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 uniqueness bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 03/29/2009, 09:35 PM 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. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 03/30/2009, 02:02 PM 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. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 03/30/2009, 03:08 PM 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. tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 03/30/2009, 11:40 PM 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 tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 03/30/2009, 11:49 PM 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 ?? tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 03/30/2009, 11:50 PM 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* tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 04/02/2009, 11:43 PM 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 bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 04/03/2009, 06:43 AM Before uniqueness, one should show however existence, because a uniqueness criterion makes no sense if there are no functions satisfying this criterion ... tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 04/03/2009, 12:22 PM 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 « Next Oldest | Next Newest »

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