Divergent Series and Analytical Continuation (LONG post)

[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)