Bourbaki

Exercise 4, § 2

Functions of a Real Variable, VI, 293

a) Prove the relation

Bn(1−X)=(−1)nBn(X) B_n(1 - X) = (-1)^n B_n(X)
(use the fact that b2n−1=0b_{2n-1} = 0 for n>1n > 1, and the relation
Bn(1−X)−Bn(−X)=(−1)nnXn−1. B_n(1 - X) - B_n(-X) = (-1)^n n X^{n-1}.
)

b) Show that

Bn(12)=bn(12n−1) B_n \left( \frac{1}{2} \right) = b_n \left( \frac{1}{2^n} - 1 \right)
(use exerc. 3).

c) Show that, for nn even, Bn(X)B_n(X) has two roots in the interval [0,1][0, 1] of R\mathbf{R}, and that for nn odd >1> 1, Bn(X)B_n(X) has a simple root at the points 0,120, \frac{1}{2} and 11 and does not vanish at any other point of [0,1][0, 1] (use bb) and the relation Bn′=nB~n−1B'_n = n \tilde{B}_{n-1}).

d) Deduce from c) that, for nn even, the maximum of ∣Bn(x)∣|B_n(x)| on the interval [0,1][0, 1] is ∣bn∣|b_n|, and that for nn odd, if ana_n is the maximum of ∣Bn(x)∣|B_n(x)| on [0,1][0, 1], then

4n+1∣bn+1∣(1−12n)≤an≤12n∣bn−1∣ \frac{4}{n+1} |b_{n+1}| \left( 1 - \frac{1}{2^n} \right) \leq a_n \leq \frac{1}{2} n |b_{n-1}|
(use the mean value theorem).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.