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+--- Thread: uniqueness (/showthread.php?tid=257)
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RE: uniqueness - bo198214 - 03/29/2009
Ansus Wrote:To be simple we have two sets of methods: those applicable for lower bases (and they are equal) and those applicable for higher bases (I suppose they also equal with each other). Probably we would not need a proof for equivalence of these two sets of methods because their areas of applicability do not overlap.
They do overlap, Matrix-Power and Intuitive Abel are applicable to all bases \( b>1 \).
And even Cauchy-Integral we saw that there are similar methods for for both base ranges.
Quote:P.S. Where I could learn about matrix powers method?
See the pdf Gottfried posted here.
RE: uniqueness - bo198214 - 03/30/2009
Ansus Wrote:Does not it use Carleman matrices whic also used for partial iteration formula?
What is partial iteration?
Indeed the matrix power method coincides with regular iteration if applied to a fixed point.
RE: uniqueness - bo198214 - 03/30/2009
Ansus Wrote:You derived your Newton formula using Carleman matrices, and they also use Carleman matrices of arbitrary iteration of exp function.
thats exactly what i said: at fixed points both methods coincide.
RE: uniqueness - tommy1729 - 03/30/2009
bo198214 Wrote:tommy1729 Wrote:f(x) = (x + q(x)) ^2 = x^2 + 2q(x)x + q(x)^2
f(f(x)) = ( x^2 + 2q(x)x + q(x)^2 + q(x^2 + 2q(x)x + q(x)^2) ) ^2
q must appear inside q.
so ??
thats no proof or disproof of anything ?
regards
tommy1729
RE: uniqueness - tommy1729 - 03/30/2009
tommy1729 Wrote:here are the equations that make half-iterate of exp(x) unique :
(under condition f(x) maps reals to reals and f(x) > x )
exp(x)
= f(f(x))
= D f(f(x)) = f ' (f(x)) * f ' (x)
= D^2 f(f(x)) = f '' (f(x)) * f ' (x)^2 + f ' (f(x)) * f '' (x)
= D^3 f(f(x)) = D^4 f(f(x))
regards
tommy1729
is there a solution where f ' (x) is not strictly rising but f(x) is Coo ??
RE: uniqueness - tommy1729 - 03/30/2009
tommy1729 Wrote:tommy1729 Wrote:here are the equations that make half-iterate of exp(x) unique :
(under condition f(x) maps reals to reals and f(x) > x )
exp(x)
= f(f(x))
= D f(f(x)) = f ' (f(x)) * f ' (x)
= D^2 f(f(x)) = f '' (f(x)) * f ' (x)^2 + f ' (f(x)) * f '' (x)
= D^3 f(f(x)) = D^4 f(f(x))
regards
tommy1729
is there a solution where f ' ' (x) is not strictly rising but f(x) is Coo ??
*corrected*
NEW uniqueness - tommy1729 - 04/02/2009
tommy1729 Wrote:tommy1729 Wrote:tommy1729 Wrote:here are the equations that make half-iterate of exp(x) unique :
(under condition f(x) maps reals to reals and f(x) > x )
exp(x)
= f(f(x))
= D f(f(x)) = f ' (f(x)) * f ' (x)
= D^2 f(f(x)) = f '' (f(x)) * f ' (x)^2 + f ' (f(x)) * f '' (x)
= D^3 f(f(x)) = D^4 f(f(x))
regards
tommy1729
is there a solution where f ' ' (x) is not strictly rising but f(x) is Coo ??
*corrected*
ok.
i see now.
i was a bit confused ...
the new uniqueness conditions , i think ,
(under condition f(x) maps reals to reals and f(x) > x )
f ' (0) > 0
f '' (0) > 0
f ''' (0) > 0
...
and f(x) is Coo
regards
tommy1729
RE: NEW uniqueness - bo198214 - 04/03/2009
Before uniqueness, one should show however existence, because a uniqueness criterion makes no sense if there are no functions satisfying this criterion ...
RE: NEW uniqueness - tommy1729 - 04/03/2009
bo198214 Wrote:Before uniqueness, one should show however existence, because a uniqueness criterion makes no sense if there are no functions satisfying this criterion ...
isnt that already done ?
by kneser ?
by robbins ?
doesnt robbins solution satisfy my uniqueness criterions ?
regards
tommy1729