A beauty function equation

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

4 users

TOPICS

No tags match your search

M

No tags match your search

M

A beauty function equation

G H J

Reveal topic tags

functional equation

real analysis unsolved

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.

417 posts

#1 h Jul 8, 2012, 10:15 AM • 2 Y

Y by Adventure10, Mango247

Let $f(x)$ satisfy the functional equation
$[f(x)]^{2}=1+x f(x+1)$
and the inequalities
$\frac{x+1}{2}\leq f(x) \leq 2(x+1)$
for all $x\geq1$ . Prove that $f(x)=x+1$ .

Z K Y

The post below has been deleted. Click to close.

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

775 posts

#2 h Mar 31, 2025, 12:01 PM

Y by

Upon inspection of the inequality $\frac{x+1}{2} \le f(x) \le 2(x+1)$ , we have $f(x) > 0$ for all $x\ge1$ . Suppose $f(x) = (x+1) \cdot g(x) \Rightarrow \frac{1}{2} \le g(x) \le 2$ (i) so that the functional equation $[f(x)]^2 = 1+xf(x+1)$ is satisfied. This leads to:

$g^{2}(x)(x+1)^2 = 1+x \cdot g(x+1)(x+2) \Rightarrow g^{2}(x) \cdot (x^2+2x+1) = 1+g(x+1) \cdot (x^2+2x)$ (ii);

and matching the coefficients of $x$ in (ii) yields:

$g^{2}(x) = g(x+1)$ (iii);

$g^{2}(x) = 1$ (iv).

From (iv) we obtain $g(x) = \pm 1$ , of which (i) and (iii) are satisfied $\Leftrightarrow g(x)=1$ . Thus, $\textcolor{red}{f(x)=x+1}$ .

This post has been edited 2 times. Last edited by Mathzeus1024, Mar 31, 2025, 12:07 PM

Z K Y

The post below has been deleted. Click to close.

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

831 posts

#3 h Mar 31, 2025, 1:40 PM

Y by

@above: Those are not polynomials to be able to match the coefficients...

Solution. We prove that if $a(x+1) \le f(x) \le b(x+1),\forall x \ge 1$ holds for some $a<1<b$ , then so does $\sqrt{a}(x+1) \le f(x) \le \sqrt{b}(x+1),\forall x \ge 1.$

The given inequalities imply $f(x)>0,\forall x \ge 1.$ Therefore, for any $x \ge 1$ we have $f(x)=\sqrt{1+xf(x+1)},$ and so $\sqrt{1+ax(x+2)} \le f(x) \le \sqrt{1+bx(x+2)}.$
It is easy to check that $\sqrt{a}(x+1) \le \sqrt{1+ax(x+2)}$ and $\sqrt{1+bx(x+2)} \le \sqrt{b}(x+1)$ hold for any $x \ge 1,$ so the claim is proved.

Consequently, if we set $a_0=\frac{1}{2}$ and $b_0=2$ we inductively get $a_n(x+1) \le f(x) \le b_n(x+1),$ for any $x \ge 1$ and any nonnegative integer $n,$ where $a_n=a_0^{1/2^n}$ and $b_n=b_0^{1/2^n}.$ Passing to the limit as $n \to \infty,$ we get $x+1 \le f(x) \le x+1,$ for any $x \ge 1$ , so $f(x)=x+1, \forall x \ge 1.$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a