+- Tetration Forum (https://math.eretrandre.org/tetrationforum)
+-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1)
+--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3)
+--- Thread: Calculating the residues of \(\beta\); Laurent series; and Mittag-Leffler (/showthread.php?tid=1367)
Calculating the residues of \(\beta\); Laurent series; and Mittag-Leffler - JmsNxn - 10/29/2021
So, this is a bit off topic, but as I haven't had any quite "eureka moments" lately. I started compiling as much information as I can about \(\beta\). And it occurred to me, I do not know the Laurent series of \(\beta\) about each singularity. And as I started evaluating it, it's actually very interesting.
We're going to work solely with \(\lambda =1\) and \(b=1\) (but the result is generalizable to all \(\lambda,\beta\)) so that,
$$
\begin{align}
\beta(s) &= \Omega_{j=1}^\infty \dfrac{e^z}{1+e^{j-s}}\,\bullet z\\
\beta(s+1) &= \dfrac{e^{\beta(s)}}{1+e^{-s}}\\
\end{align}
$$
And we're looking for the laurent series of:
$$
\beta(s+j+\pi i) = \sum_{k=-\infty}^\infty a_{jk} s^k\\
$$
For \(j\ge1\). The first surprising fact, is that the laurent series exists! Second of all, I'm going to run through an induction protocol which, well, I don't know how to explain, but it's very god damned interesting.
To begin, the first pole is at \(j=1\) and it's a simple pole. This can be derived because,
$$
\begin{align}
\beta(1+\pi i + s) &= \dfrac{\exp(\beta(\pi i + s))}{1+\exp(-\pi i - s)}\\
&= \dfrac{\exp(\beta(\pi i + s))}{1-\exp(- s)}\\
\end{align}
$$
And since \(\exp(\beta(\pi i + s))\) is holomorphic about \(s = 0\); and since the pole at \(s = 0\) for \(\frac{1}{1-\exp(- s)}\) is simple. We must have this pole is simple. Additionally we have that:
$$
\frac{1}{1-\exp(- s)} = \frac{1}{s} + g(s)\\
$$
For a holomorphic function \(g(s)\) in a neighborhood of \(s = 0\). This is derived from Cauchy's formula,
$$
\begin{align}
\text{Res}_{s=0} \frac{1}{1-\exp( - s)} &= \lim_{s \to 0} \frac{s}{1-\exp(- s)}
&= \lim_{s \to 0} \frac{1}{\exp(- s)}\,\,\text{by Hopital} = 1\\
&= 1
\end{align}
$$
From here, the first singularity of \(\beta\) looks like:
$$
\beta(1+\pi i +s) = \frac{e^{\beta(\pi i + s)}}{1 - e^{- s}}\\
$$
Which must be a simple singularity. Therefore, when we look at \(\beta(1+\pi i + s)\) we get something really nice. The residue/ laurent series is very well behaved.
$$
\int_{|s| = 1/2} \beta(1+\pi i + s)\,ds = \int_{|s| = 1/2} \frac{\exp(\beta(\pi i + s))}{1-e^{- s}}\,ds
$$
Which, by Cauchy's integral theorem, we get:
$$
\int_{|s| = 1/2} \beta(1+\pi i + s)\,ds = 2 \pi i \exp(\beta(\pi i))\\
$$
From here, we can say that:
$$
\beta(1+\pi i + s) = \frac{\exp(\beta(\pi i))}{s} + h(s)\\
$$
Where \(h\) is holomorphic in a neighborhood of \(s =0\)
Now, what's so surprising is how well the higher order singularities for \(j \ge 2\) follow this pattern. And that, each Laurent series actually follows a tetration pattern! I'm going to work through \(j=2\) which is the first induction step to enlighten.
Consider:
$$
\begin{align}
\beta(2+\pi i + s) &= \frac{e^{\beta(1+\pi i + s)}}{1-e^{-1 - s}}\\
&= \frac{e^{\displaystyle e^{\beta(\pi i)}/s + h(s)}}{1-e^{-1 - s}}\\
\end{align}
$$
Now... Quite perfectly! the only singularity that arises is in the exponent as \(s \to 0\). And, although it's an essential singularity--it's the best kind of essential singularity. Good ol' fashion \(e^{1/s}\), which certainly has a laurent series. Let's go ahead and calculate it.
$$
\beta(2 + \pi i +s) = \frac{e^{h(s)}}{1-e^{-1 -s}}\sum_{k=0}^\infty \frac{e^{\beta(\pi i) k}}{s^{k}k!}
$$
Now the real cool part happens. If we take the logarithm, we get:
$$
\log\beta(2 + \pi i +s) = \log\frac{e^{h(s)}}{1+e^{-\pi i -1 -s}} + \frac{e^{\beta(\pi i)}}{s}
$$
So, we get the recursion:
$$
\begin{align}
\log\beta(2 + \pi i +s) - \log\frac{e^{h(s)}}{1-e^{-1 -s}} &= \frac{e^{\beta(\pi i)}}{s}\\
&= \beta(1+ \pi i + s) - h(s)\\
\end{align}
$$
And now... we just apply the functional equation recursively. I'm going to write out the notation I use, which is:
$$
\beta(s + j + \pi i) = \Omega_{k=1}^{j-1} \frac{e^{z}}{1-e^{j-k-s}}\bullet \,e^{\beta(\pi i)}/s + h(s)\,\bullet z\\
$$
Which means, if:
$$
\begin{align}
a_j(s,z) &= \Omega_{k=1}^{j-1} \frac{e^{z}}{1-e^{j-k-s}}\bullet z\\
&= q_1(s,q_2(s,...,q_{j-1}(s,z)))\\
q_k(s,z) &= \frac{e^{z}}{1-e^{j-k-s}}\\
\end{align}
$$
Then:
$$
\beta(s + j + \pi i) = a_j(s,e^{\beta(\pi i)}/s + h(s))\\
$$
This is just a standard iteration of the functional equation. Now, \(a_j(s,z)\) is holomorphic in z, and in a neighborhood of zero in s. This makes the Laurent series findable; but even better a nice tetration like structure to the singularity.
Which is, in a simple terms. The value of \(\beta(s+j+\pi i) \approx \exp^{j-1} \left( e^{\beta(\pi i)}/s + h(s)\right)\). Or that, we should expect:
$$
\log^{j-1} \beta(s+j + \pi i) = \frac{e^{\beta(\pi i)}}{s} + h_j(s)\\
$$
Which means we should have a very regular structure to the singularities; and they look a lot like tetration.
Isn't that cool!
And now we can continue and find a laurent series in the neighborhood of each singularity... but that's a bit of a pain since we have a closed form expression.
Now... Onto seeing if we can find a Mittag-Leffler expansion for \(\beta\) that is valid in all of \(\mathbb{C}\). I always thought this would be impossible because the singularities would be such a mess. But with this theorem; it's probably possible. All the principle parts should look like \(f(1/(s-j-\pi i))\) for \(f\) entire. Alright!
I'll update this thread if I find anything worthy to share.
Alright, it's not quite Mittag-Leffler but it's close! Very damn close.
If you take:
$$
f_{jk}(s) = \text{Principal part of}\left(\exp^{j-1}\left(\frac{e^{\beta(\pi i)}}{s-j-(2k+1)\pi i} + h_{j}(s)\right)\right)\\
$$
This should be holomorphic on \(\mathbb{C}/\{j+(2k+1)\pi i\}\) (we've renormalized \(h_{j}\) too, just for convenience). Now the principle part, is essentially, just take the laurent series and only consider the negative exponents. So, the principle part of:
$$
\text{Principal part of}\left( \sum_{c=-\infty}^\infty a_c (z-p)^{c} \right) = \sum_{c=-\infty}^{-1} a_c (z-p)^{c}\\
$$
Now, expand the taylor series of \(f_{jk}(s)\) about \(0\). The functions \(f_{jk}\) are holomorphic about zero in increasing disks as \(j,k \to \infty\). (That's a non problem.) Take the \(2^{j+|k|}\) first terms and call this \(P_{jk}(s)\). Then,
$$
H(s) = \sum_{j=1}^\infty \sum_{k=-\infty}^\infty f_{jk}(s) - P_{jk}(s)
$$
Is a holomorphic function for \(s \neq j + (2k+1)\pi i\). This is just a slight modification of Mittag-Leffler to allow for essential singularities. Normally this doesn't work, I had to double check the literature, and the particular book I'm using: Reinhold Remmert "Classical Topics in Complex Functions Theory"--allows for the possibility of essential singularities (it's just that Mittag-Leffler's decompisition is not necessary like it is with meromorphic functions; i.e: you have to pay close care). At each singularity, it has the same principle part as \(\beta(s)\). From here, we now do something familiar to Mittag-Leffler again, and:
$$
\Pi(s) = \beta(s) - H(s)\\
$$
Which will be an entire function on \(\mathbb{C}\).
Again, this is just a sketch for the moment. But this may help... I believe we can remove the singularities of \(\beta\) and produce something like a Mittag-Leffler decomposition. This is going to take much more time though. Still not sure the best way of approach of rigorizing this. But it should look something like this.
I don't know how much this will help tetration per se. It's probably more of a curiousity with the \(\beta\) function. But you never know!