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 Post subject: functions having a certain property!Posted: Mon, 2 Jun 2008 13:59:08 UTC
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Joined: Sat, 18 Mar 2006 08:42:24 UTC
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Find all possible functions which satisfy the following condition:

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

Remark 1:
We know that for any the function

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

Remark 2:

The above problem is basically to find the endomorphisms of the group where is the

interval and is defined by:

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 Post subject: Posted: Wed, 11 Jun 2008 22:57:02 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
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Location: Austin, TX
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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 Post subject: Re: functions having a certain property!Posted: Thu, 12 Jun 2008 16:37:34 UTC
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Joined: Sat, 7 Jan 2006 18:29:24 UTC
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Location: Leeds, UK
commutative wrote:
Find all possible functions which satisfy the following condition:

This reminds me of the formula for tanh(x+y). In fact, let and , and let Then the functional equation (*) says that

Therefore and so

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 .

But then , from which it follows that g is everywhere differentiable and

That differential equation, with the initial condition g(0)=0, has the solution . Conclusion:

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 to , 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 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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 Post subject: Posted: Fri, 27 Jun 2008 16:18:17 UTC
 S.O.S. Oldtimer

Joined: Tue, 13 Jun 2006 11:57:53 UTC
Posts: 239
Location: new delhi
I think that the function is log((1-x)/(1+x))

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