Answer to Question #156225 in Real Analysis for Ange

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

$f(x) = 1 / (1 + e^x)$ is monotonously decreasing everywhere.

If $x\in I=(1/4,1/2)$ then

f(x) < f(0) = 1/2,

f(x) > f(1) = 1/(1+e) > 1/(1+3) = 1/4,

and, hence, $f(x)\in I$.

$u_0\in I$ implies $u_1=f(u_0)\in I$,

$u_n\in I$ for $n\in \mathbb{N}$ implies $u_{n+1}=f(u_n)\in I$ and, by induction, $u_n\in I$ for all $n\in \mathbb{N}$

The statement (i) is proved.

$g(x) = f(x)-x$ is a continuous function.

g(1/4) = f(1/4) - 1/4 > f(1) - 1/4 = 1/(1+e) - 1/4 > 1/(1+3) - 1/4 = 0

g(1/2) = f(1/2) - 1/2 < f(0) -1/2 = 0

Therefore, by the intermediate value theorem we have that there exists a root $s\in I$ of this function, that is, a number $s$ such that $f(s) = s$ .

By the mean value theorem

$\frac{u_{n+1} - s}{u_n-s} = \frac{f(u_{n}) - f(s)}{u_n-s} = f'(\xi_n )$

for some $\xi_n$ between $u_n$ and $s$.

$|f'(\xi_n )| = \frac{e^x}{(1+e^x)^2}\leq \frac{e^x}{(2e^{x/2})^2}\leq \frac{e^x}{4e^x} = \frac{1}{4}$ ,

where we have used the inequality of arithmetic and geometric means: $1+e^x\geq 2e^{x/2}$.

Comparing two last formulae, we have $|u_{n+1} - s|\leq |u_{n} - s|/4$, and the statement (ii) is proved.