Anthropic заявила о продвижении Claude в задаче, связанной с гипотезой Римана
Anthropic заявила, что неопубликованная исследовательская версия Claude повысила доказанную нижнюю оценку доли нулей дзета-функции Римана на критической прямой примерно с 41,6% до 67,2%. Это не является доказательством гипотезы Римана, а представленные материалы не содержат независимой проверки результата.
В официальном сообщении в социальных сетях Anthropic заявила, что неопубликованная исследовательская версия Claude работала над математической задачей, связанной с гипотезой Римана. По словам компании, система не решила саму гипотезу, но получила нижнюю оценку 67,2% для доли нулей дзета-функции Римана, расположенных на критической прямой.
Гипотеза Римана утверждает, что все нетривиальные нули имеют действительную часть 1/2, и по-прежнему не доказана. Приложенные справочные источники подтверждают прежний ориентир около 41,6–41,7%, но не позволяют независимо подтвердить новое значение 67,2%, строгость доказательства или его оценку внешними специалистами. Поэтому новость следует рассматривать как исследовательское заявление Anthropic, а не как общепринятый математический вывод.
Источники
The Riemann hypothesis isn’t as hard as you might think!aleph0.substack.com · supportingHere are some quick facts about this function. 1. The zeta function converges for complex numbers s with Re(s) > 1. 2. The zeta function can be “analytically continued” so that it is defined on the entire complex plane. 3. The zeta function vanishes at all negative even integers; that is: \(\zeta(s) = 0 \text{ for }s=-2,-4,-6, \dots \) These are called the “trivial zeroes” of the zeta function. The famous (and still unproven) Riemann Hypothesis asserts: All the zeroes of the zeta function (other than the trivial zeroes) lie on the line Re(s)=1/2. The Clay Math Institute has offered a million-dollar prize for anyone who can prove or disprove this conjecture. # Should we try solving it? [...] # Should we try solving it? The general consensus among experts seems to be that RH is simply beyond the reach of what mathematics can prove at the moment. That being said, most open problems seem impossible until they are solved. Fermat’s Last Theorem is a prime example — everyone thought
Riemann hypothesisen.wikipedia.org · supporting1. For any {\displaystyle \varepsilon >0} there exists a lower bound {\displaystyle T_{0}=T_{0}(\varepsilon )>0} such that for {\displaystyle T\geq T_{0}} and {\displaystyle H=T^{{\tfrac {1}{4}}+\varepsilon }} the interval ![{\displaystyle (T,T+H]}]( contains a zero of odd order of the function {\displaystyle \zeta {\bigl (}{\tfrac {1}{2}}+it{\bigr )}}. Let {\displaystyle N(T)} be the total number of real zeros, and {\displaystyle N_{0}(T)} be the total number of zeros of odd order of the function {\displaystyle ~\zeta \left({\tfrac {1}{2}}+it\right)~} lying on the interval ![{\displaystyle (0,T]~}]( [...] ## Zeros on the critical line [edit] Hardy (1914) and Hardy & Littlewood (1921) showed there are infinitely many zeros on the critical line, by considering moments of certain functions related to the zeta function. Selberg (1942) proved that at least a (small) positive proportion of zeros lie on the line. Levinson (1974) improved this to one-third of the zeros by relating the zero
Riemann zeta function - Wikipediaen.wikipedia.org · supportingThe Riemann hypothesis, considered one of the greatest unsolved problems in mathematics, asserts that all non-trivial zeros are on the critical line. In 1989, Conrey proved that more than 40% of the non-trivial zeros of the Riemann zeta function are on the critical line. This has since been improved to 41.7%. [...] Remove a factor of x−1/4 to make the exponents in the remainder opposites. ![{\displaystyle \xi (s)=2\int _{1}^{\infty }{\frac {d}{dx}}\left[x^{\frac {3}{2}}\psi '(x)\right]x^{-{\frac {1}{4}}}\left(x^{\frac {s-1/2}{2}}+x^{\frac {1/2-s}{2}}\right)dx}]( Using the hyperbolic functions, namely cos(x) = cosh(ix), and letting s = 1/2 + it gives ![{\displaystyle \xi (s)=4\int _{1}^{\infty }{\frac {d}{dx}}\left[x^{\frac {3}{2}}\psi '(x)\right]x^{-{\frac {1}{4}}}\cos \left({\frac {t}{2}}\log x\right)dx}]( and by separating the integral and using the power series for cos, {\displaystyle \xi (s)=\sum _{n=0}^{\infty }a_{2n}t^{2n}} which led Riemann to his famous hypothesis. ## Zeros,
Riemann Zeta Function -- from Wolfram MathWorldmathworld.wolfram.com · supporting| | | (24) | where gamma_n are the so-called Stieltjes constants. RiemannZetaFunctionGamma The Riemann zeta function can also be defined in the complex plane by the contour integral | | | (25) | for all z!=1, where the contour is illustrated above (Havil 2003, pp. 193 and 249-252). Zeros of zeta(s) come in (at least) two different types. So-called "trivial zeros" occur at all negative even integers s=-2, -4, -6, ..., and "nontrivial zeros" at certain | | | (26) | for s in the "critical strip" 0<sigma<1. The Riemann hypothesis asserts that the nontrivial Riemann zeta function zeros of zeta(s) all have real part ![sigma=R[s]=1/2](/images/equations/RiemannZetaFunction/Inline88.svg), a line called the "critical line." This is now known to be true for the first 250×10^9 roots.
Extreme values of derivatives of the Riemann zeta functionpmc.ncbi.nlm.nih.gov · supporting$$ \underset{T^{\beta} \leq t \leq T}{max} \left|\zeta^{\left(\right. ℓ \left.\right)}\left(\sigma+it\right)\right| \geq \left(\right. 1 + o \left(\right. 1 \left.\right) \left.\right) \left(\left(\right. 2 \pi \left.\right)\right)^{\sigma - \frac{1}{2}} \sqrt{\frac{\zeta \left(\right. 2 - 2 \sigma \left.\right)}{2 - 2 \sigma}} T^{\frac{1}{2} - \sigma} \left(\left(\right. log T \left.\right)\right)^{ℓ} . $$ Note that the lower bound in Theorem[1 increases when ℓ increases. So it is natural to have the following conjecture. Conjecture 1If _T_ is sufficiently large, then uniformly for all positive integers $ℓ_{1} , ℓ_{2}$ ⩽ $\left(\right. log T \left.\right)$$\cdot \left(\left(\right. log_{2} T \left.\right)\right)^{- 1}$, such that $ℓ_{1} < ℓ_{2}$, we have
Finding the nontrivial zeros of the Riemann Zeta Function using Desmosyoutube.com · supportingit turns out this is decreasing when the real part of s is bigger than zero and the limit actually is equal to Zer when the real part of s is also bigger than or equal to zero because then basically you have Infinity uh to a positive number which is infinity in the denominator that hopefully appr to zero so you can take a look at these two conditions on your own but this is basically the case when the real part of s is bigger than zero so we end up getting finally is this following thing that that ADA is actually valid for when the real part of s is bigger than zero whereas Zeta is only valid when the of s is bigger than one so now the big question is how do we actually relate these two so if you wrote them out it'll be pretty obvious to see what's going on they're very similar only Ada [...] of x = 1/2 or it intersects the red graph but if we crank K up to 1,000 have to wait a little bit for Desmos to make the change there we go just did it you can actually see the purple has actually