Arguments for the beta method not being Kneser's method

[sheldonison]

Hey Sheldon,

Do you mind if refer to your error terms as \(\rho\) as opposed to \(\tau\)...

Hey James,

That is done.  I redid all of post#51 with the changes from tau to rho, including the diagrams, and the updated pari-gp beta_tau.gp program.  In the text I added the tau/rho relationship too.  The updated pari-gp program beta_tau.gp is also in  post#51  and that beta_tau.gp code applies to this post as well.
\[\begin{align}
\tau_n(z)=\sum_{i=0}^{n}\rho_i(z)\\
\end{align}\]

This post attempts to explain how I made the graph for the approximation for a_n for the first million or so rho_4 Taylor series coefficients.  Lets start with the pari-gp program rho(z,n), where I'm most interested in rho(rr,4).  Here are some of my assumptions

• rho(rr(,4) has a nearest singularity, approximate radius of convergence 0.034681
  
• within that radius of convergence rho(rr,4) is non-zero, and has a logarithm
  
• Approximately the first 1332000 taylor series coefficients are negative for even coefficients and positive for odd coefficients
  
• it is much easier to do calculations with logrho(z,4) than rho(z,4)
  
• From Taylor series n=10...1332000 we can approximate the behavior on a circle of radius r(n) as Gaussian.  More pictures and details will follow below.
  
• above for a_n; n~>1332000, the maximum is no longer on the real axis, and the maximum switches to a complex conjugate pair of maximums that approach the singularity wall as n grows larger; these coefficients can probably still be approximated by this pair of points, maybe up to 6000000..7000000, I haven't studied the maximum yet, but as we approach the singularity other approximation techniques are required.  Eventually, you can just use the radius of convergence approximation ...
  

Lets plot logrho(rr+z,4) at about 40% of the radius of convergence using r= 0.01384115  We want to take the exponent of this function to see if we can learn more about rho_4 ...  We are plotting logrho(rr+0.01384115*exp(t*I)) where t varies from 0..2pi
   

But the real(z) is humongously negative; with a maximum value of  -2086301.8 at logrho(rr-0.01384115,4).  This is a hugely negative number that we want to take the exponent of...  So we could subtract that huge negative by plotting as follows:
exp(logrho(rr-0.0173405*exp(t*I))-logrho(rr-0.0173405,4)).  Graphing from -pi to pi, we would have a spike at 0.  We could zoom into the spike by plotting from -0.01, to +0.01.  We're making progress, but it would still be a high frequency mess.

So, we would scale the graph by exp(-1220000*t*I)!!  That's because at this radius, the graph is dominated by the coefficient for a_n*x^1220000!!  And then we get this beautiful approximately Gaussian distribution.  Here we graph from t=pi-0.01, to pi+0.01, -exp(logrho(rr+0.01384115+exp(t*I),4)*exp(-1220000*t*I)!!
   

The Gaussian approximation is so good, that it is probably accurate to several decimal digits for the envelope approximation of 0.0006076.  So how do we calculate the rn=1220000?  And how do we calculate the area of the Gaussian envelope?

This is the pari-gp equation I use.
logrho(rr+exp(log(-0.01384115)+x+O(x^3)),4)
and the output is as follows, where the polcoeff of the zero term is the magnitude reported earlier.  The first derivative is approximately 1220000.  And the 2nd derivative is the x^2 coefficient, which doubles when we take the derivative.
-2086301.83826486 + 1219999.96585252*x + 215528.202158871*x^2 + O(x^3)
\[\text{envelope} \approx \frac{1}{\sqrt{4\cdot a_2\pi}}\approx 0.0006076\]
\[\text{logcoff}(rr,rn,4)\approx\ln{|b_n|}\approx  \Re(a_0) - rn\cdot\Re(a_1)+ \ln(\text{env})\approx 3135424.03\]

The logcoeff routine uses Newton's method to quickly find the optimal radius which is dominated by the b_n Taylor series coefficient, by calling logrho iteratively until the derivative matches the desired r_n value.  The logcoeff is returned as a vector; z=logcoeff(rr,dn,4); z[1].  The envelope approximation term was 0.06076% of the value of rho(rr-0.01384115) so the other 99.94% of the value of rho comes from nearby Taylor series coefficients.  Here is the approximately Gaussian distribution r_n from 1215000 to 1225000, or 1220000+/-5000 to see each b_n's contribution at the radius of interest.  Since the infinite Fourier series of a Gaussian is ... another Gaussian, we might expect that these two graphs are both approximately Gaussian.  Notice that the b_n for n=1220000 make the approximately largest contribution, and that is the maximum at approximately 0.0006076.  By the time we get to coefficient 1215000 or 1225000, the contribution is less than 1.8*10^-16, and the sum over these 10000 coefficients is very nearly 1, also as expected.