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PostPosted: Mon, 2 Jun 2008 13:59:08 UTC 
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Find all possible functions f:(-1,1) \longrightarrow (-1,1) which satisfy the following condition:

$ f\left(\frac{x+y}{1+xy}\right)=\frac{f(x)+f(y)}{1+f(x)f(y)}, \ \ \ \forall x,y \in (-1,1). \ \ \ \ \ \ \ (*)

.......................................................................................................................................................

Remark 1:
We know that for any $ \alpha \in \mathbb{R}, the function $ \ f_{\alpha}(x)=\frac{(1+x)^{\alpha} - (1-x)^{\alpha}}{(1+x)^{\alpha} + (1-x)^{\alpha}}, \ \forall x \in (-1,1),

satisfies (*). The question is if these are all possible functions?
.......................................................................................................................................................

Remark 2:

The above problem is basically to find the endomorphisms of the group (G, *), where G is the

interval (-1,1) and (*) is defined by: $ x*y=\frac{x+y}{1+xy}, \ \ \forall x,y \in (-1,1).


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PostPosted: Wed, 11 Jun 2008 22:57:02 UTC 
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It seems this might be reformulated as a ring theory problem, since the homomorphism you've exhibited could be reformulated as a ring homomorphism.

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PostPosted: Thu, 12 Jun 2008 16:37:34 UTC 
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commutative wrote:
Find all possible functions f:(-1,1) \longrightarrow (-1,1) which satisfy the following condition:

$ f\left(\frac{x+y}{1+xy}\right)=\frac{f(x)+f(y)}{1+f(x)f(y)}, \ \ \ \forall x,y \in (-1,1). \ \ \ \ \ \ \ (*)

This reminds me of the formula for tanh(x+y). In fact, let x = \tanh s and y = \tanh t, and let g(s) = f(\tanh s). Then the functional equation (*) says that g(s+t) = \frac{g(s)+g(t)}{1+g(s)g(t)}.

Therefore g(s+t)-g(s) = \frac{g(t)(1-g(s)^2)}{1+g(s)g(t)} and so \frac{g(s+t)-g(s)}t = \frac{g(t)}t\frac{1-g(s)^2}{1+g(s)g(t)}.

It is clear from (*) (with x=y=0) that f(0)=0, and hence g(0)=0. Now suppose that f (and hence g) is differentiable at 0, with f'(0)=α (and hence by the chain rule g'(0)=α too). Therefore \lim_{t\to0}\frac{g(t)}t=\alpha.

But then \lim_{t\to0}\frac{g(s+t)-g(s)}t = \lim_{t\to0}\frac{g(t)}t\lim_{t\to0}\frac{1-g(s)^2}{1+g(s)g(t)}, from which it follows that g is everywhere differentiable and g'(s) = \alpha(1-g(s)^2).

That differential equation, with the initial condition g(0)=0, has the solution g(s) = \tanh(\alpha s). Conclusion: f(x) = \tanh(\alpha\tanh^{-1}x).

So, provided that f is differentiable at 0, there is indeed just a one-parameter family of solutions. I haven't thought about how to get from \tanh(\alpha\tanh^{-1}x) to \frac{(1+x)^{\alpha} - (1-x)^{\alpha}}{(1+x)^{\alpha} + (1-x)^{\alpha}}, but that ought to be "just algebra".

Edit. I could have done that much more simply. The map τ:x→tanh(x) is an isomorphism from the additive group (R,+) of reals to the group (G,*) (in commutative's Remark 2). So if f is an endomorphism of (G,*) then \tau^{-1}f\tau is an endomorphism of (R,+). But it is well known that if such an endomorphism is continuous (or differentiable at 0) then it is of the form x→αx for some real number α. If the map is not continuous then you have to think of it as a real-valued linear functional on R regarded as a vector space over the rationals. In that case, it can be defined arbitrarily on each element of a Hamel basis of the Q-vector space R (provided you're happy to use the Axiom of Choice).


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PostPosted: Fri, 27 Jun 2008 16:18:17 UTC 
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I think that the function is log((1-x)/(1+x))


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