(03/14/2023, 05:34 AM)JmsNxn Wrote: 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 \(0\)...I haven't read the whole post, on your early point that the sum doesn't converge-- I think this isn't actually true, and its not about precision. It IS true that the series diverges at every angle that is a rational multiple of pi. But actually, if its not a rational angle, than the convergence should depend on measure of irrationality. I think the angle I picked actually converges, but proving that is true is likely difficult without getting into detailed irrationality measurements. With this in mind, I'm thinking of taking the derivative with respsect to the direction where the function converges, and seeing whether those derivatives agree on the boundary. So, the point is it isn't discontinuous on the whole circle, and if we restrict the function to a certain ray, it might converge to a C^\infty function on that ray
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:
I suspect that this is why the 1/2^n is so much better behaved than 1/n^2. I suspect that the 1/2^n actually converges except on a set of measure 0 on the unit circle, since most numbers are probably 'irrational enough' (I'm thinking in a sense similar to the louiville irrationality measure) to converge on the boundary. On the other hand, 1/n^2 needs number to be much more irrrational to converge.
We can probably quanitfy exactly the angles that result in convergence on the unit circle by finding the irrational numbers \(\theta\) that have the property \( \frac{1}{e^{\theta i n} + 1} < c^n \) for some c<2 or something like this.
So, we want to find irrational numbers \(\theta\) so that \(\frac{1}{c^n} < e^{\theta i n} + 1\). If it turns out that most numbers satisfy this (which is what I expect), then we would find that really the natural boundary is actually really well behaved. The poles keep becoming smaller and smaller, until at irrational angles they disappear completley and we get a fully convergent series.
Okay, let me try my best to write some coherent here. You are looking at functions \( f: \mathbb{C}/U \to \mathbb{C}\). Thats all well and good-- there's lots of interesting things to look at when viewing functions that way. For instance, all of your work seems very cool and interesting, and I'm excited to see how it turns out.
HOWEVER, I think you are missing out on some important information by removing the WHOLE UNIT DISC. This is because \( f \) can sometimes still converge on the unit disc. For instance, if we have
\[
c(x) = \sum_{n=0}^\infty \frac{1}{2^n} \frac{x^n}{1+x^n}
\]
Then \( c(1) \) obviously still converges. So, one might wonder, if we restrict ourself to the positive real line (i.e., the ray \( re^{0i}, r>0 \) ), then is our function \( C^\infty \). What do the derivatives look like?
So, definitely, some angles converge on the disc. Next, we ask, for what values of \(\theta\) does \( c(e^{i \theta}) \) converge? Let's just check when it absolutely converges, which is when
\[
\sum_{n=0}^\infty \frac{1}{2^n} \frac{1}{1+e^{i \theta n}}
\]
converges. Let's rewrite the bottom part as
\[
\sum_{n=0}^\infty \frac{1}{2^n} \frac{1}{1+\cos(\theta n) + i \sin(\theta n)}
\]
To be conservative, let's take \(i \sin(\theta n) = 0\), so this assumption will cause less angles to converge than actually converge. Any, that means we want to see about a bound on how small \(1+\cos(\theta n)\) can possibly be given n. Lets again be conservative, and measure how small we can get \( |\cos(\theta n)|-1 \). This also misses some angles, for instance \( \theta = 0 \) has \( |\cos(\theta n)|-1 = 0\), but \(1+\cos(\theta n) = 2\) never becomes small. Okay, so looking for points where \( |\cos(\theta n)|-1 \) is small corresponds to measuring how close \(\theta\) gets to being a multiple of \(\pi\).
Now, if we consider \( |\cos(\theta \pi n)|-1 \), then the question reduces to, "How close can be we get \(\theta n\) to an integer?" This connects directly to irrationality measure, check out this wiki post. Now, one interesting result is that measure 1 numbers have irrationality measure of 2. Thus, for most numbers we have
\[ | x- \frac{p}{q} | \geq \frac{1}{q^2} \]
occurs at most finitely many times. Thus, for the rest of the sum, we have \( | x- \frac{p}{q} | < \frac{1}{q^2} \). Now, multiply this whole thing by \(q\). Then we have \( |qx - p| < \frac{1}{q} \). Note that for small angles we have \( \cos(x) = x\), which tells us that basically \( |\cos(\theta \pi n)|-1 = O(1/n) \) so \( \frac{1}{|\cos(\theta \pi n)|-1} = O(n) \). But \( \frac{1}{2^n} \) totally dominates this grow, so the whole thing should converge for a measure 1 set of points.
Probably, there are some small numerical mistakes in what I have above, but the point is clear. \( \frac{1}{2^n}\) is REALLY strong for convergence. Sure, the unit circle is a natural boundary, but actually, the set of points where the sum is divergent is actually very very small. I think this opens the door for doing some analysis of what \( f \) looks like on \( \mathcal{U} \).
Actually, I get something very odd when I look at plugging in \( c(x) \) on its boundary. It appears the real part is a constant value of 1, but the imaginary part varies. Actually, it looks like this type of behaviour might be more general, and I know I noticed this thing when I tried to analyze the Jacobi theta function a while back. Perhaps many series's real part actually look constant on the unit circle? Its something to look into. I think it might just be a result of it being hard to find irrational numbers that have a really good rational approximation early on-- maybe finding an irrational that has an exceptionally good approximation could break the pattern.
EDIT: it looks like
\[ \mathfrak{Re}(\frac{e^{i x}}{(1+e^{i x})}) = \frac{1}{2} \]
Thus, for most functions we will have the real part completely constant on the boundary, it will just be \( \frac{1}{2} \sum a_n \). However, the imaginary part is \( \frac{1}{2} \tan( \frac{x}{2})\), so the imaginary part actually does depend on the angle. This seems a bit weird, that should mean that if just look at the real part of these functions, they converge on the whole boundary. This almost seems to imply the natural boundary is kind of 'fake', it only shows up if you look at the function from very specific angles. For almost every angle, it looks like you are simply passing through the constant function.