Divergent Series and Analytical Continuation (LONG post)

[JmsNxn]

ALRIGHT!

I FINALLY FUCKING HAVE IT!

THIS IS THE MOST BAFFLING FUCKING FUNCTION MAN!

I AM HERE TO REDEEM MYSELF!

I AM GOING TO GO VERY SLOW!

---

Let \(f(x)\) be known as Caleb's function; it is written as:

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

The theorem we are going to present; is that for \( k \ge 1\) the value:

\[
F_k = \frac{1}{2\pi i} \int_{|z| = 2} f(z)\frac{dz}{z^{k+1}}\,dz = 0\\
\]

And that \(F_0 = 3/2\).

I have confirmed this with pari-gp; and am finally confident enough to answer this stupid annoying ass question, lmao.

We additionally, get a more complex formula for \(k \le -1\).

We start by showing that:

\[
a_{kn} = \frac{1}{2\pi i} \int_{|z| = 2} \frac{z^n}{1+z^n}\frac{dz}{z^{k+1}} = 0\\
\]

For \(k \ge 1\).

This value just equals:

\[
b_{kn} + \frac{1}{n}\sum_{q^n = -1} q^{-k}\\
\]

If \(n \not | k\) then this value is just zero. This is because \(b_{kn}\) is zero. And because the sum on the right is zero.

If \(n | k\) then something unbelievable happens; then \(k = nm\) and:

\[
(-1)^{m+1} + (-1)^m = 0\\
\]

Like holy fuck, how did I miss this.

The only time this doesn't happen is when \(k=0\), upon which the sum on the right is \(1\), and \(b_{kn} = 0\). So we end up with:

\[
F_0 = f(0) + \sum_{n=1}^\infty \frac{1}{2^n} = 3/2\\
\]

And for \(F_k\) we just get:

\[
F_k = \sum_{n=0}^\infty \frac{a_{kn}}{2^n} = 0\\
\]

This means that:

\[
B(x) = \frac{1}{2\pi i} \int_{|z| =2} f(z) \frac{dz}{z-x} = \frac{3}{2}\\
\]

Which is exactly as you saw Caleb.

---

When \(k \le -1\) we basically get the exact same thing. Except the term \(b_{kn} = 0\). So there's no cancellation and we get:

\[
F_k = -\frac{f^{(-k)}(0)}{(-k)!}\\
\]

---

Okay for fucks SAKES MAN!!!

I'm taking a break now. Been yelling at pari for the past 2 hours, lmao!!!!!

The cool formula I actually wanted is that:

\[
2-f(x) = \sum_{k\in \mathbb{Z}} F_k x^{-k}\\
\]

We should also be able to generalize this. Assume that \(\sum_{n=0}^\infty |a_n| < \infty\); then if:

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

Then:

\[
F^G_k = \frac{1}{2\pi i} \int_{|z| = 2} G(z) \frac{dz}{z^{k+1}}\\
\]

We have

\[
F^G_0 = \frac{1}{2}a_0+ \sum_{n=1}^\infty a_n\\
\]

And

\[
F^G_k = 0\,\,\text{for}\,\,k \ge 1\\
\]

And for \(k \le -1\); we have:

\[
F^G_k = - \frac{G^{(-k)}(0)}{(-k)!}\\
\]

So that the general formula is for \(|x| > 1\):

\[
C-G(1/x) = \sum_{k\in \mathbb{Z}} F^G_k x^k\\
\]

where \(C\) is an easy constant.

[Caleb]

Nice work you have here-- this gives us a unique way to connect these series outside their natural boundary into a function inside the natural boundary!

Also, based on your work, we get the 'Cauchy-like' integral formula
\[ F(x) = a_0 + \int_C \frac{F(z)}{z-\frac{1}{x}}dz\] 
Where the contour C is taken outside the natural boundary and the pole at \(z = \frac{1}{x}\) needs to outside the contour integration. Actually, we could do that, or we could only pick up the pole at \(z = \frac{1}{x}\) is pick up no other residue (but then the constant term is a bit more complciated I think). Either way, we now have an easy way to use contours only inside of the natural boundary to compute values on the inside!

[JmsNxn]

  

Also, you are very much discussing the first step of the little circle method here.

The second part is to "estimate" the half circles you have drawn, and the arcs of the circle.

For Caleb's function \(f(x)\)--we don't have to do this because we are defined outside of the unit circle. But when we are only defined within the unit disk, the only way to do this is to estimate the circle and the arcs for the partial terms.

If that makes sense

[JmsNxn]

Nice work you have here-- this gives us a unique way to connect these series outside their natural boundary into a function inside the natural boundary!

Also, based on your work, we get the 'Cauchy-like' integral formula
\[ F(x) = a_0 + \int_C \frac{F(z)}{z-\frac{1}{x}}dz\] 
Where the contour C is taken outside the natural boundary and the pole at \(z = \frac{1}{x}\) needs to outside the contour integration. Actually, we could do that, or we could only pick up the pole at \(z = \frac{1}{x}\) is pick up no other residue (but then the constant term is a bit more complciated I think). Either way, we now have an easy way to use contours only inside of the natural boundary to compute values on the inside!

Yes!

This was the kind of idea I smelled when I first saw it. Though it definitely turned out different than I originally expected. I imagine something similar will happen in more general instances.

If we let:

\[
F(z) : \mathbb{D}_{|z| < 2}/U \to \mathbb{C}\\
\]

Where there is a wall of singularities along \(|z| = 1\); but other wise everything is holomorphic, we should have something very very similar. I'm going to see if this continues to happen--or if something close to this happens. I do not think your function \(f\) is too particularly special; so something similar should pop out. I'm mostly worried if the wall of singularities needs to be simple poles or not.

I'm going to start by testing this hypothesis on:

\[
f_{\theta}(x) = \sum_{n=0}^\infty \frac{x^n}{1+e^{i\theta n} x^n} \frac{1}{2^n}\\
\]

This will scramble the poles around; but other than that, should keep the same structure as your function--when \(\theta = 0\). If a similar result follows; then we know that this result is true up to permutation of the locations of the wall of singularities.

The next step would be to take:

\[
f_\theta^K(x) = \sum_{n=0}^\infty \frac{x^n}{(1+e^{i\theta n} x^n)^{k_n}} \frac{1}{2^n}\\
\]

for a sequence of natural numbers \(K = \{k_n\}_{n=0}^\infty\). This would allow us to look at when we have second order, third order, etc... poles. This will probably create more problems, but I don't think it'll be too unreasonable.

[JmsNxn]

Okay, let's keep the same notations as before, but write:

\[
F_k^\theta = \frac{1}{2\pi i} \int_{|z|=2} f_\theta(z) \frac{dz}{z^{k+1}}\\
\]

The value:

\[
F_0^\theta = \frac{1}{2} + e^{i\theta}\sum_{n=1}^\infty \frac{e^{-i\theta n}}{2^n} = \frac{1}{2} + \frac{1}{2}\frac{1}{1-\frac{e^{-i\theta}}{2}}\\
\]

This is observed because, for \(n \ge 1\):

\[
\frac{1}{2\pi i} \int_{|z|=2} \frac{z^n}{1+e^{i\theta n} z^n} \frac{1}{z}\,dz = e^{i\theta (1-n)}\\
\]

Which is just by linear substitution.

We know that:

\[
F_k^\theta = 0\\
\]

For \(k \ge 1\) just as well. (Multiply the left hand side by \(e^{i\theta(k+1-n)}\) and we're reduced to the same integral as before). The real trouble happens for \(k \le -1\). If we recall our value \(b_{kn}\); then:

\[
\frac{1}{2\pi i} \int_{|z|=2} \frac{z^n}{1+e^{i\theta n} z^n} \frac{1}{z^{k+1}}\,dz =\frac{e^{i\theta(k+1-n)}}{2\pi i} \int_{|z|=2} \frac{e^{i\theta n}z^n}{1+e^{i\theta n} z^n} \frac{1}{e^{i\theta (k+1)}z^{k+1}}\,dz = -e^{i\theta(k+1-n)} b_{kn}\\
\]

................

Now let's sum across \(n\) with the \(1/2^n\) weight; recalling that \(k \le -1\)...

\[
F_k^\theta = -\sum_{n=0}^\infty e^{i\theta(k+1-n)} \frac{b_{-kn}}{2^n}\\
\]

This equals:

\[
-e^{i\theta(k+1)} \sum_{n=0}^\infty e^{-i\theta n} \frac{b_{-kn}}{2^n}
\]

This is definitely not so obviously related to \(-f_\theta^{(-k)}(0)/(-k)!\). This is some kind of spectral decomposition...

I'm really not sure...

But now when we write:

\[
\sum_{k\in\mathbb{Z}} F_k^{\theta}x^k = F_0^\theta -\sum_{k=1}^\infty e^{i\theta(1-k)} x^{-k}\sum_{n=0}^\infty e^{-i\theta n} \frac{b_{-kn}}{2^n}\\
\]

This still looks A LOT like \(f(1/x)\)--but there's some kind of fourier magic I can't parse....

The real magic is that:

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

So the same result seems to be popping out. I'm just not sure why yet. But I think I see why... We have to monitor the order of \(\theta\) though... this is really weird. But it's looking very very similar.
.................................................................................................................

Oh..... I'm a dumb ass it's a rotation.... This does sum to some thing close to \(f_\theta(x)\); but we apply a rotation in \(x\) before the sums line up. I'll explain in a bit. Need to double check everything....

Okay!

For example if you take \(\theta = \pi/2\), then:

\[
\sum_{k\in\mathbb{Z}}^\infty F_k^{\pi/2} x^k = C - f_{\pi/2}(-1/x)\\
\]

So the rotation is only one rotation. I'm not sure of the general rule. But i'll update as I figure it out. I'm not sure why \(i^4 = 1\) only induces a negative... So this is definitely a brain scratcher.... I thought it was going to be some \(if_{\pi/2}(-i/x)\) or something--but it appears to be much different...

The best I can say for the moment, is that:

\[
|F_k^\theta| = \left| \frac{f_\theta^{(-k)}(0)}{(-k)!}\right|\\
\]

And that these are rational rotations at worse. And they are some what related to the initial \(\theta\) value...



FUCK I'm an idiot.

Take the value:

\[
F_{-1}^{\theta}/f^{(1)}_\theta(0) = \zeta_\theta\\
\]

Then:

\[
\sum_{k\in \mathbb{Z}} F_k^{\theta} x^k = C - f_\theta(\zeta_\theta/x)\\
\]

EASY FUCKING BREEZY!!!

This actually makes perfect sense too!

All the pari code lines up, this is the answer.

So you're formula still holds, Caleb. You just need to insert a rotation depending on \(\theta\). That's it! Makes perfect sense too...

[JmsNxn]

I am going to add my hand at an entire function here.

Let:

\[
f(z,q) = \sum_{n=0}^\infty \frac{z^n}{1+q^nz^n}\frac{1}{2^n}\\
\]

Then:

\[
\frac{\partial}{\partial z}\Big{|}_{z=0} f(z,q) = 1/2\\
\]

Let:

\[
J(q) = 2F_{-1}(q) = \frac{1}{\pi i} \int_{|z| =2} f(z,q)\,dz\\
\]

---

This function looks modular as fuck.

We get that:

\[
f(0,q)-f(J(q)z,q) = \sum_{k=1}^\infty F_{-k}(q)z^k\\
\]


Oh and I am taking the modular assumption that \(q = e^{i\theta}\). I should've metioned that, lol.

---

Holy, fuck!

I've been sitting on this formula for a while.

But let's let \(b_{kn} = (-1)^{n/k+1}\) if \(k=nd\) and \(b_{00} = 1/\)--otherwise it's zero. Then Caleb's original function is written as:

\[
f(x) = \sum_{k=0}^\infty\sum_{n=0}^\infty \frac{b_{kn}}{2^n}x^k\\
\]

The thing is; we can reduce this summation.

We only have non zero values when \(k = nd\) for some value \(d \in \mathbb{N}\); where we then get:

\[
f(x) = 1/2 + \sum_{k=1}^\infty x^k \sum_{d \mid k} \frac{(-1)^{d+1}}{2^{k/d}}\\
\]

...........

When we write:

\[
f^{(k)}_\theta(0) =e^{i\theta k} k! \sum_{d \mid k} e^{i\theta k/d}\frac{(-1)^{d+1}}{2^{k/d}}
\]


We have instantly added a summative/ to /multiplicative relationship \(\sum_{d \mid k}\) and added the multiplicative/ to /summative relationship \(e^{i\theta d}\). Rather than multiplying exponentials; we've taken factors and put them in the exponentials....

I'm not saying this is modular. But this is the shit modular relationships are built from.......

I HAVE DEFINITELY MADE SOME NUMERICAL MISTAKES! I'M TRYING TO CARVE A PATH IS ALLL!!!!



EDITTTTTT:::::


\[
\frac{f^{(k)}_\theta(0)}{k!} = e^{i\theta k}\sum_{d \mid k} e^{i\theta k/d}\frac{(-1)^{d+1}}{2^{k/d}}
\]


This is the perfect formula. Only trouble is at \(k=0\); where we have to just sub in \(1/2\). But if you know how technical divisor sums are this is fine.

The function:

\[
J(e^{i\theta}) =-2\sum_{d \mid -1} e^{-i\theta/d}\frac{(-1)^{d+1}}{2^{-1/d}}
\]

Despite this expression being meaningless.......

We have given meaning to it; by writing:

\[
2F_{-1}^{\theta} = J(e^{i\theta})\\
\]

Or....

\[
J(e^{i\theta}) = \frac{1}{\pi i} \int_{|z|=2} f(z,e^{i\theta})\,dz\\
\]

The function \(J\) takes the unit circle to itself. But also takes the upper half plane to itself. There is a boundary on the unit circle; but it's different. It looks fucking modular!

[JmsNxn]

From \(t =0\) to \(t = 2 \pi\)... Then \(J(e^{it})\) graphs as:

   

This looks just like a very simple modular kind of function. We only need a single frequency to get it.
I suspect this is because it's only simple poles on the wall. If we have arbitrary order poles; we will get more complication.

Please remember the entire point is that:

\[
f(0,e^{it}) - f(J(e^{it})x,e^{it}) = \sum_{k=1}^{\infty} F_{-k}^{t}x^k\\
\]

Recalling that:

\[
F_{-k}^t = \frac{1}{2\pi i} \int_{|z| =2} f(z,e^{it})z^{k-1}\,dz\\
\]

To remind everyone:

\[
f(x,e^{it}) = \sum_{n=0}^\infty \frac{x^n}{1+e^{itn}x^n} \frac{1}{2^n}\\\
\]

[Caleb]

From \(t =0\) to \(t = 2 \pi\)... Then \(J(e^{it})\) graphs as:



This looks just like a very simple modular kind of function. We only need a single frequency to get it.
I suspect this is because it's only simple poles on the wall. If we have arbitrary order poles; we will get more complication.

Please remember the entire point is that:

\[
f(0,e^{it}) - f(J(e^{it})x,e^{it}) = \sum_{k=1}^{\infty} F_{-k}^{t}x^k\\
\]

Recalling that:

\[
F_{-k}^t = \frac{1}{2\pi i} \int_{|z| =2} f(z,e^{it})z^{k-1}\,dz\\
\]

To remind everyone:

\[
f(x,e^{it}) = \sum_{n=0}^\infty \frac{x^n}{1+e^{itn}x^n} \frac{1}{2^n}\\\
\]

I haven't read through all that you've written yet (that will take some time-- I'd like to think about all your points in detail). However, I can perhaps provide some reading if you aren't already aware of these things.

First, on your point that it looks modular, you can check out: the wiki page on lambert series. Part of my motivation for studying the series here is that they have really nice arithmetic properties. 

Second, to your point about it looking like some fourier series decomposition and the connection between \( \frac{f^{(-k)}}{(-k)!} \) and \(b_{-k}\), I have this MO post, which discusses the relationship between \( \sum a_n x^n\) and \( \sum a_{-n} x^{-n} \). Basically, I expect that these two series are always equal, up to the residues of \( a_n \) when we view \( a_n \) as a complex-analytic function defined on the plane. Since \( a_n = f^{(n)}{n!} \), then we should expect that \( \sum \frac{f^{(n)}}{n!} x^n  = -\sum \frac{f^{(-n)}}{(-n)!} x^{-n} \)

Also, a bit of a tangent, but its probably quite intersting to study the series
\[\sum \frac{x^p}{1-x^p}\]
where \(p\) is prime. If you view this as an arthemtic function, it becomes \( \sum \omega(n) x^n \) where \( \omega(n) \) counts the number of primes that divide a given n. Thus, if one found a different way to continue the series (besides just evaulting the series directly) it would provide a way to basically count the number of prime factors without actually computing the primes. The series also has the really nice property that all of its poles are 'unique' in some way, in the sense that the poles of different terms \( \frac{1}{1-x^p} \) never intersect, since to intersect they would have to share a prime factor (like, \( \frac{1}{1-x^2} \) and \( \frac{1}{1-x^4} \) have some of their poles intersect together).

[JmsNxn]

I am going to do much more reading on this.  I thought I'd add, that I posted some of the last stuff as word soup, and I may have misrepresented the function \(J(q)\). As far as I can tell \(J(q) = \frac{-1}{q^2}\). But when we start introducing \(m\) into the picture I'm expecting this to become much more complex.

I'm going to introduce the function:

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

We are getting much of the same results here. We can pretty much always assume that \(\theta =0\) without loss of generality. The results are the same upto a change of variables--which is why I think there's something modular happening here.

With this function we have:

\[
\frac{1}{2\pi i} \int_{|z| = 2} f_m(\theta;z) \frac{dz}{z^{k+1}} = 0\,\,\text{for} k \ge 1\\
\]

When \(k=0\) and \(m > 1\) we have:

\[
\frac{1}{2\pi i} \int_{|z| = 2}f_m(\theta;z) \frac{dz}{z^{k+1}} = 2^{-m}\,\,\text{for} k =0 \,\,\wedge m > 1\\
\]

The tricky part happens when \( k < 0\). Something really weird is happening here. It seems to be producing some kind of linear transformation of \(\frac{d^k}{dz^k}\Big{|}_{z=0} f _m(\theta;z)\), but it's not obvious.

This produces something very strange. For \( k > 0\) we have:

\[
b_{kn}^{(m)} = - e^{i\theta(k+1- n)}\sum_{q^n = -1} \text{Res}\left(\frac{z^{n-k-1}}{\left(1+e^{i\theta n}z^n\right)^m},z=q\right)\\
\]

where:

\[
\frac{z^n}{\left(1+e^{i\theta n}z^n\right)^m} = \sum_{k=0}^\infty b_{kn}^{(m)} z^k\\
\]

But when \(k < 0\); there is no obvious formula that appears. I thought we'd get something similar; but there appears to be something tricky going on. I think we're collecting more negatives and stuff. I'm having a little trouble finding a quick formula for the residues; because they are second order or higher residues, and for fucks sakes that's a headache, lol.

I will read your links and get around to brute forcing this stupid residue.

Also, to add to how cool this is:

\[
\sum_{m=1}^\infty f_m(\theta ; x) = \sum_{n=0}^\infty \frac{e^{-i\theta n}}{2^n} = \frac{e^{i\theta}}{e^{i\theta} - \frac{1}{2}}\\
\]

Choosing any coefficients \(a_n\) rather than \(\frac{1}{2^n}\) changes nothing from all the results I've written. I just like using this sequence because it's fast and simple. Getting a prime series will be a consequence of this; choosing a prime indicator function produces your prime series. Actually, choosing \(d_n\) where \(d_n\) is a modular sequence; then:

\[
\sum_{m=1}^\infty f_m(\theta , x) = \mathfrak{M}(\theta)\\
\]

Where \(\mathfrak{M}\) is modular.

This would mean the series:

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

Is intimately related to the modular function:

\[
\mathfrak{M}(\theta) = \sum_{n=0}^\infty d_ne^{i\theta n}\\
\]

So long as we take the modular group mod 2 \pi rather than mod 1; we should probably switch to \(e^{-2\pi i \theta}\) rather than \(e^{i\theta}\); this will clean up some of the algebra...

---

Okay! So I found the recurrence relationship for the residue. Let's call:

\[
a_{kn}^m = \frac{1}{2\pi i} \int_{|z| = 2} \frac{z^n}{(1+z^n)^m} \frac{dz}{z^{k+1}}\\
\]

Then this is equivalent to:

\[
a_{kn}^{m} = \sum_{j=1}^\infty (-1)^{j+1} a_{(k+jn)n}^{m-1}\\
\]

Because:

\[
\frac{1}{(1+z^n)^m} = \frac{1}{(1+z^n)^{m-1}} \sum_{j=1}^\infty (-1)^{j+1} z^{-j}\\
\]

When \(|z| > 1\).

This is a finite sum, because k+jn is eventually positive, and then \(a_{(k+jn)n}^m = 0\). Then our sum is written as:

\[
F_k^m = \frac{1}{2\pi i} \int_{|z| = 2} f_m(0;z) \frac{dz}{z^{k+1}}\\
\]

Where:

\[
F_k^m = \sum_{n=0}^\infty \frac{a_{kn}^m}{2^n}\\
\]

So, the function:

\[
\begin{align}
F(x) &=\sum_{k=1}^\infty F_{-k}^2 x^k\\
&= \sum_{k=1}^\infty x^k\sum_{n=0}^\infty \frac{a_{kn}^2}{2^n}\\
&= \sum_{k=1}^\infty x^k \sum_{n=0}^\infty \frac{\sum_{j=1}^\infty (-1)^{j+1}a_{(k+jn)n}}{2^n}\\
\end{align}
\]

The value \(a_{kn}\) is non-zero only when \(k = -dn\); upon which it equals \((-1)^d\). So this becomes:

\[
\sum_{j=1}^\infty (-1)^{j+1}a_{(k+jn)n} = (-1)^{d+1}d\\
\]

And Therfore!

\[
F^2(x) = \sum_{k=1}^\infty x^k \sum_{d|k} d\frac{(-1)^{d+1}}{2^{k/d}}\\
\]
..........


ALL WE DO IS ADD A POWER OF \(d\)!!!! I think this will continue and that, my conjecture is:

\[
\sum_{k=1}^\infty F_{-k}^m x^k = \sum_{k=1}^\infty x^k \sum_{d|k} d^{m-1}\frac{(-1)^{d+1}}{2^{k/d}}\\
\]

.....

NVM

\[
F_{-k}^3 = \sum_{d|k} \frac{d(d-1)}{2} \frac{(-1)^d}{2^{k/d}}\\
\]

So the answer is:

\[
h_m(d) = \sum_{j_1=1}^d \sum_{j_2 = 1}^{j} ... \sum_{j_{m-1}=1}^{j_{m-2}} 1\\
\]

Which is simply:

\[
h_m(d) = d(d-1)(d-2)\cdots (d-m+2)/(m-1)!\\
\]

which is:

\[
\begin{align}
h_1(d) &= 1\\
h_2(d) &= d\\
h_3(d) &= d(d-1)/2\\
h_4(d) &= d(d-1)(d-2)/6\\
&\vdots\\
\end{align}
\]

Or; in a clean formula:

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

And:

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

Notice we have to add in a negative based on the parity of \(m\)!

And now we get that:

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


THIS IS NOT THE COEFFICIENTS OF:

\[
f_m(0;x) = \sum_{k=0}^\infty B_k^m x^k\\
\]

They are close; but something strange is happening here... And I can't map it yet.

If we add in \(\theta\), I can still map this:

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

Then:

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

[JmsNxn]

Okay, so I'm going to call the Error function \(E^m(x)\). And I'll describe it in full detail.

Let's call the function:

\[
F^m(x) = \sum_{k=1}^\infty F_{-k}^m x^k = \sum_{k=1}^\infty x^k \sum_{d | k} h_m(d) \frac{(-1)^{m+d}}{2^{k/d}}\\
\]

Then, we will call:

\[
E^{m}(x) = f_m(x)-f_m(0) +(-1)^{m+1} F^m(x)\\
\]

For \(m=1,2\) this function is just \(0\). For \(m=3,4\) it's the same function. For \(m=4,5\) it gets a bit wilder. And as I continue shit starts to get a bit divergey; but that's because I'm doing quick calculations and avoiding large number calculations.

All the coefficients are rational of \(E^m\); and it looks pretty god damn nice, regardless of \(m\). But it also looks pretty random with the placement of these rational coefficients.

Also, it's super super suspect; that the only values which are nonzero are when \(m > 2\); which is just like modular functions--the Eisenstein series is only interesting when \(m > 2\).

The coefficients of \(E^m(x)\) also just seem to be:

\[
a_{mk}/2^k\\
\]

where \(a_{mk}\) is a natural number. So I may be missing an extra term that doesn't appear for \(m=1,2\). Jesus this shit is annoying.

---

NVM

I figured it out. And I can explain the difference.

Let's write:

\[
f_m^{k}(0) = B_k^m\\
\]

Where:

\[
f_m(x) = \sum_{k=0}^\infty B_k^m x^k\\
\]

Then:

\[
f_m(x) = \sum_{n=0}^\infty \frac{x^n}{\left(1+x^n\right)^m} \frac{1}{2^n}\\
\]

But it also equals:

\[
f_m(x) = \sum_{j=0}^\infty \sum_{n=0}^\infty (-1)^j x^{nj} \frac{x^n}{\left(1+x^n\right)^{m-1}}\frac{1}{2^n}\\
\]

This produces a SIMILAR recursive equation as with \(F_{-k}^m\), but it is not the same. We gather some extra terms