By Pearcy C. (ed.)
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Example text
Where ξ is interior to the smallest interval containing the points xi (1 (Apply exerc. 8 to the function of t f (t) − g(t) − a i p) and x. (t − x 1 )n 1 (t − x2 )n2 . . (t − x p )n p n! ) 10) Let g be an odd real function defined on a neighbourhood of 0, and 5 times differentiable on this neighbourhood. Show that x x 5 (5) g (x) + 2g (0) − g (ξ ) 3 180 g(x) (ξ θ x, 0 < θ < 1) (same method as in exerc. 9). Deduce that if f is a real function defined on [a, b]] and 5 times differentiable on this interval, then b−a 6 f (b) − f (a) f (a) + f (b) + 4 f a+b 2 − (b − a)5 (5) f (ξ ) 2880 with a < ξ < b (“Simpson’s formula”).
I DERIVATIVES c) Deduce that if f is differentiable on ]a, b[[ then the image under f of every interval contained in ]a, b[[ is itself an interval, and consequently connected (use a)). 1 , 5) Let f be the vector mapping of I [0, 1]] into R3 defined as follows: for 0 t 4 1 1 1 3 t let f(t) (−1, 4t − 1, 0); for t let f(t) f(t) (−4t, 0, 0); for 4 2 2 4 (−1, 1, 4t − 2); finally, for 34 t 1 take f(t) (4t − 4, 1, 1). Show that the convex set f(y) − f(x) generated by the set fd (I) is not identical to the closure of the set of values of y−x as (x, y) runs through the set of pairs of distinct points of I (cf.
Gn (x) x n−1 . ¶ 12) a) Let f be a vector function defined and continuous on the finite interval I [−a, +a]], taking its values in a normed space E over R and twice differentiable on I. If one puts M0 sup f(x) , M2 sup f (x) , show that for all x ∈ I one has x∈I x∈I M0 x 2 + a2 + M2 a 2a f (x) (express each of the differences f(a) − f(x), f(−a) − f(x)). b) Deduce from a) that if f is a twice differentiable function on an interval I (bounded or not), and if M0 sup f(x) and M2 sup f (x) are finite, then so is M1 sup f (x) , x∈I x∈I x∈I and one has: M1 M1 2 M0 M2 √ if I has length 2 M0 M2 if I 2 M0 M2 R.