Answer to Question #156255 in Calculus for Fletcg

Question

Given that f is the real valued function defined by f(x) = 1 / (1 + e^x). And I = (1/4, 1/2). We define the sequence (Un) by Uo = 1/4 and for all n E N where N = set of natural numbers, and E represents "an element of", U(n +1) = f(Un).

(i)Show that Un E I.
(ii) Show that for all n E N, we have ㅣU(n+1) - sㅣ <= 1/4ㅣUn - sㅣ Where s Eb(1/4, 1/2)

Answer 1

$\displaystyle (i)\\ f(x) = \frac{1}{1 + e^x}\\ f(U_n) = \frac{1}{1 + e^{U_n}}\\ \frac{2}{e^x + e^{-x}} = \frac{1}{\cosh{x}}\\ \frac{1}{1 + e^{2x}} = \frac{e^{-x}}{2\cosh{x}}\\ f(x) = \frac{1}{1 + e^{x}} = \frac{e^{-\frac{x}{2}}}{2\cosh\left(\frac{x}{2}\right)}\\ f(0) = \frac{1}{1 + e^{0}} = \frac{1}{2}\\ \textsf{If}\,\, x \in I = \left(\frac{1}{4}, \frac{1}{2}\right) \,\, \textsf{then} \\ f(x) < f(0) = \frac{1}{2}, \\ f(x) > f(1) = \frac{1}{1 + e} > \frac{1}{1 + 3} = \frac{1}{4},\\ \textsf{and hence,}\,\, f(x) \in I. \\ u_0 \in I \,\, \textsf{implies}\,\, u_1 = f(u_0) \in I. \\ u_n \in I\,\, \textsf{for} \,\, n \in \mathbb{N} \textsf{implies}\\ u_{n + 1} = f(u_n) \in I\,\, \textsf{and, by induction,} \\ u_n \in I \,\, \forall \,\, n \in \mathbb{N} \\ \textbf{\textsf{Proved}}\\ (ii) \\ h(x) = f(x) - x = \frac{1}{1 + e^x} - x\,\,\\ \textsf{is a continuous function.}\\ h\left(\frac{1}{4}\right) = f\left(\frac{1}{4}\right) - \frac{1}{4} > f(1) - \frac{1}{4} \\= \frac{1}{1 + e} - \frac{1}{4} > \frac{1}{1 + 3} - \frac{1}{4} = 0 \\ h\left(\frac{1}{2}\right) = f\left(\frac{1}{2}\right) - \frac{1}{2} < f(0) - \frac{1}{2} = 0\\ \textsf{Therefore, by the intermediate value}\\ \textsf{theorem we have that there exists a root}\,\, s \in I \\ \textsf{of this function, that is, a number}\,\, s \,\, \textsf{such that}\\ f(s) = s\\ \textsf{By MVT}\\ \frac{u_{n + 1} - s}{u_n - s} = \frac{f(u_n) - f(s)}{u_n - s} = f'(\epsilon_n) \\ \textsf{for} \,\,\epsilon_n \in (u_n, s) \\ 1 + e^x \geq 2e^{\frac{x}{2}} \,\,\,\,\, (AP \geq GP) \\ |f'(\epsilon_n)| = \frac{e^x}{(1 + e^x)^2} \leq \frac{e^x}{(2e^{\frac{x}{2}})^2} \leq \frac{e^x}{4e^x} = \frac{1}{4} \\ \textsf{Comparing the last two formulae, we have}\\ |u_{n + 1} - s| \leq \frac{|u_n - s|}{4}, \\ \textbf{\textsf{Proved}}$