functional equation

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

functional equation

G H J

Reveal topic tags

algebra proposed

functional equation

L

G H BBookmark kLocked kLocked NReply

Source: Indonesia IMO 2010 TST, Stage 1, Test 2, Problem 2

The post below has been deleted. Click to close.

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

277 posts

#1 h Nov 12, 2009, 4:32 AM • 3 Y

Y by GeorgeRP, Adventure10, Mango247

Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $f(x^3+y^3)=xf(x^2)+yf(y^2)$ for all real numbers $x$ and $y$ .
Hery Susanto, Malang

Z K Y

The post below has been deleted. Click to close.

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

68 posts

#2 h Nov 12, 2009, 8:19 AM • 1 Y

Y by Adventure10

Set $y=0$ then

$f(x^3)=xf(x^2)$ , so $f(y^3)=yf(y^2)$ , then

$f(x^3+y^3)=xf(x^2)+yf(y^2)=f(x^3)+f(y^3)$ $\rightarrow$ $f(x)+f(y)=f(x+y)$

simple induction gives : $f(x_1+x_2+...+x_n)=f(x_1)+f(x_2)+f(x_3)+...+f(x_n)$ so $\forall \ n\in\mathbb{R}$

then $f(nx)=nf(x)$ and thus by setting $x=1$ we have $f(n)=nf(1)$ , since $f(1)$ is a constant then by assuming $f(1)=a$ , gives $f(n)=an$ with $a\in\mathbb{R}$

done

Z K Y

The post below has been deleted. Click to close.

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

478 posts

math10

#3 h Nov 12, 2009, 11:03 AM • 1 Y

Y by Adventure10

matrix41 wrote:

Set $y = 0$ then

$f(x^3) = xf(x^2)$ , so $f(y^3) = yf(y^2)$ , then

$f(x^3 + y^3) = xf(x^2) + yf(y^2) = f(x^3) + f(y^3)$ $\rightarrow$ $f(x) + f(y) = f(x + y)$

simple induction gives : $f(x_1 + x_2 + ... + x_n) = f(x_1) + f(x_2) + f(x_3) + ... + f(x_n)$ so $\forall \ n\in\mathbb{R}$ then $f(nx) = nf(x)$

Why for all $n \in R$

:

matrix41 wrote:

Set $y = 0$ then

$f(x^3) = xf(x^2)$ , so $f(y^3) = yf(y^2)$ , then

$f(x^3 + y^3) = xf(x^2) + yf(y^2) = f(x^3) + f(y^3)$ $\rightarrow$ $f(x) + f(y) = f(x + y)$

simple induction gives : $f(x_1 + x_2 + ... + x_n) = f(x_1) + f(x_2) + f(x_3) + ... + f(x_n)$ so $\forall \ n\in\mathbb{R}$
then $f(nx) = nf(x)$ and thus by setting $x = 1$ we have $f(n) = nf(1)$ , since $f(1)$ is a constant then by assuming $f(1) = a$ , gives $f(n) = an$ with $a\in\mathbb{R}$

done

you only solve equation on $N$ not $R$
We have: $f(x + y) = f(x) + f(y)$ and $f(x^3) = xf(x^2)$ , $(x) = - f( - x)$ , $f(kx) = kf(x)$ for all $k \in N$
We have:
$f((x + 1)^3 + (x - 1)^3) = (x + 1)f(x^2 + 2x + 1) + (x - 1)f(x^2 - 2x + 1) = 2xf(x^2) + 2xf(1) + 4f(x)$
and $f((x + 1)^3 + (x - 1)^3) = f(2x^3 + 6x) = 2xf(x^2) + 6f(x)$
So $f(x) = xf(1)$ for all $x \in R$

Z K Y

The post below has been deleted. Click to close.

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

68 posts

#4 h Nov 13, 2009, 12:35 PM • 2 Y

Y by Adventure10, Mango247

Sorry there are some mistakes in my solution

Let me write my full solution

First from my first post, we have

$f(x_1)+f(x_2)+...+f(x_n)=f(x_1+x_2+x_3+...+x_n)$ with $n\in\mathbb{N}$ so $f(nx)=nf(x)$ $\forall n\in\mathbb{N}$

and then by setting $x=m+1$ and $y=m-1$ it gives

$f((m+1)^3+(m-1)^3)=f((m^3+3m^3+3m+1)+(m^3-3m^3+3m-1))=f(2m^3+6m)=f(2m^3)+f(6m)=2f(m^3)+6f(m)$

and

$f((m+1)^3+(m-1)^3)=(m+1)f(m^2+2x+1)+(m-1)f(x^2-2x+1)=(m+1)\left(f(m^2)+f(2m)+f(1)\right)+(m-1)\left(f(m^2)+f(-2m)+f(1)\right)=2mf(m^2)+4f(m)+2mf(1)$

since $2mf(m^2)=2f(m^3)$ , so $f((m+1)^3+(m-1)^3)=2mf(m^2)+4f(m)+2mf(1)=2f(m^3)+4f(m)+2mf(1)$

$2f(m^3)+6f(m)=f((m+1)^3+(m-1)^3)=2f(m^3)+4f(m)+2mf(1)$

$2f(m)=2mf(1)$ $\rightarrow$ $f(m)=mf(1)$ $\forall m\in\mathbb{R}$

please check....

EDITED : Sorry I didn't know that my solution is similar to math10's solution

Z K Y

The post below has been deleted. Click to close.

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

761 posts

#5 h Oct 30, 2011, 5:27 AM • 1 Y

Y by Adventure10

not so hard.
let $y=0$ ,then
$f(x^3)=xf(x^2)$
so $f(x^3+y^3)=f(x^3)+f(y^3)$
hence f satisfies Cauchy's function,so for rational x, $f(x)=x$
hence $f(1)=1$
by letting $x=x+1$ we get
$f((x+1)^3)=(x+1)f((x+1)^2)$
hence $2f(x^2)=(2x-1)f(x)+x$
let $x=x+1$
we get $2(f(x^2)+2f(x)+1)=(2x+1)(f(x)+1)+x+1$
hence $f(x)=x$ .

Z K Y

The post below has been deleted. Click to close.

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

23437 posts

pco

#6 h Oct 30, 2011, 6:07 AM • 2 Y

Y by Adventure10, Mango247

littletush wrote:

... hence f satisfies Cauchy's function,

Right

littletush wrote:

so for rational x, $f(x)=x$

Wrong : $f(x)=ax$

Z K Y

The post below has been deleted. Click to close.

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

761 posts

#7 h Oct 30, 2011, 6:17 AM • 2 Y

Y by Adventure10, Mango247

pco wrote:

littletush wrote:

... hence f satisfies Cauchy's function,

Right

littletush wrote:

so for rational x, $f(x)=x$

Wrong : $f(x)=ax$

oh gush!such a terrible mistake!
a can be any real number.

Z K Y

The post below has been deleted. Click to close.

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

204 posts

#8 h Jul 14, 2014, 7:46 AM • 1 Y

Y by Adventure10

So which solution is correct?????

Z K Y

The post below has been deleted. Click to close.

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

769 posts

#9 h Sep 5, 2016, 9:18 AM • 1 Y

Y by Adventure10

It is easy to show that ∃ $a\in \mathbb R$ s.t. $f(x)=ax(\forall x\in \mathbb Q)$ .

If we show that $f$ is
-continuous at some point
-monotonous
-either upperbounded or lowerbounded on some open interval

$f(x)=ax(\forall x\in \mathbb R)$

Anyone has idea?

Z K Y

The post below has been deleted. Click to close.

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

23437 posts

pco

#10 h Sep 5, 2016, 9:44 AM • 4 Y

Y by Takeya.O, bgn, Wizard_32, Adventure10

Raja Oktovin wrote:

Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $f(x^3+y^3)=xf(x^2)+yf(y^2)$ for all real numbers $x$ and $y$ .
Hery Susanto, Malang

Let $P(x,y)$ be the assertion $f(x^3+y^3)=xf(x^2)+yf(y^2)$
Let $a=f(1)$

$P(0,0)$ $\implies$ $f(0)=0$
$P(x,0)$ $\implies$ $f(x^3)=xf(x^2)$
And so $f(x^3+y^3)=f(x^3)+f(y^3)$ and so $f(x)$ is additive.

So $f(px)=pf(x)$ $\forall x$ and $\forall p\in\mathbb Q$

Let $x\in\mathbb R$ and $k\in\mathbb Q$
$P(x+k,0)$ $\implies$ $f(x^3+3kx^2+3k^2x+k^3)=(x+k)f(x^2+2kx+k^2)$

Which may be written $k^2(f(x)-ax)+2k(f(x^2)-xf(x))=0$

Considering this as a polynomial in $k$ with infinitely many roots (any rational), we get thet it must be the zero polynomial and so, looking at coefficient of $k^2$ :

$\boxed{f(x)=ax\text{ }\forall x}$ which indeed is a solution, whatever is $a\in\mathbb R$

Z K Y

The post below has been deleted. Click to close.

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

363 posts

#11 h Sep 5, 2016, 9:45 AM • 3 Y

Y by Takeya.O, Wizard_32, Adventure10

My solution:Plug x=y=0 so f(0)=0.
Plug x=0 so f(x^3)=xf(x^2).(1)
So f(x^3+y^3)=f(x^3)+f(y^3) for every x;y real.
So f is addictive.
Now we will caculate f( (x+1)^3+(x-1)^3) in 2 ways.
f( (x+1)^3+(x-1)^3) = (x+1)f((x+1)^2)+(x-1)f((x-1)^2).
=(x+1)[f(x^2)+2f(x)+f(1)]+(x-1)(f(x^2)-2f(x)+f(1)].
=2xf(x^2)+2xf(1)+4f(x).(2)
On the other hand,
f( (x+1)^3+(x-1)^3) =f(2x^3+6x)=2f(x^3)+6f(x).(3)
By (1);(2) and (3) we have :
f(x)=cx(c=f(1)).

Z K Y

The post below has been deleted. Click to close.

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

769 posts

#12 h Sep 5, 2016, 10:02 AM • 2 Y

Y by Adventure10, Mango247

@pco,@anhtaitran
What a nice solution!

Are you a FE master?

This post has been edited 1 time. Last edited by Takeya.O, Sep 5, 2016, 10:03 AM
Reason: Latex miss

Z K Y

The post below has been deleted. Click to close.

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

104 posts

#13 h Feb 2, 2018, 4:32 PM • 2 Y

Y by Adventure10, Mango247

pco wrote:

And so $f(x^3+y^3)=f(x^3)+f(y^3)$ and so $f(x)$ is additive.

So $f(px)=pf(x)$ $\forall x$ and $\forall p\in\mathbb Q$

How do I conclude this? Induction would work on $\mathbb N$ but what can I do here?

This post has been edited 1 time. Last edited by Evenprime123, Feb 2, 2018, 4:32 PM

Z K Y

The post below has been deleted. Click to close.

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

23437 posts

pco

#14 h Feb 2, 2018, 4:52 PM • 2 Y

Y by Adventure10, Mango247

Evenprime123 wrote:

pco wrote:

And so $f(x^3+y^3)=f(x^3)+f(y^3)$ and so $f(x)$ is additive.

So $f(px)=pf(x)$ $\forall x$ and $\forall p\in\mathbb Q$

How do I conclude this? Induction would work on $\mathbb N$ but what can I do here?

You should consider this as a well-known property of additive functions.

If you want to prove it :
$f(2x)=f(x)+f(x)=2f(x)$
$f(3x)=f(2x)+f(x)=3f(x)$
And, with induction : $f(nx)=nf(x)$ $\forall x$ , $\forall n\in\mathbb N$

So $f(x)=f(q\frac xq)=qf(\frac xq)$ and so $f(\frac xq)=\frac 1qf(x)$ $\forall x$ , $\forall q\in\mathbb N$

So $f(\frac pqx)=pf(\frac xq)=\frac pqf(x)$ and so $f(px)=pf(x)$ $\forall x$ , $\forall p\in\mathbb Q^+$

It remains to remember that $f(-x)=-f(x)$ and $f(0)=0$ and so
$f(px)=pf(x)$ $\forall x$ , $\forall p\in\mathbb Q$

Z K Y

The post below has been deleted. Click to close.

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

23437 posts

pco

#16 h Feb 3, 2018, 3:20 PM • 1 Y

Y by Adventure10

GeronimoStilton wrote:

Some progress:
...
... so I can't finish the problem.

Sorry, but what is the interest of this post ?
Just publicly claim that you are working on this problem ? And what about your progress about your sister's garden ? and what is the last film you liked ?
Just request for some help ? : read the thread ! there is already a full solution (see post #10)
Just bring some new informations to the community ? but this contribution already has been given at least thrice in the current thread (and btw is the beginning of the full solution in the post #10)

Z K Y

The post below has been deleted. Click to close.

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

464 posts

#17 h Mar 23, 2021, 8:27 AM

Y by

This is a folklore:
$\bigstar \color{blue}{\textit{\textbf{ANS:}}}$ $f(x)=cx \quad x\in \mathbb{R}$ where $c\in \mathbb{R}$ .
$\spadesuit \color{red}{\textit{\textbf{Proof:}}}$ It's easy to see that this is a solution to the given FE. Let $P(x,y)$ denote the given assertion, $P(x,-x): f(0)=0$ $P(x,0): f(x^3)=xf(x^2).$ So, let $u=x^3, v=y^3$ , $f(u+v)=f(x^3+y^3)=xf(x^2)+yf(y^2)=f(x^3)+f(y^3)=f(u)+f(v)$ and $f$ is additive. We use a standard trick: $P(x+1,x-1): \boxed{f(x)=f(1)x \quad \forall x\in \mathbb{R}}$ as many terms cancel nicely due to additivity and it yields the desired result. $\quad \blacksquare$

This post has been edited 3 times. Last edited by Keith50, Mar 23, 2021, 8:30 AM

Z K Y

The post below has been deleted. Click to close.

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

338 posts

#18 h Jun 19, 2023, 10:49 PM

Y by

Raja Oktovin wrote:

Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying $f(x^3+y^3)=xf(x^2)+yf(y^2)$ for all real numbers $x$ and $y$ .
Hery Susanto, Malang

$\color{blue} \boxed{\textbf{Answer: f(x)=cx}}$
$\color{blue} \boxed{\textbf{Proof:}}$
$\color{blue} \rule{24cm}{0.3pt}$
$f(x^3+y^3)=xf(x^2)+yf(y^2)...(\alpha)$ In $(\alpha) y=0:$
$\Rightarrow f(x^3)=xf(x^2)...(\beta)$ $(\beta)$ in $(\alpha):$
$\Rightarrow f(x^3+y^3)=f(x^3)+f(y^3)$ $\Rightarrow f(x+y)=f(x)+f(y)...(I)$ In $(\beta) x=x+1:$
$\Rightarrow f(x^3+3x^2+3x+1)=(x+1)f(x^2+2x+1)$ By $(I):$
$\Rightarrow f(x^3)+f(3x^2)+f(3x)+f(1)=xf(x^2)+xf(2x)+xf(1)+f(x^2)+f(2x)+f(1)$ $\Rightarrow f(x)=xf(1)$ $\Rightarrow \boxed{\textbf{f(x)=cx}}_\blacksquare$ $\color{blue} \rule{24cm}{0.3pt}$

This post has been edited 1 time. Last edited by hectorleo123, Jun 19, 2023, 10:49 PM

Z K Y

The post below has been deleted. Click to close.

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

F10tothepowerof34

195 posts

F10tothepowerof34

#19 h Jun 19, 2023, 11:01 PM

Y by

This problem is very similar to Macedonian Mathematical Olympiad 2007 P4

Z K Y

The post below has been deleted. Click to close.

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

338 posts

#20 h Jun 19, 2023, 11:33 PM

Y by

F10tothepowerof34 wrote:

This problem is very similar to Macedonian Mathematical Olympiad 2007 P4

It's the same problem...

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a