Divergent Series and Analytical Continuation (LONG post)

[tommy1729]

Alright guys time to slighty debunk most of this 

f(1/x) = f(x) + C

does not make sense since

iterating 2 times gives 

f(x) = f(x) + 2 C

so C must be 0.

Although if the sum of coeff is indeed C = 0 it works.

( caleb was ahead of me when describing the coef sum , i wanted to post it )

but then it need to be consistant with plug in.
and it probably is.

Notice if C = 0 then f(1) is defined and continu just like I mentioned earlier.

But this is not super interesting.

it gives f(x) = f(x) and f(1/x) = f(1/x).


since 1/x is 2 cyclic our other function should also be 2 cylclic.

So more interesting is

f(1/x) = 1 - f(x).

since 1-x iterated 2 times is x.

this is equivalent to

f(1/x) + f(x) = 1.

another way is

f(1/x) + f(x) = g(x + 1/x).

And 

f(x) - f(1/x) = log(x).

or

f(x) + f(1/x) = log(x + 1/x) = log(x) + log(1+1/x^2)

it is clear that 1/x or conj(1/x) is key here together with the 2 cycle iterations.


***

I want to point out that not all of them have the same natural boundary or the unit boundary , it is just a way of thinking.

In particular if 

f(1/x) = g( f(x) ) 

and g maps cross the unit circle , we have analytic continuation there by g , so the unit circle is not the boundary.


see also the related topic :



regards

tommy1729