functions having a certain property!

commutative · **Joined:** Sat, 18 Mar 2006 08:42:24 UTC **Posts:** 834

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). \ \ \ \ \ \ \ (*)$

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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?
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Remark 2:

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

where

is the

interval (-1,1)

and

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

Shadow · **Posted:** Wed, 11 Jun 2008 22:57:02 UTC

It seems this might be reformulated as a ring theory problem, since the homomorphism you've exhibited could be reformulated as a ring homomorphism.

Opalg · **Posted:** Thu, 12 Jun 2008 16:37:34 UTC

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).

pankaj · **Posted:** Fri, 27 Jun 2008 16:18:17 UTC

I think that the function is log((1-x)/(1+x))

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functions having a certain property!

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