[MO] Residue at ∞ and ∑(-1)^n x^(2^(2^n))

[Caleb]

I wrote this post yesterday on MO, I wonder if anyone has some ideas on how to solve. Here is the link: https://mathoverflow.net/questions/44373...s-at-infty

I'm also curious if anyone has run into these 'residue at ∞' objects before. In some sense, I view them as a bit of a generalization of the residue theorem. Also, residues at ∞ can have lots of complexity, for instance, you can write the jacobi theta function completely in terms of a single residue at ∞. Likewise, one can take an arbitrary infinite series and use the 'reverse' borel transform (i.e. we do (-n)!/(-n)! instead of (n!)/(n!)) to push all the behaviour of the series into a residue at infinity, so computing residues at infinity is, in theory, at least as hard as computing infinite series. These are just some idle thoughts, they might be meaningless.

[JmsNxn]

Hey, Caleb

Just came here to say your statement that this may be "idle thoughts" is entirely correct; and that's its strongest point. I PM'd you the rough draft of the paper I'm working on. And early on I talk about residues at \(\infty\) and how they relate to residues at \(0\). The entire thesis of "Lambert's reflection formula" is that we can use the residues at \(\infty\) to write the residues at \(0\). So, sure, it's idle thoughts; but you're on to something! You are definitely on to something. I'm going to reserve comments on the specific problem in this post; as I'm still stuck on a couple things with more manufactured cases. But "idle thoughts" are what makes math math . I'm not sure how to answer your MO question; but I imagine it'll follow similar rules to what I've constructed so far.

Sincere regards, James

[tommy1729]

I agree with james.

The residue at oo is linked to those at 0.

However in case our function is more complicated or we do not have a reflection formula ...
Im not sure.

Maybe something like

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

Can be used and considered a reflection kinda.

regards

tommy1729

[JmsNxn]

I agree with james.

The residue at oo is linked to those at 0.

However in case our function is more complicated or we do not have a reflection formula ...
Im not sure.

Maybe something like

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

Can be used and considered a reflection kinda.

regards

tommy1729

Hey, Tommy;

I sent Caleb the PDF, not you yet; but "reflection formula" is more a term I'm using loosely. It's not literally \(f(x) = f(1/x)\) or something like this. It's more so we can use \(f(x)\) to construct \(f(1/x)\) and vice versa. Which is why I refer to it as a reflection formula. Essentially; we can reconstruct \(f(x)\) about \(x =0\) using the taylor coefficients of \(f(x)\) about \(x =\infty\)--I just like using "reflection formula" as a term; because it relates to Lambert series; and generalizes the simple reflection formula you see. I'm not actually just saying \(f(1/x) = h(f(x))\) or something!

[Caleb]

I agree with james.

The residue at oo is linked to those at 0.

However in case our function is more complicated or we do not have a reflection formula ...
Im not sure.

Maybe something like

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

Can be used and considered a reflection kinda.

regards

tommy1729

Hey, Tommy;

I sent Caleb the PDF, not you yet; but "reflection formula" is more a term I'm using loosely. It's not literally \(f(x) = f(1/x)\) or something like this. It's more so we can use \(f(x)\) to construct \(f(1/x)\) and vice versa. Which is why I refer to it as a reflection formula. Essentially; we can reconstruct \(f(x)\) about \(x =0\) using the taylor coefficients of \(f(x)\) about \(x =\infty\)--I just like using "reflection formula" as a term; because it relates to Lambert series; and generalizes the simple reflection formula you see. I'm not actually just saying \(f(1/x) = h(f(x))\) or something!

I haven't read all the way through the paper yet, but I'm guessing that the type of residues I'm considering here aren't quite covered under the cased you consider. These aren't your usual residue at ∞, they are more like parts or sections of a residue. Here's what it looks like if you map from z -> 1/z, so that the residues at ∞ get moved to zero. Here, for example, is what a countour looks like that picks up the residue
   
You can't use any simple for the the Cauchy integral formula here, because you can't close the contour. Its a bit interesting, because your integrating over a function that is analytic inside the whole region, except on only a single point at 0, and there's no way around that point, i.e. every contour needs to include 0 to be closed and integrate over the right region.

[JmsNxn]

Hey, Caleb!

You're right; I don't cover shit like this til' close to the end; and there I void all possible contours like this.

If you take:

\[
z \mapsto e^{-1/z}\\
\]

You will notice very similar behaviour; and that is where I cap the lambert reflection formula.

Where you are seeing here is more like:

\[
z \mapsto e^{-1/\sqrt{z}}\\
\]

Which has a branching singularity at \(0\) (coincidentally similarly at \(z = \infty\)). The rough draft I sent you is intended solely for singularities which aren't branching. Your graph above is clearly a branching singularity.

That's why I don't want to comment on this new question too much; but it deeply relates to "residue at infinity" as I'm talking about. It's just a branching residue rather than a meromorphic, or essential residue. I'm focusing on \(z \mapsto 1/z^k\) or \(z \mapsto e^{-1/z}\), I'm entirely ignoring \(z \mapsto e^{-1/\sqrt{z}}\). Not because I can't speak on it, but because I'm less confident, and more unsure what happens here.

Awesome job still!! I love these graphs! Keep em coming!

Sincere Regards, James

EDIT!

Also the graph you just posted is a Riemann surface defined off of a second order singularity. This means that the dimension of the surface is \(2\). This is just like \(z = y^2\) produces a Riemann Surface \(S\) in \(2\) complex dimension. So that, the "branching point" of the graph you just took, is only 2nd degree. It looks like \(z = y^2\).  But the base singularity is essential... we can see this just how the function behaves near \(z=0\). It gets arbitrarily large the closer to \(z =0\) (which means it's essential). From there \(e^{-1/\sqrt{z}}\), is a second order Riemann Surface; or rather; there are two copies \(e^{\pm \frac{1}{y}}\) which are projections from the Riemann surface \(S\) in \(z\) to a singular complex variable \(y\).

I apologize if I've appeared a tad daft, Caleb. I'm just trying to do baby steps first; and build a slow theory. But I still believe I can answer your questions. Just remember that professional math is a marathon. And if you do a marathon, you'll come in 15th place. If you try to sprint your way through, you'll end up in last place...

[Caleb]

Hey, Caleb!

You're right; I don't cover shit like this til' close to the end; and there I void all possible contours like this.

If you take:

\[
z \mapsto e^{1/z}\\
\]

You will notice very similar behaviour; and that is where I cap the lambert reflection formula.

Where you are seeing here is more like:

\[
z \mapsto e^{1/\sqrt{z}}\\
\]

Which has a branching singularity at \(0\) (coincidentally similarly at \(z = \infty\)). The rough draft I sent you is intended solely for singularities which aren't branching. Your graph above is clearly a branching singularity.

That's why I don't want to comment on this new question too much; but it deeply relates to "residue at infinity" as I'm talking about. It's just a branching residue rather than a meromorphic, or essential residue. I'm focusing on \(z \mapsto 1/z^k\) or \(z \mapsto e^{1/z}\), I'm entirely ignoring \(z \mapsto e^{1/\sqrt{z}}\). Not because I can't speak on it, but because I'm less confident, and more unsure what happens here.

Awesome job still!! I love these graphs! Keep em coming!

Sincere Regards, Jame

I'm not sure what your definition of branching singularity is, but I'm not sure the object I have shown fits into that categorization. In particular, it only has a single branch, defined everywhere, unlike \( e^{\frac{1}{\sqrt{z}} \). In particular, if you don't consider \(e^\frac{1}{z}\) to have a branching singularity, then what do you thinking about \(e^\frac{1}{z^2}\). This has the same type of behaviour as this function. In particular, consider a contour of this shape
   
This has the same type of behaviour as the function in the original post, and you can't apply Cauchy's integral theorem to compute it. The behaviour gets more extreme when we increase the power, for instance, at \( e^{\frac{1}{z^10}} \) it looks like this
   
Actually, with these type of contours, you can do something like Cauchy's theorem, but weaker. In particular, there is actually a Cauchy's theorem that I devoloped a while ago that applies to open circles, instead of closed curves. Take \( f(z) = \sum_{n=-\infty}^\infty a_n x^n \). In the case where you take a closed curve (a closed circle in particular), you cancel out every term except \( a_{-1} \) (this is the Cauchy integral formula). But, if you take a semi circle, you cancel out every other term (except a_{-1}). A third of a circle cancels out every 3rd term, and a fourth of a circle cancels out every 4th, etc. You can use this to obtain a divergent series representation for the integral over these badly defined contours, but this series representation has generally seemed useless to me since the series is divergent.

I've found some old notes where I derived the precise formula for half circles (its actually easy to compute for other ones as well, but I haven't gone through the algebra yet), I took \(f(z)=\sum_{n=1}^\infty \frac{a_n}{z^n}\) and then we obtain \(\int_C f(z)dz = a_1\pi i-2ir\sum_{n=1}^\infty \frac{(-1)^n a_{2n}}{(2n-1)r^{2n}}\) if \(C\) is a half-circle from \(\theta=-\frac{\pi}{2}\) to \(\theta = \frac{\pi}{2}\) with radius \(r\). Thus, we may compute the residue as \(\lim_{r \to 0} a_1\pi i-2ir\sum_{n=1}^\infty \frac{(-1)^n a_{2n}}{(2n-1)r^{2n}}\), which allows us to obtain that, in general, the residue picked up by the contour for \( \frac{1}{z^2} t^{\frac{1}{z^2}}\) is equal to \(- \frac{\sqrt{\pi}}{\sqrt{-\ln(t)}}\).

[JmsNxn]

Hmmmmm.


Now I'm confused. I think you may be missing some background knowledge on complex analysis.

\[
\int_{|z| = 1} e^{1/z}\,dz = 2\pi i\\
\]

If I take \(k>1\):

\[
\int_{|z| = 1} e^{1/z^k}\,dz = 0\\
\]


This is not numerically verifiable; but it's mathematically true. Additionally we can substitute \(\xi = 1/z\); and take taylor coefficients at \(\xi\); which relate directly to the above values; but as "residue at infinity" relates to "residue at zero".


When we introduce:

\[
\int_{|z| = 1} e^{1/\sqrt{z}}\,dz\\
\]

It is impossible to take this integral without considering it in an "improper manner". Which was my point.


I apologize if I'm not making any sense. But from that point; could you perhaps elaborate further from this perspective? I'm not trying to be a jack ass, I'd just love to get a better grasp of what you mean???

Again, sorry.

James

[Caleb]

Hmmmmm.


Now I'm confused. I think you may be missing some background knowledge on complex analysis.

\[
\int_{|z| = 1} e^{1/z}\,dz = 2\pi i\\
\]

If I take \(k>1\):

\[
\int_{|z| = 1} e^{1/z^k}\,dz = 0\\
\]


This is not numerically verifiable; but it's mathematically true. Additionally we can substitute \(\xi = 1/z\); and take taylor coefficients at \(\xi\); which relate directly to the above values; but as "residue at infinity" relates to "residue at zero".


When we introduce:

\[
\int_{|z| = 1} e^{1/\sqrt{z}}\,dz\\
\]

It is impossible to take this integral without considering it in an "improper manner". Which was my point.


I apologize if I'm not making any sense. But from that point; could you perhaps elaborate further from this perspective? I'm not trying to be a jack ass, I'd just love to get a better grasp of what you mean???

Again, sorry.

James

Fair point-- I didn't explain the shape of my contours well enough. Look closely again at the shape of the contours I am choosing, for instance, the one for \( e^{\frac{1}{z^2}} \) or \( e^\frac{1}{z^10} \) are easy ones to see. Also, to be specific, the contour I have is drawn in white, I'm not sure if I mentioned that before. 

All the contours are going THROUGH THE NON-ANALYTIC POINT AT 0! That means you can't apply Cauchy's theorem for residues! The integrals you have shown are regular circles-- I know how to deal with these cases. For instance, the contour I'm looking at for \( e^\frac{1}{z^2} \) is over the contour C where C is the path
\[ -i \to 0 \to i \to i+1 \to -i + 1 \to -i \]
Actually, if we want to be precise, we need to remove \(0\) from this path, since \( e^\frac{1}{z^2} \) is not defined there.


Okay, so I've introduced a weird object, why do I care about it? I care about such weird path as these because they allow me to pick up, for isntace, the residue at \( + \infty \) without picking up the residue at at other angles. For instance, I might wonder-- what is the integral \( \int_{-\infty}^\infty e^{-z^2} dz\) equal to? I could draw this as a contour over the imaginary line of \( e^{z^2} \), it looks like this
   

BUT, notice that I want to pick up the residue at \( + \infty \) WITHOUT ALSO PICKING UP THE RESIDUE AT \( -\infty \). In particular, I make the change \(dz \to d\frac{1}{z}\) and now I'm looking at \( \frac{1}{z^2} e^{\frac{1}{z^2}} \). Then, the contour integral around the full circle is zero-- but this isn't what we want, the integral is clearly now zero so thats not helpful. The contour we actually want is this one
   


In particular, notice that 
\[\int_{-\infty}^\infty e^{-z^2} dz = \sqrt{\pi}\]
But 
\[\int_{|r| = 1} \frac{1}{z^2} e^{\frac{1}{z^2}}dz = 0 \]
Whereas (where C is the contour I drew in the picture)
\[\frac{1}{i} \int_{C} \frac{1}{z^2} e^{\frac{1}{z^2}}dz =\sqrt{\pi}  \]
As expected

[tommy1729]

This is all very interesting !