Ecommerce web host - 348 ARITHMETIC 4.5.3 Continuing, we see that if
348 ARITHMETIC 4.5.3 Continuing, we see that if g has a bounded first derivative, we can differen- tiate term by term to show that Tg does also: (TgY(x) x)2 &) = - kFl ((k + 1 + - (L;x;Jg( - -+k;x,2g (i&)) =- (g(&)-g(k+:+x)) + (k + x):(:; 1 + 2) There is consequently a third linear operator, U, such that (Tg) = -U(g ), namely = dt UP(X)c k v(t) (k + 1 + x) k>l + (k+x)i(:;l+x) . (29) What is the relevance of all this to our problem? Well, if we set Fn(x) = w + x) + q@ + xl), (30) fn(x) = cl+ x) F:,(x) = &Cl + qdl+ x,)), (31) we have f;(x) = RE(Ml + x))/((ln 2) P + x)); (32) the effect of the lg(1 + x) term disappears, after these transformations. F urther- more since Fn = PFo we have fn = Tnfo and f; = (-l)nU fb. Both F, and fn have bounded derivatives, by induction on 72. Thus (32) becomes (-l)nRK(lg(l + x)) = (1 + x)(ln 2)2 U fb(x). (33) Now Fe(x) = x, fo(x) = 1 + x, and fb(x) is the constant function 1. We are going to show that the operator U takes the constant function into a function with very small values, hence (Rx(x)1 must be very small for 0 5 x 5 1. Finally we can clinch the argument by showing that R,(x) itself is small: Since we have R,(O) = Rn(l) = 1, it follows from a well-known interpolation formula (cf. exercise 4.6.4-15 with x0 = 0, x1 = x, x2 = 1) that R,(x) = - ( ; ) R;@(x)) (34) for some function c(x), where 0 5 E(x) 5 1 when 0 2 x 2 1.