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 the distributive property tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 05/27/2014, 10:49 PM (05/27/2014, 12:46 PM)MphLee Wrote: Please.. can you give me a soft explaination? $f$ is meant to be a fractional (half) iterate of exp? How do you arrive at $f ( f^{\circ -1}(a) + f^{\circ -1}(b) ) = a + b$? is the starting Hypotesis? After that i can't follow the differentiation argument (my fault)...seems me that you are proving that whenerever f is an addition automorphism it is linear but is obvious that if $f(x)$ is $Cx$ for some $C$ it is an automorphism of the addition... in other words I don't get why we need such automorphism $f$...and how it helps us. f is a function we look for that is invertible and C^3 or higher. SO we solve for f such that $f ( f^{\circ -1}(a) + f^{\circ -1}(b) ) = a + b$ So this f is very general hence I proved more than I had too. But sometimes a generalization is easier to prove ! Its really essential that you understand why this equation occurs. Perhaps an example : lets say we switch base e with base B. then B^x = exp(ln B * ln (x)) and B_ln(x) = ln(x)/ln B. As said before we wrote : exp(f(x)) and f^[-1]ln(x). Hence we get f(x) = ln B * x and f^[-1](x) = x / ln B. IF we plug in we get : exp(ln(a) + ln(b)) = B^(ln(a)/ln B + ln(b)/ln B) = exp( f ( f^[-1](ln(a)) + f^[-1](ln(b)) ) Now its Obvious ( take ln on both sides if not yet Obvious for you ) to see that we require f to satisfy : $f ( f^{\circ -1}(a) + f^{\circ -1}(b) ) = a + b$ For this particular example its clear that our f as defined above works ... BUT if we solve this equation in a more general setting we get all valid f's and our proof. Hence we try to solve for ALL f satisfying the equation $f ( f^{\circ -1}(a) + f^{\circ -1}(b) ) = a + b$? such that f is invertible and C^3 or better ( such as C^oo or holomorphic ). Of course f(x) = C x works fine ! But we wanted to prove ONLY f(x) = C x works !! We continue to do so by using the property of differentiability. And then we end up with f(x) must be of the form C x. Hence we arrive at f ( f^[-1](a) + f^[-1](b) ) = a + b now differentiate with respect to a : f ' (f^[-1](a) + f^[-1](b)) / f ' (f^[-1](a)) = 1 ( use the chain rule ! ) substitute X = f(a) , Y = f^[-1](b) : f ' ( a + Y) = f ' (a)^2 Now rewrite f ' = G and differentiate with respect to Y : G ' (Y + a) = 0 Hence G is constant and thus f = lineair. Since f(0) = 0 we conclude f(x) = C x. thus it only holds for r = 0,1 QED So we actually proved the more general : exp( f ( f^[-1](ln a) + f^[-1](ln b) ) )= exp(ln a + ln b) ( = ab ) only holds for f = C x. let a" = ln^[Q](a) , b" = ln^[Q](b) then we can write : exp^[Q]( f ( f^[-1](ln^[Q] a) + f^[-1](ln^[Q] b) ) )= exp^[Q](ln^[Q] a + ln^[Q] b) only holds for all f = C x. QED Possibly not the most elegant proof. On the other hand quite direct and general. Im not into (homo)morphisms for uncountable sets. I rather use that for group theory ,set theory or sometimes topology. I hesitated to try and use some Cauchy tricks such as his functional equation , but it seems a bit unnatural. Still looking for a more elegant proof... But this proof should convince you. regards tommy1729 « Next Oldest | Next Newest »

 Messages In This Thread the distributive property - by tommy1729 - 09/25/2010, 12:05 AM RE: the distributive property - by tommy1729 - 09/28/2010, 12:22 PM RE: the distributive property - by tommy1729 - 10/13/2010, 11:07 PM RE: the distributive property - by tommy1729 - 05/23/2014, 11:12 PM RE: the distributive property - by MphLee - 05/24/2014, 09:46 AM RE: the distributive property - by tommy1729 - 05/25/2014, 12:16 AM RE: the distributive property - by MphLee - 05/25/2014, 09:35 AM RE: the distributive property - by tommy1729 - 05/26/2014, 11:37 PM RE: the distributive property - by MphLee - 05/27/2014, 07:43 AM RE: the distributive property - by tommy1729 - 05/27/2014, 12:23 PM RE: the distributive property - by MphLee - 05/27/2014, 12:46 PM RE: the distributive property - by tommy1729 - 05/27/2014, 10:49 PM RE: the distributive property - by andydude - 05/27/2014, 07:45 PM RE: the distributive property - by MphLee - 05/27/2014, 08:22 PM RE: the distributive property - by tommy1729 - 05/27/2014, 10:55 PM RE: the distributive property - by MphLee - 06/06/2014, 09:56 AM RE: the distributive property - by tommy1729 - 06/06/2014, 11:56 AM RE: the distributive property - by MphLee - 06/06/2014, 12:09 PM

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