07/25/2021, 11:20 PM
i would like to clarify a bit on why at most one branch jump.
apart from log(1/x) = - log(x) and other sign arguments or " small number " arguments ( who only give at most 2 pi branch jump or 2 pi abs change anyways - as desired ! - * which is essentially the same !! * ) there is also this ;
consider a sequence of continu ( on the complex plane , thus 2D continu ) functions f1 , f2 , ... , then jumping 2 branches would be impossible !
If we would jump 2* branches at a point s , ( *with respect to its neighbourhood ) then we must have a " created " discontinu point.
IT would mean manually creating a disconu jump of (at least in the subcomputation ) 2 pi i.
this is because this point s is no longer connected by an angle ( at most 2 pi i ) to its neighbourhood.
this is similar to ln(exp(s)) , you jump at most 1 branch for the ln with respect the neighbourhood , otherwise it is not a smooth path on the riemann surface and hence not analytic.
In other words you cannot naturally get a sudden jump of 2 branches , and you are not " trying to interpret it analytically , rather deliberately trying to suggest it is not analytic ".
since if for FINITE n : f_n(s) is analytic , so is f_(n+1)(s+1) since it is just a log of an analytic function that is not 0 anywhere.
We cannot use induction here to imply it for n = oo but we can say that for n+1 we also get a continu function(*).
( A minor detail is I LEFT OUT THE CONDITION THAT THE DERIVATIVES ARE NOT GOING TO oo , but that is easy to overcome ! ( in particular for finite n and n + 1 )
Notice the log is a very smooth function , not having many bumps or anything.
regards
tommy1729
apart from log(1/x) = - log(x) and other sign arguments or " small number " arguments ( who only give at most 2 pi branch jump or 2 pi abs change anyways - as desired ! - * which is essentially the same !! * ) there is also this ;
consider a sequence of continu ( on the complex plane , thus 2D continu ) functions f1 , f2 , ... , then jumping 2 branches would be impossible !
If we would jump 2* branches at a point s , ( *with respect to its neighbourhood ) then we must have a " created " discontinu point.
IT would mean manually creating a disconu jump of (at least in the subcomputation ) 2 pi i.
this is because this point s is no longer connected by an angle ( at most 2 pi i ) to its neighbourhood.
this is similar to ln(exp(s)) , you jump at most 1 branch for the ln with respect the neighbourhood , otherwise it is not a smooth path on the riemann surface and hence not analytic.
In other words you cannot naturally get a sudden jump of 2 branches , and you are not " trying to interpret it analytically , rather deliberately trying to suggest it is not analytic ".
since if for FINITE n : f_n(s) is analytic , so is f_(n+1)(s+1) since it is just a log of an analytic function that is not 0 anywhere.
We cannot use induction here to imply it for n = oo but we can say that for n+1 we also get a continu function(*).
( A minor detail is I LEFT OUT THE CONDITION THAT THE DERIVATIVES ARE NOT GOING TO oo , but that is easy to overcome ! ( in particular for finite n and n + 1 )
Notice the log is a very smooth function , not having many bumps or anything.
regards
tommy1729