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Source: IISc Pravega, Enumeration 2023-24 Finals P1

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#1 h Jan 27, 2024, 2:57 PM

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Find all functions $f\colon \mathbb R^+ \mapsto \mathbb R^+$ , such that for all positive reals $x,y$ , the following is true:

$xf(1+xf(y))= f\left(f(x) + \frac 1y\right)$
Kazi Aryan Amin

This post has been edited 1 time. Last edited by Aryan-23, Jan 27, 2024, 2:58 PM

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pco

#2 h Jan 27, 2024, 6:26 PM • 2 Y

Y by math_comb01, Aryan-23

Aryan-23 wrote:

Find all functions $f\colon \mathbb R^+ \mapsto \mathbb R^+$ , such that for all positive reals $x,y$ , the following is true:

$xf(1+xf(y))= f\left(f(x) + \frac 1y\right)$

Let $P(x,y)$ be the assertion $xf(1+xf(y))=f(f(x)+\frac 1y)$

$P(\frac x{f(x+1)},x+1)$ $\implies$ $x=f(f(\frac x{f(x+1)})+\frac 1{x+1})$ and so $f(x)$ is surjective

If $f(a)=f(b)$ for some $a\ge b$ , comparaison of $P(x,a)$ and $P(x,b)$ implies $f(f(x)+\frac 1a)=f(f(x)+\frac 1b)$ and so, since surjective, $f(x+\frac 1a)=f(x+\frac 1b)$
And so $f(x+\Delta)=f(x)$ $\forall x>\frac 1a$ and for $\Delta=\frac 1b-\frac 1a$
Choosing then $u$ such that $f(u)=1$ and $x>\frac 1a$ , comparaison of $P(x,u)$ and $P(x+\Delta,u)$ gives $\Delta f(1+x)=0$ , and so $\Delta=0$ and $f(x)$ is injective.

Then $P(1,x)$ implies, using injectivity, $f(x)=\frac 1x+f(1)-1$ and so, since $f(x)$ is positive and surjective, $f(1)=1$ and

$\boxed{f(x)=\frac 1x\quad\forall x>0}$ , which indeed fits.

This post has been edited 1 time. Last edited by pco, Jan 27, 2024, 6:27 PM
Reason: Typos

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#3 h Jan 28, 2024, 2:34 AM

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The answer is $f(x)=\tfrac{1}{x}$ only, which clearly works. Let $P(x,y)$ denote the assertion. From $P(\tfrac{x}{f(x+1)},x+1)$ we find that $f$ is surjective.

I now claim that $f$ is injective. Indeed, if $f(a)=f(b)$ with $a>b$ then $f(t_1+\tfrac{1}{a})=f(t_1+\tfrac{1}{b})$ for all $t_1>0$ by surjectivity. Then
$f\left(t_2+\frac{1}{t_1+\frac{1}{a}}\right)=f\left(t_2+\frac{1}{t_1+\frac{1}{b}}\right).$ As $t_1 \in \mathbb{R}^+$ varies, $\tfrac{1}{t_1+\frac{1}{a}}-\tfrac{1}{t_1+\frac{1}{b}}=\tfrac{a-b}{(at_1+1)(bt_1+1)}$ varies and hits all values in the interval $(0,a-b)$ . Furthermore, $\tfrac{1}{t_1+\tfrac{1}{a}} \leq a$ for all $t_1$ , so by repeatedly applying this we get $f$ eventually constant, say equal to $C>0$ . But then by fixing $y$ sufficiently close to $0$ , and then sending $x \to \infty$ implies $Cx=C$ , which is absurd, implying the desired.

Now plug in $x=1$ to get $1+f(y)=f(1)+\tfrac{1}{y}$ for all $y$ . By surjectivity (and the fact that $f$ is positive), we have $f(y)=\tfrac{1}{y}$ for all $y$ , as desired. $\blacksquare$

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#6 h Jan 28, 2024, 4:07 AM • 2 Y

Y by Aryan-23, Deadline

very tricky problem

here is the solution by my team "60 Dipole"
solution

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#9 h Jan 28, 2024, 9:39 AM

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I was not having high hopes of solving this... But look what happened!

Solution:The answer is $f(x) = \frac{1}{x}$ for all $x \in \mathbb{R}^+$ . It clearly works and we will now show its the only possible solution. Denote $P(x,y)$ as the assertion to the given functional equation.

Claim: $f$ is bijective from $\mathbb{R}^+ \to \mathbb{R}^+$ .

Proof: Start by $P(x/f(y), y)$ which yields
$xf(1+x) = f(y)f\left( f\left( \frac{x}{f(y)} \right) + \frac{1}{y} \right).$ Putting $x+1 = y$ in above equation yields the surjectivity.

For the other part assume $f(a) = f(b)$ for $b>a$ . $P(x,a)$ and $P(x,b)$ gives
$f\left( f(x) + \frac{1}{a} \right) = f\left( f(x) + \frac{1}{b} \right).$ As $f$ is a surjection, we can conclude that $f$ is eventually periodic with period $T \coloneqq \frac{1}{a} - \frac{1}{b}$ . Pick $k$ such that $f(k)$ is an integer. For large enough $x$ , consider $P(x+T,k)$ and $P(x,k)$ . This would yield
$(x+T)f(1+xf(k)) = xf(1+xf(k)) \iff T = 0$ which is a contradiction since $b>a$ . This proves the claim. $\square$

Finally, consider $P(1,y)$ to get
$f(1+f(y)) = f\left(f(1) + \frac{1}{y}\right).$ By injectivity, we can get $f(y) = 1/y + c$ for some positive constant $c$ . Putting this back into the original equation, its not hard to get $c = 0$ . This completes solution. $\blacksquare$

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#10 h Jan 28, 2024, 11:29 AM • 1 Y

Y by GeoKing

Very standard, I think such FE's are really common.

First note that $P\left(\frac{x}{f(x+1)}, x+1\right)$ shows that $f$ is surjective. Now, I claim that $f$ is injective. Indeed, assume not, let $f(a)=f(b)$ for some distinct positive reals $a,b$ . Then, $P(x,a)-P(x,b)$ reveals that $f$ is eventually periodic with period $T$ . Now, choose $u$ so that $f(u)=N$ where $N$ is a large positive integer. Then, $P(x+T,y)-P(x,y)$ just yields that $(x+T)f(1+xN+TN) = xf(1+xN)$ but as $f$ is eventually periodic, we have that $(x+T)f(1+xN)=xf(1+xN)$ for $x$ large enough, a contradiction, so $f$ is indeed injective. Now, $P(1,x)$ finishes.

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#11 h Apr 30, 2025, 3:40 AM

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Aryan-23 wrote:

Find all functions $f\colon \mathbb R^+ \mapsto \mathbb R^+$ , such that for all positive reals $x,y$ , the following is true:

$xf(1+xf(y))= f\left(f(x) + \frac 1y\right)$
Kazi Aryan Amin

$P\left(\frac x{f(x+1)},x+1\right)\Rightarrow f\left(f\left(\frac x{f(x+1)}\right)+\frac1{x+1}\right)=x$ so $f$ is surjective. Let $f(k)=1$ .

Suppose $f(a)=f(b)$ for some $a,b$ .
$P(x,a)\Rightarrow xf(1+xf(a))=f\left(f(x)+\frac1a\right)$
$P(x,a)\Rightarrow xf(1+xf(a))=f\left(f(x)+\frac1b\right)$
Comparing, we get $f\left(f(x)+\frac1a\right)=f\left(f(x)+\frac1b\right)$ . By surjectivity, $f\left(x+\frac1a\right)=f\left(x+\frac1b\right)$ .
$P\left(x+\frac1a,k\right)\Rightarrow\left(x+\frac1a\right)f\left(1+x+\frac1a\right)=f\left(f\left(x+\frac1a\right)+\frac1k\right)$
$P\left(x+\frac1b,k\right)\Rightarrow\left(x+\frac1b\right)f\left(1+x+\frac1a\right)=f\left(f\left(x+\frac1a\right)+\frac1k\right)$
Comparing, we get $a=b$ , so $f$ is injective.

$P(1,x)\Rightarrow f(1+f(x))=f\left(f(1)+\frac1x\right)\Rightarrow f(x)=\frac1x+f(1)-1$ , testing all functions of this form gives that only $\boxed{f(x)=\frac1x}$ works.

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