Bourbaki

Exercise 1, § 4

General Topology, V

a) Let (xn,yn)n∈Z(x_n, y_n)_{n \in \mathbf{Z}} be a family of points of R2\mathbf{R}^2 such that for all n∈Zn \in \mathbf{Z} we have xn<xn+1x_n < x_{n+1} and

yn+1−ynxn+1−xn<yn+2−yn+1xn+2−xn+1. \frac{y_{n+1} - y_n}{x_{n+1} - x_n} < \frac{y_{n+2} - y_{n+1}}{x_{n+2} - x_{n+1}}.
Show that, for all integers n∈Z,m>0,p>0n \in \mathbf{Z}, m > 0, p > 0, we have
yn+m−ynxn+m−xn<yn+m+p−ynxn+m+p−xn. \frac{y_{n+m} - y_n}{x_{n+m} - x_n} < \frac{y_{n+m+p} - y_n}{x_{n+m+p} - x_n}.

b) If a>0a > 0 and x≠0x \neq 0, put

fa(x)=ax−1x. f_a(x) = \frac{a^x - 1}{x}.
Show that, if x<y,x≠0,y≠0x < y, x \neq 0, y \neq 0 and a≠1a \neq 1, we have fa(x)<fa(y)f_a(x) < f_a(y) [prove the relation first for x,yx, y integral, by means of aa], then for x,yx, y rational, and finally in general].

c) Deduce that, for all a>0a > 0, the function fa(x)f_a(x), defined for x≠0x \neq 0, has a limit on the right and a limit on the left as x→0x \to 0; show that these two limits have a common value φ(a)\varphi(a), and that φ(a)≠0\varphi(a) \neq 0 if a≠1a \neq 1. Show that if a≠1a \neq 1, the function log⁡ax/x−1\log_a x/x - 1, defined for x≠1x \neq 1, tends to the limit 1/φ(a)1/\varphi(a) as x→1x \to 1.

d) Show that, for all a>0a > 0 and b>0b > 0, we have φ(b)=φ(a)log⁡ab\varphi(b) = \varphi(a) \log_a b (if a≠1a \neq 1). Deduce that there exists a number ee between 2 and 4 such that φ(e)=1\varphi(e) = 1, and that φ(a)=log⁡ea\varphi(a) = \log_e a for all a>0a > 0.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.