10/29/2021, 11:44 PM

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!

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!