02/28/2023, 07:18 PM
(02/28/2023, 07:03 PM)JmsNxn Wrote: 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.
