Ok I am a bit confused. So here comes a question,
Consider a maclaurin series for $f(z)$
$$f(z) = \sum_{n=0}^{\infty} f_n z^n$$
where $f(z)$ has a radius of exactly $1$.
$f(z)$ may or may not have a natural boundary at $|z| = 1$.
The next step - especially for lacunary series - :
Use a Lambert series expansion or one of its variants :
$$f(z) = \sum_{n=0}^{\infty} f_n z^n = \sum_{n=0}^{\infty} l_n \frac{z^n}{1-z^n}$$
and assuming it converges everywhere apart from the natural boundary or singularities/poles we have done a continuation !
However we need to be careful, even with convergeance.
If there is no natural boundary, is the result correct ; is it consistent with analytic continuation ?
Consider the functional equation
$$f(z) = f(1/z) $$
for $z$ not on the unit circle.
We can force this !
$$f(z) = \sum_{n=0}^{\infty} f_n z^n = \sum_{n=0}^{\infty} c_n \frac{z^n}{1+z^{2n}}$$
Now we arrived at the functional equation $f(z) = f(1/z)$. Even if there was no natural boundary !
- Another example :
Lambert series like this can lead to non standard continuations. The book generalized analytic continuation by Ross, and Shapiro explores these ideas and gives an example in page 1 of chapter 1 of a lambert series that is $\frac{x}{1-x}$ inside the unit disk and $-\frac{x}{1-x}$ outside the unit disk reproduced below:
$$ \frac{x}{1-x^2} + \frac{x^2}{1-x^4} + ... + \frac{x^{2^{n-1}}}{1-x^{2^n}} + ... $$
But even then I think there is a space of ideas here worth exploring. I also have examples (In particular cubic theta series via sieving) where the standard continuation seems like the only possible continuation via lambert series.
Also interesting to note is perhaps this here :
Analytic continuation of a lambert series $f(z)=\sum_{n=1}^\infty \frac{2^n z^n}{1-(z/2)^n}$?
Some go a bit further
use polynomial division : $$\exp(x) = "(1+x)(1+ ..x^2 + ..x^3 + ..x^4) = (1+x)(1+..x^2)(1+..x^3 + ..x^4)= ..."$$
etcetera to arrive at "a product expansion" for the exp valid within the unit circle.
Or they do it with $f(z)$. And then taking the derivative of log on both sides
$$\frac{f'(z)}{f(z)} = \sum \frac{g(n) n x^{n-1}}{1 + g(n) x^n} $$
And finally using that to compute outside radius $1$.
So far the intro.
Lets generalize:
We want (for our continuation) to put
$$f(z) = \sum_{n=0}^{\infty} f_n z^n = \sum_{n=0}^{\infty} l_n \frac{z^n}{T_k(z^n)}$$
Where $T_k(z)$ is a polynomial with roots only on the unit circle of degree $k$.
The cyclotomic polynomials $C_k(z)$ are probably the most famous, but notice the case $C_2(z)$ is not well suited due to the forced functional equation (see previous). (Btw Cauchy officially came first with that forced functional equation or so I heard and read)
The case $C_1(z)$ is just the lambert series in essense.
It seems the cyclotomic polynomials of odd prime degree are ideal.
For computation, we use the generalized moebius inversion ofcourse.
So I was wondering about
$$f(z) = \sum_{n=0}^{\infty} f_n z^n = \sum_{n=0}^{\infty} w_n \frac{z^n}{1+z^n+z^{2n}+z^{3n}+z^{4n}}$$
$$f(z) = \sum_{n=0}^{\infty} f_n z^n = \sum_{n=0}^{\infty} u_n \frac{z^n}{1+z^n+z^{2n}}$$
And now ofcourse we want both to hold everywhere for $|z| \neq 1$ and
$$\sum_{n=0}^{\infty} w_n \frac{z^n}{1+z^n+z^{2n}+z^{3n}+z^{4n}} = \sum_{n=0}^{\infty} u_n \frac{z^n}{1+z^n+z^{2n}}$$
But when does that happen ??
Does that only happen when there is analytic continuation ?
Does that always hold when there is analytic continuation ?
Does that even care about analytic continuation ?
Does this imply a functional equation such as say $ f(z) = b - f(1/z)$ ?
So in essence :
MAIN QUESTION
we want for all complex $|z| \neq 1$
$$\sum_{n=0}^{\infty} w_n \frac{z^n}{1+z^n+z^{2n}+z^{3n}+z^{4n}} = \sum_{n=0}^{\infty} u_n \frac{z^n}{1+z^n+z^{2n}}$$
But when does that happen ??
It seems one decays faster than the other, so is that even possible for nonconstant functions ?