a functional equation on positive reals

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a functional equation on positive reals

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Source: Czech and Slovak third round,2004,p6

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#1 h Mar 3, 2012, 5:49 AM • 4 Y

Y by jhu08, HWenslawski, Adventure10, Mango247

Find all functions $f:\mathbb R^+ \rightarrow \mathbb R^+$ such that for all positive real numbers $x,y$ ,
$x^2[f(x)+f(y)]=(x+y)f(yf(x)).$

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pco

#2 h Mar 3, 2012, 6:23 AM • 5 Y

Y by utlight, jhu08, HWenslawski, Adventure10, Mango247

littletush wrote:

Find all functions $f:R^+ \rightarrow R^+$ such that for all positive real numbers $x,y$ ,
$x^2[f(x)+f(y)]=(x+y)f(yf(x))$ .

Let $P(x,y)$ be the assertion $x^2(f(x)+f(y))=(x+y)f(yf(x))$

$P(x,x)$ $\implies$ $f(xf(x))=xf(x)$ and so $f(f(1))=f(1)$

$P(xf(x),f(1))$ $\implies$ $x^2f(x)^2=f(xf(x)f(1))$
$P(f(1),xf(x))$ $\implies$ $f(1)^2=f(xf(x)f(1))$

And so $f(x)=\frac{f(1)}x$ and, plugging this in original equation, we get $f(1)=1$

Hence the answer : $\boxed{f(x)=\frac 1x}$

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#3 h Oct 5, 2018, 6:03 PM • 3 Y

Y by jhu08, Adventure10, Mango247

$P(1,1)\implies f(1)=f(f(1))$
$P(1,x)\implies f(1)+f(x)=(1+x)f(xf(1))$
Using the first equation in this post we get
$P(f(1),1)\implies f(1)^2\cdot 2f(1)=f(1)^2\cdot(f(f(1))+f(1))=(f(1)+1)(f(f(1))=(f(1)+1)f(1)\iff 2f(1)^2=f(1)+1\iff f(1)\in\lbrace -0.5, 1\rbrace\iff f(1)=1$
Returning to the second $1+f(x)=(1+x)f(x)\iff f(x)=\frac{1}{x}$

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#4 h Oct 5, 2018, 6:24 PM • 3 Y

Y by jhu08, Adventure10, Mango247

Here is a different one. Let $P(x,y)$ be the given property. We claim that $f(\cdot)$ is injective. For this, suppose that $f(x_1)=f(x_2)$ . We have,
$P(x,1)\implies x^2(f(x)+f(1))=(x+1)f(f(x))\implies \frac{x_1^2}{x_1+1}=\frac{x_2^2}{x_2+1}\implies (x_1-x_2)(x_1x_2+x_1+x_2)=0.$ In particular, if $x_1\neq x_2$ , we must have $x_1x_2+x_1+x_2=0$ , a contradiction.

Now, $P(x,x)$ gives $xf(x)=f(xf(x))$ , in particular, $xf(x)$ is a fixed point of $f(\cdot)$ . We will now show that, as a consequence of injectivity, the only fixed point of $f(\cdot)$ is 1, which will prove that $f(x)=1/x$ . Now, suppose $t\neq 1$ is a fixed point. Then, $f(tf(t))=tf(t)$ yields that $t^2$ is also a fixed point, Similarly, $t^4,t^8,\dots$ are all fixed points, in particular, there is a strictly monotonic sequence $t_n$ for which, $f(t_n)=t_n$ for every $n$ .

Finally, $P(1,x)$ gives us, $f(x)+1 = (x+1)f(xf(1))$ , and for each $t_n$ , $f(t_n)+1=(t_n+1)=(t_n+1)f(t_nf(1))$ , and thus, $f(t_nf(1))=1$ . Using injectivity, we conclude that $t_nf(1)$ is constant, thus, $t=1$ is the only fixed point, proving the claim.

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RayThroughSpace

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RayThroughSpace

#5 h Oct 16, 2020, 9:54 PM • 1 Y

Y by jhu08

Plugging in $x=1,y =1$ : we get $f(f(1)) = f(1)$
Pluging $y=1,x=f(1)$ : we get $f(1)^2 /f(1)+1 = f(f(f(1)))/2f(1)$ . The RHS is $1/2$ from step 1 and we get $2f(1)^2 -f(1) -1 = 0 \implies f(1) =1$ .
Now plugging $x=1$ gives $f(y) +1 = (1+y)f(y)$ so $f(y) = 1/y$ , which is indeed a solution.

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Project_Donkey_into_M4

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Project_Donkey_into_M4

#6 h Nov 23, 2021, 1:39 PM • 1 Y

Y by jhu08

Czech-Slovak 2004 P6 wrote:

Find all functions $f:\mathbb R^+ \rightarrow \mathbb R^+$ such that for all positive real numbers $x,y$ ,
$x^2[f(x)+f(y)]=(x+y)f(yf(x)).$

Say $P$ be the assertion.
$P(1,1) \implies 2f(1)=2f(f(1)) \implies f^{n}(1)=f(1)$
for all $n$ belonging to $\mathbb{N}$ .
$P(f(1),1) \implies f(1)^{2}(f(f(1))+f(1))=(f(1)+1)(f(f(1).1)$ or
$f(1)^{2}.2f(1)=(f(1)+1)f(1) \implies (2f(1)+1)(f(1)-1)=0$ .
So either $f(1)=1$ or $f(1)=-\frac{1}{2}$ .
But $f : \mathbb{R}+ \rightarrow \mathbb{R}+$ ,
So $f(1)=1$ .
Now $P(1,x) \implies f(1)+f(x)=(1+x)f(f(1).x)$
$f(x)+1=xf(x)+f(x) \implies xf(x)=1 \implies f(x)=1/x$
And we are done $\blacksquare$

This post has been edited 1 time. Last edited by Project_Donkey_into_M4, Nov 23, 2021, 1:39 PM

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#7 h Nov 23, 2021, 2:50 PM • 2 Y

Y by jhu08, megarnie

Let $f(u)=u$ :
$P(u,1)\Rightarrow u^2+u(f(1)-1)-1=0$
This equation has at most one positive root, so $f$ has at most one fixed point.

$P(x,x)\Rightarrow f(xf(x))=xf(x)\Rightarrow xf(x)=u\Rightarrow\boxed{f(x)=\frac 1x}$ after testing all solutions $f(x)=\frac ux$ .

This post has been edited 2 times. Last edited by jasperE3, Apr 8, 2025, 8:26 PM

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Project_Donkey_into_M4

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Project_Donkey_into_M4

#8 h Nov 24, 2021, 7:10 AM • 1 Y

Y by jhu08

jasperE3 wrote:

Let $f(u)=u$ :
$P(u,u)\Rightarrow f(u^2)=u^2$
$P(u,1)\Rightarrow u^2+u(f(1)-1)-1=0$
This equation has at most one positive root, so $f$ has at most one fixed point.

$P(x,x)\Rightarrow f(xf(x))=xf(x)\Rightarrow xf(x)=u\Rightarrow\boxed{f(x)=\frac ux},u\in\mathbb R^+$ which are solutions.

:huh:

:wacko:

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pco

#9 h Nov 24, 2021, 7:29 AM

Y by

jasperE3 wrote:

... $\Rightarrow\boxed{f(x)=\frac ux},u\in\mathbb R^+$ which are solutions.

No. Only $u=1$ fits.

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#10 h Nov 24, 2021, 3:40 PM • 1 Y

Y by megarnie

pco wrote:

jasperE3 wrote:

... $\Rightarrow\boxed{f(x)=\frac ux},u\in\mathbb R^+$ which are solutions.

No. Only $u=1$ fits.

Thanks. $~~~~~$

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#12 h Apr 8, 2025, 6:39 PM

Y by

Let $P(x , y)$ be the assertion.

$P(1 , 1)$ gives $f(f(1)) = f(1)$ .
Let's prove $f$ is injective.
Assume that there is $a , b$ positive reals such that $f(a) = f(b)$ but $a \neq b$ .

$P(a , 1)$ gives $a^{2}(f(a) + f(1)) = (a+1)f(f(a))$ .
$P(b , 1)$ gives $b^{2}(f(b) + f(1)) = (b+1)f(f(b))$ .
This means $a^{2}/(a+1) = b^{2}/(b+1)$ .

$a^2b+a^2=b^2a+b^2$ and $(ab+a+b)(a-b)=0$ soo since $a , b$ are positive reals then $a = b$ contradiction. $\square$ .

After knowing $f$ is injective it is easy to see that $f(1) = 1$ .

$P(1 , x)$ gives $1 + f(x) = (x + 1)f(x)$ this means $xf(x) = 1$ .Soo $\boxed{f(x)=1/x}$ for all positive reals $x$ .Soo we are done.

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This post has been edited 2 times. Last edited by Frd_19_Hsnzde, Apr 8, 2025, 6:41 PM

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