By Azoulay E., Avignant Jean

Azoulay E., Avignant J. Mathematiques four. Algebre (MGH, 1984)(fr)(ISBN 2704210896)

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**Sample text**

The last two questions of the example show that this is not the case. 3) We get by diﬀerentiation fn (x) = n(1 − x)n − n2 x(1 − x)n−1 = n(1 − x)n−1 (1 − x − nx) = n(1 − x)n−1 {1 − (n + 1)x}. Since fn (0) = fn (1) = 0, and fn (x) > 0 for 0 < x < 1, the continuous function fn (x) must have a maximum. 1 in ]0, 1[, this value corresponds Since fn ∈ C ∞ , and since fn (x) = 0 is only fulﬁlled for x = n+1 to the unique maximum. The value of the function here is fn 1 n+1 =n· 1 n+1 1− n 1 n+1 1− = 1 n+1 n+1 .

A more careful solution is the following. 1) To every ε > 0 we choose N = N (ε), such that 1 ≤ε 2N choose N ≥ ln(1/ε) ﬁxed . ln 2 2) Since (1/2n ) is decreasing for increasing n, we have 1 1 ≤ N ≤ε 2n 2 for every n ≥ N (ε) [independently of x]. 3) For every x ≥ 1 we have 1 1 ≤ n ≤ε (1 + x2 )n 2 for every n ≥ N (ε) [independently of x]. Finally we summarize the above: To every ε > 0 there exists an N = N (ε) [which does not depend on x], such that for every n ≥ N and every x ≥ 1, |fn (x) − 0| ≤ ε, which means that we have uniform convergence on [1, ∞[.

NNE Pharmaplan offers me freedom with responsibility as well as the opportunity to plan my own time. com NNE Pharmaplan is the world’s leading engineering and consultancy company focused exclusively on the pharma and biotech industries. NNE Pharmaplan is a company in the Novo Group. 1 Prove that the sequence of functions fn (x) = x2 x x x + 2 + ··· + 2 , +1 x +4 x + n2 where fn : [−1, 1] → R, is pointwise convergent on the interval [−1, 1]. Hint. Apply the General principle of convergence. Apply a programmable pocket calculator to sketch the graph of fn (x) for some large n.