Cycle length of (x^2-1)/2x derivative relation power of two crazy

Art of Problem Solving

AoPS Online

College Math Topics in undergraduate and graduate studies

Topics in undergraduate and graduate studies

3 M G

BBookmark VNew Topic kLocked

College Math Topics in undergraduate and graduate studies

Topics in undergraduate and graduate studies

3 M G

BBookmark VNew Topic kLocked

w

1 user

TOPICS

No tags match your search

M

No tags match your search

M

Cycle length of (x^2-1)/2x derivative relation power of two crazy

G H J

Reveal topic tags

newton s method

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

762 posts

#1 h Feb 28, 2025, 7:26 AM

Y by

Let

$f(x) = \frac{x^2 - 1}{2x}$

(this is the approximation using Newton’s method for $x^2 + 1 = 0$ . Consider the sequence given by $a_{n+1} = f(a_n)$ . Suppose for some $a_0$ , we have the relationship $a_0 = a_\ell$ , where $a_0 \neq a_i$ for $0<i<\ell$ (in other words, a cycle of length $\ell$ ). Show that

$\frac{d}{dx}\, f^{(\ell)}(a_0) = 2^{\ell}$ ,

where $f^{(\ell)}(x)$ denotes $\ell$ function compositions of $f(x)$ (i.e. $f^{(\ell + 1)}(x) = f(f^{\ell}(x))$ , $f^{(1)}(x) = f(x))$ .)

—————

For example, when $x = \frac{1}{\sqrt{3}}$ , it produces a cycle of length $2$ , and $\left. \frac{d}{dx} f(f(x))\, \right|_{x=1/\sqrt{3}} = 2^2 = 4$

This post has been edited 1 time. Last edited by ehz2701, Feb 28, 2025, 7:26 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

44 posts

#2 h Feb 28, 2025, 9:25 AM

Y by

wrong forum,i think

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

2947 posts

#3 h Mar 1, 2025, 3:33 AM • 1 Y

Y by ehz2701

Note that there exists a $\theta_0$ such that $a_0=\cot\theta_0$ . Then using the trigonometric identity $\cot 2\theta=\frac{\cot^2\theta-1}{2\cot\theta}$ gives $a_n=\cot 2^n\theta_0$

$x=\cot\theta\implies f^{(\ell)}(x)=\cot 2^\ell\theta\implies\frac{df^{(\ell)}}{dx}=\frac{df^{(\ell)}}{d\theta}\cdot\frac{d\theta}{dx}=\frac{2^\ell\csc^2 2^\ell\theta}{x^2+1}=\frac{2^\ell\csc^2 2^\ell\theta}{\cot^2\theta+1}=\frac{2^\ell\csc^2 2^\ell\theta}{\csc^2\theta}$

If the sequence $\{f^{(k)}(a_0)\}_{k=0}^\infty$ has period $\ell$ then we have $\cot 2^\ell\theta_0=\cot\theta_0$ and so $\csc^2 2^\ell\theta_0=\csc^2\theta_0$ which makes $\frac{df^{(\ell)}}{dx}(a_0)=2^\ell$ .

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a