uniqueness
#4
tommy1729 Wrote:NO

of course not.

for example : the first case :

exp(x) = f ' (f(x)) * f ' (x)

now consider a solution that satisfies f(f(x)) = exp(x)

and let assume f ' (x) = 0 has a finite real zero at x = r1.

thus f ' (r1) = 0

exp( r1 ) = 0 * f ' (f(r1)) => ??????

you see , contradiction.

That just shows that there is no halfiterate with f'(x)=0 for some x.
The equation \( \exp(x)=f'(f(x))*f'(x) \) is just a consequence of \( \exp(x)=f(f(x)) \), so it is valid for *every* halfiterate f (which of course must be differentiable).
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Messages In This Thread
uniqueness - by tommy1729 - 03/24/2009, 02:50 AM
RE: uniqueness - by bo198214 - 03/24/2009, 09:39 AM
RE: uniqueness - by tommy1729 - 03/26/2009, 07:34 PM
RE: uniqueness - by bo198214 - 03/26/2009, 08:11 PM
RE: uniqueness - by tommy1729 - 03/27/2009, 12:57 PM
RE: uniqueness - by bo198214 - 03/29/2009, 07:50 PM
RE: uniqueness - by tommy1729 - 03/29/2009, 08:51 PM
RE: uniqueness - by bo198214 - 03/29/2009, 09:06 PM
RE: uniqueness - by tommy1729 - 03/30/2009, 11:40 PM
RE: uniqueness - by tommy1729 - 03/30/2009, 11:49 PM
RE: uniqueness - by tommy1729 - 03/30/2009, 11:50 PM
NEW uniqueness - by tommy1729 - 04/02/2009, 11:43 PM
RE: NEW uniqueness - by bo198214 - 04/03/2009, 06:43 AM
RE: NEW uniqueness - by tommy1729 - 04/03/2009, 12:22 PM

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