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Arguments for the beta method not being Kneser's method
#55
(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;

.gp   beta_tau.gp (Size: 8.79 KB / Downloads: 21)
- Sheldon
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Messages In This Thread
RE: Arguments for the beta method not being Kneser's method - by sheldonison - 10/23/2021, 03:13 AM

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