Hey, Sheldon.
So if I'm reading you correctly, we are trying to look at solutions of;
\(
\phi(s) + s = (2k+1)\pi i\\
\)
Then, for \( |x| < \delta \),
\(
\phi(s+x) + s+x = 0\\
\)
Therefore these points cluster when doing the iteration,
\(
\tau \mapsto s + \log(1+\frac{\tau(s+1)}{\phi(s+1)})\\
\)
Because we get too close to \( \log(0) \) at the real line!
Fascinating! Fascinating! I have made a grave mistake; and I think I can prove it!
Quick question though; do you mind looking at the iteration:
\(
\tau(x+i\pi/2) = x+i\pi/2 +\log(1+ \frac{\tau(x+1+i\pi/2)}{\phi(x+1+i\pi/2)})\\
\)
For \( \tau_0(x + i\pi/2) = 0 \) and \( \tau_1(x+i\pi/2) = x+i\pi/2 \). If this converges there may be something else I can say. And I may not be entirely wrong. What if it's just \( C^{\infty} \) on the real line (the singularities cluster there); but in the complex plane we don't necessarily always suffer from this problem.
What if \( (-\infty,-2] \) is not the branch cut but \( \mathbb{R} \) is, and so is \( \mathbb{R} + 2\pi i k \)? Naturally it's nowhere analytic on the branch-cut and it can't converge uniformly there (what you see as clustering of singularities) (but in the complex plane the singularities cluster towards the real line; away from from a compact set excluding \( 2\pi i \) multiples of \( \mathbb{R} \)!).
I think it may still be holomorphic on \( \mathbb{C} \) minus branch cuts; just \( \mathbb{R} \) is a branch cut...
(LAST EDIT I SWEAR!)
Think: the log's can't force the imaginary part to zero because we keep on adding in an imaginary part... not sure how to explain it.
Long story short: I was so focused on \( 0 < \Im(s) < 2\pi \) I forgot about the end points.
So if I'm reading you correctly, we are trying to look at solutions of;
\(
\phi(s) + s = (2k+1)\pi i\\
\)
Then, for \( |x| < \delta \),
\(
\phi(s+x) + s+x = 0\\
\)
Therefore these points cluster when doing the iteration,
\(
\tau \mapsto s + \log(1+\frac{\tau(s+1)}{\phi(s+1)})\\
\)
Because we get too close to \( \log(0) \) at the real line!
Fascinating! Fascinating! I have made a grave mistake; and I think I can prove it!
Quick question though; do you mind looking at the iteration:
\(
\tau(x+i\pi/2) = x+i\pi/2 +\log(1+ \frac{\tau(x+1+i\pi/2)}{\phi(x+1+i\pi/2)})\\
\)
For \( \tau_0(x + i\pi/2) = 0 \) and \( \tau_1(x+i\pi/2) = x+i\pi/2 \). If this converges there may be something else I can say. And I may not be entirely wrong. What if it's just \( C^{\infty} \) on the real line (the singularities cluster there); but in the complex plane we don't necessarily always suffer from this problem.
What if \( (-\infty,-2] \) is not the branch cut but \( \mathbb{R} \) is, and so is \( \mathbb{R} + 2\pi i k \)? Naturally it's nowhere analytic on the branch-cut and it can't converge uniformly there (what you see as clustering of singularities) (but in the complex plane the singularities cluster towards the real line; away from from a compact set excluding \( 2\pi i \) multiples of \( \mathbb{R} \)!).
I think it may still be holomorphic on \( \mathbb{C} \) minus branch cuts; just \( \mathbb{R} \) is a branch cut...
(LAST EDIT I SWEAR!)
Think: the log's can't force the imaginary part to zero because we keep on adding in an imaginary part... not sure how to explain it.
Long story short: I was so focused on \( 0 < \Im(s) < 2\pi \) I forgot about the end points.