By Kenneth S. Miller.

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**Sample text**

194). b) Show that the set of points of I where fd is continuous is the complement of a meagre subset of I (cf. Gen. , IX, p. 255, exerc. 21). 6) Let (rn ) be the sequence formed by the rational numbers in [0, 1]], arranged in a certain ∞ order. Show that the function f (x) 2−n (x − rn )1/3 is continuous and differentiable n 0 at every point of R, and has an inﬁnite derivative at every point rn . (To see that f is differentiable at a point x distinct from the rn , distinguish two cases, according to whether the series with general term 2−n (x − rn )−2/3 has sum +∞ or converges; in the second case, note for all x 0 and all y x, one has 0 (y 1/3 − x 1/3 )/(y − x) 4/3x 2/3 ).

Let us show that we must have c b. So suppose that c < b and that moreover c∈ / A : then fd (c) and gr (c) exist and satisfy (6); suppose in the ﬁrst place that gr (c) (which is necessarily 0) is ﬁnite; then one can always write fd (c) ugr (c), with u 1; since the function f(x) − ug(x) has zero right derivative at the point c there must exist a y such that c < y b and such that for c x y one has f(x) − f(c) − u(g(x) − g(c)) ε(x − c) from which f(x) − f(c) g(x) − g(c) + ε(x − c) and, taking account of (8), in which x is replaced by c, f(x) − f(a) g(x) − g(a) + ε(x − a) + ε an

When f has a derivative of order n + 1 on I, one can estimate rn (x) in terms of this n + 1th derivative, on all of I, and not just on an unspeciﬁed neighbourhood of a : PROPOSITION 3. If f (n+1) (x) rn (x) M on I, then we have M |x − a|n+1 (n + 1)! (9) on I. Indeed, the formula holds for n 0, by I, p. 15, th. 2. Let us prove it by induction on n : by the induction hypothesis applied to f , one has |y − a|n n! from which the formula (9) follows by the mean value theorem (I, p. 23, th. 2). rn (y) M COROLLARY.