Bourbaki

Exercise 4, § 2

Functions of a Real Variable, VI, p. 23

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)nXn−1. B_n(1-X) - B_n(-X) = (-1)^n X^{n-1}.

b) Prove that one has

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

(use Exercise 3).

c) Prove 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 b) and the relation Bn′=nBn−1B'_n = n B_{n-1}).

d) Deduce from c) that, for nn even, the maximum of ∣Bn(x)∣|B_n(x)| in the interval (0,1)(0,1) is equal to ∣bn∣|b_n|, and that for nn odd, if ana_n is the maximum of ∣Bn(x)∣|B_n(x)| in (0,1)(0,1), one has

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

(use the theorem of finite increments).

Solution

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Machine translation of the English, by gpt-5-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.