Divergent Series and Analytical Continuation (LONG post)

[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.........