Pictures of some generalized analytical continuations

[Caleb]

I want to share some illustrations to give others some context about what generalized analytical continuation looks like.

First, let me mention my general philosophy in studying an object like generalized analytical continuation. At the moment, math research is only a hobby, so my main motivation in studying these objects is purely aesthetic. Thus, my main goal is really to find the "most beautiful" way to continue a function beyond its boundary. Thus, some of the analysis in the following post will be motivated by what type of options seem most mathemtically elegent. 

This first function
\[ f(x)= \sum_{n=1}^\infty \frac{x^n}{1+x^n} \frac{1}{n^2}\] 
and it is continued in the natural way (i.e. by plugging in)

To me, this graph seems a bit incoherent between the two parts, but this is just a feeling. 

Here is a graph of the natural continuation of 
\[f(x) = \sum_{n=0}^\infty \frac{1}{2^n} \frac{1}{1+x^n}\]
I've graphed this one in a slighty different way to bring out the structure better. I think the natural continuation is perfect for this function-- its hard to tell whats inside the boundary and what's not. 

However, the continuation seems to get worse when \(2^n\) gets closer to \(1^n\). For instance, looking at 
\[f(x) = \sum \frac{1}{1.5^n}{1}{1+x^n}\]
We get the graph below, which doesn't seem as nice.   


One idea I have been considering is that there are cases where you already know the location of all the poles. For instance, consider Mick's function
\[f(x) = \sum_{k=0}^\infty \left(\frac{1}{1-\frac{x^{k+1}}{2^{k-1}}}-1\right) \]
We know that the poles have location \(x = 2 * \sqrt[k+1]{\frac{1}{2^2}}\). For instance, on the upper half of the disc we get the poles


However, if we changed 2 to something less than 1, then we would know the location of the poles "should be" \( \alpha \cdot \sqrt[k+1]{\frac{1}{\alpha^2}}\), however we wouldn't be able to sum the series. This makes gives a sanity check to make sure that certain continuation techniques work with what we expect.

Also, another interesting thing to note is that in all of these graphs the poles and zeroes seem to always be paired. In particular, there always seems to be a natural way to pair up a zero inside the boundary with one on the outside. I guess this probably follows from the 1/x thing, but I still find it interesting. 

Anyway, I'm thinking I'll just use this thread to keep track of my drawing and make them available to others in case it happens to be useful.

[tommy1729]

However, the continuation seems to get worse when \(2^n\) gets closer to \(1^n\). For instance, looking at 
\[f(x) = \sum \frac{1}{1.5^n}{1}{1+x^n}\]
We get the graph below, which doesn't seem as nice.   

I guess you made a typo here.

[tommy1729]

Make sure the functions actually have no fake poles/boundary such as the lambert series for id(z).
Lambert series always have poles dense to eachother. But they can be removable as for id(z).

If you feel something is odd , maybe the boundary is removable.

I have not checked for your functions just a remark.

regards

tommy1729

[tommy1729]

Now hold on a minute.
 If I am not mistaken you took micks function and provided an alternative series expansion that converges in a larger domain.

But now you use the alternative series expansion and plug in 1/2 instead of 2.

But if you took the original series for mick's function it converges better if you pick 1/2 instead of 2.

So you got this one backwards ?

You only needed to replace 2 with 1/2 in micks original series, and that converges.

So if you do nothing or go backwards from your alternative series to the double sum , back to micks definition , then your problem is solved.

Then it looks like one of your simple "plug-ins".

Or did I make a mistake ?


The idea of using geometric series seems to work so often , apart from anyting that resembles ( with multiple sums , internal or external ) 
1+1+1+1+...

which means here that 1/2 and 2 work but 1 does not extend ?
Just an idea or the philosophy of the geometric series in my viewpoint.



regards

tommy1729

[Caleb]

Now hold on a minute.
 If I am not mistaken you took micks function and provided an alternative series expansion that converges in a larger domain.

But now you use the alternative series expansion and plug in 1/2 instead of 2.

But if you took the original series for mick's function it converges better if you pick 1/2 instead of 2.

So you got this one backwards ?

You only needed to replace 2 with 1/2 in micks original series, and that converges.

So if you do nothing or go backwards from your alternative series to the double sum , back to micks definition , then your problem is solved.

Then it looks like one of your simple "plug-ins".

Or did I make a mistake ?


The idea of using geometric series seems to work so often , apart from anyting that resembles ( with multiple sums , internal or external ) 
1+1+1+1+...

which means here that 1/2 and 2 work but 1 does not extend ?
Just an idea or the philosophy of the geometric series in my viewpoint.



regards

tommy1729

I'll edit the original post when I have better internet. For now, see this desmos graph: https://www.desmos.com/calculator/wulciifdwn. The main point is that when \(\alpha > 1 \), (like is the case in micks function), then all of the poles are INSIDE the unit circle. Thus, the regular meromorphic (not generalized analytical continuation), is able to pick up all of these poles. In the graph, I draw in the complex plane the location of where all the poles "should be," even for \( \alpha < 1 \). The problem is, that for \( \alpha < 1 \) you don't get to see any of these poles, since they are outside the natural boundary.

So, my suggestion was that one criteria for determining whether a generalized analytical continuation is good is to check whether it preverses where the poles are supposed to be. In particular, when we look at the slight variation of micks function where \( \alpha < 1 \), then we should be interested in checking whether the poles are at the locatoin they should be.

[JmsNxn]

I'm going to hold off on talking about this too much at the moment. But I believe I have a way to relate "the most beautiful" extension--as you so called it. And the manner I refer to this as, is we can reconstruct \(f(z)\) for \(|z|>1\) using \(|z| <1\), and vice versa, using nothing but Cauchy's integral theorem...

I don't want to muddy the pond too much by discussing details I haven't worked out yet. I'm about 8 pages in to a well written write up (expecting about 30-35 pages); but for restricted cases: \(L(z) : \mathbb{C}/\mathcal{U} \to \mathbb{C}\) where \(\mathcal{U}\) is the unit disk, I believe I've found your "beautiful" extension. I cannot speak of the more modular functions which have poles within the disk. Though I imagine it won't take much to generalize the work.

Also, super nice pictures   I love graphing complex functions! I wish Pari/gp had better onboard graphing protocols. Though, I've made mike3's graphing program fairly similar in philosophy to the desmos protocol you are using.

Super exciting topic, Caleb. Very happy you've started to discuss this! I'm learning a lot and having the time of my life. Reminding me of my number theory days!  

Sincere, Regards

[JmsNxn]

Here are two graphs of:

\[
f_2(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^2} \frac{1}{2^n}\\
\]

And:

\[
f_3(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^3} \frac{1}{2^n}\\
\]

Over the box domain \(- 2 \le \Re(z) \le 2\) and \(-2 \le \Im(z) \le 2\). This graphing protocol is done fairly "low res", as I didn't want to spend a day compiling it. But this is still very accurate. Every singularity is mapped to a black pixel; but the area near a singularity is mapped well enough. I can make these graphs "higher res", it just takes more time.

These functions have a reflection formula about \(z \mapsto 1/z\).

Taking \(f_2(z)\) we get:

   

Please note, that this will be more accurate if you increase terms used. I only used about 16 terms, and 20 digits. This will fill out more with more of a boundary as you progress further.

Taking \(f_3(z)\) we get:

   

I took a fairly lower precision version of this picture, because I don't want these to compile all night. But it's still what it should look like.

Lastly, I am going to take Caleb's function, when \(m =1\). This is the same function as your first post. It is just compiled through my code; and graphed in a low res manner:

   

I want to point out, that all three of these functions have a reflection formula...

Ultimately these graphs are only using sample points along \(z^{16} = -1\), which is pretty low demanding. But it is very accurate as a calculator at this point--about 5 digit accuracy. So,you won't see much change in the graph if, say, we take \(z^{100} = -1\) as our sample points. It will look more "hi-res", but it'll be the same function.

This is a symptom of just how well behaved these Lambert/Caleb functions are....

[JmsNxn]

I'm hesitant to post these pictures, as people might not understand. We are taking \(z^{16} =-1\) sample points. And we are writing:

\[
H(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^n} \frac{1}{2^n}\\
\]

And,

\[
G(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^{n^2}} \frac{1}{2^n}\\
\]

I opted for higher res pictures of these objects; just because they encompass more chaos. But they are graphed over the same box domain: \(|\Re(z)| < 2\) and \(|\Im(z)|<2\). 

Here is \(H\):

   

Here is \(G\):

   

---

These functions have a reflection formula... And that's my main point....

I'm going to take \(z^{40} = -1\) amount of sample points; and make a much bigger graph, with even higher res; but just for \(H(z)\). This will take 5-6 hours at best...

In the mean time, here is \(z^{40} =-1\) and a hi res graph of Caleb's original function...

   

[Caleb]

I'm hesitant to post these pictures, as people might not understand. We are taking \(z^{16} =-1\) sample points. And we are writing:

\[
H(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^n} \frac{1}{2^n}\\
\]

And,

\[
G(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^{n^2}} \frac{1}{2^n}\\
\]

I opted for higher res pictures of these objects; just because they encompass more chaos. But they are graphed over the same box domain: \(|\Re(z)| < 2\) and \(|\Im(z)|<2\). 

Here is \(H\):



Here is \(G\):

---

These functions have a reflection formula... And that's my main point....

I'm going to take \(z^{40} = -1\) amount of sample points; and make a much bigger graph, with even higher res; but just for \(H(z)\). This will take 5-6 hours at best...

In the mean time, here is \(z^{40} =-1\) and a hi res graph of Caleb's original function...

What function are you refering to as "Caleb's" function? Is this
\[
G(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)} \frac{1}{2^n}\\
\]
When I graph this function I get a bit more constrast between the inner and outer function. For instance, if I plot the series of the first 200 terms in Mathemtica, I get this graph

Which definitely seems to have that kind of unnatural jump. I see this same thing when I graph the first 100 terms in my plotter

In fact, here is a plot of the real part of first 1000 terms at the angle \( \theta = \frac{\pi}{5.5} \) (i.e. I'm graphing \( f(r*e^{i \theta}) \) from r = -2 to 2)

And you can kind of notice the unnatural increasing on the right side. Something similarly weird happens at other angles, for instance at \( \theta = \frac{\sqrt{2}}{2} \pi \) we get the graph 

You can see the weirdness if we plot the derivative of the function above. You get an odd and sudden switch in the behaviour of the function near 1. For instance, a plot of the derivative of the first 2000 terms of the sum at the angle  \( \theta = \frac{\sqrt{2}}{2} \pi \) gives the following graph

This of course doesn't mean anything definitive, but it suggests that even matching all derivatives on the boundary might not fully resolve the issue of finding the right continuation.

[JmsNxn]

We are not matching the derivatives on the boundary!!!!!!!!!!!! I know it might look like that. But that's not what we want. I mean, \(1/z^2\) isn't continuous at \(0\)...

We are talking about a function \(L(z) : \mathbb{C}/\mathcal{U} \to \mathbb{C}\) where \(\mathcal{U}\) is the unit circle. It is discontinuous on this circle. That's not what's in question. The graphs you showed should behave like that. Also, the reason that your graph looks different--I have done a super precision version. You can't get these with MatLab, or other graphing software. I did the maximal amount of poles my computer could take. And I graphed it raw. Which means it'll look slightly different than the result you graphed. Because these can be computer intensive results to calculate. And something like Desmos/Matlab may default to a lower precision, to save CPU exhaustion. I'm making my CPU run marathons to graph this. Additionally; I am using mike3's colour schema--which is a tad different than a standard colour schema.

You have turned me on to these results; and perhaps I haven't made it clear that I am proving things you are trying to prove.

Let's define:

\[
c_m^{k} = \frac{1}{(m-1)!}\sum_{d \vert\ k} (-1)^{d+1} \frac{\prod_{i=0}^{m-2}(d+i)}{2^{k/d}}\\
\]

Where \(k \in \mathbb{Z}\); and when we take \(k=dn\), the value \(d\) is negative if \(k\) is negative. This can be written more clearly that \(k/d\) is always positive. This is how we take this divisor sum.

Then,

\[
\begin{align*}
\sum_{k=1}^\infty c_m^{k} z^k &= \sum_{n=1}^\infty \frac{z^n}{(1+z^n)^m}\frac{1}{2^n}\,\,\text{while}\,\,|z| < 1\\
-\sum_{k=1}^\infty c_m^{-k} z^{-k} &= \sum_{n=1}^\infty \frac{z^n}{(1+z^n)^m}\frac{1}{2^n}\,\,\text{while}\,\,|z| > 1\\
\end{align*}
\]

This is a perfectly, one-off, defined arithmetic function. It satisfies your curious functional relationship:

\[
\sum_{k=0}^\infty f(k)z^k = -\sum_{k=1}^\infty f(-k)z^{-k}\\
\]

It does so without referencing Ramanujan. While still; being written as an analytic expression.

This is a very deep result, Caleb. I am calling the function:

\[
c(z) = \sum_{n=0}^\infty \frac{z^n}{1+z^n} \frac{1}{2^n}\\
\]

The Caleb function. And it's a very good jumping off point, especially because the coefficients are geometrically converging. In my head I think of m'th order Caleb functions; which are just:

\[
c^m(z) = \sum_{n=0}^\infty \frac{z^n}{(1+z^n)^m} \frac{1}{2^n}\\
\]

There is so much more going on here; that if we restrict:

\[
L(z) = \infty\,\,\text{only when}\,\,z^n = -1\,\,\text{for some}\,\, n\\
\]

Then \(L(z)\) suffers this same reflection formula. We just have to express the residues at \(z^n = -1\) using a superposition of Lambert functions...

I plan to write this much more detailed. I apologize if I'm not making any sense. I'm prioritizing writing the paper now. But I have it figured out; such that we can generalize the behaviour of Caleb's function as \(z \mapsto 1/z\) to any function \(L(z) : \mathbb{C}/\mathcal{U} \to \mathbb{C}\)... At this point in time I am confident with my result; depending on two conditions.

1.) The function \(L(z)\) solely has singularities when \(|z| = 1\). These singularities are poles, just that, poles--not essential singularities/branching points/logarithmic singularities. If \(L(q) = \infty\) then, \(L(q+h) = O(1/h^m)\).

2.) The singularities can only occur at \(z =q\) where: \(|q| = 1\), and \(q^n = -1\). This is for any \(n > 0\).

Given these two conditions, we have the answer to the question you we're hinting at:

\[
-\sum_{k=1}^\infty L^{(-k)}(0)z^{k} = L(1/z) - L(\infty)\\
\]

While also satisfying:

\[
\sum_{k=1}^\infty L^{(k)}(0)z^k = L(z) - L(0)\\
\]

We can freely swap these variables; and every function with a dense amount of singularities on \(|z| =1\)--where the only singularities are at \(z^n =-1\)..... then there's your reflection formula.

Sincere regards, James

BONUS!

Here is the function:

\[
F(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^n} \frac{1}{2^n}\\
\]

Which satisfies this reflection formula...

Graphed in a super hi res manner: