05/01/2023, 11:12 AM
Hey everyone!
So I'm stuck on something pretty frustrating. But before I get into what I'm frustrated on, because it's a little involved to explain. There are quite a few functions in this post; and each play their own part. I'm going to start with defining the \(\beta\) function as:
\[
\beta(s) = \Omega_{j=1}^\infty \frac{e^z}{1+e^{j - s - \upsilon}}\,\bullet z\\
\]
The value \(\upsilon = 1.519081744458...\); and is the unique real constant to make it so that \(\beta(0) = 1\). Which means that:
\[
\beta(s+1) = \frac{e^{\beta(s)}}{1+e^{-s-\upsilon}}\\
\]
And is the unique holomorphic function to satisfy this this equation.
This function is \(2\pi i\) periodic, and real valued on the real line, and on \(\Im(s) = \pi\). This beta function is super easy to construct, and I can do so with incredible efficiency. I've posted this picture many times, but to remind everyone, this is what this function looks like:
The next function is our famed \(\text{slog}\). For this function, I only need a generated taylor series:
\[
\text{slog}(1+z)= \sum_{j=1}^\infty a_j z^j\\
\]
So give me the coefficients \(a_j\), and I can construct the super logarithm everywhere. I use Sheldon's program to do this; which I store as a file that we can keep. This constructs kneser's superlogarithm which looks like:
This of course has branch cuts; and the program I wrote chooses the branch cuts for us. I particularly like these branch cuts because they draw the chi star \(\Psi(\mathbb{R})\) about the fixed point \(L\) (where \(\Psi\) is the Schroder function for \(\exp\)). This function is also \(2 \pi i\) periodic.
This is a perfectly valid abel function, and always satisfies \(\text{slog}(e^z) = \text{slog}(z) + 1\); it does get a little wacky as we go further out, but is always analytic (except at the branch cuts of course). Branch cuts appear at the preimages of \(L\) and along the line \(\Im(z) = \pi\). If \(\exp^{\circ n}(z_0) = L\) and \(|\arg(z_0)| < \pi\), then \(\text{slog}\) has a branching singularity at \(z_0\)... it'll look like the chi star Sheldon talked about.
The next function I introduce is what I've dubbed the \(\chi\) function--this is not Sheldon's \(\chi\) function, but it does a similar function. This is much more related to Kneser's construction. This function is exactly:
\[
\chi(z) = \beta(\text{slog}(z))
\]
A graph of which looks like the below:
The function \(\chi\) has quite a few interesting properties. Firstly it has a fixed point at \(1\). This is seen as:
\[
\chi(1) = \beta(\text{slog}(1)) = \beta(0) = 1\\
\]
This fixed point is also attracting, and fairly well attracting:
\[
\chi(1+z) = 1+ \beta'(0)\text{slog}'(1) z + O(z^2) = 1 + (0.682403718747...)z + O(z^2)
\]
This function is just as holomorphic as \(\text{slog}\) is, subtracting further points because \(\beta\) has singularities at \((2k+1) \pi i + j - \upsilon\). It is also \(2 \pi i\) periodic. This will be very important as we progress.
The philosophy I am presenting is simple. We can interchange Kneser's super exponential and \(\beta\) upto \(2 \pi i\). This is done by the following trick:
\[
\chi(e^z) = \frac{e^{\chi(z)}}{1+e^{-\text{slog}(z)}}\\
\]
Where we get the identity:
\[
\chi(\text{sexp}(z)) = \beta(z)\\
\]
Which is valid for \(|\Im(z)| < \pi\). Going further out is a problem for another day, but I think I see how to do it. The trouble happens when we try to invert \(\chi\) and discuss \(\chi^{-1}(z)\). I can handle this for the real line; where I've written:
\[
\text{sexp}(z) = \chi^{-1}(\beta(z))\\
\]
Here's a graph of kneser for \([-2,2]\):
This agrees with Kneser to 40 digits; and does so with polynomials. I remind the reader that the only thing we use from Sheldon is the construction of \(\{a_j\}_{j=1}^\infty\). And I've now constructed Kneser from this sequence, and the beta function.
........The trouble happens about \(\Im(z) = 1\)........
The function \(\chi^{-1}(z)\) is calculable; but it's a fucking headache. So far, I can only get it near \(z=0\) or \(z=1\), to make a closed neighborhood of \((0,1) \subset \mathbb{C}\). This is enough to get \(\text{sexp}(z)\) for \(|\Im(z)| < 1\), but it starts to fail as we exit this area.
I'm just posting this hoping someone may have a suggestion!
The math seems to suggest that this should construct Kneser for \(|\Im(z)| < \pi\), but I'm fucking something up so that it's only working for \(|\Im(z)| < 1\)...
Regards, James
So I'm stuck on something pretty frustrating. But before I get into what I'm frustrated on, because it's a little involved to explain. There are quite a few functions in this post; and each play their own part. I'm going to start with defining the \(\beta\) function as:
\[
\beta(s) = \Omega_{j=1}^\infty \frac{e^z}{1+e^{j - s - \upsilon}}\,\bullet z\\
\]
The value \(\upsilon = 1.519081744458...\); and is the unique real constant to make it so that \(\beta(0) = 1\). Which means that:
\[
\beta(s+1) = \frac{e^{\beta(s)}}{1+e^{-s-\upsilon}}\\
\]
And is the unique holomorphic function to satisfy this this equation.
This function is \(2\pi i\) periodic, and real valued on the real line, and on \(\Im(s) = \pi\). This beta function is super easy to construct, and I can do so with incredible efficiency. I've posted this picture many times, but to remind everyone, this is what this function looks like:
The next function is our famed \(\text{slog}\). For this function, I only need a generated taylor series:
\[
\text{slog}(1+z)= \sum_{j=1}^\infty a_j z^j\\
\]
So give me the coefficients \(a_j\), and I can construct the super logarithm everywhere. I use Sheldon's program to do this; which I store as a file that we can keep. This constructs kneser's superlogarithm which looks like:
This of course has branch cuts; and the program I wrote chooses the branch cuts for us. I particularly like these branch cuts because they draw the chi star \(\Psi(\mathbb{R})\) about the fixed point \(L\) (where \(\Psi\) is the Schroder function for \(\exp\)). This function is also \(2 \pi i\) periodic.
This is a perfectly valid abel function, and always satisfies \(\text{slog}(e^z) = \text{slog}(z) + 1\); it does get a little wacky as we go further out, but is always analytic (except at the branch cuts of course). Branch cuts appear at the preimages of \(L\) and along the line \(\Im(z) = \pi\). If \(\exp^{\circ n}(z_0) = L\) and \(|\arg(z_0)| < \pi\), then \(\text{slog}\) has a branching singularity at \(z_0\)... it'll look like the chi star Sheldon talked about.
The next function I introduce is what I've dubbed the \(\chi\) function--this is not Sheldon's \(\chi\) function, but it does a similar function. This is much more related to Kneser's construction. This function is exactly:
\[
\chi(z) = \beta(\text{slog}(z))
\]
A graph of which looks like the below:
The function \(\chi\) has quite a few interesting properties. Firstly it has a fixed point at \(1\). This is seen as:
\[
\chi(1) = \beta(\text{slog}(1)) = \beta(0) = 1\\
\]
This fixed point is also attracting, and fairly well attracting:
\[
\chi(1+z) = 1+ \beta'(0)\text{slog}'(1) z + O(z^2) = 1 + (0.682403718747...)z + O(z^2)
\]
This function is just as holomorphic as \(\text{slog}\) is, subtracting further points because \(\beta\) has singularities at \((2k+1) \pi i + j - \upsilon\). It is also \(2 \pi i\) periodic. This will be very important as we progress.
The philosophy I am presenting is simple. We can interchange Kneser's super exponential and \(\beta\) upto \(2 \pi i\). This is done by the following trick:
\[
\chi(e^z) = \frac{e^{\chi(z)}}{1+e^{-\text{slog}(z)}}\\
\]
Where we get the identity:
\[
\chi(\text{sexp}(z)) = \beta(z)\\
\]
Which is valid for \(|\Im(z)| < \pi\). Going further out is a problem for another day, but I think I see how to do it. The trouble happens when we try to invert \(\chi\) and discuss \(\chi^{-1}(z)\). I can handle this for the real line; where I've written:
\[
\text{sexp}(z) = \chi^{-1}(\beta(z))\\
\]
Here's a graph of kneser for \([-2,2]\):
This agrees with Kneser to 40 digits; and does so with polynomials. I remind the reader that the only thing we use from Sheldon is the construction of \(\{a_j\}_{j=1}^\infty\). And I've now constructed Kneser from this sequence, and the beta function.
........The trouble happens about \(\Im(z) = 1\)........
The function \(\chi^{-1}(z)\) is calculable; but it's a fucking headache. So far, I can only get it near \(z=0\) or \(z=1\), to make a closed neighborhood of \((0,1) \subset \mathbb{C}\). This is enough to get \(\text{sexp}(z)\) for \(|\Im(z)| < 1\), but it starts to fail as we exit this area.
I'm just posting this hoping someone may have a suggestion!
The math seems to suggest that this should construct Kneser for \(|\Im(z)| < \pi\), but I'm fucking something up so that it's only working for \(|\Im(z)| < 1\)...
Regards, James
