Example. Knowledge renewal depends on useful yield [ftip-00OM]
Example. Knowledge renewal depends on useful yield [ftip-00OM]
Let \(D(t)\) denote effective task-relevant coverage, rather than raw token count. Suppose useful new human input arrives at rate \(h\geq 0\), synthetic generation at rate \(v\geq 0\) has effective yield \(a\geq 0\), and coverage depreciates at rate \(\delta _D>0\). Under the stipulated law
\[\dot D=h+av-\delta _DD,\qquad D(t)=D_*+(D_0-D_*)e^{-\delta _Dt},\qquad D_*=(h+av)/\delta _D.\]Direct differentiation verifies the solution. For a required coverage \(D_{\min }>0\), if \(D_0\geq D_{\min }\) and \(h+av\geq \delta _DD_{\min }\), then \(D(t)\geq D_{\min }\) at every time. If \(h+av<\delta _DD_{\min }\) and \(D_0\) is finite, the path eventually falls below the threshold. The necessary synthetic contribution is exactly \(av\geq \max \{0,\delta _DD_{\min }-h\}\). Positive synthetic yield is necessary when human inflow leaves a deficit; an affordable schedule must also produce and check the generated material.
The scalar law omits distributional coverage and estimation error. The contrast between recursive replacement in Shumailov et al., accumulation in Gerstgrasser et al., and consistency conditions in Barzilai and Shamir is a reason to specify \(a\) and the learning process, not to assume that every generated token has fixed positive knowledge value.