Find f

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Redriver

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Redriver

#1 h Jun 25, 2006, 5:36 PM • 2 Y

Y by Adventure10, Mango247

Find all $: R \to R : \ \ f(x^2+f(y))=y+f^2(x)$

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cefer

294 posts

cefer

#2 h Jun 25, 2006, 7:44 PM • 2 Y

Y by Adventure10, Mango247

Firstly if $f(x)=f(y)$ then $f(a^2+f(x))=f(a^2+f(y))$
From here $x+f(a)^2=y+f(a)^2 \Longrightarrow x=y$
We get that $f(x)$ is injective.
Taking $x=0$ we get $f(f(y))=y+f(0)^2$ $(1)$
So there exists $t\in R$ such that $f(t)=0$
Taking $x=y=t$ we get $f(t^2)=t \Longrightarrow f(f(t^2))=f(t) \Longrightarrow t^2+f(0)^2=0$
So $t=f(0)=0$
Taking $y=0$ we get $f(x^2)=f(x)^2=f(-x)^2$
From injectivity of $f$ we get $f(-x)=-f(x)$
Taking $y=0$ we get $f(x^2)=f(x)^2$ .So $f(x)\ge 0$ if and only if $x\ge 0$ $(2)$
Taking $y=-x^2$ we have $f(x^2+f(-x^2))=-x^2+f(x)^2$
$f(x^2-f(x^2))=f(x)^2-x^2$
From $(2)$ we get that $f(x)=x$
That's all

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Blackbeam999

32 posts

Blackbeam999

#3 h Apr 30, 2025, 5:04 AM

Y by

cefer wrote:

Firstly if $f(x)=f(y)$ then $f(a^2+f(x))=f(a^2+f(y))$
From here $x+f(a)^2=y+f(a)^2 \Longrightarrow x=y$
We get that $f(x)$ is injective.
Taking $x=0$ we get $f(f(y))=y+f(0)^2$ $(1)$
So there exists $t\in R$ such that $f(t)=0$
Taking $x=y=t$ we get $f(t^2)=t \Longrightarrow f(f(t^2))=f(t) \Longrightarrow t^2+f(0)^2=0$
So $t=f(0)=0$
Taking $y=0$ we get $f(x^2)=f(x)^2=f(-x)^2$
From injectivity of $f$ we get $f(-x)=-f(x)$
Taking $y=0$ we get $f(x^2)=f(x)^2$ .So $f(x)\ge 0$ if and only if $x\ge 0$ $(2)$
Taking $y=-x^2$ we have $f(x^2+f(-x^2))=-x^2+f(x)^2$
$f(x^2-f(x^2))=f(x)^2-x^2$
From $(2)$ we get that $f(x)=x$
That's all

Why f(x)=x? I am confused

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jasperE3

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jasperE3

#4 h Apr 30, 2025, 5:14 AM • 1 Y

Y by Blackbeam999

Blackbeam999 wrote:

.

Why f(x)=x? I am confused

it is kinda confusing yeah

First consider the case $x\ge0$ , we use the equation $f\left(x^2-f\left(x^2\right)\right)=f(x)^2-x^2~(1)$ .
We got earlier that $f$ is odd, so:
$f\left(f\left(x^2\right)-x^2\right)=-f\left(x^2-f\left(x^2\right)\right)=-\left(f(x)^2-x^2\right)=x^2-f(x)^2~(2).$ So
$f\left\lvert f\left(x^2\right)-x^2\right\rvert=-\left|f(x)^2-x^2\right|,$ since if $\left\lvert f\left(x^2\right)-x^2\right\rvert=f\left(x^2\right)-x^2$ this is equivalent to $(2)$ and if $\left\lvert f\left(x^2\right)-x^2\right\rvert=x^2-f\left(x^2\right)$ this is equivalent to $(1)$ .
Then, recall that $f(x)\ge0$ for all $x\ge0$ , so:
$0\ge-\left|f(x)^2-x^2\right|=f\left|f\left(x^2\right)-x^2\right|\ge0,$ and equality must hold everywhere so $f\left(x^2\right)=x^2$ . Using oddness, $f(x)=x$ everywhere.

This post has been edited 2 times. Last edited by jasperE3, Apr 30, 2025, 5:18 AM

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Blackbeam999

32 posts

Blackbeam999

#5 h Apr 30, 2025, 5:59 AM

Y by

jasperE3 wrote:

Blackbeam999 wrote:

.

Why f(x)=x? I am confused

it is kinda confusing yeah

First consider the case $x\ge0$ , we use the equation $f\left(x^2-f\left(x^2\right)\right)=f(x)^2-x^2~(1)$ .
We got earlier that $f$ is odd, so:
$f\left(f\left(x^2\right)-x^2\right)=-f\left(x^2-f\left(x^2\right)\right)=-\left(f(x)^2-x^2\right)=x^2-f(x)^2~(2).$ So
$f\left\lvert f\left(x^2\right)-x^2\right\rvert=-\left|f(x)^2-x^2\right|,$ since if $\left\lvert f\left(x^2\right)-x^2\right\rvert=f\left(x^2\right)-x^2$ this is equivalent to $(2)$ and if $\left\lvert f\left(x^2\right)-x^2\right\rvert=x^2-f\left(x^2\right)$ this is equivalent to $(1)$ .
Then, recall that $f(x)\ge0$ for all $x\ge0$ , so:
$0\ge-\left|f(x)^2-x^2\right|=f\left|f\left(x^2\right)-x^2\right|\ge0,$ and equality must hold everywhere so $f\left(x^2\right)=x^2$ . Using oddness, $f(x)=x$ everywhere.

Thank!

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cefer

294 posts

cefer

#6 h Apr 30, 2025, 6:26 PM • 1 Y

Y by Blackbeam999

Blackbeam999 wrote:

cefer wrote:

Firstly if $f(x)=f(y)$ then $f(a^2+f(x))=f(a^2+f(y))$
From here $x+f(a)^2=y+f(a)^2 \Longrightarrow x=y$
We get that $f(x)$ is injective.
Taking $x=0$ we get $f(f(y))=y+f(0)^2$ $(1)$
So there exists $t\in R$ such that $f(t)=0$
Taking $x=y=t$ we get $f(t^2)=t \Longrightarrow f(f(t^2))=f(t) \Longrightarrow t^2+f(0)^2=0$
So $t=f(0)=0$
Taking $y=0$ we get $f(x^2)=f(x)^2=f(-x)^2$
From injectivity of $f$ we get $f(-x)=-f(x)$
Taking $y=0$ we get $f(x^2)=f(x)^2$ .So $f(x)\ge 0$ if and only if $x\ge 0$ $(2)$
Taking $y=-x^2$ we have $f(x^2+f(-x^2))=-x^2+f(x)^2$
$f(x^2-f(x^2))=f(x)^2-x^2$
From $(2)$ we get that $f(x)=x$
That's all

Why f(x)=x? I am confused

We have
$f(x^2-f(x)^2)=f(x)^2-x^2$ and we also know that $f(x) \geq 0$ if and only if $x \geq 0$ . If $x^2 > f(x)^2$ , then we get that $f$ gets negative value at positive input. Similarly, if $x^2 < f(x)^2$ , then $f$ gets positive value at negative input. So $x^2 = f(x)^2$ and this gives us $f(x) = x$ .