(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