02/07/2021, 11:11 PM
Hey, so I wanted to try and approach from the path of proving NON-convergence. When you can't prove it does, try and prove it can't. We're going to simplify to the case;
\(
-\phi(t+\pi i) = \psi(t)\\
\)
Now, I had boiled it down into one thing or another--which is \( \frac{\psi(t)}{t} \to 1 \) or we get precisely my worst fear; the graphs Sheldon is showing. I was hoping it would level out and not grow too fast, so that it couldn't shrink. But lo and behold, it looks like it grows too fast, causing it to shrink; causing the violent oscillation we see in this graph. I tried running my own pari-gp just to double check (maybe just maybe there was an artifact in Sheldon's computations (not likely though)), and I do notice this weird hump we see in this graph right before the oscillation. Which is truly the death of the method. It has to stay below \( t \) but approach \( t \) for this thing to work. The moment it breaks this pattern, we get chaos. (And we can notice this in Sheldon's graph, where the tiny hump happens where it just dips below \( -t \) causing the insane breakdown.) Now I can't prove anything exactly, but I thought I'd try and frame this post from the worst case scenario.
Define the sequence of numbers \( \delta_n = \psi(t+2n) \) for \( 0 < a < t < b < 1 \). Assume for some \( N>2 \) that \( \delta_N < 1 \) (which is assuming the worst possible case is going to happen.)
Then,
\(
\delta_{N+1} = e^{t+2N+1 - e^{t+2N - \delta_N}}\\
0 \le e^{2N+1 - e^{2N+1}} \le |\delta_{N+1}| \le e^{2N+2 - e^{2N-1}} < 1\\
\)
which implies that, \( \delta_n \to 0 \) and it looks double exponential. Now define the sequence of numbers, \( \mu_n = \psi(t+2n+1) \).
\(
\mu_n = e^{t+2n - \delta_n}\\
0 \le e^{2n - 1} \le \mu_n \le e^{2n+1}\\
\)
Now it's no hard fact to sandwich,
\(
0 \le e^{2n-1}e^{2n+1 - e^{2n+1}} \le \mu_n \delta_n \le e^{2n+1}e^{2n+2 - e^{2n-1}}\\
0 \le \mu_n \delta_n \le e^{4n+3-e^{2n-1}} \to 0\\
\)
Now, the reason this is so important, and why this is my biggest fear, is that we can now confidently say that, for even \( m=2k \)
\(
\prod_{j=1}^m \psi(t+j) =\prod_{j=1}^k \mu_j \delta_j \to 0\\
\)
This kaputs the entire construction of \( \tau(s) \) along the line \( \Im(s) = \pi \). Since the crucial philosophy of the proof is to use Banach's Fixed Point Theorem,
\(
|\tau_{m+1}(t+\pi i) - \tau_m(t+ \pi i)| \le \frac{|t+m + \pi i|}{|\prod_{j=1}^m \psi(t+j)|} \to \infty\\
\)
So unless Sheldon's code is making a rounding error causing us to just dip below \( -t \), and the nested exponentials magnify that error causing oscillation... It looks like this construction will fail on the line \( \Im(s)=\pi \). Which is very unfortunate. I, for one, have faith in Sheldon's code. So I am officially in the not holomorphic camp--it's just \( C^{\infty} \) on \( (-2,\infty) \).
Regards, James
\(
-\phi(t+\pi i) = \psi(t)\\
\)
Now, I had boiled it down into one thing or another--which is \( \frac{\psi(t)}{t} \to 1 \) or we get precisely my worst fear; the graphs Sheldon is showing. I was hoping it would level out and not grow too fast, so that it couldn't shrink. But lo and behold, it looks like it grows too fast, causing it to shrink; causing the violent oscillation we see in this graph. I tried running my own pari-gp just to double check (maybe just maybe there was an artifact in Sheldon's computations (not likely though)), and I do notice this weird hump we see in this graph right before the oscillation. Which is truly the death of the method. It has to stay below \( t \) but approach \( t \) for this thing to work. The moment it breaks this pattern, we get chaos. (And we can notice this in Sheldon's graph, where the tiny hump happens where it just dips below \( -t \) causing the insane breakdown.) Now I can't prove anything exactly, but I thought I'd try and frame this post from the worst case scenario.
Define the sequence of numbers \( \delta_n = \psi(t+2n) \) for \( 0 < a < t < b < 1 \). Assume for some \( N>2 \) that \( \delta_N < 1 \) (which is assuming the worst possible case is going to happen.)
Then,
\(
\delta_{N+1} = e^{t+2N+1 - e^{t+2N - \delta_N}}\\
0 \le e^{2N+1 - e^{2N+1}} \le |\delta_{N+1}| \le e^{2N+2 - e^{2N-1}} < 1\\
\)
which implies that, \( \delta_n \to 0 \) and it looks double exponential. Now define the sequence of numbers, \( \mu_n = \psi(t+2n+1) \).
\(
\mu_n = e^{t+2n - \delta_n}\\
0 \le e^{2n - 1} \le \mu_n \le e^{2n+1}\\
\)
Now it's no hard fact to sandwich,
\(
0 \le e^{2n-1}e^{2n+1 - e^{2n+1}} \le \mu_n \delta_n \le e^{2n+1}e^{2n+2 - e^{2n-1}}\\
0 \le \mu_n \delta_n \le e^{4n+3-e^{2n-1}} \to 0\\
\)
Now, the reason this is so important, and why this is my biggest fear, is that we can now confidently say that, for even \( m=2k \)
\(
\prod_{j=1}^m \psi(t+j) =\prod_{j=1}^k \mu_j \delta_j \to 0\\
\)
This kaputs the entire construction of \( \tau(s) \) along the line \( \Im(s) = \pi \). Since the crucial philosophy of the proof is to use Banach's Fixed Point Theorem,
\(
|\tau_{m+1}(t+\pi i) - \tau_m(t+ \pi i)| \le \frac{|t+m + \pi i|}{|\prod_{j=1}^m \psi(t+j)|} \to \infty\\
\)
So unless Sheldon's code is making a rounding error causing us to just dip below \( -t \), and the nested exponentials magnify that error causing oscillation... It looks like this construction will fail on the line \( \Im(s)=\pi \). Which is very unfortunate. I, for one, have faith in Sheldon's code. So I am officially in the not holomorphic camp--it's just \( C^{\infty} \) on \( (-2,\infty) \).
Regards, James