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 Post subject: Functional Equation
PostPosted: Thu, 26 Apr 2012 19:53:37 UTC 
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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 x and y 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 y 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.

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 Post subject: Re: Functional Equation
PostPosted: Thu, 26 Apr 2012 20:32:32 UTC 
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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 x and y 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 y 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?

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 Post subject: Re: Functional Equation
PostPosted: Fri, 27 Apr 2012 01:52:05 UTC 
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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?

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 Post subject: Re: Functional Equation
PostPosted: Fri, 27 Apr 2012 05:46:07 UTC 
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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.

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 Post subject: Re: Functional Equation
PostPosted: Fri, 27 Apr 2012 12:58:26 UTC 
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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 x and y 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 y 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...

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\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: Functional Equation
PostPosted: Fri, 27 Apr 2012 15:34:16 UTC 
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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 f(t)=0

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

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 Post subject: Re: Functional Equation
PostPosted: Fri, 27 Apr 2012 15:47:27 UTC 
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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 f(x^2+y^2), 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 f(t)=0

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


Why does t=xyf(x^2+y^2) mean f(t)=xyf(t)?

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: Functional Equation
PostPosted: Fri, 27 Apr 2012 16:21:29 UTC 
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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 f(x^2+y^2), 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 xyf(x^2+y^2), not f(x^2+y^2), 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 f(t)=0

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


Why does t=xyf(x^2+y^2) mean f(t)=xyf(t)?


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 :P

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) on both sides yields:
xy'=1 or f(t)=0.
The result quickly follows if we can show that f(t) isn't zero.

EDIT: t is defined as x^2+y'^2 I didn't feel like writing x^2+y'^2 throughout the solution.

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 Post subject: Re: Functional Equation
PostPosted: Mon, 14 May 2012 20:05:32 UTC 
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outermeasure wrote:
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.


Didn't see this until just awhile ago, the problem is solved by plugging in 1's and 0's. There may be a few small typos.~

My Solution:
Spoiler:
Let P(x,y,z) be the assertion P(x,y,z) \rightarrow g(1+x^2)g(y^2+z^2)=g(z+xy)g(y-xz)

Claim 1: g(1)g(y^2+z^2)=g(y)g(z)
P(0,y,z) \rightarrow g(1)g(y^2+z^2)=g(y)g(z)

Claim 2: g(1)g(y^2)=g(y)g(0)
Take z=0 for Claim 1.

Claim 3: [g(x^2+1)]^2=g(0)g(1+x^2)
P(x,x,1) \rightarrow [g(x^2+1)]^2=g(0)g(1+x^2)

Claim 4: g(1+x^2)g(0)=[g(0)]^2
P(x,0,0) \rightarrow g(1+x^2)g(0)=[g(0)]^2

Claim 5: g(1+x^2)g(y^2)=g(y)g(yx)
P(x,y,0) \rightarrow g(1+x^2)g(y^2)=g(y)g(yx)

Claim 6: g(1)g(2y^2)=[g(y)]^2
P(0,y,y) \rightarrow g(1)g(2y^2)=[g(y)]^2

Claim 7: g(x^2+1)=g(0) for all real x.
From Claims 3 and 4,
[g(x^2+1)]^2=g(0)g(1+x^2) and g(1+x^2)g(0)=[g(0)]^2.
Or by subtracting g(1+x^2)g(0) from both sides of each equation,
[g(x^2+1)]^2-g(0)g(1+x^2)=0 and -g(1+x^2)g(0) + [g(0)]^2 =0.
Adding these 2 equations together yields:
[g(x^2+1)-g(0)]^2=0
which is true iff
g(x^2+1)=g(0).

Claim 8: g(1)=g(0)
Observe that by Claim 7 when x=0, g(1)=g(0)

Now, we have two cases.

Case #1: g(0)=0.
This implies that g(1)=0, so by Claim 5, g(x)=0 for all real x.

Case #2: g(0) \ne 0
This implies that g(1) \ne 0 by Claim 8.

We can actually prove something a little stronger - that g(x) is equal to g(0).

Observe that if z=1 in Claim 3, then
g(1)g(1+y^2)=g(y)g(1)
But, g(1+y^2)=g(1)=g(0), and since g(0) \ne 0 g(y)=g(0).



Hence for all real x, g(x)=c, where c is a fixed yet arbitrary number. Plugging this back in, we find that it is indeed a solution.


A somewhat similar problem: Day #2 Problem #5

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