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 Calculating the residues of $$\beta$$; Laurent series; and Mittag-Leffler JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 10/29/2021, 11:44 PM (This post was last modified: 10/30/2021, 09:25 AM by JmsNxn.) 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! « Next Oldest | Next Newest »

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