Functional Equation

rdj5933mile5math64 · **Posted:** Thu, 26 Apr 2012 19:53:37 UTC

Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that f(1)=1

and
$f\left(f(x)y+\frac{x}{y}\right)=xyf\left(x^2+y^2\right)$ for all real numbers

and

with $y \neq 0$

My Work:

$P(0,y) \rightarrow f(0)=0.$
Note that the equation $x^2+y^2=f(x)y+\frac{x}{y}$ can be rewritten as a cubic in terms of

with 1 real root. So, we can have them set equal to each other, but until we prove that f(x)=0

iff x is zero, we can't use this info. Additionally, we can't wlog anything... And as a conjecture I think the function is $f(x)=\frac{1}{x}$ when x is nonzero and f(0)=0

. But I can't prove it.

Edit (Shadow): Spoiler tags removed.

Shadow · **Posted:** Thu, 26 Apr 2012 20:32:32 UTC

rdj5933mile5math64 wrote:

Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that f(1)=1

and
$f\left(f(x)y+\frac{x}{y}\right)=xyf\left(x^2+y^2\right)$ for all real numbers

and

with $y \neq 0$

My Work:

$P(0,y) \rightarrow f(0)=0.$
Note that the equation $x^2+y^2=f(x)y+\frac{x}{y}$ can be rewritten as a cubic in terms of

with 1 real root. So, we can have them set equal to each other, but until we prove that f(x)=0

iff x is zero, we can't use this info. Additionally, we can't wlog anything... And as a conjecture I think the function is $f(x)=\frac{1}{x}$ when x is nonzero and f(0)=0

. But I can't prove it.

Edit (Shadow): Spoiler tags removed.

What is this P?

rdj5933mile5math64 · **Posted:** Fri, 27 Apr 2012 01:52:05 UTC

Sorry, let P(x,y) be the assertion $f\left(f(x)y+\frac{x}{y}\right)=xyf\left(x^2+y^2\right)$

EDIT: Should I refrain from hiding my work in general?

Shadow · **Posted:** Fri, 27 Apr 2012 05:46:07 UTC

rdj5933mile5math64 wrote:

Sorry, let P(x,y) be the assertion $f\left(f(x)y+\frac{x}{y}\right)=xyf\left(x^2+y^2\right)$

EDIT: Should I refrain from hiding my work in general?

Yes you should.

Spoilers are for cases when you have hints which more or less give things all the way away and might ruin the problem, but also may be necessary for the poster to understand things. With your own work, there's never any reason to hide it.

outermeasure · **Posted:** Fri, 27 Apr 2012 12:58:26 UTC

rdj5933mile5math64 wrote:

Find all functions $f:\mathbb{R}\to\mathbb{R}$ such that f(1)=1

and
$f\left(f(x)y+\frac{x}{y}\right)=xyf\left(x^2+y^2\right)$ for all real numbers

and

with $y \neq 0$

My Work:

$P(0,y) \rightarrow f(0)=0.$
Note that the equation $x^2+y^2=f(x)y+\frac{x}{y}$ can be rewritten as a cubic in terms of

with 1 real root. So, we can have them set equal to each other, but until we prove that f(x)=0

iff x is zero, we can't use this info. Additionally, we can't wlog anything... And as a conjecture I think the function is $f(x)=\frac{1}{x}$ when x is nonzero and f(0)=0

. But I can't prove it.

Edit (Shadow): Spoiler tags removed.

Where does $x^2+y^2=f(x)y+\frac{x}{y}$ come from? Are you assuming xy=1 and playing with consequence of (unproven) injectivity? But then there is no need to invoke the cubic in y, and it is not sufficient to prove $f(x)=0\iff x=0$ ...

rdj5933mile5math64 · **Posted:** Fri, 27 Apr 2012 15:34:16 UTC

outermeasure wrote:

Where does $x^2+y^2=f(x)y+\frac{x}{y}$ come from?

A common strategy that I use on olympiad fuctional equation problems is to set the things that we are taking "f" of equal to each other and see what happens. I wasn't necessarily assuming xy=1, or injectivity (kinda wishing I could though).~

outermeasure wrote:

Are you assuming xy=1 and playing with consequence of (unproven) injectivity? But then there is no need to invoke the cubic in y, and it is not sufficient to prove $f(x)=0\iff x=0$ ...

No I'm playing with what happens if $x^2+y^2=f(x)y+\frac{x}{y}$ is true (intuition is above). The reason for invoking the cubic is to show that for any x there exists a real y such that $x^2+y^2=f(x)y+\frac{x}{y}$ is true.

I'll show my work a little more formally this time

My Work:

Let P(x,y)

be the assertion $f(f(x)y+\frac{x}{y})=xyf(x^2+y^2)$ .

Claim 1: f(0)=0

Assume for the sake of contradiction that this is not the case.
$P(0,\frac{1}{f(0)}) \rightarrow f(1)=0$
which is a contradiction.

Claim 2: The problem is finished if we can prove $f(x) = 0 \iff x=0$ .
Consider the polynomial:
y^3-y^2f(x)+yx^2-x=0

Observe that this is a cubic and therefore it must have one real root for any given x.
But this is equivalent to there being at least one real y satisfying:
$f(f(x)y+\frac{x}{y})=xyf(x^2+y^2)=t$ .
Hence, we have
f(t)=xyf(t)

Subtracting f(t)

on both sides yields:
xy=1

or

This feels like progress, but I don't know how I could show f(t) is nonzero... :/

outermeasure · **Posted:** Fri, 27 Apr 2012 15:47:27 UTC

rdj5933mile5math64 wrote:

outermeasure wrote:

Where does $x^2+y^2=f(x)y+\frac{x}{y}$ come from?

A common strategy that I use on olympiad fuctional equation problems is to set the things that we are taking "f" of equal to each other and see what happens. I wasn't necessarily assuming xy=1, or injectivity (kinda wishing I could though).~

But that can't be true. RHS is

, not

, so it is not a even a remotely good strategy. For example, $g\colon\mathbb{R}\to\mathbb{R};\quad g(1+x^2)g(z^2+y^2)=g(z+xy)g(y-xz)$ for all real x,y,z.

rdj5933mile5math64 wrote:

outermeasure wrote:

Are you assuming xy=1 and playing with consequence of (unproven) injectivity? But then there is no need to invoke the cubic in y, and it is not sufficient to prove $f(x)=0\iff x=0$ ...

No I'm playing with what happens if $x^2+y^2=f(x)y+\frac{x}{y}$ is true (intuition is above). The reason for invoking the cubic is to show that for any x there exists a real y such that $x^2+y^2=f(x)y+\frac{x}{y}$ is true.

I'll show my work a little more formally this time

My Work:

Let P(x,y)

be the assertion $f(f(x)y+\frac{x}{y})=xyf(x^2+y^2)$ .

Claim 1: f(0)=0

Assume for the sake of contradiction that this is not the case.
$P(0,\frac{1}{f(0)}) \rightarrow f(1)=0$
which is a contradiction.

Claim 2: The problem is finished if we can prove $f(x) = 0 \iff x=0$ .
Consider the polynomial:
y^3-y^2f(x)+yx^2-x=0

Observe that this is a cubic and therefore it must have one real root for any given x.
But this is equivalent to there being at least one real y satisfying:
$f(f(x)y+\frac{x}{y})=xyf(x^2+y^2)=t$ .
Hence, we have
f(t)=xyf(t)

Subtracting f(t)

on both sides yields:
xy=1

or

This feels like progress, but I don't know how I could show f(t) is nonzero... :/

Why does

mean

?

rdj5933mile5math64 · **Posted:** Fri, 27 Apr 2012 16:21:29 UTC

outermeasure wrote:

rdj5933mile5math64 wrote:

outermeasure wrote:

Where does $x^2+y^2=f(x)y+\frac{x}{y}$ come from?

A common strategy that I use on olympiad fuctional equation problems is to set the things that we are taking "f" of equal to each other and see what happens. I wasn't necessarily assuming xy=1, or injectivity (kinda wishing I could though).~

But that can't be true. RHS is xyf(x^2+y^2)

, not

, so it is not a even a remotely good strategy.

There are 2 types of functional equations. Those that you can intuitively solve and those that involve plugging somewhat random things in. This is the latter from what I can tell.

outermeasure wrote:

rdj5933mile5math64 wrote:

outermeasure wrote:

Where does $x^2+y^2=f(x)y+\frac{x}{y}$ come from?

A common strategy that I use on olympiad fuctional equation problems is to set the things that we are taking "f" of equal to each other and see what happens. I wasn't necessarily assuming xy=1, or injectivity (kinda wishing I could though).~

But that can't be true. RHS is

, not

, so it is not a even a remotely good strategy.

rdj5933mile5math64 wrote:

outermeasure wrote:

Are you assuming xy=1 and playing with consequence of (unproven) injectivity? But then there is no need to invoke the cubic in y, and it is not sufficient to prove $f(x)=0\iff x=0$ ...

No I'm playing with what happens if $x^2+y^2=f(x)y+\frac{x}{y}$ is true (intuition is above). The reason for invoking the cubic is to show that for any x there exists a real y such that $x^2+y^2=f(x)y+\frac{x}{y}$ is true.

I'll show my work a little more formally this time

My Work:

Let P(x,y)

be the assertion $f(f(x)y+\frac{x}{y})=xyf(x^2+y^2)$ .

Claim 1: f(0)=0

Assume for the sake of contradiction that this is not the case.
$P(0,\frac{1}{f(0)}) \rightarrow f(1)=0$
which is a contradiction.

Claim 2: The problem is finished if we can prove $f(x) = 0 \iff x=0$ .
Consider the polynomial:
y^3-y^2f(x)+yx^2-x=0

Observe that this is a cubic and therefore it must have one real root for any given x.
But this is equivalent to there being at least one real y satisfying:
$f(f(x)y+\frac{x}{y})=xyf(x^2+y^2)=t$ .
Hence, we have
f(t)=xyf(t)

Subtracting f(t)

on both sides yields:
xy=1

or

This feels like progress, but I don't know how I could show f(t) is nonzero... :/

Why does

mean

?

Yeah, lol I'm very sorry for any confusion I may have caused. I am really tired and and as a result I'm not thinking the clearest today.~

Let me try a Take 3 on Claim 2

Claim 2: The problem is finished if we can prove $f(x) = 0 \iff x=0$ .
Consider the polynomial:
y^3-y^2f(x)+yx^2-x=0

Observe that this is a cubic and therefore it must have one real root for any given x.
But this is equivalent to there being at least one real y (let's say y') satisfying:
$f(x)y'+\frac{x}{y'}=x^2+y'^2=t$ .
Hence, we have
$f(f(x)y'+\frac{x}{y'})=f(x^2+y'^2)=f(t)$
But, note that
$P(x,y') \rightarrow f(t)=xy'f(t)$
Subtracting f(t)