Proposition 10. Define x_5=.01 and x_ j+1 =10x_j^2 for j≥5 . Then, for every j≥5 , u_j≤x_j, e_ k_j (j)≤x_j, 1-A(M_j,j)≤x_j. In particular x_j=10^ -1-2^ j-5 .

The table proves the two boundary inequalities at j=5 , and u_5<.000015<.01 . Suppose holds, and write x=x_j≤.01 . Here k_j=M_j/2 and u_ j+1 =6u_j^2≤6x^2 . The first argument of F in satisfies √ d(u_ j+1 ) ≤ √(6x^2)/(2-18x^2)≤2x. The first output is therefore bounded by F(2x,x,1-x) . For x>0 , division by x^2 gives (F(2x,x,1-x))/(x^2) =(2)/(1-2x)((1-x)^ -1 +(1-x)^ -2 +(1-x)^ -3 ). This expression is increasing on [0,.01] after its continuous extension at zero. Its endpoint value is 42430000/6792093<10 . Thus we may take E_ j+1 =10x^2 . For factors in [0,1] , the defect of their product is at most the sum of their defects. Thus the four-factor product in has defect at most x+x+6x^2+10x^2=2x+16x^2 . Moreover (f(2x+16x^2))/(x^2) =((2+16x)^2)/(2(1-2x-16x^2)^2) is increasing on [0,.01] and has endpoint value 3645000/1495729<10 . Thus we may also take η_ j+1 =10x^2 . Finally u_ j+1 ≤6x^2≤10x^2 . This proves at the next level; the sequence remains in [0,.01] . Solving its recurrence gives.