Sorry, Tommy, I made a typo.
You are assuming this is a single holomorphic function. IT is not a holomorphic function, it is a generalized holomorphic function. I've been doing more reading on this.
Yes; if: \(f\) is holomorphic you are correct; excusing my typo, you are wrong. But these are not holomorphic functions. We are taking a function holomorphic on two disconnected domains. Where it has different values on both domains.
Perhaps to help you out I should write it out more explicitly.
Let:
\[
\begin{align}
f_1(x) &= \sum_{n=0}^\infty \frac{x^n}{1+x^n}\frac{1}{2^n}\,\,\text{for}\,\,|x|<1\\
f_2(x) &= \sum_{n=0}^\infty \frac{x^n}{1+x^n}\frac{1}{2^n}\,\,\text{for}\,\,|x|>1\\
\end{align}\\
\]
Then let's rewrite my equation as:
\[
f_1(x) = 2 - f_2(1/x)\\
\]
This is actually very easy to numerically verify, as Caleb pointed out.
\[
\frac{(1/x)^n}{1+(1/x)^n} = \frac{1}{x^n +1}\\
\]
Then:
\[
1-\frac{1}{x^n +1} = \frac{x^n+1}{x^n+1} - \frac{1}{x^n + 1} = \frac{x^n}{x^n + 1}\\
\]
Therefore:
\[
2 - f_2(1/x) = \sum_{n=0}^\infty \frac{1}{2^n}-\frac{1}{x^n+1} \frac{1}{2^n}\\
\]
Because: \(2 = \sum_{n=0}^\infty \frac{1}{2^n}\). And this just equals:
\[
f_1(x) = \sum_{n=0}^\infty \frac{x^n}{1+x^n} \frac{1}{2^n}\\
\]
This is not nonsense. We are working with two functions; which are "natural extensions of each other" to disconnected domains. This is known as a "generalized analytic function"; which is holomorphic on disjointed domains!
What I am saying, is that I am observing this phenomena for more exotic functions than this simple rendition. I can't prove it yet; but calculators don't lie... I should have written that the formula turns out like:
\[
L(1/x) = C - L(x)\\
\]
Which DOES NOT suffer your duplication formula. Sorry, I fucked up the signs before. I was just note dumping too fast!
EDIT!
I am referring to this as Lambert's Reflection formula. But it doesn't just follow for:
\[
\frac{x^n}{1+x^n}\\
\]
In the simple way Caleb presented it; or how I just explained. It also follows for:
\[
\frac{x^n}{(1+x^n)^2}\\
\]
But you have to be a bit more careful on how you rearrange the objects. And it also follows for higher powers. It also follows for mixing the powers. IT FUCKING FOLLOWS EVERYWHERE. The identity:
\[
L(x) = C - L(1/x)\\
\]
Is super fucking general, Tommy!!!!!!!
We can take \(L(x)\) for \(|x| >1\)--take a Cauchy integral, and reconstruct \(L(x)\) for \(|x| <1\). And it looks like a Lambert Reflection Formula.
I'm trying to map out the exact mathematics right now. I don't have it yet. But I am pretty fucking close.
You are assuming this is a single holomorphic function. IT is not a holomorphic function, it is a generalized holomorphic function. I've been doing more reading on this.
Yes; if: \(f\) is holomorphic you are correct; excusing my typo, you are wrong. But these are not holomorphic functions. We are taking a function holomorphic on two disconnected domains. Where it has different values on both domains.
Perhaps to help you out I should write it out more explicitly.
Let:
\[
\begin{align}
f_1(x) &= \sum_{n=0}^\infty \frac{x^n}{1+x^n}\frac{1}{2^n}\,\,\text{for}\,\,|x|<1\\
f_2(x) &= \sum_{n=0}^\infty \frac{x^n}{1+x^n}\frac{1}{2^n}\,\,\text{for}\,\,|x|>1\\
\end{align}\\
\]
Then let's rewrite my equation as:
\[
f_1(x) = 2 - f_2(1/x)\\
\]
This is actually very easy to numerically verify, as Caleb pointed out.
\[
\frac{(1/x)^n}{1+(1/x)^n} = \frac{1}{x^n +1}\\
\]
Then:
\[
1-\frac{1}{x^n +1} = \frac{x^n+1}{x^n+1} - \frac{1}{x^n + 1} = \frac{x^n}{x^n + 1}\\
\]
Therefore:
\[
2 - f_2(1/x) = \sum_{n=0}^\infty \frac{1}{2^n}-\frac{1}{x^n+1} \frac{1}{2^n}\\
\]
Because: \(2 = \sum_{n=0}^\infty \frac{1}{2^n}\). And this just equals:
\[
f_1(x) = \sum_{n=0}^\infty \frac{x^n}{1+x^n} \frac{1}{2^n}\\
\]
This is not nonsense. We are working with two functions; which are "natural extensions of each other" to disconnected domains. This is known as a "generalized analytic function"; which is holomorphic on disjointed domains!
What I am saying, is that I am observing this phenomena for more exotic functions than this simple rendition. I can't prove it yet; but calculators don't lie... I should have written that the formula turns out like:
\[
L(1/x) = C - L(x)\\
\]
Which DOES NOT suffer your duplication formula. Sorry, I fucked up the signs before. I was just note dumping too fast!
EDIT!
I am referring to this as Lambert's Reflection formula. But it doesn't just follow for:
\[
\frac{x^n}{1+x^n}\\
\]
In the simple way Caleb presented it; or how I just explained. It also follows for:
\[
\frac{x^n}{(1+x^n)^2}\\
\]
But you have to be a bit more careful on how you rearrange the objects. And it also follows for higher powers. It also follows for mixing the powers. IT FUCKING FOLLOWS EVERYWHERE. The identity:
\[
L(x) = C - L(1/x)\\
\]
Is super fucking general, Tommy!!!!!!!
We can take \(L(x)\) for \(|x| >1\)--take a Cauchy integral, and reconstruct \(L(x)\) for \(|x| <1\). And it looks like a Lambert Reflection Formula.
I'm trying to map out the exact mathematics right now. I don't have it yet. But I am pretty fucking close.
