Hey Sheldon,
Do you mind if refer to your error terms as \(\rho\) as opposed to \(\tau\). The reason being, I had considered a similar \(\rho\) beforehand; yours is algebraically more clever though. I tried to reduce it into a sum of error terms; and I had tried it with the use of the variable \(\rho\) as opposed to \(\tau\). This is much more consistent with my notation when doing infinite compositions. Where we compound errors as \(\sum_j \rho_j\). In such a sense, I've reserved \(\rho\) for compositions mapped to additions. Which keeps, in tone, with a lot of my previous papers.
That is to say:
\[
\begin{align}
\rho^0_\lambda(s) &= -\log(1+\exp(-\lambda s))\\
\tau^n_\lambda(s) &= \sum_{j=0}^{n-1} \rho_\lambda^j(s)\\
\end{align}
\]
I had considered these earlier; but couldn't make heads or tails of it. I never thought:
\[
\begin{align}
\rho_\lambda^n(s) = \log\left(1+\frac{\rho_\lambda^{n-1}(s+1)}{\beta_\lambda(s+1) + \tau_\lambda^{n-1}(s+1)}\right)\\
= \log\left(1+\frac{\rho_\lambda^{n-1}(s+1)}{\beta_\lambda(s+1) + \sum_{j=0}^{n-2} \rho_\lambda^j(s+1)}\right)
\end{align}
\]
Which is the quintessential speed up you are employing.
Nonetheless, is it okay if we refer to these as \(\rho\) as opposed to \(\tau\)? Because I have some good asymptotics of \(\rho\) if we talk about it like this. To me, \(\tau\) is the direct recursion and \(\rho\) is reducing it into a summation. Upon which; I have many tools to handle this sum asymptotically. And it fits very well with the notation I used to prove \(\beta\) is holomorphic in the first place. It makes the notation more consistent.
To me, notationally, \(\rho\) implies we are creating a summative bound of a sequence of compositions. To me \(\rho\) means a bounding map from \(\Omega \to \sum\).
Regards, James
Do you mind if refer to your error terms as \(\rho\) as opposed to \(\tau\). The reason being, I had considered a similar \(\rho\) beforehand; yours is algebraically more clever though. I tried to reduce it into a sum of error terms; and I had tried it with the use of the variable \(\rho\) as opposed to \(\tau\). This is much more consistent with my notation when doing infinite compositions. Where we compound errors as \(\sum_j \rho_j\). In such a sense, I've reserved \(\rho\) for compositions mapped to additions. Which keeps, in tone, with a lot of my previous papers.
That is to say:
\[
\begin{align}
\rho^0_\lambda(s) &= -\log(1+\exp(-\lambda s))\\
\tau^n_\lambda(s) &= \sum_{j=0}^{n-1} \rho_\lambda^j(s)\\
\end{align}
\]
I had considered these earlier; but couldn't make heads or tails of it. I never thought:
\[
\begin{align}
\rho_\lambda^n(s) = \log\left(1+\frac{\rho_\lambda^{n-1}(s+1)}{\beta_\lambda(s+1) + \tau_\lambda^{n-1}(s+1)}\right)\\
= \log\left(1+\frac{\rho_\lambda^{n-1}(s+1)}{\beta_\lambda(s+1) + \sum_{j=0}^{n-2} \rho_\lambda^j(s+1)}\right)
\end{align}
\]
Which is the quintessential speed up you are employing.
Nonetheless, is it okay if we refer to these as \(\rho\) as opposed to \(\tau\)? Because I have some good asymptotics of \(\rho\) if we talk about it like this. To me, \(\tau\) is the direct recursion and \(\rho\) is reducing it into a summation. Upon which; I have many tools to handle this sum asymptotically. And it fits very well with the notation I used to prove \(\beta\) is holomorphic in the first place. It makes the notation more consistent.
To me, notationally, \(\rho\) implies we are creating a summative bound of a sequence of compositions. To me \(\rho\) means a bounding map from \(\Omega \to \sum\).
Regards, James