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Let $1<a<e^{1/e}$, and define $f(x)=a^x$

(b) Show that if $x_1$ is any point in the interval $(1,e)$ and $p$ is a fixed point of $f(x)$ in the interval $(1,e)$, then

$$|f(x_1) - p | <|x_1 - p |$$

(c) If $x_1$ is as in (b), define a sequence $\{x_n\}_{n=1}^{\infty}$ as follows:

Let $x_2 = f(x_1),\:x_3=f(x_2),\ldots x_{n+1}=f(x_n), \ldots$

Show that the sequence $\{x_n\}$ converges to $p$

Solutions(?):

(b)

From the mean value theorem it follows that there exists some $x_2$ between $x_1$ and $p$ where $x_1 \neq p$ such

that

$f'(x_2)=|\frac{f(x_1)-p}{x_1-p}|$ then it follows:

$$|\frac{f(x_1)-p}{x_1-p}|=f'(x_2)<|\frac{\max\{f(e)=a^e:a\in(1,e^{1/e})\}-\min\{f(1)=a^1:a\in (1,e^{1/e})\}}{e-1}|<|\frac{e-1}{e-1}|=1$$

From this it follows that $|\frac{f(x_1)-p}{x_1-p}|<1 \Rightarrow |f(x_1) - p | <|x_1 - p | $

(c) I feel like it would have something to do with the Banach fixed-point but that would mean I did something wrong with (b) since it needs to be Lipschitz continuous with its constant being less than $1$ but I feel I did something wrong in (b) that makes (c) impossible or it is some constraint I have left unaccounted for.

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Your upper bound for $f'(x)$ can be improved. For $1\leq x \leq e$ we have $$0<f'(x)=a^x\log a<a^x/e<(e^{1/e})^x/e=e^{x/e-1}\leq 1.$$So $x\in [1,e]\implies 0<f'(x)<1$. Since $f'$ is continuous and $ 0<f'<1$ on the closed interval $[1,e]$, there exists $c\in (0,1)$ such that $$\forall x\in [1,e] (|f'(x)|\leq c).$$ Taking $p\in (1,e)$ with $(\log p)/p=\log a $, we have $f(p)=p$ . Your use of the mean value theorem of calculus shows that $$\forall x\in [1,e](|f(x)-p|\leq c|x-p|).$$ Therefore if $x_0\in [1,e]$ and $x_{n+1}=f(x_n)$ then $$|x_n-p|\leq c^n|x_0-p|.$$ Footnotes (1) :The existence of $p\in (1,e)$ satisfying $(\log p)/p=\log a$ can be shown by considering the continuous function $g(x)=((\log x)/x)-\log a.$ We have $g(0)<0$ and $g(e)>0.$....(2)We can show that if $b=e^{1/e}$ and $b_0=b$ and $b_{n+1}=b^{b_n}$ then $(b_n)_n$ converges to $e$.

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