Is this FE solvable?

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Is this FE solvable?

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: Original

The post below has been deleted. Click to close.

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

149 posts

#1 h Apr 21, 2025, 3:02 AM

Y by

Let $c_1,c_2 \in \mathbb{R^+}$ . Find all $f : \mathbb{R^+} \rightarrow \mathbb{R^+}$ such that for all $x,y \in \mathbb{R^+}$ $f(x+c_1f(y))=f(x)+c_2f(y)$

Z K Y

The post below has been deleted. Click to close.

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

23515 posts

pco

#2 h Apr 21, 2025, 9:34 AM • 2 Y

Y by internationalnick123456, ItzsleepyXD

ItzsleepyXD wrote:

Let $c_1,c_2 \in \mathbb{R^+}$ . Find all $f : \mathbb{R^+} \rightarrow \mathbb{R^+}$ such that for all $x,y \in \mathbb{R^+}$ $f(x+c_1f(y))=f(x)+c_2f(y)$

There are infinitely many solutions and I'm not sure we can find a general form for all of them.

Here is a huge family of solutions (I'm not sure these are all of them) :

Let $U$ any additive subgroup of $\mathbb R$ not limited at $\{0\}$ and stable by multiplication by $c_2$
Let $\sim$ the equivalence relation over $\mathbb R$ defined as $x\sim y$ $\iff$ $x-y\in U$
Let $r(x)$ from $\mathbb R\to\mathbb R$ a function which associates to any real a representent (unique per class) of its equivalence class.

Let $u(x)$ any function from $r(\mathbb R)\to U$ such that $u(x)\ge c_2x$ (infinitemy many such functions exist since $U$ is not overbounded)

Define $f(x)$ as $f(x)=\frac{c_2x+u(r(x))-c_2r(x)}{c_1}$

Examples :
Whatever is $c_2$ , the basic $U=\mathbb R$ gives trivial solution :
$\boxed{f(x)=\frac{c_2}{c_1}x+a\quad\forall x>0}$ , xhich indeed fits, whatever is $a\ge 0$

If $c_2\in\mathbb Z_{>0}$ , we can choose for example $U=\mathbb Z$ , $r(x)=\{x\}$ and $u(x)=2c_2\left\lfloor\frac 1{1-x}\right\rfloor$ and we get the solution
$\boxed{f(x)=\frac{c_2}{c_1}\left(\lfloor x\rfloor+2\left\lfloor\frac1{1-\{x\}}\right\rfloor\right)}$
And a lot lot of others

If $c_2=\sqrt 2$ , we can choose $U=Q[\sqrt 2]$ and a lot of solutions there (no kind example since there are no explicit form for an $r(x)$ ) ...

...

Z K Y

The post below has been deleted. Click to close.

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

149 posts

#3 h Apr 21, 2025, 11:15 AM

Y by

pco wrote:

ItzsleepyXD wrote:

Let $c_1,c_2 \in \mathbb{R^+}$ . Find all $f : \mathbb{R^+} \rightarrow \mathbb{R^+}$ such that for all $x,y \in \mathbb{R^+}$ $f(x+c_1f(y))=f(x)+c_2f(y)$

There are infinitely many solutions and I'm not sure we can find a general form for all of them.

Here is a huge family of solutions (I'm not sure these are all of them) :

Let $U$ any additive subgroup of $\mathbb R$ not limited at $\{0\}$ and stable by multiplication by $c_2$
Let $\sim$ the equivalence relation over $\mathbb R$ defined as $x\sim y$ $\iff$ $x-y\in U$
Let $r(x)$ from $\mathbb R\to\mathbb R$ a function which associates to any real a representent (unique per class) of its equivalence class.

Let $u(x)$ any function from $r(\mathbb R)\to U$ such that $u(x)\ge c_2x$ (infinitemy many such functions exist since $U$ is not overbounded)

Define $f(x)$ as $f(x)=\frac{c_2x+u(r(x))-c_2r(x)}{c_1}$

Examples :
Whatever is $c_2$ , the basic $U=\mathbb R$ gives trivial solution :
$\boxed{f(x)=\frac{c_2}{c_1}x+a\quad\forall x>0}$ , xhich indeed fits, whatever is $a\ge 0$

If $c_2\in\mathbb Z_{>0}$ , we can choose for example $U=\mathbb Z$ , $r(x)=\{x\}$ and $u(x)=2c_2\left\lfloor\frac 1{1-x}\right\rfloor$ and we get the solution
$\boxed{f(x)=\frac{c_2}{c_1}\left(\lfloor x\rfloor+2\left\lfloor\frac1{1-\{x\}}\right\rfloor\right)}$
And a lot lot of others

If $c_2=\sqrt 2$ , we can choose $U=Q[\sqrt 2]$ and a lot of solutions there (no kind example since there are no explicit form for an $r(x)$ ) ...

...

Thanks a lot PCO :omighty:

:omighty:

:omighty:

But I wonder if add injective to the condition of function will make it solvable or not?

Z K Y

The post below has been deleted. Click to close.

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

11337 posts

#4 h Apr 22, 2025, 4:31 AM

Y by

It's solvable with nondecreasing condition I believe.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a