the distributive property MphLee Long Time Fellow Posts: 321 Threads: 25 Joined: May 2013 05/27/2014, 12:46 PM (This post was last modified: 05/27/2014, 03:58 PM by MphLee.) 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. MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 05/27/2014, 07:45 PM @MphLee Hyperoperations, in the general sense, are any sequence of binary operations that includes addition and multiplication. The commutative hyperoperations satisfy this property because $\exp^0(\ln^0(a) + \ln^0(b)) = a + b$ and $\exp^1(\ln^1(a) + \ln^1(b)) = e^{\ln(a) + \ln(b)} = e^{\ln(a)}e^{\ln(b)} = a \times b$. That formula is the starting point, it is the definition of commutative hyperoperations. The fact that it contains addition and multiplication can be discussed and proved from the definition. MphLee Long Time Fellow Posts: 321 Threads: 25 Joined: May 2013 05/27/2014, 08:22 PM (05/27/2014, 07:45 PM)andydude Wrote: @MphLee Hyperoperations, in the general sense, are any sequence of binary operations that includes addition and multiplication. The commutative hyperoperations satisfy this property because $\exp^0(\ln^0(a) + \ln^0(b)) = a + b$ and $\exp^1(\ln^1(a) + \ln^1(b)) = e^{\ln(a) + \ln(b)} = e^{\ln(a)}e^{\ln(b)} = a \times b$. That formula is the starting point, it is the definition of commutative hyperoperations. The fact that it contains addition and multiplication can be discussed and proved from the definition. I'm aware of this, of the Bennet's definition and I'm aware of the Rubtsov's generalizzation using different bases values (reflexive operations sequence). I'm even aware that the term Hyperoperations usually means (can be formalized as) an indexed family of binary operations $\{*_i\}_{i \in I}$ whith addition, multiplication and exponentiation belonging to the image of the indexed family (the image of the family is defined to be the image of the set of indexes- set of ranks- via the indicization function). This definition is the one I found on Wikipedia and is very smart even if it cuts the Commutative hyperoperations out of the game (Maybe we can make a weaker concept of Hyperoperations Family without the exponentiation requirement, I would call them Weak Hyperoperations Families)... Anyways I'm very courious...I was not able to find references about this terminology and I did not even find who introduced this formal definition. MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ tommy1729 Ultimate Fellow Posts: 1,668 Threads: 368 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 tommy1729 Ultimate Fellow Posts: 1,668 Threads: 368 Joined: Feb 2009 05/27/2014, 10:55 PM (05/27/2014, 07:45 PM)andydude Wrote: @MphLee Hyperoperations, in the general sense, are any sequence of binary operations that includes addition and multiplication. The commutative hyperoperations satisfy this property because $\exp^0(\ln^0(a) + \ln^0(b)) = a + b$ and $\exp^1(\ln^1(a) + \ln^1(b)) = e^{\ln(a) + \ln(b)} = e^{\ln(a)}e^{\ln(b)} = a \times b$. That formula is the starting point, it is the definition of commutative hyperoperations. The fact that it contains addition and multiplication can be discussed and proved from the definition. Andy ! Im honored too see your return at my thread ! However I think your post will not remove MphLee's confusion. Hope Im more Lucky with my own post. Maybe you can improve the proof. I dont like it too much now. regards tommy1729 MphLee Long Time Fellow Posts: 321 Threads: 25 Joined: May 2013 06/06/2014, 09:56 AM (05/27/2014, 10:55 PM)tommy1729 Wrote: Andy ! Im honored too see your return at my thread ! However I think your post will not remove MphLee's confusion.In fact because he wn't talking about the same thing but about the term Hyperoperation family Quote:Hope Im more Lucky with my own post. Maybe you can improve the proof. I dont like it too much now. regards tommy1729 I think I've got it, You were playing with the change-of-base formula right? Thanks, now is more clear I guess. MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ tommy1729 Ultimate Fellow Posts: 1,668 Threads: 368 Joined: Feb 2009 06/06/2014, 11:56 AM I did not consider the change of base. Maybe its related , maybe not. regards tommy1729 MphLee Long Time Fellow Posts: 321 Threads: 25 Joined: May 2013 06/06/2014, 12:09 PM (06/06/2014, 11:56 AM)tommy1729 Wrote: I did not consider the change of base. Maybe its related , maybe not. regards tommy1729 That confuses me again ! xD MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ « Next Oldest | Next Newest »