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not easy functional system
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Diogene
Yang-Mills Theory
Yang-Mills Theory

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#1
not easy functional system
C.B.D
Find the functions f,g: R^+\rightarrow R^+( positive real numbers) such that :
f(x)=\frac 1x + \frac 1{f(g(x))},\ g(x)=\frac 1x + \frac 1{g(f(x))}
Cool
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PostPosted: Sun Nov 08, 2009 3:13 pm  Back to top 
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pco
Navier-Stokes Equations
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#2
Re: not easy functional system
C.B.D
Diogene wrote:
Find the functions f,g: R^ + \rightarrow R^ +( positive real numbers) such that :
f(x) = \frac 1x + \frac 1{f(g(x))},\ g(x) = \frac 1x + \frac 1{g(f(x))}
Cool


I think this problem is very difficult.
Beside the trivial solution f(x) = g(x) = \frac 2x, there exist infinitely many solutions.

========================== example of solutions =====================
For example, considering the cases where f(x) = g(x) and f(x) = \frac 1x + \frac 1{f(f(x))}, we have solutions like :

Let p > 1\in\mathbb N
Let A = \{x = 2^u(p + 1)^{\frac v2}p^{\frac w2}, \forall u,v,w\in\mathbb Z\}
Let B = \mathbb R^ + \backslash A

Consider the relation \sim defined in B\times B by x\sim y \iff either xy = (p + 1)p^n, either \frac xy = p^n for some n\in\mathbb Z
This clearly is an equivalence relation.
Let then h(x): B\to B any choice function associating to x\in B a representant for it's equivalence class.

Let x\in B. Since x\sim h(x) :
either \exists n\in\mathbb Z such that xh(x) = (p + 1)p^n
either \exists n\in\mathbb Z such that \frac {h(x)}x = p^n
and , for a given x, it's impossible to have two different integers n_1 and n_2 verifying such equalities :
xh(x) = (p + 1)p^{n_1} = (p + 1)p^{n_2} would imply n_1 = n_2

\frac {h(x)}x = p^{n_1} = p^{n_2} would imply n_1 = n_2

xh(x) = (p + 1)p^{n_1} and \frac {h(x)}x = p^{n_2} would imply x^2 = (p + 1)p^{n_1 - n_2} and so x\in A and so x\notin B

So there is a well defined function n(x): B\to\mathbb Z such that either xh(x) = (p + 1)p^{n(x)}, either \frac {h(x)}x = p^{n(x)}

Then we can define f(x) = g(x) as :

\forall x\in A : f(x) = \frac 2x

\forall x\in B such that xh(x) = (p + 1)p^{n(x)} : f(x) = h(x)p^{ - n(x) - 1} =\frac{p+1}{px}

\forall x\in B such that \frac {h(x)}x = p^{n(x)} : f(x) = \frac {(p + 1)p^{n(x)}}{h(x)} =\frac{p+1}x


verification that this function fits the requirements

1) x\in A
Then f(x) = \frac 2x and f(x)\in A. So f(f(x)) = \frac 2{f(x)} = x and \frac 1x + \frac 1{f(f(x))} = \frac 1x + \frac 1x = f(x)
And so f(x) = \frac 1x + \frac 1{f(f(x))}

2) x\in B such that xh(x) = (p + 1)p^{n(x)}
Then f(x)=h(x)p^{ - n(x) - 1} and so f(x)\in B and f(x)\sim h(x) and so h(f(x)) = h(x)

So f(x) = h(f(x))p^{ - n(x) - 1} and \frac {h(f(x))}{f(x)} = p^{n(x) + 1} and so n(f(x)) = n(x) + 1

So \frac {h(f(x))}{f(x)} = p^{n(f(x))} and f(f(x)) = \frac {(p + 1)p^{n(f(x))}}{h(f(x))} = \frac {(p + 1)p^{n(x) + 1}}h(x)}

Then \frac 1x + \frac 1{f(f(x))} = \frac {h(x)}{(p + 1)p^{n(x)}} + \frac {h(x)}{(p + 1)p^{n(x) + 1}} = \frac {(p + 1)h(x)}{(p + 1)p^{n(x) + 1}} = \frac {h(x)}{p^{n(x) + 1}} = f(x)

And so f(x) = \frac 1x + \frac 1{f(f(x))}

3) x\in B such that \frac {h(x)}x = p^{n(x)}
Then f(x) = \frac {(p + 1)p^{n(x)}}{h(x)} and so f(x)\in B and f(x)h(x) = (p + 1)p^{n(x)} so f(x)\sim h(x) and so h(f(x)) = h(x)

So f(x)h(f(x)) = (p + 1)p^{n(x)} and n(f(x)) = n(x)

Then f(x)h(f(x)) = (p + 1)p^{n(f(x))} and so f(f(x)) = h(f(x))p^{ - n(f(x)) - 1} = h(x)p^{ - n(x) - 1}

So \frac 1x + \frac 1{f(f(x))} = \frac {p^{n(x)}}{h(x)} + \frac {p^{n(x) + 1}}{h(x)} = \frac {(p + 1)p^{n(x)}}{h(x)} = f(x)

And so f(x) = \frac 1x + \frac 1{f(f(x))}


Q.E.D





========================== end of example ================================

I would be very interested, Diogene, on informations about the origin of the problem (it's a crazy problem for olympiad, or olympiad preparation, in my opinion).

And I would be very interested in your own general solution.

Thanks in advance.
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Patrick

PostPosted: Mon Nov 09, 2009 10:21 am  Back to top 
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Diogene
Yang-Mills Theory
Yang-Mills Theory

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#3
Hello pco,
I was inspired by http://www.artofproblemsolving.com/Forum/viewtopic.php?t=307928
The functional system f(x) = \frac ax + \frac 1{f(g(x))},\ g(x) = \frac ax + \frac 1{g(f(x))} has only the solution f(x) = g(x) = \frac {a + 1}x if a > 1
If a \leq 1 the probleme is very hard, and I am not much more advanced than you .

Cool
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D.. to A.. : "Stand out of my light."

PostPosted: Mon Nov 09, 2009 7:56 pm  Back to top 
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