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Source: Oral ENS

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#1 h Oct 28, 2024, 11:34 AM

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Hello guys, here is a fun functional equation !
Find all functions from $\mathbb{R^*_+}$ to $\mathbb{R^*_+}$ such that for all $x$ : $f(f(x))=-f(x)+6x$
Hint : Click to reveal hidden text

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#2 h Oct 28, 2024, 11:37 AM

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What is $\mathbb{R}_+^*$ ?

This post has been edited 1 time. Last edited by AshAuktober, Oct 28, 2024, 11:38 AM

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#4 h Oct 28, 2024, 12:16 PM

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AshAuktober wrote:

What is $\mathbb{R}_+^*$ ?

It's the set of positive reals.

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#5 h Oct 28, 2024, 12:25 PM

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We observe that the RHS has a linear term, and the LHS contains a self-composition of $f$ . Knowing that a self-composition of a linear function is itself linear, let $f(x) = Ax+B$ such that:

$A(Ax+B) + B = -(Ax+B) + 6x \Rightarrow A^{2}x + AB + B = (6-A)x - B$ ; (i)

which gives us the system (after matching the coefficients in (i)):

$A^2 = 6 - A$ (ii);

$B(A+1) = -B$ (iii).

We obtain $A^2 + A - 6 = (A+3)(A-2) = 0 \Rightarrow A = -3, 2$ from (ii). Substitution of these values into (iii) produces:

$B(-2) = -B \Rightarrow B = 0$ and $B(3) = - B \Rightarrow B = 0$ ;

or $f(x) = -3x$ and $f(x) = 2x$ . Since we require $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ , the only solution is $\fbox{f(x) = 2x}$ .

This post has been edited 2 times. Last edited by Mathzeus1024, Oct 28, 2024, 12:31 PM
Reason: e

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#6 h Mar 29, 2025, 4:16 AM • 2 Y

Y by MS_asdfgzxcvb, Filipjack

Mathzeus1024 wrote:

We observe that the RHS has a linear term, and the LHS contains a self-composition of $f$ . Knowing that a self-composition of a linear function is itself linear, let $f(x) = Ax+B$ such that:

$A(Ax+B) + B = -(Ax+B) + 6x \Rightarrow A^{2}x + AB + B = (6-A)x - B$ ; (i)

which gives us the system (after matching the coefficients in (i)):

$A^2 = 6 - A$ (ii);

$B(A+1) = -B$ (iii).

We obtain $A^2 + A - 6 = (A+3)(A-2) = 0 \Rightarrow A = -3, 2$ from (ii). Substitution of these values into (iii) produces:

$B(-2) = -B \Rightarrow B = 0$ and $B(3) = - B \Rightarrow B = 0$ ;

or $f(x) = -3x$ and $f(x) = 2x$ . Since we require $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ , the only solution is $\fbox{f(x) = 2x}$ .

This is incorrect, if you apply such a technique to functional equation $f:\mathbb R^+\to\mathbb R^+$ such that $f(f(x))=x$ you won't find the solution $f(x)=\frac1x$ .

My solution:
Fix $x$ , let $a_n=f^n(x)$ (denoting iteration). Notice that $a_{n+2}=-a_{n+1}+6a_n$ , with characteristic equation $r^2=-r+6$ . This has roots $r=-3$ and $r=2$ , so we can write $a_n=A\cdot(-3)^n+B\cdot2^n$ for some $A,B\in\mathbb R$ and all $n\in\mathbb N$ (provable by simple induction where we choose $A$ and $B$ such that the base cases $n=1$ and $n=2$ are satisfied).

Rewriting as $A\cdot(-1.5)^n+B=2^{-n}a_n>0$ , we can see that if $A<0$ the LHS will be arbitrarily small for large even values of $n$ , and likewise if $A>0$ the LHS will be arbitrarily small for large odd values of $n$ . This contradicts $\text{LHS}>0$ , so we need $A=0$ , hence $a_n=B\cdot2^n$ . In particular, we get $a_2=4B$ and $a_1=2B$ , so $f(f(x))=-f(x)+6x$ rewrites as $4B=-2B+6x$ , so $B=x$ . Then $f(x)=a_1=2B=2x$ , so $\boxed{f(x)=2x}$ for all $x$ which fits, we're done.

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ektorasmiliotis

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ektorasmiliotis

#7 h Mar 29, 2025, 3:02 PM

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sefatod628 wrote:

Hello guys, here is a fun functional equation !
Find all functions from $\mathbb{R^*_+}$ to $\mathbb{R^*_+}$ such that for all $x$ : $f(f(x))=-f(x)+6x$
Hint : Click to reveal hidden text

Wrong solution.....

This post has been edited 1 time. Last edited by ektorasmiliotis, Mar 29, 2025, 5:04 PM

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#8 h Mar 29, 2025, 3:19 PM • 1 Y

Y by ektorasmiliotis

ektorasmiliotis wrote:

f(f(x))+f(x)=6x
if f(x)>2x for some x, then for these x:
6x=f(f(x))+f(x)>2f(x)+2x>4x+2x=6x

No. You can not conclude from $f(x)>2x$ for some $x$ that $f(f(x))>2f(x)$ also (except if you have proved before that $f(x)$ is monotonous, which you did not)

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