Divergent Series and Analytical Continuation (LONG post)

[JmsNxn]

Okay........

I have figured out the formula for:

\[
f_m(\theta ; x) = \sum_{k=0}^\infty B^m_k(\theta) x^k\\
\]

And the formula for:

\[
F^m(\theta;x) = \sum_{k=1}^{\infty} F_{-k}^m(\theta) x^k
\]

The values \(B^m_k\) are only slightly different than \(F_{-k}^m(\theta)\)...

I need to sleep on this a bit, just to make sure everything is okay

My code is fucking perfect though!


EDIT:

Okay, so the way I see it is that:

\[
B^m_k(\theta) = e^{i\theta k} \sum_{d|k} h_m(d)e^{-i\theta k/d} \frac{(-1)^{d+1}}{2^{k/d}}\\
\]

Where:

\[
h_m(d) =  \frac{1}{m-1!}\prod_{i=1}^{m-1} (d+i-1)\\
\]

Then for \(F_{-k}^m\) we have the exact same formula; but we have to take a slightly different divisor sum. It is technically the exact same formula; but we have to use \(-k\) and we are taking divisors \(d\) which are negative; so that:

\[
F_{-k}^m = e^{-i\theta k} \sum_{d|-k} h_m(d)e^{i\theta k/d} \frac{(-1)^{d+1}}{2^{-k/d}}\\
\]

Or if you prefer to write this with positive divisors:

\[
F_{-k}^m = -e^{-i \theta k} \sum_{d|k} h_m(-d)e^{-i\theta k/d} \frac{(-1)^{d+1}}{2^{k/d}}
\]

So that actually; TECHNICALLY:

\[
F_{-k}^m(\theta) = -B_{-k}^m(\theta)
\]

But the term \(h_m(-d)\) produces a lot of chaos in this formula; because it is zero for \(d < m-1\). So I believe your intuition is definitely correct. I think we will get that:

\[
\sum_{k=0}^\infty B_k^m(\theta) x^k = -\sum_{k=1}^\infty F_{-k}^m(\theta) x^{-k}\\
\]

Or something close to this........

Alright; I'm off to study the holy grail now. Let \(\mathcal{M} = \{m_n\}_{n=0}^\infty\)  be a sequence of natural numbers \(m_n \in \mathbb{N}\) for \(m_n \ge 0\). And look at the generalized Caleb function:

\[
f_{\mathcal{M}}(\theta ; x) = \sum_{n=0}^\infty \frac{z^n}{\left(1+e^{i\theta n}z^n\right)^{m_n}} \frac{1}{2^n}\\
\]

If it works for this function; I think it'll work for every function. And by what I'm seeing so far; I see no reason this shouldn't produce the exact same result. I think it will be important to introduce arbitrary sequences \(a_n\) now, rather than \(1/2^n\)...


WAIT!

I think it's just:

\[
B^{\mathcal{M}}_k(\theta) = e^{i\theta k} \sum_{d|k} h_{m(k/d)}(d)e^{-i\theta k/d} \frac{(-1)^{d+1}}{2^{k/d}}\\
\]

I'll double check this though. The same formula should pop out for \(F_{-k}^{\mathcal{M}}\)....

UPDATE!!!!!!!!!

Okay, lets let \(m(n) : \mathbb{N} \to \mathbb{N}\). Let's let \(a_n : \mathbb{N} \to \mathbb{C}\), but additionally let's assume that:

\[
\sum_{n=0}^\infty |a_n| < \infty\\
\]

Let's define the generalized Lambert function as:

\[
L(\theta; z) = \sum_{n=0}^\infty a_n \frac{z^n}{\left(1+e^{i\theta n}z^n\right)^{m(n)}}\\
\]

Let:

\[
\ell_k = \frac{L^{(k)}(\theta ; 0)}{k!} = e^{i\theta k} \sum_{d | k} e^{-i\theta k/d}a_{k/d} h_{m(k/d)}(d)(-1)^{d+1}
\]

Then:

\[
L(\theta ; z) = \sum_{k=0}^\infty \ell_k z^k\\
\]

Let:

\[
P_k = \frac{1}{2\pi i} \int_{|z| = 2} L(\theta ; z) z^{k-1}\,dz\\
\]

for \(k > 0\). Then:

\[
P_k = e^{-i\theta k} \sum_{d | k} a_{k/d} e^{-i\theta k/d}h_{m(k/d)}(-d)(-1)^{d} = \ell_{-k}\\
\]

This is about how general I can make it. I will now work on showing that your identity that we can reconstruct \(L\) using only the coefficients \(P_k\)--as the coefficients of a Laurent series. We'll see how it goes!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!


This is super cool; because we don't need Ramanujan at all. But it's doing everything Ramanujan does! AWesome!

[JmsNxn]

Okay, I'll stop blowing up this thread after this last identification.

Let:

\[
L(z) : \mathbb{C}/ U \to \mathbb{C}\\
\]

Where \(U\) is the unit circle. Let's assume that:

\[
\int_{|\zeta| = R} L(\zeta)\frac{d\zeta}{\zeta^{k-1}} = 0\\
\]

For all \(R > 1\), and for all \(k > 0\). Then call the sum:

\[
F(z) = -\sum_{k\in \mathbb{Z}} z^{-k}\int_{|\zeta| = R} L(\zeta)\frac{d\zeta}{\zeta^{k-1}}\\
\]

This just reduces to:

\[
F(z) = -A - \sum_{k=1}^\infty \frac{1}{2\pi i}\int_{|\zeta| = R} L(\zeta)z^k \zeta^{k-1}\,d\zeta\\
\]

Which is just:

\[
F(z) = -A + \frac{1}{2\pi i} \int_{|\zeta| = R} \frac{L(\zeta)}{\zeta - \frac{1}{z}}\,d\zeta\\
\]

The conjecture by Caleb; is that:

\[
F(z) = L(z) + C\\
\]

Where the constant \(C\) is easily discoverable. Where it can be written more compactly as:

\[
L(z) = L(0) + \frac{1}{2\pi i} \int_{|\zeta| = R} \frac{L(\zeta)}{\zeta - \frac{1}{z}}\,d\zeta\\
\]

For all \(|z| < 1\). This actually creates a reflection formula between \(L(1/z)\) and \(L(z)\). Where we can simply evaluate this integral as:

\[
\frac{1}{2\pi i} \int_{|\zeta| = R} \frac{L(\zeta)}{\zeta - \frac{1}{z}}\,d\zeta\\ = L(1/z) + A\\
\]

Where:

\[
A = \frac{1}{2\pi i} \int_{|z| = R} \frac{L(\zeta)}{\zeta}\,d\zeta\\
\]

Because:

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

When \(|z\zeta| > 1\); which is always true if \(|\zeta| = R\) is large enough. And the only non-zero term, is when \(k = 1\).

I HAVE CONFIRMED THIS NUMERICALLY FOR ALL GENERALIZED LAMBERT FUNCTIONS!!!!!!!

I think this might be confirmation of a much much deeper identity; but we have to prove Caleb's conjecture first. Which is that:

\[
L(z) = L(0) + \frac{1}{2\pi i} \int_{|\zeta| = R} \frac{L(\zeta)}{\zeta-\frac{1}{z}}\,d\zeta\\
\]

For all \(1 < \left|\frac{1}{z}\right| < R\), \(\frac{1}{R} < |z| < 1\).

This would intimately relate both functions \(L(z)\) and \(L(1/z)\) that I can't even believe is possible. This is like striking straight up gold!

BEAUTIFUL INTUITION, CALEB! This is a reflection formula for the books! And I think it'll go a long way in doing what you want to do! But do it without needing to fall back on Fourier Analysis/ Ramanujan shit!

Also, for you modular nerds! If \(L(0) = 0\), then we have shown that:

\[
L(1/z) = L(z) + A\\
\]

If we also have that \(A =0\)... then \(L(1/z) = L(z)\)! So if we assume that \(L(\infty) =0\) and \(L(0) = 0\); then these functions are carbon copies.

Which is the first step in identifying a modular relationship. The next would be a periodic condition... But I can imagine this as "analytically continuing modular functions"

YOU ARE ONTO SOMETHING HERE CALEB!!!!!!!!!!!!!!!!

Sincere regards, SUPER PUMPED, James

[Caleb]

Okay, I'll stop blowing up this thread after this last identification.

Let:

\[
L(z) : \mathbb{C}/ U \to \mathbb{C}\\
\]

Where \(U\) is the unit circle. Let's assume that:

\[
\int_{|\zeta| = R} L(\zeta)\frac{d\zeta}{\zeta^{k-1}} = 0\\
\]

For all \(R > 1\), and for all \(k > 0\). Then call the sum:

\[
F(z) = -\sum_{k\in \mathbb{Z}} z^{-k}\int_{|\zeta| = R} L(\zeta)\frac{d\zeta}{\zeta^{k-1}}\\
\]

This just reduces to:

\[
F(z) = -A - \sum_{k=1}^\infty \frac{1}{2\pi i}\int_{|\zeta| = R} L(\zeta)z^k \zeta^{k-1}\,d\zeta\\
\]

Which is just:

\[
F(z) = -A + \frac{1}{2\pi i} \int_{|\zeta| = R} \frac{L(\zeta)}{\zeta - \frac{1}{z}}\,d\zeta\\
\]

The conjecture by Caleb; is that:

\[
F(z) = L(z) + C\\
\]

Where the constant \(C\) is easily discoverable. Where it can be written more compactly as:

\[
L(z) = L(0) + \frac{1}{2\pi i} \int_{|\zeta| = R} \frac{L(\zeta)}{\zeta - \frac{1}{z}}\,d\zeta\\
\]

For all \(|z| < 1\). This actually creates a reflection formula between \(L(1/z)\) and \(L(z)\). Where we can simply evaluate this integral as:

\[
\frac{1}{2\pi i} \int_{|\zeta| = R} \frac{L(\zeta)}{\zeta - \frac{1}{z}}\,d\zeta\\ = L(1/z) + A\\
\]

Where:

\[
A = \frac{1}{2\pi i} \int_{|z| = R} \frac{L(\zeta)}{\zeta}\,d\zeta\\
\]

Because:

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

When \(|z\zeta| > 1\); which is always true if \(|\zeta| = R\) is large enough. And the only non-zero term, is when \(k = 1\).

I HAVE CONFIRMED THIS NUMERICALLY FOR ALL GENERALIZED LAMBERT FUNCTIONS!!!!!!!

I think this might be confirmation of a much much deeper identity; but we have to prove Caleb's conjecture first. Which is that:

\[
L(z) = L(0) + \frac{1}{2\pi i} \int_{|\zeta| = R} \frac{L(\zeta)}{\zeta-\frac{1}{z}}\,d\zeta\\
\]

For all \(1 < \left|\frac{1}{z}\right| < R\), \(\frac{1}{R} < |z| < 1\).

This would intimately relate both functions \(L(z)\) and \(L(1/z)\) that I can't even believe is possible. This is like striking straight up gold!

BEAUTIFUL INTUITION, CALEB! This is a reflection formula for the books! And I think it'll go a long way in doing what you want to do! But do it without needing to fall back on Fourier Analysis/ Ramanujan shit!

Also, for you modular nerds! If \(L(0) = 0\), then we have shown that:

\[
L(1/z) = L(z) + A\\
\]

If we also have that \(A =0\)... then \(L(1/z) = L(z)\)! So if we assume that \(L(\infty) =0\) and \(L(0) = 0\); then these functions are carbon copies.

Which is the first step in identifying a modular relationship. The next would be a periodic condition... But I can imagine this as "analytically continuing modular functions"

YOU ARE ONTO SOMETHING HERE CALEB!!!!!!!!!!!!!!!!

Sincere regards, SUPER PUMPED, James

I apoligise for the late response, I should have mentioned this earlier- but I belive we have only stubled upon a trivial property  . In particular, its not that \( L(z) \) and \( L(1/z) \) are related in some interesting way-- they are actually just the same evaulation. In paritcular, notice that
\[ f(x) = \sum \frac{x^n}{1+x^n} a_n \]
Using the geometric series formula for \( |x|<1 \) gives
\[ f(x) = \sum_n a_n  \sum_k x^{n(k+1)}\]
Now take a look at 
\[f(1/x) = \sum \frac{(\frac{1}{x})^n}{1+(\frac{1}{x})^n} a_n = \sum a_n \frac{1}{x^n +1} \]
But now, for \(|x|<1\), we have that 
\[\sum a_n \frac{1}{x^n +1} = \sum_n a_n \sum_{k=0}^\infty (x)^{nk}\]
But now the only missing term is \( k=0\), so 
\[f(1/x) - f(x) = \sum_n a_n \sum_{k=0}^0 = \sum_n a_n \]
So, unforunately I don't think anything interseting is actually going on here, at least for this relationship.

[JmsNxn]

I apoligise for the late response, I should have mentioned this earlier- but I belive we have only stubled upon a trivial property  . In particular, its not that \( L(z) \) and \( L(1/z) \) are related in some interesting way-- they are actually just the same evaulation. In paritcular, notice that
\[ f(x) = \sum \frac{x^n}{1+x^n} a_n \]
Using the geometric series formula for \( |x|<1 \) gives
\[ f(x) = \sum_n a_n  \sum_k x^{n(k+1)}\]
Now take a look at 
\[f(1/x) = \sum \frac{(\frac{1}{x})^n}{1+(\frac{1}{x})^n} a_n = \sum a_n \frac{1}{x^n +1} \]
But now, for \(|x|<1\), we have that 
\[\sum a_n \frac{1}{x^n +1} = \sum_n a_n \sum_{k=0}^\infty (x)^{nk}\]
But now the only missing term is \( k=0\), so 
\[f(1/x) - f(x) = \sum_n a_n \sum_{k=0}^0 = \sum_n a_n \]
So, unforunately I don't think anything interseting is actually going on here, at least for this relationship.

I appreciate your candor, Caleb! This is definitely a trivial Fourier type identity. The point I am trying to make is this follows without this expansion! And that is definitely non-trivial!!!!

So, this result follows pretty trivially if we follow the coefficients on a generalized Lambert function. The real trouble would be to generalize this to arbitrary functions:

\[
\begin{align}
L(z) &: \mathbb{C} / U \to \mathbb{C}\\
L(z) &= L(0) + \frac{1}{2\pi i} \int_{|\zeta| =R} \frac{L(\zeta)}{\zeta - \frac{1}{z}}\,d\zeta\\
\end{align}
\]

Where we are given, if:

\[
\int_{|\zeta| = R} L(\zeta)\frac{d \zeta}{\zeta^{k-1}} = 0\\
\]

For \( k \ge 1\); then:

\[
L(1/z) = L(z) - L(0) + \frac{1}{2\pi i} \int_{|\zeta| = R} \frac{L(\zeta)}{\zeta}\,d\zeta\\
\]

This is HIGHLY non-trivial!

Yes for the Lambert function, this reduces to some fourier rearrangements and is trivial. But When we write this as arbitrary functions, this is absolutely not trivial!

This is a fucking crazy result, Caleb!

It is a generalized function result; for functions holomorphic on \(\mathbb{D}\) and on \(\mathbb{C}/\overline{\mathbb{D}}\). If they satisfy the above integral condition, they must satisfy this reflection formula.

For things like Lambert's functions, it is very obvious; just by rearranging the series. It is not obvious for general functions. We have gold here, bro

For example; if we have \(L(z) : \mathbb{D} \to \mathbb{C}\) with a natural boundary on \(|z| =1 \). There exists a function \(G(z) : \mathbb{C}/\mathbb{D} \to \mathbb{C}\) such that:

\[
G(z) = L(1/z) + C\\
\]

For a constant \(C\). This constant can be chosen uniquely such that:

\[
\int_{|\zeta| = R} \frac{G(\zeta)}{\zeta^{k+1}}\,d\zeta = 0\\
\]

For all \(k \ge 1\). And this gives our analytic continuation!!!!!!!!

Usually, finding \(C\) will be very hard; and in most cases won't exist. But if it does exist; we have continued a function \(L:\mathbb{D} \to\mathbb{D}\) to \(L:\mathbb{C}/U \to \mathbb{C}\). This is super fucking coool111

[JmsNxn]

Let's say we have a function:

\[
E(z) : \mathbb{H} \to \mathbb{C}\\
\]

Where \(\mathbb{H} = \{\Im(z) > 0\}\)...

Let's write the value:

\[
E(-\infty) = E_0 = \lim_{z\to -\infty} E(z)\\
\]

Now let's write the value:

\[
A = \int_{-\infty}^{\infty} E(t) \,d\mu\\
\]

Where \(d\mu\) is the measure induced on a circle mapped to the real line with the weight \(d\zeta/\zeta\).

Let's assume both of these things converge.

\[
E(\overline{z}) = E(z) -E_0 + A\\
\]

We have defined this function on the lower half plane, solely from its value on the upper half plane!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

EDIT!!!

To explain what I mean, let's take:

\[
F(z) = \prod_{n=1}^\infty \left(1-x^n\right)\\
\]

The value \(F(0) = 1\). Now there is a natural boundary at \(|z| = 1\). But this boundary is integrable. We have that:

\[
\int_{|z| =1} F(z)\frac{dz}{z} = 0\\
\]

We also have that:

\[
\int_{|z| =1} F(z)\frac{dz}{z^{k+1}} = 0\\
\]

For all \(k \ge 0\).

Using this Lambert identity we have that:

\[
F(1/z) = F(z) - 1+ 0\\
\]

And not only is this a viable function; it satisfies our uniqueness conditions:

\[
\int_{|z| = R} \frac{F(z)}{z^{k+1}}\,dz = \int_{|z| = 1/R} \left(F(z)-1\right)z^{k-1}\,dz\\
\]

If \(k \ge 1\)... And therefore this is a natural analytic continuation!

This shit is highly non-trivial BRO!

[Caleb]

I apoligise for the late response, I should have mentioned this earlier- but I belive we have only stubled upon a trivial property  . In particular, its not that \( L(z) \) and \( L(1/z) \) are related in some interesting way-- they are actually just the same evaulation. In paritcular, notice that
\[ f(x) = \sum \frac{x^n}{1+x^n} a_n \]
Using the geometric series formula for \( |x|<1 \) gives
\[ f(x) = \sum_n a_n  \sum_k x^{n(k+1)}\]
Now take a look at 
\[f(1/x) = \sum \frac{(\frac{1}{x})^n}{1+(\frac{1}{x})^n} a_n = \sum a_n \frac{1}{x^n +1} \]
But now, for \(|x|<1\), we have that 
\[\sum a_n \frac{1}{x^n +1} = \sum_n a_n \sum_{k=0}^\infty (x)^{nk}\]
But now the only missing term is \( k=0\), so 
\[f(1/x) - f(x) = \sum_n a_n \sum_{k=0}^0 = \sum_n a_n \]
So, unforunately I don't think anything interseting is actually going on here, at least for this relationship.

I appreciate your candor, Caleb! This is definitely a trivial Fourier type identity. The point I am trying to make is this follows without this expansion! And that is definitely non-trivial!!!!

So, this result follows pretty trivially if we follow the coefficients on a generalized Lambert function. The real trouble would be to generalize this to arbitrary functions:

\[
\begin{align}
L(z) &: \mathbb{C} / U \to \mathbb{C}\\
L(z) &= L(0) + \frac{1}{2\pi i} \int_{|\zeta| =R} \frac{L(\zeta)}{\zeta - \frac{1}{z}}\,d\zeta\\
\end{align}
\]

Where we are given, if:

\[
\int_{|\zeta| = R} L(\zeta)\frac{d \zeta}{\zeta^{k-1}} = 0\\
\]

For \( k \ge 1\); then:

\[
L(1/z) = L(z) - L(0) + \frac{1}{2\pi i} \int_{|\zeta| = R} \frac{L(\zeta)}{\zeta}\,d\zeta\\
\]

This is HIGHLY non-trivial!

Yes for the Lambert function, this reduces to some fourier rearrangements and is trivial. But When we write this as arbitrary functions, this is absolutely not trivial!

This is a fucking crazy result, Caleb!

It is a generalized function result; for functions holomorphic on \(\mathbb{D}\) and on \(\mathbb{C}/\overline{\mathbb{D}}\). If they satisfy the above integral condition, they must satisfy this reflection formula.

For things like Lambert's functions, it is very obvious; just by rearranging the series. It is not obvious for general functions. We have gold here, bro

OOHH I see, your right that this is much more interesting for arbitrary functions! 

So... a simple condition on the fourier coefficents, implies a strong coherence between the two components? That IS insane! I really have to go back and read all the work you've done so far-- I've been busy so I'm far behind on all the progress you've made but soon I'll have time to read it.

Anyway, it looks like you've done great work so far I'm excited to read up on all you've done! This is very exciting stuff! Also, my recent post on micks function makes me super super curious about what a function like his looks like outside its natural boundary, it will be interseting to test it out in that case.

[tommy1729]

Alright guys time to slighty debunk most of this 

f(1/x) = f(x) + C

does not make sense since

iterating 2 times gives 

f(x) = f(x) + 2 C

so C must be 0.

Although if the sum of coeff is indeed C = 0 it works.

( caleb was ahead of me when describing the coef sum , i wanted to post it )

but then it need to be consistant with plug in.
and it probably is.

Notice if C = 0 then f(1) is defined and continu just like I mentioned earlier.

But this is not super interesting.

it gives f(x) = f(x) and f(1/x) = f(1/x).


since 1/x is 2 cyclic our other function should also be 2 cylclic.

So more interesting is

f(1/x) = 1 - f(x).

since 1-x iterated 2 times is x.

this is equivalent to

f(1/x) + f(x) = 1.

another way is

f(1/x) + f(x) = g(x + 1/x).

And 

f(x) - f(1/x) = log(x).

or

f(x) + f(1/x) = log(x + 1/x) = log(x) + log(1+1/x^2)

it is clear that 1/x or conj(1/x) is key here together with the 2 cycle iterations.


***

I want to point out that not all of them have the same natural boundary or the unit boundary , it is just a way of thinking.

In particular if 

f(1/x) = g( f(x) ) 

and g maps cross the unit circle , we have analytic continuation there by g , so the unit circle is not the boundary.


see also the related topic :



regards

tommy1729

[JmsNxn]

Sorry, Tommy, I made a typo.

You are assuming this is a single holomorphic function. IT is not a holomorphic function, it is a generalized holomorphic function. I've been doing more reading on this.

Yes; if: \(f\) is holomorphic you are correct; excusing my typo, you are wrong. But these are not holomorphic functions. We are taking a function holomorphic on two disconnected domains. Where it has different values on both domains.

Perhaps to help you out I should write it out more explicitly.

Let:

\[
\begin{align}
f_1(x) &= \sum_{n=0}^\infty \frac{x^n}{1+x^n}\frac{1}{2^n}\,\,\text{for}\,\,|x|<1\\
f_2(x) &= \sum_{n=0}^\infty \frac{x^n}{1+x^n}\frac{1}{2^n}\,\,\text{for}\,\,|x|>1\\
\end{align}\\
\]

Then let's rewrite my equation as:

\[
f_1(x) = 2 - f_2(1/x)\\
\]

This is actually very easy to numerically verify, as Caleb pointed out.

\[
\frac{(1/x)^n}{1+(1/x)^n} = \frac{1}{x^n +1}\\
\]

Then:

\[
1-\frac{1}{x^n +1} = \frac{x^n+1}{x^n+1} - \frac{1}{x^n + 1} = \frac{x^n}{x^n + 1}\\
\]

Therefore:

\[
2 - f_2(1/x) = \sum_{n=0}^\infty \frac{1}{2^n}-\frac{1}{x^n+1} \frac{1}{2^n}\\
\]

Because: \(2 = \sum_{n=0}^\infty \frac{1}{2^n}\). And this just equals:

\[
f_1(x) = \sum_{n=0}^\infty \frac{x^n}{1+x^n} \frac{1}{2^n}\\
\]

---

This is not nonsense. We are working with two functions; which are "natural extensions of each other" to disconnected domains. This is known as a "generalized analytic function"; which is holomorphic on disjointed domains!

What I am saying, is that I am observing this phenomena for more exotic functions than this simple rendition. I can't prove it yet; but calculators don't lie... I should have written that the formula turns out like:

\[
L(1/x) = C - L(x)\\
\]

Which DOES NOT suffer your duplication formula. Sorry, I fucked up the signs before. I was just note dumping too fast!

---

EDIT!

I am referring to this as Lambert's Reflection formula. But it doesn't just follow for:

\[
\frac{x^n}{1+x^n}\\
\]

In the simple way Caleb presented it; or how I just explained. It also follows for:

\[
\frac{x^n}{(1+x^n)^2}\\
\]

But you have to be a bit more careful on how you rearrange the objects. And it also follows for higher powers. It also follows for mixing the powers. IT FUCKING FOLLOWS EVERYWHERE. The identity:

\[
L(x) = C - L(1/x)\\
\]

Is super fucking general, Tommy!!!!!!!

We can take \(L(x)\) for \(|x| >1\)--take a Cauchy integral, and reconstruct \(L(x)\) for \(|x| <1\). And it looks like a Lambert Reflection Formula.

I'm trying to map out the exact mathematics right now. I don't have it yet. But I am pretty fucking close.

[tommy1729]

Ok consider we have only 1 natural boundary.

Now let A be the interior of the boundary and B the exterior.

Then A union B = C the complex sphere without the boundary.

And A intersect B is empty.

Now we want a function L(z) such that 

1) L(z) is meromorphic on the whole complex sphere.

2) L(A) = B and L(B) = A , L(A) =/= A , L(B) =/= B

to be clear every element x of A maps to some L(x) in B , AND every element y of B maps to some L(y) in A. 

3) L© maps bijectively to C. ( f: C arrow C )

A consequence is that L(L(A)) = A , L(L(B) = B 

In fact it follows that L(L(z)) = z.

Not many functions have that property.

In fact the only possible ones are these

L(z) = (a z + b)/(c z + d)

such that one of them is true :

1) a + d = 0
2) L(z) = b - z
3) L(z) = b/z

 

Then every reflection formula equation is of the type 

L_1 ( f(z) ) = f( L_2(z) )

( And this even holds without the existance of a boundary )

(   in fact almost every reflection formula in math books is of the form L_1 ( f(z) ) = f( L_2(z) ) or f( L_1 (z) ) = f( L_2(z) )   )

***

now I mentioned ( here and else ) equations like

f(c/x) + f(x) = W

f(c/x) - f(x) = W - 2W log(x)/ln©

The first one is of the type L_1 ( f(z) ) = f( L_2(z) )

Since L_1 is entire this implies that A and B are entire regions.
So that is even better if you do not likes poles outside the boundary.


f(c/x) - f(x) = W - 2W log(x)/ln©

is a bit different.

one solution is f(x) = W * ln(x)/ln©


f(c/x) = W ln(c/x)/ln© = W (ln© - ln(x)) / ln© = W * ( 1- ln(x)/ln© )

so  f(c/x) - f(x) = W * ( 1 - ln(x)/ln© - ln(x)/ln© ) = W - 2W log(x)/ln©


However 

if f(x) = W * log(x)/ln©

f(c/x) = W ln(c/x)/ln© = W (ln© - ln(x)) / ln© = W * ( 1- ln(x)/ln© )

so  f(c/x) + f(x) = W * ( 1 - ln(x)/ln© + ln(x)/ln© ) = W 


so we solved ( at least one solution )

f(c/x) + f(x) = W

without any boundaries.

But also notice there may be other solutions.

And also notice the log© ; if c = 1 then

f(1/x) + f(x) = W

f(x) = f(1/x) = W/2 

is a solution and others ??

Does this mean we need expansions in terms of log to solve 


L_1 ( f(z) ) = f( L_2(z) )

?

What about the equation I mentioned earlier 

f(x) - f(c/x) = k ln(x)

It seems similar.



I know I only used "algebraic" ideas.
And I only used ideas based on the boundary , not even considering functions usually.

But as I show here , the options are already limited by just considering basic properties.

I did not consider other properties such as the determinant ( ad - bc ) and such ( the theory is rich ).


We use different approaches but seem to get to the same idea.
As often in math.

btw i see no reason to write f as two functions based on f(A) or f(B) but I guess that is a matter of taste.


@james : calm down

so I use only f and not 2 f's , never said it was analytic everywhere ofcourse so no contradiction just a function defined for A and B.

Also I am aware of the terminology generalized holomorphic function.

I did not say nonsense.

I feel you are not calm and rather offensive or defensive.

easy on the fu.. and !!!!!

I also want to point out I was more replying to caleb then to you.

And Im not saying you guys are fundamentally wrong or anything.

Just minor things.


Maybe I misinterpret your reaction.
Maybe I sounded hostile or disagreeing more than I intended.
( debunk was a big word )


@caleb ; what are you thoughts on my comment to your C needing to be 0 ? Or did you made a sign mistake ? I did not double check.


More investigation is needed.



regards

tommy1729

[JmsNxn]

@james : calm down

so I use only f and not 2 f's , never said it was analytic everywhere ofcourse so no contradiction just a function defined for A and B.

Also I am aware of the terminology generalized holomorphic function.

I did not say nonsense.

I feel you are not calm and rather offensive or defensive.

easy on the fu.. and !!!!!

I also want to point out I was more replying to caleb then to you.

And Im not saying you guys are fundamentally wrong or anything.

Just minor things.


Maybe I misinterpret your reaction.
Maybe I sounded hostile or disagreeing more than I intended.
( debunk was a big word )

Hey, Tommy

I apologize. I wrote my answer, realized it was nonsense, and then I retyped my answer. I think my first answer was much more hostile, which appears to be the one you read. I deleted it pretty quickly and rewrote it to be less aggressive. I apologize but emotional context can be hard to read from posts. Plus I was up all night, and irritated, but that's no excuse to not be civil. I apologize for being insensitive and rude. Just know it only came from a place of heated debate, not anger or anything.

I'd like to give a little justification on what I am talking about though; and trust me, I am just as confused as you, about why this reflection formula seems to be happening in ways that don't really make sense. If you can spot flaws in this deduction it's only going to help. But this reflection formula seems to be happening with all of these "quasi Lambert" series.

I'm going to take the simple function:

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

This function is holomorphic on \(\mathbb{C} / \mathcal{U}\) for the unit disk \(\mathcal{U}\)

The values:

\[
\frac{1}{2\pi i}\int_{|\zeta| = 2} \frac{f_2(\zeta)}{\zeta^{k+1}} \,d \zeta = 0\\
\]

For all \(k \ge 1\), this is a pretty easy check. Additionally, the coefficients of \(f_2(z)\) look like:

\[
f_2(z) = \sum_{k=0}^\infty z^k \sum_{d \vert k} d \frac{(-1)^{d+1}}{2^{k/d}}\\
\]

*********
EDIT to reflect the main point:

Theorem:
This function satisfies the reflection formula: \(f_2(z) = f_2(1/z)\)...
*********

The values of:

\[
\frac{1}{2\pi i}\int_{|\zeta| = 2} \frac{f_2(\zeta)}{\zeta^{k+1}} \,d \zeta= -\sum_{d \vert k} d \frac{(-1)^{d+1}}{2^{k/d}}\\
\]

For \(k \le - 1\). This divisor sum is done for \(- k \le d \le -1\), so we get another negative, and:

\[
\frac{1}{2\pi i}\int_{|\zeta| = 2} \frac{f_2(\zeta)}{\zeta^{k+1}} \,d \zeta= f_2^{(-k)}(0)/(-k)!\\
\]

***I fucked up earlier, but the value at \(k=0\) is just \(f_2(0)\)....

Therefore; if we take the sum:

\[
\sum_{k=0}^\infty \frac{z^k}{2\pi i}\int_{|\zeta| = 2} f_2(\zeta)\zeta^{k-1}\,d \zeta = f_2(z)\\
\]

You'll note, for \(0 < |z| < 1/2\) so that \(|z \zeta| < 1\), and the series converges; we have the expression:

\[
\frac{1}{2\pi i} \int_{|\zeta| = 2} \frac{f_2(\zeta)}{1/\zeta - z}\,d\zeta = f_2(z)\\
\]

BUT! we can also just evaluate this integral using residue calculus, which comes out as:

\[
f_2(1/z)\\
\]

so long as \(1 < |z| < 2\)...

VOILA!

This is happening and it's fucking weird, and I don't understand why... I cannot prove this, and obviously this statement of equations is not a proof, because I have hand waved a good amount of things, but kinda, not really at all. We still get this reflection--or something that's looking like a reflection. I'm planning on doing a full write up, but I have numerically confirmed this reflection, and it's confusing as hell. I'm doing flip it on its head too many times...

I apologize for note dumping too much, but this is just beyond fascinating to me. There's some kind of reflection formula happening, and it's odd...

I am seeing a similar reflection formula happening for all:

\[
f_m(\theta; z) = \sum_{n=0}^\infty \frac{z^n}{\left(1 + e^{i\theta n}z^n\right)^m} \frac{1}{2^n}\\
\]

And it's even happening when \(m = m(n)\) is non constant--takes \(\mathbb{N} \to \mathbb{N}\)... It also happens for arbitrary \(\ell^1\) sequences \(d_n\) rather than the simple \(\frac{1}{2^n}\)...

I can't prove it yet, but this reflection formula is looking very very universal.

Sincere regards. Sorry for the outburst of       . Should be a good reminder not to stay up for 24 hours and start posting math, lmao!

EDIT!

OMG! I think you're right tommy. The reflection formula gets a tad more complicated for this case. As I was just checking Taylor coefficients; they appeared to be right. But the signs seem to be changing. You're right. It isn't \(L(1/z) = C-L(z)\); but it looks VERY VERY close to this; but there seems to be an extra term that I lost! Man this problem is giving me a fucking headache!

EDIT2!

****I edited this post to make the math correct*****

GODDAMN IT! I FORGOT A MAIN POINT! When \(m=2\), the constant term is neglected. So the answer is:

\[
f_2(1/z) = f_2(z)\\
\]

Which is in line with what you are saying! When \(m=3\) and etc... things are going to get really spicy, though! I see my dumb mistake again. I was forgetting constant terms; and ignoring them too much. But in the end it does look kind of like \(L(1/z) = C - L(z)\), but there's something extra going on I can't map yet. It probably looks more like \(L(1/z) = (-1)^m L(z) + O(z^m)\) or something like this.........