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03/14/2023, 05:45 AM
(This post was last modified: 03/14/2023, 07:09 AM by Caleb.)
(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 \(0\)...
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 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
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.
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03/14/2023, 07:11 AM
(This post was last modified: 03/14/2023, 07:18 AM by JmsNxn.)
THANK YOU, CALEB!
I AM AWARE OF EVERYTHING IN YOUR POST. IT MAY BE CONTINUOUS AT A BOUNDARY POINT! IT MAY EVEN BE COMPLEX DIFFERENTIABLE!
BUT!!!!!!!!!!!!!!!!!!!!!! IT IS NOT COMPLEX DIFFERENTIABLE ON AN OPEN DOMAIN!
Funny you mention \(1/n^2\). Let me take the function:
\[
f(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^n} \frac{1}{(n+1)^2}\\
\]
Then MY FORMULA STILL WORKS!!!!!
My formula doesn't care that \(1/2^n\) goes to zero fast; All I care is that \(G(n) \to 0\) and:
\[
\sum_{n=0}^\infty |G(n)| < \infty\\
\]
Then:
\[
f_G(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^n} G(n)\\
\]
Satisfies the reflection formula. Let's make it even more complex. Let's say \(m(n) : \mathbb{N} \to \mathbb{N}\), so instead of \(n \mapsto n\), we have \(n \mapsto m(n)\). So we write:
\[
f_G^m(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^{m(n)}} G(n)\\
\]
THIS SATISFIES THE SAME REFLECTION FORMULA!
The fact you are noticing values in which this object is finite/differentiable along \(|z| = 1\), is entirely expected. At every value \(|z| = 1\) such that \(z \neq e^{\pi i/ n}e^{2\pi i \frac{k}{n}}\) for all \(n \ge 1\) and \(0 \le k < n\)-- this object is differentiable.
The point, and maybe, I'll chop it up to you not understanding some subtleties. Just because \(f(z)\) is complex differentiable at \(z_0\), does not mean it is holomorphic at \(z_0\). For \(f(z)\) to be holomorphic at \(z_0\), we need \(f(z)\) to be complex differentiable for \(|z-z_0| < \delta\) for some \(\delta> 0\). This is a SUPER SUBTLE DIFFERENCE!
What you are detailing is that \(f(z_0)\) and \(f'(z_0)\) exist! Which, all the power to you! That's correct as fuck! But HOLOMORPHY, is a stricter requirement. So when I say it is NOWHERE HOLOMORPHIC on \(|z| = 1\), I am correct. BUT IT ALLOWS, for \(f(z_0)\) and \(f'(z_0)\) to still exist!
I am entirely on board with you, Caleb! No where in this post am I trying to disagree. I am just clarifying terminology. This function is NOWHERE HOLOMORPHIC for \(|z| = 1\). It can be pointwise differentiable! BUT NOT HOLOMORPHIC! At least, assuming that there are dense poles about \(z^n = -1\)....
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03/14/2023, 07:23 AM
(This post was last modified: 03/14/2023, 07:40 AM by Caleb.)
(03/14/2023, 07:11 AM)JmsNxn Wrote: THANK YOU, CALEB!
I AM AWARE OF EVERYTHING IN YOUR POST. IT MAY BE CONTINUOUS AT A BOUNDARY POINT! IT MAY EVEN BE COMPLEX DIFFERENTIABLE!
BUT!!!!!!!!!!!!!!!!!!!!!! IT IS NOT COMPLEX DIFFERENTIABLE ON AN OPEN DOMAIN!
Funny you mention \(1/n^2\). Let me take the function:
\[
f(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^n} \frac{1}{(n+1)^2}\\
\]
Then MY FORMULA STILL WORKS!!!!!
My formula doesn't care that \(1/2^n\) goes to zero fast; All I care is that \(G(n) \to 0\) and:
\[
\sum_{n=0}^\infty |G(n)| < \infty\\
\]
Then:
\[
f_G(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^n} G(n)\\
\]
Satisfies the reflection formula. Let's make it even more complex. Let's say \(m(n) : \mathbb{N} \to \mathbb{N}\), so instead of \(n \mapsto n\), we have \(n \mapsto m(n)\). So we write:
\[
f_G^m(z) = \sum_{n=0}^\infty \frac{z^n}{\left(1+z^n\right)^{m(n)}} G(n)\\
\]
THIS SATISFIES THE SAME REFLECTION FORMULA!
The fact you are noticing values in which this object is finite/differentiable along \(|z| = 1\), is entirely expected. At every value \(|z| = 1\) such that \(z \neq e^{\pi i/ n}e^{2\pi i \frac{k}{n}}\) for all \(n \ge 1\) and \(0 \le k \le n\)-- this object is differentiable.
The point, and maybe, I'll chop it up to you not understanding some subtleties. Just because \(f(z)\) is complex differentiable at \(z_0\), does not mean it is holomorphic at \(z_0\). For \(f(z)\) to be holomorphic at \(z_0\), we need \(f(z)\) to be complex differentiable for \(|z-z_0| < \delta\) for some \(\delta> 0\). This is a SUPER SUBTLE DIFFERENCE!
What you are detailing is that \(f(z_0)\) and \(f'(z_0)\) exist! Which, all the power to you! That's correct as fuck! But HOLOMORPHY, is a stricter requirement. So when I say it is NOWHERE HOLOMORPHIC on \(|z| = 1\), I am correct. BUT IT ALLOWS, for \(f(z_0)\) and \(f'(z_0)\) to still exist!
I am entirely on board with you, Caleb! No where in this post am I trying to disagree. I am just clarifying terminology. This function is NOWHERE HOLOMORPHIC for \(|z| = 1\). It can be pointwise differentiable! BUT NOT HOLOMORPHIC! At least, assuming that there are dense poles about \(z^n = -1\)....
Yes, I agree with what you have here, thank you for clarifying the terminology. Of course the function is nowhere holomorphic at the boundary, the boundary has a dense set of singularity.
But, if we want to study function beyond natural boundary, I think there are some regular notions we must let go of. In particular, I thinking about what happens when we reduce the definition of holomorphic from "F is holomorphic on an open subset" TO "F is holomorphic on an open subset minus a set of measure zero".
My point is just that the functions we are dealing with here only have a measure 0 set of bad points-- the rest of the points are well defined. So, I'm interested if your extension will recover any familar theorems when we weaken the definition of holomorphic in the way I have suggested.
Also, I don't mean to disagree with the work you have done. I'm not imply anything you have is wrong. I'm just suggesting that we look at some details of coherence on the unit circle to see if a given extension is 'good'. I.e., we might be able to classify 'bad' extensions by their bad behaviour on the unit circle. My cursory look at generalized analytical continuation suggests they do something like this-- they look for matching boundary values as one condition for continuation. I think complex differentiablity might be another condition to look for, to shift out bad extensions from the good ones
Lets call my defintion of holomorphic weakly holomorphic. We might wonder if there are any functions that have a natural boundary but a weakly holmorphic extension. Is the natural extension of c(x) weakly holomorphic? Weakly holomorphic seems like a pretty strong condition, so I'm thinking that showing c(x) satisfies it would be a pretty strong step towards full uniqueness.
EDIT: Sorry-- I now realize "holomorphic except on a set of measure zero" isn't a coherent idea. I think there is something there-- but thats not the right idea. I'll have to think about it a bit more to figure out what I really want to say. Anyway, I still think that having such good convergence on the boundary (convergence except on a set of measure 0) definitely means something, and it means that a lot of analysis can be done with a sufficent weakening of definitions
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(03/14/2023, 07:23 AM)Caleb Wrote: My point is just that the functions we are dealing with here only have a measure 0 set of bad points-- the rest of the points are well defined. So, I'm interested if your extension will recover any familar theorems when we weaken the definition of holomorphic in the way I have suggested.
Also, I don't mean to disagree with the work you have done. I'm not imply anything you have is wrong. I'm just suggesting that we look at some details of coherence on the unit circle to see if a given extension is 'good'. I.e., we might be able to classify 'bad' extensions by their bad behaviour on the unit circle. My cursory look at generalized analytical continuation suggests they do something like this-- they look for matching boundary values as one condition for continuation. I think complex differentiablity might be another condition to look for, to shift out bad extensions from the good ones
This is my point also, Caleb. Because we described the singularities at \(|z| = 1\) as the same thing; we have the same result for non-singular points on \(|z| = 1\). My equation will describe the point wise differential properties; but the only data it needs to produce this, is singular data for \(|z| =1\).
I am loving everything you are saying. I'm not going to prove this result for you. But all the things you are worried about; it happens. Every thing works exactly as your imagining. The math is turning out to be incredibly beautiful. But it's very technical. And I want to get everything right. So I'm focusing on writing at 25-30 page paper, which covers all bases.
But, what you see as differential points; can be covered by only looking at the singular points. That's kind of the thesis of what I am saying!
I'm super excited to go deeper into this. But, I should probably shut up. I should just finish the 25-30 pages of work that goes into it. This is a paper; not a post on the tetration forum... I should treat this knowledge with the respect it deserves.
Nonetheless! Keep posting, Caleb! I'm always weirdly angry and upset about how your right, but it doesn't make sense, lmfao.
Sincere Regards, James
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(03/14/2023, 07:43 AM)JmsNxn Wrote: (03/14/2023, 07:23 AM)Caleb Wrote: My point is just that the functions we are dealing with here only have a measure 0 set of bad points-- the rest of the points are well defined. So, I'm interested if your extension will recover any familar theorems when we weaken the definition of holomorphic in the way I have suggested.
Also, I don't mean to disagree with the work you have done. I'm not imply anything you have is wrong. I'm just suggesting that we look at some details of coherence on the unit circle to see if a given extension is 'good'. I.e., we might be able to classify 'bad' extensions by their bad behaviour on the unit circle. My cursory look at generalized analytical continuation suggests they do something like this-- they look for matching boundary values as one condition for continuation. I think complex differentiablity might be another condition to look for, to shift out bad extensions from the good ones
This is my point also, Caleb. Because we described the singularities at \(|z| = 1\) as the same thing; we have the same result for non-singular points on \(|z| = 1\). My equation will describe the point wise differential properties; but the only data it needs to produce this, is singular data for \(|z| =1\).
I am loving everything you are saying. I'm not going to prove this result for you. But all the things you are worried about; it happens. Every thing works exactly as your imagining. The math is turning out to be incredibly beautiful. But it's very technical. And I want to get everything right. So I'm focusing on writing at 25-30 page paper, which covers all bases.
But, what you see as differential points; can be covered by only looking at the singular points. That's kind of the thesis of what I am saying!
I'm super excited to go deeper into this. But, I should probably shut up. I should just finish the 25-30 pages of work that goes into it. This is a paper; not a post on the tetration forum... I should treat this knowledge with the respect it deserves.
Nonetheless! Keep posting, Caleb! I'm always weirdly angry and upset about how your right, but it doesn't make sense, lmfao.
Sincere Regards, James
Yes sorry-- I will shut up for now to let you work on the paper. I'm super excited to read it, so I'll bide my time doing something else for a few days. Good luck on the writing-- I can't wait to read it!
Thanks again for your work,
Caleb
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03/14/2023, 08:16 AM
(This post was last modified: 03/14/2023, 08:27 AM by JmsNxn.)
(03/14/2023, 07:51 AM)Caleb Wrote: Yes sorry-- I will shut up for now to let you work on the paper. I'm super excited to read it, so I'll bide my time doing something else for a few days. Good luck on the writing-- I can't wait to read it!
Thanks again for your work,
Caleb
Don't be sorry! This is some of the weirdest math I've ever seen! You have sparked Ramanujan's identity:
\[
-\sum_{k=1}^\infty f(-k)z^{-k} = \sum_{k=1}^\infty f(k) z^k\\
\]
Where:
\[
\begin{align*}
\sum_{k=1}^\infty f(k) z^k &= c(z) - c(0)\\
-\sum_{k=1}^\infty f(-k)z^{k} &= c(1/z) - c(\infty)\\
\end{align*}
\]
You typically prove this by showing that \(f(k)\) is subject to Carlson's Theorem for \(k \in \mathbb{C}\).
We are dodging all of these requirements! I don't quite get how, yet. But we are avoiding this entirely!
My suspicion is because we are writing this as a "zeta" function, because we are writing it as a divisor sum; rather than a normal sum! So, some "zeta-like" sums pop out. And the average zeta-function is subject to Carlson's theorem. But this still doesn't make sense! Either way though; these functions satisfy Ramanujan's identity.
Beautiful work, Caleb! Absolutely fucking beautiful!
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(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 \(0\)...
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:
that last picture seems to have an analytic opening ?
truncated result or optical illusion ?
regards
tommy1729
Posts: 1,214
Threads: 126
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03/16/2023, 01:29 AM
(This post was last modified: 03/16/2023, 08:15 AM by JmsNxn.)
(03/14/2023, 11:16 PM)tommy1729 Wrote: that last picture seems to have an analytic opening ?
truncated result or optical illusion ?
regards
tommy1729
That opening is a bit of an optical illusion. What's happening is that at \(z=1\) this function is essentially infinitely differentiable. So it is \(\mathcal{C}^\infty\) here. This causes a kind of regularity near here. It is still absolutely not holomorphic here, or near here. But the singularities are measure zero on the unit disk; so because it is smooth at this point; it looks like things are "cancelling out" and the poles are almost negligible.
It's definitely not a truncation error because I took 50 digit precision, and 100 terms to the sum:
\[
L(z) = \sum_{n=0}^{100} \frac{z^n}{(1+z^n)^n} \frac{1}{2^n}\\
\]
So any truncation error would be absolutely invisible to the naked eye. My hypothesis is the fact that this function is smooth at \(1\). There's probably an asymptotic series here; which I'm sure you could figure out; doesn't look too hard to manipulate out.
Also additionally, the values \(|q-1| < \delta\) is small and \(q^n = -1\) implies that \(n\) is very very large (how well we are rationally approximating \(\pi\)); which then implies that it is a miniscule term that is added to the value. I think all functions:
\[
L(z) = \sum_{n=0}^\infty P(n) \frac{z^n}{(1+z^n)^{M(n)}}\\
\]
Should have a similar result near \(1\)--a kind of "gateway" near \(z= 1\). Provided that:
\[
\limsup_{n\to\infty} |P(n)|^{1/n} \le 1\\
\]
But the "gateway" can look worse or better depending on how fast \(M(n)\) and \(P(n)\) grows.
I'm not certain this is the reasoning; but it's my best guess at the moment
So, to be to the point. There are still a bunch of poles near this gateway; they are just \(\frac{1}{2^N}* \frac{1}{z}\) where \(N\) is so ridiculously large, that the naked eye can't see the pole...
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(03/16/2023, 01:29 AM)JmsNxn Wrote: (03/14/2023, 11:16 PM)tommy1729 Wrote: that last picture seems to have an analytic opening ?
truncated result or optical illusion ?
regards
tommy1729
That opening is a bit of an optical illusion. What's happening is that at \(z=1\) this function is essentially infinitely differentiable. So it is \(\mathcal{C}^\infty\) here. This causes a kind of regularity near here. It is still absolutely not holomorphic here, or near here. But the singularities are measure zero on the unit disk; so because it is smooth at this point; it looks like things are "cancelling out" and the poles are almost negligible.
It's definitely not a truncation error because I took 50 digit precision, and 100 terms to the sum:
\[
L(z) = \sum_{n=0}^{100} \frac{z^n}{(1+z^n)^n} \frac{1}{2^n}\\
\]
So any truncation error would be absolutely invisible to the naked eye. My hypothesis is the fact that this function is smooth at \(1\). There's probably an asymptotic series here; which I'm sure you could figure out; doesn't look too hard to manipulate out.
Also additionally, the values \(|q-1| < \delta\) is small and \(q^n = -1\) implies that \(n\) is very very large (how well we are rationally approximating \(\pi\)); which then implies that it is a miniscule term that is added to the value. I think all functions:
\[
L(z) = \sum_{n=0}^\infty P(n) \frac{z^n}{(1+z^n)^{M(n)}}\\
\]
Should have a similar result near \(1\)--a kind of "gateway" near \(z= 1\). Provided that:
\[
\limsup_{n\to\infty} |P(n)|^{1/n} \le 1\\
\]
But the "gateway" can look worse or better depending on how fast \(M(n)\) and \(P(n)\) grows.
I'm not certain this is the reasoning; but it's my best guess at the moment
So, to be to the point. There are still a bunch of poles near this gateway; they are just \(\frac{1}{2^N}* \frac{1}{z}\) where \(N\) is so ridiculously large, that the naked eye can't see the pole...
thanks
tommy1729
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