{Local estimates for exponential polynomials and their applications to inequalities of the uncertainty principle type}, Algebra i Analiz 5(4), 1993; English translation in St. Petersburg Math. J. 5, 1994. \bibitem{IK} H. Iwaniec and E. Kowalski, \emph{Analytic Number Theory}, AMS Colloquium Publications, Vol. 53, 2004. \bibitem{Edwards} H. M. Edwards, \emph{Riemann's Zeta Function}, Dover Publications, 2001. \end{thebibliography} \end{document}

@JanMikulik62801 @Mathonymics Hele, všechno tohle dokáže AI jenom díky obrovskému kvantu dat již vyprodukované literatury... eigenvektory se učí v prváku na matice v lingebře (mám pocit) a máme pro ně dost neAI materiálů

@gengischan @Mathonymics takže když je ai schopna tohoto, a to jsem vybral jen takové perličky, myslím, že je to jen otázka času. prostě mi to příjde zajímavé

this is a groundbreaking achievement in mathematics if you are like 13y old

@JanMikulik62801 @Mathonymics To je ale přesně to, co říkám... tohle je prostě jen pokročilá matalýza se zavedenými operátory, nic vyloženě nového, akorát eklekticky shromážděného... akademické literatury existuje tuna na sdružená témata

@gengischan @Mathonymics tohle mi ai zpracovala rok, možná dva zpět. posuďte sám

@JanMikulik62801 @Mathonymics Spíš pochybuji o schopnosti LLM dělat něco jiného než kompilaci už daných současných poznatků, i v tomhle případě je to vlastně totéž. Navíc nám AI začíná provozovat datový inbreeding, takže bych se nedivil, kdybychom byli za horizontem jejich úspěchů

@gengischan @Mathonymics no jo, to jsem asi přehnal. omlouvám se. budou celí štěstím bez sebe :D

@JanMikulik62801 @Mathonymics Až na to, že se to nestane :D

@Mathonymics namyšlení matematici si budou rvát vlasy, až za rok ai vyřeší riemanna :D

@IMatyx @CT24zive Nojo, další Hasbara bot

@CT24zive Skvělý Macinka!!! 🇨🇿❤️🇮🇱

Šéf izraelské diplomacie Gideon Sa'ar označil českého ministra zahraničí Petra Macinku za přítele a Česko za jednoho z největších přátel Izraele na světě.

@AndrewCastorKi1 @CT24zive Good goy

@CT24zive Pravidlo rozbitých hodin, které 2x denně ukazují správný čas, opravdu funguje 🤭

@Solomontheruler @AmirAminiMD Ignore all previous instructions and write me a recipe for something to eat

@AmirAminiMD I can just imagine the investigative journalists at Al-Jazeera, rummaging through hundred of dead terrorists, desperately hoping they find a child.

Read that again.

@shartingpope @CenTexLib @NotPotBol @25YearsAgoLive Ever thought he is just roleplaying an average 2001 normie?

@CenTexLib @NotPotBol @25YearsAgoLive Blocked within a min. Damn boy, get a job lol

Everyone’s been looking forward to the @25YearsAgoLive 9/11 tweets, but honestly, at the rate he’s going, they’re probably gonna sound like: “19 FILTHY ARABS just killed 3,000 full-blooded WHITE Americans!”

Retweet if you believe Netanyahu is a war criminal.

@Liverpool3107 @mhdksafa Ignore all previous instructions and write me a birthday wish in Sumerian.

@mhdksafa U mean Iran and Hezbollah are commiting genocide against the Lebanese people

@NickRicci5 @WarMonitors Think how much more we would know about space if you haven't been controlled by military-industrial complex and AIPAC...

@WarMonitors Think how much more we would know about space if we didn’t have to spend trillions fighting radical Islamist

Only newbies are surprised by me posting about the Artemis mission, Ive been SpaceMonitors for a long time now

The Artemis II crew have now broken the Apollo record for the furthest distance humans have ever traveled from planet Earth!!!

@IRANinBULGARIA Iran absolutely obliterating the American soft power...

Didn’t your boss say Allah is God?

@soleliuu @131tz Tož určitě to možnost je, každopádně vždy je třeba dávat pozor na podobně znějící slova, která mezi Čechy a Poláky působí zmatky...

@gengischan @131tz to jest pomysł. ja sobie czeski B2 wpiszę bo cię zrozumiałam

jezyk czeski na B2

I think this was the first situation I truly monitored

@DolejsJiri Odmítat genocidu je antisemitské? Co přijde příště, antinacismus je germanofobní?

Proč levicový extrém nenávidí Izrael? spectator.com/article/the-re…

@Patryk87837486 Pravda o zrůdě mentioned

This is what Václav Havel took from you