or equivalently ogSββ On the other hand, from()we have A(S) A(2)<A(s+1 aA(s) BA(t)<(a+1A(S) or 段A+ AG罗 Thus A(S) 10gS/ A(t) log t When B is large enough, we have A(t)=k log ta b log t log S < b a + 1 b or equivalently On the other hand, from (*) we have A(S ) A(t ) < A(S ) a a b a+1 or A(s) bA(t) < (a+1)A(S) (**) b a A(t) A(S) < a b + 1 b (***) Thus A(t) A(S) - log t log S < 1 b When b is large enough, we have A(t) = k log t