07/05/2022, 12:06 AM
I understand that, Daniel. I apologize if my response seemed hostile. Wasn't my intention, lol.
All I'm saying is that the Taylor series approach, is inherently Schroder's construction. I'm not doubting that it works in any way shape or form. But I suggest you observe your iteration \(f^{\circ 1/2}(z)\) of \(f = \sqrt{2}^z\), and trace \(z\) from \(2 \to 4\). Somewhere along that path there is a singularity (probably a tiny discontinuity/jump at about 1E-10 height). If not, you've done something wrong.
All I'm saying is that the Taylor series approach, is inherently Schroder's construction. I'm not doubting that it works in any way shape or form. But I suggest you observe your iteration \(f^{\circ 1/2}(z)\) of \(f = \sqrt{2}^z\), and trace \(z\) from \(2 \to 4\). Somewhere along that path there is a singularity (probably a tiny discontinuity/jump at about 1E-10 height). If not, you've done something wrong.