double functional equation , continuum sum and analytic continuation

[tommy1729]

...

So in this case * if you accept it *

we get

for Re(z) < 1

f(z^2) = f(z) - z

f(z^4) = f(z) - z - z^2

and for Re(z) > 1

f(z^2) = f(z) - 1/z

f(z^4) = f(z) - 1/z - 1/z^2

and for all z not on the unit circle

f(z) = f(1/z)

well assuming all paradoxes have now been removed.

regards

tommy1729

notice that 

- 1/z - 1/z^2 - 1/ z^3 = -z - z^2 - z^3

has only got solutions on the unit circle !!

However

1/z + 1/z^2 + 1/z^4 = z + z^2 + z^4

has solutions off the unit circle !!

This is both an argument for and against the reflection

f(1/z) = f(z)


notice that x + x^2 + x^3 + ... occured in the zeta topics !



regards

tommy1729