02/26/2023, 09:33 PM
(02/26/2023, 11:52 AM)JmsNxn Wrote: Okay, so take this post with a Grain of salt. I wanted to look deeper into this question. Because my suspicion is that we can say that sometimes, these are natural analytic continuations. I apologize if this is dumb. I'm just spit balling
I'd like to add my own two cents on:
\[
f(x) = \sum_{n=0}^\infty \frac{x^n}{1+x^n} \frac{1}{2^n}\\
\]
First of all, the word I used to use for these functions is a function that maps \(\mathbb{C} \to \mathbb{C}\) upto a measure zero set in \(\mathbb{C}\). In this case, the measure zero set is the unit circle \(U\). So that \(f : \mathbb{C}/U \to \mathbb{C}\). There is absolutely zero way to analytically continue these functions, or refer to them as analytic continuations of each other. By standard literature; they are two functions which are holomorphic on disjointed domains of \(\mathbb{C}\).
But yes, ONE is more natural; obviously the one which is just plugging in the number. This is a fallacy that Cauchy called the Generality of Algebra. And it was a stark criticism of Euler. Who upon which, Euler would use the Generality of Algebra to get correct results. But he also got incorrect results.
...
Lets define
\[
f(x) = \sum_{n=0}^\infty \frac{x^n}{1+x^n} \frac{1}{2^n}\\
\]
and
g(x)
\[
g(x) = \sum_{n=0}^\infty \frac{x^n}{1+x^n} \frac{1}{3^n}\\
\]
Now there probably exist entire functions A(z),B(z) such that
A(f(z)) = B(g(z))
The idea was that taylor series of f or g would still have the same natural boundary because their powers do.
(although the multiplicity changes)
This leads to the idea that if we accept the continuation by plug-in for f and g then
Q(z) as a taylor expansion in two variables (a(z),b(z))
where a(z) = f(z) , b(z) = g(z)
starts to make sense.
Also
\[
t(x) = \sum_{n=0}^\infty \frac{x^n}{1+x^n} 2^n\\
\]
by using 1 + 2 + 4 + 8 + ... = -1
might make sense.
t(1) =
\[
t(1) = \sum_{n=0}^\infty \frac{1}{1+1} 2^n\\
\]
so t(1) = - 1/2.
But we must be careful , we do not know if using such ideas is self-consistant with eachother.
I think this is somewhat in the spirit of what Caleb wanted to do.
regards
tommy1729


