Divergent Series and Analytical Continuation (LONG post)

[Gottfried]

1) I like vrey much the attempt of Caleb to find contoures of a "greater picture" of this type of divergence and the hope of finding a way to extend the function(s) beyond its/their natural boundary.

2) I didn't see in the above discussin (overlooked?) links to two Q&A in MO which Caleb and James surely know, but most other tetforum-members possibly not. The following links seem to me much fruitful for more analytical and heuristic properties of this style of functions and their evaluations as sums/series:

 - https://mathoverflow.net/q/198665/7710  Properties of summation of an eventually oscillating series. A couple of interesting answers about exactly Caleb's series can be found, I liked (and like) always the contribution(s) of Robert Israel as very often much enlightening.

- https://mathoverflow.net/a/198871/7710  my answer to the above, showing some specific analysis/heuristics using the reordering of the double-series like Caleb has shown it here.

- https://mathoverflow.net/q/201098/7710  a follow up question of mine, trying to put that specific series *in a greater context* of what I called here sometimes "(alternating) iteration series [AIS]" on which I had the similar property of oscillation, but having no answer so far. 


(Unfortunately, after pandemy years and own general health/age issues I can no more undertake reasonable attacks on this matter myself, so I likely cannot be of much help in this issue.)

[Caleb]

1) I like vrey much the attempt of Caleb to find contoures of a "greater picture" of this type of divergence and the hope of finding a way to extend the function(s) beyond its/their natural boundary.

2) I didn't see in the above discussin (overlooked?)  links to two Q&A in MO which Caleb and James surely know, but most other tetforum-members possibly not. The following links seem to me much fruitful for more analytical and heuristic properties of this style of functions and their evaluations as sums/series:

 - https://mathoverflow.net/q/198665/7710  Properties of summation of an eventually oscillating series. A couple of interesting answers about exactly Caleb's series can be found, I liked (and like) always the contribution(s) of Robert Israel as very often much enlightening.

- https://mathoverflow.net/a/198871/7710  my answer to the above, showing some specific analysis/heuristics using the reordering of the double-series like Caleb has shown it here.

- https://mathoverflow.net/q/201098/7710  a follow up question of mine, trying to put that specific series *in a greater context* of what I called here sometimes "(alternating) iteration series [AIS]" on which I had the similar property of oscillation, but having no answer so far. 


(Unfortunately, after pandemy years and own general health/age issues I can no more undertake reasonable attacks on this matter myself, so I likely cannot be of much help in this issue.)

Gottfried! Its great to see you here! I'll surely check out your MO quesiton, it looks to have some deep and interesting content. 

Also, as to your second link, the similarity between my approach and your is not a coincidence. The methods I devoloped were inspired by seeing your answer! In fact, your last sentence on your answer was

Possibly we have here something like in the Ramanujan-summation of divergent series where we have to add some integral to complete the divergent sums, but I really don't know.

And, this is essentially exactly what I have done in this post (the main difference being that the integral in this case goes in the complex plane as opposed to the Ramanjuan integral which is on the real line).

[JmsNxn]

https://mathoverflow.net/questions/19866...g-function

Oh wow, that is a beautiful thread, Gottfried!

And the answer is through Fourier Transform...and what is the Mellin transform but a variable changed Fourier Transform...

But I digress.

---

I spent the whole morning--scribbling in a notebook--looking at Caleb's function:

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

This function really fascinates me, because simply evaluating \(|x|>1\) should be the analytic continuation of \(f(x)\) for \(|x| < 1\). But what constitutes an "analytic continuation" is not what this is. It is in no way anything that fits in standard complex analytic theory. There's something special going on with this function.

I realized I made some dumb mistakes in my original evaluation--but I was on the right track. So I'm going to start over. Sorry, I was up all night and then wrote an analysis in the wee hours of the morning. So I was all over the place. I'm going to start over again with a fresh mind--after a pot of coffee and tea. Forgive me, Caleb; if this isn't what you are looking for. I may be going out on a tangent--but this function really fascinates me!

---

We start by defining:

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

If \(f(x)\) were holomorphic for \(|x| < 2+\delta\), then this value would simply be \(\frac{f^{(k)}(0)}{k!}=  B_k\). So we can write Caleb's function for \(|x| < 1\) as:

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

We will also call:

\[
\frac{x^n}{1+x^n} = \sum_{k=0}^\infty b_{kn} x^k = \sum_{j=0}^\infty (-1)^{j} x^{n(j+1)}\\
\]

So that:

\[
\sum_{n=0}^\infty \frac{b_{kn}}{2^n} = B_{k}\\
\]

The values \(b_{kn}\) are fairly elementary. If \(k \equiv 0\pmod{n}\) then \(b_{kn} = (-1)^{\frac{n}{k}-1}\) and \(k > 0\). And otherwise it is just \(0\).

---

I am now going to assume that \(k \in \mathbb{Z}\) for \(F_k\). So I am assuming that "perhaps" there is a nontrivial Laurent part being added to this. I am now going to do a similar thing, but for the poles of \(\frac{x^n}{1+x^n}\). And by such, I will call:

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

The function \(\frac{z^n}{1+z^n}\frac{1}{z^{k+1}}\) has poles at the \(n\)'th roots of unity of \(-1\); and for \(k\ge n\), it has a pole at \(0\). The pole at \(0\) is easy to handle; it's just the value \(b_{kn}\). The poles at the roots of unity are a bit different. But they are all simple poles; and therefore the standard formula for the formula of a simple pole is given as:

\[
\text{Res}\left( \frac{g(z)}{h(z)},q\right) = \frac{g(q)}{h'(q)}\\
\]

By which we get that:

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

We are only going to care about the second part; as the second part is the interesting part. We already know what \(b_{kn}\) behaves like when we sum it; we care about how the poles react when we sum it. So to begin; we're going to rewrite this strange sum:

\[
\sum_{q^n = -1} \frac{1}{q^k} = \sqrt[n]{-1}\sum_{j=0}^{n-1} e^{2\pi i \frac{k j}{n}} = \sqrt[n]{-1}\frac{1- e^{2\pi i k}}{1-e^{2\pi i \frac{k}{n}}}\\
\]

The value \(\sqrt[n]{-1}\) is any representative from this set; any will do just fine...

Which is just re-indexing; and the standard geometric series. For some reason I thought this was a Gaussian sum before; I should've stopped posting; I was way too tired! I fucked my calculations up, but I ended up with the right answer--because I smelled the right answer...

But... This is just \(0\) for any integer \(k \in \mathbb{Z}\)...

This let's us grab the value \(F_k\) rather painlessly now. Just sum over \(n\) with a geometric weight:...

\[
\begin{align}
F_k &= \sum_{n=0}^\infty \frac{a_{kn}}{2^n}\\
&=B_k + 0\\
\end{align}
\]

---

So the first thing we should identify, is the value of \(F_0\). This value is given as:

\[
F_0 = \frac{1}{2\pi i} \int_{|z| = 2} f(z)\frac{dz}{z}\\
\]

This value is the value \(B_0 = f(0)\).... This may seem kind of stupid, but it means that for all \(R \neq 1\); we know that:

\[
F_0 = \frac{1}{2\pi i} \int_{|z| = R} f(z) \frac{dz}{z} = f(0)\\
\]

The residues of the poles on the boundary, will just cancel out... The second thing to note is that this happens for every \(k\). The sum of the roots of \(-1\) always sum to zero... And therefore:

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

This takes care of the point I was trying to make!

---

We can recover Caleb's function \(f(x)\) for \(|x| < 1\) by writing:

\[
f(x) = \sum_{k=0}^\infty F_k x^k\\
\]

Where, \(F_k\) is given as:

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

Where \(R \neq 1\)...

This means that the "analytic continuation" of \(f(x)\) for \(|x| >1\), basically acts like there is no wall of singularities. IT COMPLETELY IGNORES THE WALL OF SINGULARITIES!!!!!

So, what this means is much much deeper than Caleb is perhaps letting on. The standard manner of saying something a function \(G\) is an analytic continuation of \(g\); is by saying they are holomorphic on intersecting domains, and they agree there.

Here we have a function \(f_1\) holomorphic for \(|x| < 1\), and a function \(f_2\) holomorphic for \(|x|>1\). And both functions have a natural boundary at \(|x| = 1\). But \(f_2\) acts as an analytic continuation of \(f_1\). BECAUSE CAUCHY'S EQUATIONS ARE SATISFIED!

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

For all \(|x| < 1\)!!!!!!!!!!!!!!!

I don't know a word for this. I'm going to call this a "removable wall of singularities".

Fuck this is a super cool function  

---

Okay, I hopefully redeemed my dumb ass Gaussian sum comment, lmaooo

Regards, James

[Caleb]

https://mathoverflow.net/questions/19866...g-function

Oh wow, that is a beautiful thread, Gottfried!

And the answer is through Fourier Transform...and what is the Mellin transform but a variable changed Fourier Transform...

But I digress.

---

I spent the whole morning--scribbling in a notebook--looking at Caleb's function:

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

This function really fascinates me, because simply evaluating \(|x|>1\) should be the analytic continuation of \(f(x)\) for \(|x| < 1\). But what constitutes an "analytic continuation" is not what this is. It is in no way anything that fits in standard complex analytic theory. There's something special going on with this function.

I realized I made some dumb mistakes in my original evaluation--but I was on the right track. So I'm going to start over. Sorry, I was up all night and then wrote an analysis in the wee hours of the morning. So I was all over the place. I'm going to start over again with a fresh mind--after a pot of coffee and tea. Forgive me, Caleb; if this isn't what you are looking for. I may be going out on a tangent--but this function really fascinates me!

---

We start by defining:

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

If \(f(x)\) were holomorphic for \(|x| < 2+\delta\), then this value would simply be \(\frac{f^{(k)}(0)}{k!}=  B_k\). So we can write Caleb's function for \(|x| < 1\) as:

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

We will also call:

\[
\frac{x^n}{1+x^n} = \sum_{k=0}^\infty b_{kn} x^k = \sum_{j=0}^\infty (-1)^{j} x^{n(j+1)}\\
\]

So that:

\[
\sum_{n=0}^\infty \frac{b_{kn}}{2^n} = B_{k}\\
\]

The values \(b_{kn}\) are fairly elementary. If \(k \equiv 0\pmod{n}\) then \(b_{kn} = (-1)^{\frac{n}{k}-1}\) and \(k > 0\). And otherwise it is just \(0\).

---

I am now going to assume that \(k \in \mathbb{Z}\) for \(F_k\). So I am assuming that "perhaps" there is a nontrivial Laurent part being added to this. I am now going to do a similar thing, but for the poles of \(\frac{x^n}{1+x^n}\). And by such, I will call:

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

The function \(\frac{z^n}{1+z^n}\frac{1}{z^{k+1}}\) has poles at the \(n\)'th roots of unity of \(-1\); and for \(k\ge n\), it has a pole at \(0\). The pole at \(0\) is easy to handle; it's just the value \(b_{kn}\). The poles at the roots of unity are a bit different. But they are all simple poles; and therefore the standard formula for the formula of a simple pole is given as:

\[
\text{Res}\left( \frac{g(z)}{h(z)},q\right) = \frac{g(q)}{h'(q)}\\
\]

By which we get that:

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

We are only going to care about the second part; as the second part is the interesting part. We already know what \(b_{kn}\) behaves like when we sum it; we care about how the poles react when we sum it. So to begin; we're going to rewrite this strange sum:

\[
\sum_{q^n = -1} \frac{1}{q^k} = \sqrt[n]{-1}\sum_{j=0}^{n-1} e^{2\pi i \frac{k j}{n}} = \sqrt[n]{-1}\frac{1- e^{2\pi i k}}{1-e^{2\pi i \frac{k}{n}}}\\
\]

The value \(\sqrt[n]{-1}\) is any representative from this set; any will do just fine...

Which is just re-indexing; and the standard geometric series. For some reason I thought this was a Gaussian sum before; I should've stopped posting; I was way too tired! I fucked my calculations up, but I ended up with the right answer--because I smelled the right answer...

But... This is just \(0\) for any integer \(k \in \mathbb{Z}\)...

This let's us grab the value \(F_k\) rather painlessly now. Just sum over \(n\) with a geometric weight:...

\[
\begin{align}
F_k &= \sum_{n=0}^\infty \frac{a_{kn}}{2^n}\\
&=B_k + 0\\
\end{align}
\]

---

So the first thing we should identify, is the value of \(F_0\). This value is given as:

\[
F_0 = \frac{1}{2\pi i} \int_{|z| = 2} f(z)\frac{dz}{z}\\
\]

This value is the value \(B_0 = f(0)\).... This may seem kind of stupid, but it means that for all \(R \neq 1\); we know that:

\[
F_0 = \frac{1}{2\pi i} \int_{|z| = R} f(z) \frac{dz}{z} = f(0)\\
\]

The residues of the poles on the boundary, will just cancel out... The second thing to note is that this happens for every \(k\). The sum of the roots of \(-1\) always sum to zero... And therefore:

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

This takes care of the point I was trying to make!

---

We can recover Caleb's function \(f(x)\) for \(|x| < 1\) by writing:

\[
f(x) = \sum_{k=0}^\infty F_k x^k\\
\]

Where, \(F_k\) is given as:

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

Where \(R \neq 1\)...

This means that the "analytic continuation" of \(f(x)\) for \(|x| >1\), basically acts like there is no wall of singularities. IT COMPLETELY IGNORES THE WALL OF SINGULARITIES!!!!!

So, what this means is much much deeper than Caleb is perhaps letting on. The standard manner of saying something a function \(G\) is an analytic continuation of \(g\); is by saying they are holomorphic on intersecting domains, and they agree there.

Here we have a function \(f_1\) holomorphic for \(|x| < 1\), and a function \(f_2\) holomorphic for \(|x|>1\). And both functions have a natural boundary at \(|x| = 1\). But \(f_2\) acts as an analytic continuation of \(f_1\). BECAUSE CAUCHY'S EQUATIONS ARE SATISFIED!

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

For all \(|x| < 1\)!!!!!!!!!!!!!!!

I don't know a word for this. I'm going to call this a "removable wall of singularities".

Fuck this is a super cool function  

---

Okay, I hopefully redeemed my dumb ass Gaussian sum comment, lmaooo

Regards, James

This is an interesting analysis! I'll look over in detail later today-- but for now, I should mention that my intuition is that \(f(x)\) has a 'fake' natural boundary, but the other representation of \(f(x)\), where \(F(x) = \sum_{n=0}^\infty (-1)^n x^{2^n}\) has a 'real' natural boundary. The difference between a fake natural boundary and a real one is what I think drives the difference between the sums outside the natural boundary. I think if you try to do your analysis with the residues for \(F(x) = \sum_{n=0}^\infty (-1)^n (1/e)^{a^n}\) you should find that extra residues get picked up, and that they don't sum up to zero. Just a thought-- I'll definintely try this out myself later today and see what happens.

[JmsNxn]

This is an interesting analysis! I'll look over in detail later today-- but for now, I should mention that my intuition is that \(f(x)\) has a 'fake' natural boundary, but the other representation of \(f(x)\), where \(F(x) = \sum_{n=0}^\infty (-1)^n x^{2^n}\) has a 'real' natural boundary. The difference between a fake natural boundary and a real one is what I think drives the difference between the sums outside the natural boundary. I think if you try to do your analysis with the residues for \(F(x) = \sum_{n=0}^\infty (-1)^n (1/e)^{a^n}\) you should find that extra residues get picked up, and that they don't sum up to zero. Just a thought-- I'll definintely try this out myself later today and see what happens.

Oh you are definitely right. I don't think this works for \(g(x) = \sum_n (-1)^n x^{2^n}\). And for all wall of singularities situations. I think this function just happens to have a "removable" wall of singularities. Why I was so interested in this function. So, it's a bit of a tangent. But I'm sure, still somewhat relevant. I don't expect to find a formula for \(g\) from this, lmao. I'm just perplexed by this weird ass fucking relationship.

I made a post on MO, asking for references on similar phenomena of this \(f\) situation:

https://mathoverflow.net/questions/44183...nt-domains

Please upvote it; because it's the only way anyone of stature ever fucking reads it, lmao!

But this is definitely a bit of a tangent...

But still, this is the Kernel situation. When it goes to zero under the integral. Which can't hurt to know right?

Sincere regards.

[JmsNxn]

Okay, I want to add a stronger result to my last result.

Assume \(|x| \neq 1\). Call the contour \(C\) the parameterization of \(|z| =R\) for \(R\neq 1\). Then for all \(|x| < R\) and \(|x|\neq 1\), we have that:

\[
f(x) = \frac{1}{2\pi i} \int_C \frac{f(z)}{z-x}\,dz\\
\]

At this point I have only shown this result for \(|x| < \text{min}(R,1)\). I want to show it for \(|x| < R\).

This is actually pretty simple...

Let's take the summand \(\frac{z^n}{(1+z^n)(z-x)}\)... The value of all the residues of this summand are:

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

The sum:

\[
\sum_{q^n = -1}\frac{q}{q-x} = \sum_{j = 0}^{n-1} \frac{\sqrt[n]{-1}e^{2\pi i \frac{kj}{n}}}{\sqrt[n]{-1}e^{2\pi i \frac{kj}{n}}-x}\\
\]

So that:

\[
\sum_{j=0}^{n-1} \sum_{m=0}^\infty \sqrt[n]{-1}^{-m} e^{-2\pi i \frac{kjm}{n}}x^m=0\\
\]

Because:

\[
\sum_{j=0}^{n-1} e^{-2\pi i \frac{kjm}{n}} = 0\\
\]

And that, all we end up with is that:

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

So that the wall of singularities disappear even better than I originally thought.

---

My conclusion is that Caleb's function \(f(x)\) is holomorphic for \(\mathbb{C}/U\) where \(U\) is the unit circle. And that for all disks \(|z| = R\) where \(R \neq 1\)-- we have Cauchy's integral Formula:

For all \(|x| < R\) and \(|x| \neq 1\) and \(R \neq 1\)--where \(C\) is the parameterization of \(|z| =R\):

\[
f(x) = \frac{1}{2\pi i} \int_C \frac{f(z)}{z-x}\,dz\\
\]

For all \(|x| < R\)...

Fuck this function is cool as fuck!

This becomes not a question of "analytic continuation"--but whether it satisfies Cauchy's integral formula... This has to do with functional analysis (where advanced interpretations of Cauchy's integral formula exist). I'm super excited by your work, Caleb!

[Caleb]

Okay, I want to add a stronger result to my last result.

Assume \(|x| \neq 1\). Call the contour \(C\) the parameterization of \(|z| =R\) for \(R\neq 1\). Then for all \(|x| < R\) and \(|x|\neq 1\), we have that:

\[
f(x) = \frac{1}{2\pi i} \int_C \frac{f(z)}{z-x}\,dz\\
\]

At this point I have only shown this result for \(|x| < \text{min}(R,1)\). I want to show it for \(|x| < R\).

This is actually pretty simple...

Let's take the summand \(\frac{z^n}{(1+z^n)(z-x)}\)... The value of all the residues of this summand are:

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

The sum:

\[
\sum_{q^n = -1}\frac{q}{q-x} = \sum_{j = 0}^{n-1} \frac{\sqrt[n]{-1}e^{2\pi i \frac{kj}{n}}}{\sqrt[n]{-1}e^{2\pi i \frac{kj}{n}}-x}\\
\]

So that:

\[
\sum_{j=0}^{n-1} \sum_{m=0}^\infty \sqrt[n]{-1}^{-m} e^{-2\pi i \frac{kjm}{n}}x^m=0\\
\]

Because:

\[
\sum_{j=0}^{n-1} e^{-2\pi i \frac{kjm}{n}} = 0\\
\]

And that, all we end up with is that:

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

So that the wall of singularities disappear even better than I originally thought.

---

My conclusion is that Caleb's function \(f(x)\) is holomorphic for \(\mathbb{C}/U\) where \(U\) is the unit circle. And that for all disks \(|z| = R\) where \(R \neq 1\)-- we have Cauchy's integral Formula:

For all \(|x| < R\) and \(|x| \neq 1\) and \(R \neq 1\)--where \(C\) is the parameterization of \(|z| =R\):

\[
f(x) = \frac{1}{2\pi i} \int_C \frac{f(z)}{z-x}\,dz\\
\]

For all \(|x| < R\)...

Fuck this function is cool as fuck!

This becomes not a question of "analytic continuation"--but whether it satisfies Cauchy's integral formula... This has to do with functional analysis (where advanced interpretations of Cauchy's integral formula exist). I'm super excited by your work, Caleb!

Hmm, I did a little bit of numerical testing and I've found that
\[\frac{1}{2\pi i} \int_C \frac{f(z)}{z-x} dz = -\frac{3}{2}\]
if the contour \(C\) encloses the whole natural boundary. In particular, I think we actually have that 
\[\frac{1}{2\pi i} \int_C \frac{x^n}{x^n+1} \frac{1}{z-x} = 1 \]
As long as \(C\) encloses all the poles of \(x^n+1\).

Let me know if any of these computations are wrong, I did these in a rush and I'll check them later

[JmsNxn]

Hmm, I did a little bit of numerical testing and I've found that
\[\frac{1}{2\pi i} \int_C \frac{f(z)}{z-x} dz = -\frac{3}{2}\]
if the contour \(C\) encloses the whole natural boundary. In particular, I think we actually have that 
\[\frac{1}{2\pi i} \int_C \frac{x^n}{x^n+1} \frac{1}{z-x} = 1 \]
As long as \(C\) encloses all the poles of \(x^n+1\).

Let me know if any of these computations are wrong, I did these in a rush and I'll check them later

Oh My GOD!

The plot thickens!

No, you are right!

I wrote:

\[
\frac{1}{n}\sum_{j=0}^{n-1} \sum_{m=0}^\infty \sqrt[n]{-1}^{-m} e^{2\pi i \frac{jkm}{n}} = 0\\
\]

But the actual formula is:

\[
\frac{1}{n}\sum_{j=0}^{n-1} \sum_{m=1}^\infty \sqrt[n]{-1}^{-m} e^{2\pi i \frac{jkm}{n}} = 0\\
\]

When \(m=0\), the formula produces a \(1\)!

\[
\frac{1}{n}\sum_{j=0}^{n-1} 1 = 1\\
\]

....Okay so for \(|x| > 1\) the Cauchy integral formula doesn't hold.

But it does hold for \(|x| < 1\)...

God this shit makes my head spin, lmao!

I apologize, Caleb

[JmsNxn]

Christian Remling has put me to shame.

He's expressed your qualms, Caleb. And done so, to show that my Cauchy formula is not correct!!!

There is a residue factor which gets added in. He uses the \(f(0) = 1/2\) and my formula equals \(3/2\) as an example.

I definitely screwed up some part of my formula. But it's probably just a couple indicator values which get compounded...

In the case of \(f(0) = 1/2\); my formula is \(f(0) + 1\). And for further derivatives \(f^{(k)}(0)\)..

There's some small error I forgot. And it's probably \(n=k\) kind of \(\delta[n=k] = 1\) value.

This'll probably point even more towards your work!

There's some small rational function that:

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

Where now:

\[
\frac{1}{2\pi i} \int_{|z| = 2} \frac{f(z)}{z-x}\,dz = f(x) + \sum_{k=0}^\infty d_k x^k\\
\]

Where \(d_k\) is a small collection of \(1\)'s and \(-1\)'s I missed out on...

Fuckkkkkkkk

[Caleb]

Christian Remling has put me to shame.

He's expressed your qualms, Caleb. And done so, to show that my Cauchy formula is not correct!!!

There is a residue factor which gets added in. He uses the \(f(0) = 1/2\) and my formula equals \(3/2\) as an example.

I definitely screwed up some part of my formula. But it's probably just a couple indicator values which get compounded...

In the case of \(f(0) = 1/2\); my formula is \(f(0) + 1\). And for further derivatives \(f^{(k)}(0)\)..

There's some small error I forgot. And it's probably \(n=k\) kind of \(\delta[n=k] = 1\) value.

This'll probably point even more towards your work!

There's some small rational function that:

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

Where now:

\[
\frac{1}{2\pi i} \int_{|z| = 2} \frac{f(z)}{z-x}\,dz = f(x) + \sum_{k=0}^\infty d_k x^k\\
\]

Where \(d_k\) is a small collection of \(1\)'s and \(-1\)'s I missed out on...

Fuckkkkkkkk

The computations I have suggest that there is really no relation between the Cauchy Integral formula when we take contours outside the natural boundary, and the function inside. For all higher derivatives (i.e. k>0), we have that the Cauchy Integral formula says \( f^{(k)} = 0 \)

This shoud follow because 
\[ \frac{1}{2 \pi i} \int_C \frac{x^n}{1+x^n} \frac{1}{x-z} dx = 1 \]
FOR ALL \( z \). Since this doesn't depend on z at all, its derivatives must be zero. Thus, the only term that acatually contributes something is the term where n=0 in the original sum. (ASIDE: There's a really good website for doing some quick complex analysis testing stuff HERE. It has support for contour integration, which is amazing since there are lots of case where I like to be able to draw in my own contours without needing to pull out mathemtica and define everything) 

Now this to me is really bizarre. Somehow, the natural boundary is HIDING all the information about f(z). Basically, any integration that takes place outside the natural boundary tells you nothing about the function inside.

However, your note a few days ago about the little circle method got me think-- YOU CAN ACTUALLY STILL RECOVER THE INFORMATION WITH THE LITTLE CIRCLE TECHNIQUE. Or, at least, I think you can-- I don't really understand exactly how the little circle method works, so this might be something different, but I kind of think its in the same kind of spirit as the little circle technique thing you mentioned before. 

I'm bad at explaining stuff, so instead I'll draw a bad picture. We can actually recover the information for the n=3 term by taking a integration path where we place little indentations around each of the 3rd roots of (-1). If we did this, the natural boundary would hide all the information about all the terms where n != 3, but for n=3, we woudl be able to recover the coefficent since we removed the 3rd roots which were hiding the information for the 3rd term.   
My point is, to recover the n=3 term, all we have to do is subtract out the 3rd root of unity poles

Generally, since all the poles should be simple poles, we can figure out the residue at a given nth root of unity by taking a limit FROM OUTSIDE THE NATURAL BOUNDARY as we move towards those poles. Here's another really bad drawing (I'm doing this with a trackpad on a computer, so excuse its crudeness)
So, we could get the information about the residues completely by information about f on the outside. Anyway, let me know if what I'm saying here make any sense at all-- I'm probably too tired to explain anything well. Also, this only works if we look at the derivative, since the contour integral of  (where the contour is outside the natural boundary)
\[ \int \frac{x^n}{x^n+1} \frac{1}{x-z} dz \neq 0\]
but 
\[ \int \frac{x^n}{x^n+1} \frac{1}{(x-z)^2} dz = 0\]
Okay, so the formula should look like this. If we have
\[f(x) = \sum_{n=0}^\infty a_n \frac{x^n}{1+x^n} \]
Then
\[ a_k = \frac{d}{dx} \frac{1}{\frac{x^k}{1+x^k}} \bigg|_{x =t} \sum_{n=0}^k \text{Residue}(\frac{f(z)}{(z-t)^2}, z=e^{\pi i + \frac{2 \pi i}{k}}) \]

Or something like that, IDK I guess the idea is just that the residues at the roots of unity tell you about the nth term in the series expansion, but you have to make sure \( t \neq 0 \). I'm basically just ranting at this point-- I'll try to do something series with this tommorow. The main point is just that the inner series is uniquely determined if I claim its of the form
\[f(x) = \sum_{n=0}^\infty a_n \frac{x^n}{1+x^n} \]
So I guess that uniqueness, we have that series of the form \( \sum_{n=0}^\infty a_n \frac{x^n}{1+x^n}\) are uniquely determined inside the disc by their values on the outside (probably up to a constant or something like that)