Divergent Series and Analytical Continuation (LONG post) tommy1729 Ultimate Fellow     Posts: 1,906 Threads: 409 Joined: Feb 2009 03/14/2023, 10:52 PM (This post was last modified: 03/14/2023, 11:30 PM by tommy1729.) This is mainly aimed at Caleb because it might be new to him , but ofcourse to all people Although it is not a natural boundary at all I can not help think about this  pxp(z) = (1 + a_0 z)(1 + a_1 z^2)(1 + a_2 z^3)... where pxp(z) clearly has the unit circle as " boundary " since it has a dense set of zero's there. But pxp(z) is actually just exp(z). This is a product expansion for exp(z) valid for |z| < 1. I will put the pxp paper by Gottfried in an attachement. Now what is fun is the analogue ; we have a unit boundary ( kinda ) , an infinite expression and even a functional equation : exp(- x) = 1/exp(x) hence pxp(- z) = 1/pxp(z).   Then again it does not map the inside to the outside of the boundary and vice versa. Just a comment ... And a reminder that we have not even considered products yet. *** I want to say (again) f(x) = x + x^2  + x^4 + x^8 + ... now x^(2^n) is the super of the function g(x) := x^2. and x^(2^(c i)) = x^(-1) is thus a function that commutes with it ( iterates of g(x) )  ( for appropriate c ) Notice how the imaginary direction of the super is linked to the reflection. more general :  IF a function is an infinite sum of iterates ,  and it has a natural boundary that is equal to the edge of its filled julia , and the filled julia set has only one component. and the filled julia set has only one fixed point where the dynamics is analytically conjugate to an irrational rotation ( siegel disk) or an attracting c z or c z^n (for constants c and n ).[url=https://en.wikipedia.org/wiki/Irrational_rotation][/url] then the imaginary direction of the super is linked to the reflection if a reflection formula exists. Then again , here I have so many restrictions that makes a nice reflection formula unlikely. Remember In previous posts I wrote the reflection formulas that have nice properties and make most sense. But say moebius transforms can not map to any edge of some julia. However maybe a riemann mapping can map things to a circle and save the day. Notice there are two reflection functions : L_1( f(z) ) = f( L_2 (z) ) after a riemann mapping we might get L_1 ( f(z) ) = f*( 1/z ) where f* relates to f by the riemann mapping. or something like that. Just some thoughts. regards tommy1729 Attached Files dreamofasequence.pdf (Size: 626.14 KB / Downloads: 18) tommy1729 Ultimate Fellow     Posts: 1,906 Threads: 409 Joined: Feb 2009 03/14/2023, 11:11 PM i keep mentioning  x + x^2  + x^4  + x^8 + ... because unlike most with reflection formula's this one does not have a simple plug-in that works outside the boundary. Maybe some series expanion will turn it into a function such that we can just use plug-in values. regards tommy1729 JmsNxn Ultimate Fellow     Posts: 1,214 Threads: 126 Joined: Dec 2010 03/16/2023, 02:11 AM (This post was last modified: 03/16/2023, 04:18 AM by JmsNxn.) (03/14/2023, 10:52 PM)tommy1729 Wrote: This is mainly aimed at Caleb because it might be new to him , but ofcourse to all people Although it is not a natural boundary at all I can not help think about this  pxp(z) = (1 + a_0 z)(1 + a_1 z^2)(1 + a_2 z^3)... where pxp(z) clearly has the unit circle as " boundary " since it has a dense set of zero's there. But pxp(z) is actually just exp(z). This is a product expansion for exp(z) valid for |z| < 1. I will put the pxp paper by Gottfried in an attachement. Now what is fun is the analogue ; we have a unit boundary ( kinda ) , an infinite expression and even a functional equation : exp(- x) = 1/exp(x) hence pxp(- z) = 1/pxp(z).   Then again it does not map the inside to the outside of the boundary and vice versa. Just a comment ... And a reminder that we have not even considered products yet. I haven't posted much on products yet; but I have considered them in my work upto now. Essentially; from my perspective; this becomes a problem of where there are FINITE singularities along $|z| = 1$. Where the finite number happens to be zero. A large part of what I am trying to do is to extend products with natural boundaries to a larger domain by considering a reflection formula that is universal. So when you write: $\text{pxp}(-z) = \prod_{n=0}^\infty \left(1+ a_n z^n\right)^{-1}\\$ I want to use the residues (which are zero in this case) at the poles, to construct the Taylor series or the Laurent Series; or both of them as one. And in order to do that, I want to write them as a Lambert series first. Another point to observe is that if: $\frac{(-1)^k}{k!} = \sum_{d | k} b_d\\$ Where then: $b_n = \sum_{k|n} \mu(k/n) \frac{(-1)^k}{k!}\\$ Then: $\exp(-z) = \sum_{n=0}^\infty b_n \frac{z^n}{1-z^n}\\$ Things get more complicated though; once we allow for second order poles; rather than just first order poles. This switch becomes more intricate. Not to mention, when we allow arbitrary order poles on the boundary. Now, we can still express it as a Lambert series; but it hides the actual complexity a good amount. My goal so far; has been to relate the sequence $b_n$, To the values at $q^n = -1$: $L(z) = \sum_{j=1}^m \frac{r_j}{(z-q)^j} + O(1)\\$ This has been unbelievably effective! But the key is to include more general Lambert series; so that we can talk about arbitrary order poles. For example; let's take Euler's series: $E(z) = \sum_{n=0}^\infty \#n z^n = \prod_{n=1}^\infty \frac{1}{(1-z^n)}\\$ with the partition function $\# n$. It's been long known that the residues; and singular part of the poles at $z^n = 1$ create an exact equality to $\# n$--a la circle method. I'm basically trying to reprove this same result; but in turn, by trying to reprove it, I am constructing $E(1/z)$. Ultimately the goal is: If you give me the singular part of $L(z)$ at each $q^n = -1$, I can construct a generalized Lambert series, and vice versa. The amount of number theory that I'm using is minimal for the moment; but I am seeing huge applications to Caleb's original idea of "analytically continuing" a modular function. I'm still a little spooked by this. And I'm not trying to overstep the deepness of this. But this is looking really really fucking cool! There is definitely something very deep going on behind the scenes! I'm sticking for the moment to the case where the singularities happen at $q^n = -1$ (Rather than $q^n = 1$ because it makes the "gateway" at $z= 1$); but this can be altered to $q^n = e^{i\theta}$ with little change to the fundamentals. I do not think this will generalize to when the singularities aren't measure zero on the boundary. I think the singularities need to be measure zero... Additionally, I'm only working on rationally dense sequences $q^n = e^{i\theta}$... I'm still wishy washy though. I'm at page 15 of a rough draft; and I'm only starting to get to the deep stuff! Caleb Fellow   Posts: 51 Threads: 6 Joined: Feb 2023 03/17/2023, 12:18 AM (03/16/2023, 02:11 AM)JmsNxn Wrote: (03/14/2023, 10:52 PM)tommy1729 Wrote: This is mainly aimed at Caleb because it might be new to him , but ofcourse to all people Although it is not a natural boundary at all I can not help think about this  pxp(z) = (1 + a_0 z)(1 + a_1 z^2)(1 + a_2 z^3)... where pxp(z) clearly has the unit circle as " boundary " since it has a dense set of zero's there. But pxp(z) is actually just exp(z). This is a product expansion for exp(z) valid for |z| < 1. I will put the pxp paper by Gottfried in an attachement. Now what is fun is the analogue ; we have a unit boundary ( kinda ) , an infinite expression and even a functional equation : exp(- x) = 1/exp(x) hence pxp(- z) = 1/pxp(z).   Then again it does not map the inside to the outside of the boundary and vice versa. Just a comment ... And a reminder that we have not even considered products yet. I haven't posted much on products yet; but I have considered them in my work upto now. Essentially; from my perspective; this becomes a problem of where there are FINITE singularities along $|z| = 1$. Where the finite number happens to be zero. A large part of what I am trying to do is to extend products with natural boundaries to a larger domain by considering a reflection formula that is universal. So when you write: $\text{pxp}(-z) = \prod_{n=0}^\infty \left(1+ a_n z^n\right)^{-1}\\$ I want to use the residues (which are zero in this case) at the poles, to construct the Taylor series or the Laurent Series; or both of them as one. And in order to do that, I want to write them as a Lambert series first. Another point to observe is that if: $\frac{(-1)^k}{k!} = \sum_{d | k} b_d\\$ Where then: $b_n = \sum_{k|n} \mu(k/n) \frac{(-1)^k}{k!}\\$ Then: $\exp(-z) = \sum_{n=0}^\infty b_n \frac{z^n}{1-z^n}\\$ Things get more complicated though; once we allow for second order poles; rather than just first order poles. This switch becomes more intricate. Not to mention, when we allow arbitrary order poles on the boundary. Now, we can still express it as a Lambert series; but it hides the actual complexity a good amount. My goal so far; has been to relate the sequence $b_n$, To the values at $q^n = -1$: $L(z) = \sum_{j=1}^m \frac{r_j}{(z-q)^j} + O(1)\\$ This has been unbelievably effective! But the key is to include more general Lambert series; so that we can talk about arbitrary order poles. For example; let's take Euler's series: $E(z) = \sum_{n=0}^\infty \#n z^n = \prod_{n=1}^\infty \frac{1}{(1-z^n)}\\$ with the partition function $\# n$. It's been long known that the residues; and singular part of the poles at $z^n = 1$ create an exact equality to $\# n$--a la circle method. I'm basically trying to reprove this same result; but in turn, by trying to reprove it, I am constructing $E(1/z)$. Ultimately the goal is: If you give me the singular part of $L(z)$ at each $q^n = -1$, I can construct a generalized Lambert series, and vice versa. The amount of number theory that I'm using is minimal for the moment; but I am seeing huge applications to Caleb's original idea of "analytically continuing" a modular function. I'm still a little spooked by this. And I'm not trying to overstep the deepness of this. But this is looking really really fucking cool! There is definitely something very deep going on behind the scenes! I'm sticking for the moment to the case where the singularities happen at $q^n = -1$ (Rather than $q^n = 1$ because it makes the "gateway" at $z= 1$); but this can be altered to $q^n = e^{i\theta}$ with little change to the fundamentals. I do not think this will generalize to when the singularities aren't measure zero on the boundary. I think the singularities need to be measure zero... Additionally, I'm only working on rationally dense sequences $q^n = e^{i\theta}$... I'm still wishy washy though. I'm at page 15 of a rough draft; and I'm only starting to get to the deep stuff!Here are some brief undevoloped thoughts. On tommy's point, the functional equation can break down with finite singularities. Consider  $f(x) = \sum_{n=1}^\infty \mu(n) \frac{x^n}{1-x^n} = x$ But then of course $f(x) + f(\frac{1}{x})$ is not constant. Similar problems occur when $\sum_{n=1}^\infty \mu(n) \frac{x^n}{1+x^n}$. However, this means that equations with an infinite number of poles also have problems, since given some $g(x)$ we can just consider $g(x) + f(x)$, and now we have a function with the same poles but $g(1/x) + f(1/x)+f(x) + g(x) \neq C$. An argument might be made that the issue is converge on the unit disc, since $f(x)$ doesn't become small on the disc, so perhaps in some sense the poles don't have measure 0?? However, I think this would be too limiting. I've been thinking about the series  $F(x) = \sum_{n=0}^\infty \frac{x^n}{1+x^n}$ Which has the 'natural' continuation $\tilde F(x) = \sum_{k=0}^\infty \frac{(-1)^k}{1-x^{k+1}}$ If we cesaro sum, these two series agree inside the natural boundary, but the second one $\tilde F$ converges on the outside as well. This one seems to be an interesting example, since it doesn't converge on the unit disc, but it still has a nice continuous. It's also related to the arithmetic function $\sum_{k=0}^\infty x^m \left(\sum_{k | m} (-1)^k\right)$ I think it would be much more interesting to look at $F(x) = \sum_{n=1}^\infty \frac{x^n}{1-x^n}$, but I'm not confident that the natural extension that emerges is the 'right' one.  On this point of continuing beyond natural boundaries series that don't absolutely converge, I'm wondering if there is some way to use $L(x) + L(1/x) = C$ to provide a way to compute certain infinite series. In particular, we have that $L(x) = \sum_{n=0}^\infty a_n (\frac{x^n}{1+x^n}-1+1) = -\sum_{n=0}^\infty a_n \frac{1}{x^n + 1} + \sum a_n = -L(\frac{1}{x}) + C$ So, for these lambert series, $C = \sum_{n=0}^\infty a_n$. I've always been very curious about evaluating a few specific arithmetic infinite series. In particular, I'd love to figure out the evaluations of  $\sum_{n=0}^\infty \mu(n) \quad \sum_{n=0}^\infty \lambda(n) \quad \sum_{n=0}^\infty \varphi(n) \quad \sum_{n=0}^\infty \Lambda(n)$   And many of the other common number theory functions. In particular, using the method I have outlined, we have that $F(x) + \tilde F(1/x) = \sum_{n=0}^\infty 1$, and the constant turns out to be $\frac{1}{2}$, which is exactly what you would expect (i.e. its $1 + \zeta(0)$).  On a completely unrelated note, I've noticed through graphing an intereting pairing between poles and zeroes. For instance, with a weird sort of domain coloring, I created harsh lines where the function is completely real (i.e. has no imaginary part). For some reason, with lots of different functions, I notice that it creates these odd egg-shaped regions. They always go through 2 zeroes and 2 poles. This is true for the so-called "Caleb's function" $c(x)$, and it even seemed to be true for micks function which has many poles inside the disc. For instance, the below image is a graph of $F(x), \tilde F(x)$ after a conformal mapping that maps the inside of the circle into the upper half plane, and the outside of the circle into the lower half plane. You'll notice that it is almost periodic Here is another image, with a more usual domain coloring (for some reason, the attachment isn't working on this one) I'm not sure if this weird behaviour leads to anything. JmsNxn Ultimate Fellow     Posts: 1,214 Threads: 126 Joined: Dec 2010 03/18/2023, 04:05 AM (This post was last modified: 03/18/2023, 08:38 AM by JmsNxn.) Caleb! You are absolutely correct! I'm not trying to show my hand too much right now; but this is what's so confusing! $p(z) = \sum_{k=0}^\infty p_k z^k\\$ And let's assume that: $b_k = \sum_{ d \vert k} (-1)^{k/d} p_d \mu(k/d)$ Then $p(z)$ is also written as: $p(z) = \sum_{k=0}^\infty b_k \frac{z^k}{1+z^k}\\$ BUT! this tells us nothing about $p(z)$ for $|z| > 1$... Just take $p(z)$ to be a polynomial and you can identify this. My theory is evolving constantly; but we have to make two additional claims to the above statement. We must ensure that $p(z)$ only has poles at $q^n = -1$; and we must use a generalized Lambert series... If I write: $p_k = \sum_{d \vert k} H(d, M(k/d)) P(k/d)\\$ Where: $H(d,m) = \frac{(-1)^{m+1}}{(m-1)!} \prod_{i=0}^{m-2} (d+i)\\$ Then the expression $p(z)$ is just: $p(z) = \sum_{n=0}^\infty \frac{P(n)z^n}{(1+z^n)^{M(n)}}\\$ BUT NOW THE REFLECTION FORMULA HOLDS! The trouble becomes; how do we find $P(n)$ and $M(n)$, as arithmetic functions, solely using the function $p(z)$ and it's taylor coefficients $p_k$???? Ironically enough; this is solvable for polynomials... And the "generalized Lambert series" is just a polynomial. Which satisfies the reflection. For example, take $p(z) = z$. Then $P(1) = 1$ and $P(n) = 0$ everywhere else; and $M(1) = 0$ and $M(n)$ is arbitrary... Voila! We've done our reflection formula.... $M(n) : \mathbb{N} \to \mathbb{N}$ and $P(n) : \mathbb{N} \to \mathbb{C}$; so there really is no problem. And the mechanics I use to write the reflection formula still follow; even if we choose the trivial $M(n) = 0$--but it can be a bit of a fucking headache to write everything perfectly.... I'm excited to release my program soon... I'm still developing some of the ideas; but finding a flawless reflection for arbitrary functions is appearing clearer and clearer. EDIT: I'm going to explain the polynomial interpolation; because this has to do with extending near the boundary. Additionally it is something you have mentioned. Let's define: \begin{align} P(1) &= 1\\ P(n) &= \epsilon\,\,\text{if}\,\, n \ge 0\,\,n\neq 1\\ M(1) &= 0\\ \end{align} Then the function: $\sum_{n=0}^\infty \frac{P(n)z^n}{\left(1+z^n\right)^{M(n)}} = z + O(\epsilon)\\$ We can do this for all polynomials. And my number theoretic tricks will solve this generally... The reflection formula becomes: $L(z) = 1/\left(1/z + O(\epsilon)\right) = z + O(\epsilon)\\$ « Next Oldest | Next Newest »

 Possibly Related Threads… Thread Author Replies Views Last Post Pictures of some generalized analytical continuations Caleb 18 1,166 03/17/2023, 12:56 AM Last Post: tommy1729 double functional equation , continuum sum and analytic continuation tommy1729 6 469 03/05/2023, 12:36 AM Last Post: tommy1729 Discussion on "tetra-eta-series" (2007) in MO Gottfried 40 2,563 02/22/2023, 08:58 PM Last Post: tommy1729 continuation of fix A to fix B ? tommy1729 22 1,727 02/06/2023, 11:59 PM Last Post: tommy1729 Tetration Asymptotic Series Catullus 18 3,006 07/05/2022, 01:29 AM Last Post: JmsNxn Formula for the Taylor Series for Tetration Catullus 8 2,595 06/12/2022, 07:32 AM Last Post: JmsNxn Calculating the residues of $\beta$; Laurent series; and Mittag-Leffler JmsNxn 0 967 10/29/2021, 11:44 PM Last Post: JmsNxn Trying to find a fast converging series of normalization constants; plus a recap JmsNxn 0 895 10/26/2021, 02:12 AM Last Post: JmsNxn Reducing beta tetration to an asymptotic series, and a pull back JmsNxn 2 1,947 07/22/2021, 03:37 AM Last Post: JmsNxn Hey, Everyone; been a long time... JmsNxn 17 10,857 01/28/2021, 09:53 AM Last Post: MphLee

Users browsing this thread: 1 Guest(s) 