10/23/2021, 03:13 AM
(This post was last modified: 10/23/2021, 10:33 PM by sheldonison.)

(10/22/2021, 03:54 AM)JmsNxn Wrote: ... fascinating, Sheldon.... I'm a little dumbfounded by how you are calculating logrho so fast about the singularity--but it makes sense for the most part.

Hey James,

Now lets define a function \(\text{logrho}(z)=\ln(-\rho(z))\) where I'll use the shorthand notation \(l\rho(z)\) for the remainder of this post. Lets start with the following from my previous post, again this is for the 2pii periodic beta(z,1).

$$\begin{align}

f_0(z)=\beta(z)-\ln(1+\exp(-z));\;\;\; f_n(z) = \ln^{\circ n}f(z+n)\\

\rho_0(z)=-\ln(1+\exp(-z))\\

\rho_n(z)=\ln\left(1+\frac{\rho_{n-1}(z+1)}{f_{n-2}(z+1)}\right)\\

\end{align}$$

Now lets change the recursive equation for \(\rho\) to a recursive equation for \(l\rho\)

$$\begin{align}

l\rho_0(z)=\ln\Big(\ln\big(1+\exp(-z)\big)\Big)\\

l\rho_n(z)=\ln\left(-\ln\left(1+\frac{\rho_{n-1}(z+1)}{f_{n-2}(z+1)}\right)\right)\\

l\rho_n(z)=\ln\left(-\ln\left(1+\frac{-\exp(l\rho_{n-1}(z+1))}{\exp(f_{n-1}(z))}\right)\right)\\

l\rho_n(z)=\ln\bigg(-\ln\Big(1-\exp\big( l\rho_{n-1}(z+1) - f_{n-1}(z) \big) \Big) \bigg)\\

\end{align} $$

Next I implemented in pari-gp a routine I called loglogmexp(z) which implements the following:

$$\begin{align}

\text{loglogmexp}(y)=\ln\Big(-\ln\big(1-\exp(y)\big)\Big)\\

l\rho_n(z)=\text{loglogmexp}\big( \rho_{n-1}(z+1) - f_{n-1}(z)\big);\;\;\; y=\rho_{n-1}(z+1)-f_{n-1}(z)\\

\end{align}$$

Now, often times \(\Re(y)\) is large enough negative, that we can replace the inner most \(-\ln\big(1-\exp(y) \big)\) with the approximation of: \(\exp(y)\)!! If we are closer to the singularity then I implemented either a more exact series, or else directly implemented the exponents and logarithms. But for n=4, for most cases this is an extremely accurate approximation. This approximation is accurate to >=~60 decimal digits at a radius of less than 99.998% of the radius of convergence!

$$\begin{align}

l\rho_n(z) \approx l\rho_{n-1}(z+1) - f_{n-1}(z) \\

l\rho_n(z) \approx \ln\Big(\ln\big(1+\exp(-z-n)\big)\Big)-\sum_{i=1}^{n}f_{i-1}(z+n-i)\\

\end{align} $$

edit and update: The equation above is dominated by \(f_0(z+n-1)\) or if centering at Tet(0), \(e\uparrow\uparrow(z+n-1)\). In my program, I call f(z,n), beta_tau(z,n). You can see the individual contributions, by running "logrho_n(rr,4)" instead of logrho(rr,4).

Code:

`z=logrho_n(rr,4);`

-5.74639913386489 log(log(1+exp(-z-4)))

-3814279.10476022 -beta_tau(z+3,0)

-15.1542622414793 -beta_tau(z+2,1)

-2.71828182845905 -beta_tau(z+1,2)

-1.00000000000000 -beta_tau(z+0,3)

z=-3814303.72370342;

beta_tau.gp (Size: 8.79 KB / Downloads: 110)

- Sheldon