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 Post subject: Functional EquationPosted: Thu, 26 Apr 2012 19:53:37 UTC
 S.O.S. Oldtimer

Joined: Wed, 4 Apr 2012 03:51:40 UTC
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Find all functions such that and
for all real numbers and with

My Work:

Note that the equation 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 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 when x is nonzero and . But I can't prove it.

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 Post subject: Re: Functional EquationPosted: Thu, 26 Apr 2012 20:32:32 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 13973
Location: Austin, TX
rdj5933mile5math64 wrote:
Find all functions such that and
for all real numbers and with

My Work:

Note that the equation 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 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 when x is nonzero and . But I can't prove it.

What is this P?

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 Post subject: Re: Functional EquationPosted: Fri, 27 Apr 2012 01:52:05 UTC
 S.O.S. Oldtimer

Joined: Wed, 4 Apr 2012 03:51:40 UTC
Posts: 145
Location: Hockeytown
Sorry, let P(x,y) be the assertion

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

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 Post subject: Re: Functional EquationPosted: Fri, 27 Apr 2012 05:46:07 UTC
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rdj5933mile5math64 wrote:
Sorry, let P(x,y) be the assertion

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 EquationPosted: Fri, 27 Apr 2012 12:58:26 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
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rdj5933mile5math64 wrote:
Find all functions such that and
for all real numbers and with

My Work:

Note that the equation 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 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 when x is nonzero and . But I can't prove it.

Where does 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 ...

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 Post subject: Re: Functional EquationPosted: Fri, 27 Apr 2012 15:34:16 UTC
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Joined: Wed, 4 Apr 2012 03:51:40 UTC
Posts: 145
Location: Hockeytown
outermeasure wrote:
Where does 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 ...

No I'm playing with what happens if 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 is true.

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

My Work:

Let be the assertion .

Claim 1:
Assume for the sake of contradiction that this is not the case.

Claim 2: The problem is finished if we can prove .
Consider the polynomial:

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:
.
Hence, we have

Subtracting on both sides yields:
or

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 EquationPosted: Fri, 27 Apr 2012 15:47:27 UTC
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Joined: Mon, 29 Dec 2008 17:49:32 UTC
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Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
rdj5933mile5math64 wrote:
outermeasure wrote:
Where does 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, 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 ...

No I'm playing with what happens if 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 is true.

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

My Work:

Let be the assertion .

Claim 1:
Assume for the sake of contradiction that this is not the case.

Claim 2: The problem is finished if we can prove .
Consider the polynomial:

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:
.
Hence, we have

Subtracting on both sides yields:
or

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

Why does mean ?

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 Post subject: Re: Functional EquationPosted: Fri, 27 Apr 2012 16:21:29 UTC
 S.O.S. Oldtimer

Joined: Wed, 4 Apr 2012 03:51:40 UTC
Posts: 145
Location: Hockeytown
outermeasure wrote:
rdj5933mile5math64 wrote:
outermeasure wrote:
Where does 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.

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 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 ...

No I'm playing with what happens if 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 is true.

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

My Work:

Let be the assertion .

Claim 1:
Assume for the sake of contradiction that this is not the case.

Claim 2: The problem is finished if we can prove .
Consider the polynomial:

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:
.
Hence, we have

Subtracting on both sides yields:
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 .
Consider the polynomial:

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:
.
Hence, we have

But, note that

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

EDIT: is defined as I didn't feel like writing throughout the solution.

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 Post subject: Re: Functional EquationPosted: Mon, 14 May 2012 20:05:32 UTC
 S.O.S. Oldtimer

Joined: Wed, 4 Apr 2012 03:51:40 UTC
Posts: 145
Location: Hockeytown
outermeasure wrote:
for all real x,y,z.

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

My Solution:
Spoiler:
Let be the assertion

Claim 1:

Claim 2:
Take for Claim 1.

Claim 3:

Claim 4:

Claim 5:

Claim 6:

Claim 7: for all real x.
From Claims 3 and 4,
and .
Or by subtracting from both sides of each equation,
and .

which is true iff
.

Claim 8:
Observe that by Claim 7 when ,

Now, we have two cases.

Case #1: .
This implies that , so by Claim 5, for all real x.

Case #2:
This implies that by Claim 8.

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

Observe that if in Claim 3, then

But, , and since .

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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