(05/27/2014, 12:46 PM)MphLee Wrote: Please.. can you give me a soft explaination?

is meant to be a fractional (half) iterate of exp?

How do you arrive at ? 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 is for some it is an automorphism of the addition...

in other words I don't get why we need such automorphism ...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

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 :

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

?

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