Another way or at least idea is this
Reflection idea.
Since f(x^2) - f(x) + x = 0 for all x
,it is possible to extend this idea to all z not on the unit circle.
It even makes that equation differentiable.
Lets say
f(x^2) - f(x) = - 2.
then we try to find x.
A solution is garanteed to exist by the fact that it is analytic when not on the unit circle.
And now we compute f(x) = y
Then we simply set
f(x) = y = f(2)
f(x^2) = f(2^2) = f(4) = - 2 + y.
If this is consistant with the previous post is another matter.
also
Lets say
f(x^2) - f(x) = - 2.
then we find x.
might be an issue because it might not give a unique x , so more care is needed.
deciding which x might use the stuff from the other post though.
It feels a bit like going from the reals to the complex.
just like x^2 = -1 had no solution before the complex numbers,
f(x^2) - f(x) = -2
has no evident solution in the complex.
Just like finding the real numbers as solutions to the cubic equation required complex numbers ,
maybe finding complex solutions to this
f(z^2) - f(z) = -2
requires new numbers.
Just an idea.
Ok in this case the analogue is a bit weird :
afterall f(z^2) - f(z) = - id(z) when considered literal and f(4) = f(2) = oo ( diverges )
so
oo - oo = -2 ??
maybe with limits ...
Maybe a better analogue is this
sqrt(2)^sqrt(2)^... = 2.
x^(1/x) = sqrt(2)
x= 2
however we have on the other side
x = 4 works as well !!
On the other hand solving f(x^2) - f(x) might not be easier than computing f(x)
so the benefit is not proven.
***
I also suggested another reflection idea in the past and recently again.
The reflection formula is simply based on the shape of the boundary :
f( z ) = f ( conjugate (1/z) )
or
f( z ) = f ( 1 / z )
This might give the conditions
f(z^2) - f(z) = - z
f(z^4) = f(z) - z - z^2
f(z^(-1)) = f(z)
a triple functional equation ??
but does this not give contradictions somewhere ???
LETS SEE
if
f(z) = f(1/z)
f( 1/z^2 ) = f(z^2)
so
f(z) = 1/z + 1/z^2 + 1/z^4 + 1/z^8 + ...
for |z| > 1
f(z^2) = f(z) - z
should hold
but now for
f(z^2) = f(z) - 1/z !!!
if |z| > 1.
These lead to conditional functional equations rather than new numbers !
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
Reflection idea.
Since f(x^2) - f(x) + x = 0 for all x
,it is possible to extend this idea to all z not on the unit circle.
It even makes that equation differentiable.
Lets say
f(x^2) - f(x) = - 2.
then we try to find x.
A solution is garanteed to exist by the fact that it is analytic when not on the unit circle.
And now we compute f(x) = y
Then we simply set
f(x) = y = f(2)
f(x^2) = f(2^2) = f(4) = - 2 + y.
If this is consistant with the previous post is another matter.
also
Lets say
f(x^2) - f(x) = - 2.
then we find x.
might be an issue because it might not give a unique x , so more care is needed.
deciding which x might use the stuff from the other post though.
It feels a bit like going from the reals to the complex.
just like x^2 = -1 had no solution before the complex numbers,
f(x^2) - f(x) = -2
has no evident solution in the complex.
Just like finding the real numbers as solutions to the cubic equation required complex numbers ,
maybe finding complex solutions to this
f(z^2) - f(z) = -2
requires new numbers.
Just an idea.
Ok in this case the analogue is a bit weird :
afterall f(z^2) - f(z) = - id(z) when considered literal and f(4) = f(2) = oo ( diverges )
so
oo - oo = -2 ??
maybe with limits ...
Maybe a better analogue is this
sqrt(2)^sqrt(2)^... = 2.
x^(1/x) = sqrt(2)
x= 2
however we have on the other side
x = 4 works as well !!
On the other hand solving f(x^2) - f(x) might not be easier than computing f(x)
so the benefit is not proven.
***
I also suggested another reflection idea in the past and recently again.
The reflection formula is simply based on the shape of the boundary :
f( z ) = f ( conjugate (1/z) )
or
f( z ) = f ( 1 / z )
This might give the conditions
f(z^2) - f(z) = - z
f(z^4) = f(z) - z - z^2
f(z^(-1)) = f(z)
a triple functional equation ??
but does this not give contradictions somewhere ???
LETS SEE
if
f(z) = f(1/z)
f( 1/z^2 ) = f(z^2)
so
f(z) = 1/z + 1/z^2 + 1/z^4 + 1/z^8 + ...
for |z| > 1
f(z^2) = f(z) - z
should hold
but now for
f(z^2) = f(z) - 1/z !!!
if |z| > 1.
These lead to conditional functional equations rather than new numbers !
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