Math Olympiads problem that I couldn't do.

[mtayab1994]

Homework Statement

for every (x,y) in ℝ^2:

f(x^2+f(y))= y-x^2

Homework Equations

Find all functions.

The Attempt at a Solution

I was wondering if rewriting it as x^2+f(y)=f^-1(y-x^2)?

 

[micromass]

So, what is the question?? And what is f??

 

[deluks917]

I believe the idea is to find all functions satisfying that condition.

 

[mtayab1994]

I believe the idea is to find all functions satisfying that condition.

Yes that's what it is. Find all functions satisfying that condition.

 

[micromass]

Note that

$$x^2+f(x^2+f(y))=y$$

So if we define

$$g_x(y):=x^2+f(y)$$

Then what can you say about $g_x\circ g_x$?

 

[mtayab1994]

Note that

$$x^2+f(x^2+f(y))=y$$

So if we define

$$g_x(y):=x^2+f(y)$$

Then what can you say about $g_x\circ g_x$?

I don't get what you're trying to tell me with gx∘gx.

I understand that its a category from gx to gx, so u can say: gx(g(x)). right?

 

[Quinzio]

I'm sorry for my intrusion, but after having spent some time on this I would like to add my comment.
From intuition and some trials I can say the $f(x) = -x$ works.
So that:
$f(y) = -y$
$f(y) + x^2= -y+x^2$
$f(f(y) + x^2)= y-x^2$

But is there an analytical or mechanical way to solve these problems ?

Then what can you say about $g_x\circ g_x$?

I can't get this too. Maybe one should apply the derivative ?

 

[Dick]

I'd try just differentiating both sides with respect to x.

 

[micromass]

I'd try just differentiating both sides with respect to x.

Yes, but what if f is not differentiable??

 

[Quinzio]

I don't get what you're trying to tell me with gx∘gx.

I understand that its a category from gx to gx, so u can say: gx(g(x)). right?

I'd try just differentiating both sides with respect to x.

But it's not a function of 2 variables. There's only one "input".

 

[mtayab1994]

I'm sorry for my intrusion, but after having spent some time on this I would like to add my comment.
From intuition and some trials I can say the $f(x) = -x$ works.
So that:
$f(y) = -y$
$f(y) + x^2= -y+x^2$
$f(f(y) + x^2)= y-x^2$

But is there an analytical or mechanical way to solve these problems ?


I can't get this too. Maybe one should apply the derivative ?

f(f(y)+x^x)=y−x^2 can also be written as fof(y)+f(x^2)=y-x^2

 

[Quinzio]

But the problem says "find all the solutions". So there are more than 1 ? What the others look like then ?

 

[Dick]

Yes, but what if f is not differentiable??

I'd say find all the differentiable ones first, then try thinking about if there could be more.

 

[Dick]

But it's not a function of 2 variables. There's only one "input".

Treat y as a constant.

 

[mtayab1994]

But the problem says "find all the solutions". So there are more than 1 ? What the others look like then ?

I don't know that's what I'm trying to reach but i also got this one:

y=f^-1(f^-1(y-x^2)-x^2)

 

[micromass]

I don't know that's what I'm trying to reach but i also got this one:

y=f^-1(f^-1(y-x^2)-x^2)

Yes, but you got to watch out. f is not necessarily invertible. So the function $f^{-1}$ might not exist!

 

[mtayab1994]

Yes, but you got to watch out. f is not necessarily invertible. So the function $f^{-1}$ might not exist!

Yes but is it correct?

 

[mtayab1994]

That's the question, its the second one if you understand some french micromass.

 

[micromass]

Yes but is it correct?

Not until you show f to be invertible. What happens if x=0 ?

 

[mtayab1994]

for x=0 you get fof(y)=y and f(y)=f^-1(y)

 

[micromass]

for x=0 you get fof(y)=y and f(y)=f^-1(y)

Indeed! So f is its now inverse!

Now, take your original relationship again

$$f(x^2+f(y))=y-x^2$$

Now take f of both sides. What happens if y=0?

 

[mtayab1994]

Indeed! So f is its now inverse!

Now, take your original relationship again

$$f(x^2+f(y))=y-x^2$$

Now take f of both sides. What happens if y=0?

x^2=f(-x^2)

 

[micromass]

x^2=f(-x^2)

No, that's not what I get. How did you get that?

 

[mtayab1994]

No, that's not what I get. How did you get that?

did u get f(y)=f(-x^2)-x^2?

 

[micromass]

did u get f(y)=f(-x^2)-x^2?

No. Remember that we put y=0!

 

[mtayab1994]

No. Remember that we put y=0!

idk but i keep getting x^2=f(-x^2)

 

[micromass]

idk but i keep getting x^2=f(-x^2)

Can you post how you get there??

 

[mtayab1994]

Can you post how you get there??

Case: y=0:

f(x^2+f(y))=y-x^2 <=> f(x^2)+f(f(0))=-x^2 <=> x^2+fof(0) = f^-1(-x^2)

<=> x^2=f^-1(-x^2).

 

[micromass]

In your first line you use f(a+b)=f(a)+f(b). But you don't know whether that property holds! So you can't do that!

 

[mtayab1994]

In your first line you use f(a+b)=f(a)+f(b). But you don't know whether that property holds! So you can't do that!

What solution did u get for Y=0 just curious to know.

 

[micromass]

What solution did u get for Y=0 just curious to know.

Hold on, let me guide you there step by step.

So, you have

$$f(x^2+f(y))=y-x^2$$

What happens if you take f of both sides??

 

[mtayab1994]

Hold on, let me guide you there step by step.

So, you have

$$f(x^2+f(y))=y-x^2$$

What happens if you take f of both sides??

I can't grasp what you mean by if you take the f of both sides?

 

[micromass]

I can't grasp what you mean by if you take the f of both sides?

For example, if we have a=b. If we take f of both sides, then we have f(a)=f(b).

 

[mtayab1994]

For example, if we have a=b. If we take f of both sides, then we have f(a)=f(b).

ooohh alright i got: fof(x^2+f(y))=f(y-x^2)

 

[micromass]

ooohh alright i got: fof(x^2+f(y))=f(y-x^2)

OK, and fof=... ??