Analytic continuation of a lambert series $f(z)=\sum_{n=1}^\infty \frac{2^n z^n}{1-(z/2)^n}$?

Question

Let's define a function for complex $z$ :

$$f(z)=\sum_{n=1}^\infty \frac{2^n z^n}{1-(z/2)^n}$$

Lambert series often have an analytic continuation and can even be entire functions so the poles can be removable.

As typical example we have $\sum_{n=1}^\infty \frac{\mu(n) z^n}{1-z^n}=z$ where $\mu(n)$ is the moebius function.

For our function $f(z)$ we have a dense set of poles with $|z| = 2$ although they might be removable.

But our function starts to diverge as soon as $|z|>\frac{1}{2}$.

So the question is how to do analytic continuation of $f(z)$.

Perhaps

$$f(z)=\sum_{n=1}^\infty \frac{2^n z^n}{1-(z/2)^n}=\sum_{n=1}^\infty \frac{4^n (z/2)^n}{1-(z/2)^n}$$

can help ?

We could maybe use $4^n=\sum_{m=0}^\infty \frac{\ln(4)^m n^m}{m!}$ and use a double sum ?

We have a closed form of $f(z,t) = \sum_{n=1}^\infty \frac{n^t z^n}{1-z^n}$ for every positive integer $t$ afterall in term of taylor series with coefficients from the generalized divisors function.

But I am not sure if that helps.

We could try to take taylor series expansions at various points but Im not sure how to find the radius.

Can we compute $f(3)$ ?? Does $f(z)$ have any natural boundaries, and if so where are they ? Is $f(z)$ a meromorphic function ?

Does it help to define $1 + 2 + 4 + 8 + 16 + ... = -1$ ? What summability methods should we or could we use here ?

Answer 1

I'll come back and write a more formal answer later. For now, I'll just give a brief argument and then show some pretty pictures ;)

Using the geometric series, we have that (for $|x|<2$) $$ \sum_{n=1}^\infty \frac{2^nx^n}{1-(\frac{x}{2})^n} = \sum_{n=1}^\infty 2^nx^n \sum_{k=0}^\infty \frac{x^{nk}}{2^{nk}}$$ The sum absolutely converges for $|x|<1/2$, so we can switch the order of summation to obtain $$ \sum_{n=1}^\infty 2^nx^n \sum_{k=0}^\infty \frac{x^{nk}}{2^{nk}}= \sum_{k=0}^\infty \sum_{n=1}^\infty \frac{x^{n(k+1)}}{2^{n(k-1)}}$$ Now that the inner sum coverges when $|x|<2$, and we get $$\sum_{k=0}^\infty \sum_{n=1}^\infty \frac{x^{n(k+1)}}{2^{n(k-1)}} = \sum_{k=0}^\infty \left(\frac{1}{1-\frac{x^{k+1}}{2^{k-1}}}-1\right)$$ Define $f(z) =\sum_{k=0}^\infty \left(\frac{1}{1-\frac{z^{k+1}}{2^{k-1}}}-1\right)$. Then $f(z)$ is meromorphic for $|z|<2$ and agrees with the original series for $|z|<1/2$. By the identity theorem, it is an analytical continuation up to $|z|<2$. At that point, it has a natural boundary.

Now to the interesting part-- pictures! This is a graph of the original function, before continuation. And this is what the continuation looks like:

There are some really striking patterns in the locations of the zeroes and poles. For instance, if I change some of the domain coloring properties, I get this beautiful picture of the function (where only the top half is graphed, since the rest is symmetric)

Now, the real interesting question is 'Can we continue this function beyond its natural boundary? What does the function look like there?' I don't know the answer to these questions, but I hope to make some progress on them soon.

Explanation of the Pictures

Picture 1 and Picture 2 share the same coloring method. In HSB (Hue, Saturation, Brightness), the function I'm using is (with 255 as the highest value) $$F(re^{i \theta}) = \left[ 255\frac{\theta}{2\pi},250,125 + mod(155|\ln(r)|,255) \right]$$ So, the hue corresponds to the angle. The brightness corresponds to the log of the size. The breaks in color you appear to see come from the mod(x,255) term. This means the color resets backward when it becomes too large. This makes it easy to detect poles and zeroes, since their will be lots of these breaks close together near poles and zeroes, since there the log of the size becomes infinite.

In the second picture, I make the hue correspond to the angle. This is why you see those cuts-- the angle isn't a continuous function. In particular, if you go around a circle, the angle goes from $0$ to $2 \pi$, but then jumps back to $0$. So the lines correspond to points where the angle is jumping from $0$ to $2 \pi$. Thus, along every cut, the function is purely real with a positive real part.

The cuts connect together the poles and zeroes in this function. In particular, if I draw the poles in white, then I obtain this graph: (Ignore the white near the boundary, that is just because it becomes too hard to compute the function accurately there). I think that's pretty interesting, the poles and zeroes seem to come in pairs, and those pairs seem to be connected

Answer 2

This is not an answer(yet?) but more of a comment and in particular aimed at Caleb.

To extend beyond Caleb answer for $2<|z|$ I talked to my mentor tommy and he mentioned the generalized q-gamma function. This might work.

Here is the basic idea or at least my interpretation.

Say $x=10$.

Then the sum is

$$f(10) = \sum \frac{20^n}{1 - 5^n}$$

Now lets use taylor series , say

$$g(x) = \sum \frac{x^n}{1 - 5^n}$$

This implies

$$g(x) - g(5x) = \frac{1}{1-x}$$

And that has the solution for a suitable constant $C$ :

$$g(x) - C = \frac{\psi_5(\frac{\ln(x)}{\ln(5)}) - \psi_5(1)}{\ln(5)} - \frac{\ln(x)}{\ln(5)}$$

Where $\psi$ is the q-gamma function.

The q-gamma function satisfies recursion equations and reflection equations so this might just work.

So

$$f(10) = g(20) = C + \frac{\psi_5(\frac{\ln(20)}{\ln(5)}) - \psi_5(1)}{\ln(5)} - \frac{\ln(20)}{\ln(5)}$$

I have little time so might improve later. Comments please ?

edit

ps :

See

https://en.wikipedia.org/wiki/Q-gamma_function

where many formulas are explained.

---