Anthropic称研究版Claude推进黎曼猜想相关问题
Anthropic表示,一款未发布的Claude研究版本将黎曼ζ函数临界线上零点所占比例的已证明下界,从约41.6%提高到67.2%。该成果并非黎曼猜想的证明,且所给材料未包含独立同行评议验证。
Anthropic在一则官方社交媒体公告中称,未发布的Claude研究版本参与了与黎曼猜想有关的数学研究。该公司表示,模型没有解决黎曼猜想,但为“位于临界线上的黎曼ζ函数零点比例”给出了67.2%的新下界。
黎曼猜想断言所有非平凡零点都位于实部为1/2的临界线上,仍是未解决的数学难题。所附背景资料支持此前约41.6%至41.7%的下界这一历史基线,但不能独立核实67.2%的新证明、其严格性或学界审查情况。因此,该消息应视为Anthropic的研究声明,而非已获广泛验证的定论。
来源证据
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