08/11/2022, 07:03 PM
(08/11/2022, 03:28 AM)JmsNxn Wrote:(08/10/2022, 02:12 PM)Leo.W Wrote: ...
Ironically, Leo. I wanted to approach this problem using work from Remmert's two part textbooks on complex ....
I should add, that Remmert has a section on summing the very function you mention: \(\sum_{n=0}^\infty z^{2^n}\).
Not to judge, but any sum with terms with \(O(n!^k)\) can converge after at most (k+1)-times Borel summation, bro, so it's not the big deal.
This is how we deal with generalized Hypergeometric functions like, \(f(z)=\sum_{n=0}^\infty{n!^2z^n}\), can converge after 2 borel summation procedure.
About the function \(\sum_{n=0}^\infty z^{2^n}\), it's hadamard's problem and called lacunary functions, I mentioned it because I thought it'd be a great example for a damn fast growing func, because \(O(a^{2^n})=O(e^{ln(a)2^n})>>O(e^{knln(n)})>O(n^{kn})>O(n!^k)\) for any k since \(O(2^n)>>O(knln(n))\)
And then \(O(a^{2^n})\) is not Borel summable even after k times, but still summable by contour integrals with Residue theorem. And hence we can sum \(O(^n10)\) like \(1-10+10^{10}-10^{10^{10}}+\cdot\)
My focus was on the summation and at least now have no idea bout coefficients'.
Regards, Leo

