02/19/2023, 09:13 AM
I believe you are confused, Caleb. To ask for an analytic continuation of \(\lambda(n)\)--is very unnatural. To ask for an analytic continuation, you must specify some additional condition, to what the continuation looks like. In this case we are very unnatural.
Let's say we write:
\[
\lambda(n) =(-1)^{\Omega(n)}\\
\]
Where \(\Omega(n)\)--counts how many prime factors \(n\) has. When we write:
\[
\Omega(s)\\
\]
For \(\Re(s) > 1\); where \(\Omega(s) \Big{|}_{\mathbb{N}} = \Omega(n)\). We have to make an additional distinction claim on \(\Omega(s)\). Let's say that \(\Omega(s)\) is in our Fractional Calculus space. So let's say:
\[
\Omega(s+1) = \frac{d^{s}}{dw^s}\Big{|}_{w=0} \sum_{n=0}^\infty \Omega(n+1) \frac{(-w)^n}{n!}\\
\]
In order to call this an interpolation, you have to show that:
\[
\begin{align}
f(w) &= \sum_{n=0}^\infty \Omega(n+1) \frac{(-w)^n}{n!}\\
\Omega(s+1) &= \frac{1}{\Gamma(1-s)} \int_0^\infty f(w) w^{-s}\,dw\\
\int_0^\infty |f(w)|\,dw &< \infty\\
\end{align}
\]
Now this would perfectly solve your problem. BUT, AND THIS IS A HUGE FUCKING BUT!
It's on you to show that:
\[
\int_0^\infty |f(w)|\,dw < \infty
\]
In order to say that:
\[
\lambda(s)\\
\]
Is holomorphic. Or it's on you to say that:
\[
\lambda(s+1) = \frac{d^s}{dw^s}\Big{|}_{w=0} \sum_{n=0}^\infty \lambda(n+1) \frac{(-w)^n}{n!}\\
\]
And test that:
\[
\int_0^\infty \left| \sum_{n=0}^\infty \lambda(n+1) \frac{(-w)^n}{n!}\right| \,dw < \infty\\
\]
BUT YOU HAVE TO PROVE THIS. AND IT IS SUPER NONTRIVIAL!!!!!!!
Let's instead look at your problem at face value:
\[
\sum_{n=0}^\infty \frac{\lambda(n) z^n}{1-z^n} = \sum_{n=0}^\infty z^{n^{2}}\\
\]
Setting \(\lambda(0) = 1\) and \(\Omega(0) = 0\).
The question, and area of research you are broaching on, Caleb. Is an old avenue of research. I am not that well versed in it; but it's referred to as: THE CIRCLE METHOD. Or, Ramanujan's and Hardy's CIRCLE METHOD. What happens is close to what you are writing. But it gets very harder.
\[
\sum_{n=0}^\infty z^{n^2} = \sum_{n=0}^M \frac{\lambda(n)z^n}{1-z^n} + \sum_{j=0}^M H_j(z)\\
\]
Where as \(M\to \infty\), the sum \(H_j(z)\) turns into asymptotic series. Which allow for estimates on growth. END OF STORY. Your quest to analytically continue this sum is futile.
The function:
\[
F(z) = \sum_{n=0}^\infty z^{n^2}\\
\]
Has a dense amount of singularities on \(|z| =1\). This function is reducible into:
\[
F(e^{2\pi i x}) = \sum_{n=0}^\infty e^{2\pi i x n^2}\\
\]
Which is holomorphic for \(\Im(x) > 0\). Where a natural boundary appears along \(x \in \mathbb{R}\). There is no analytic continuation. But for; \(\Im(x) < 0\) we have:
\[
\begin{align}
F(e^{2\pi i x}) &= \overline{\sum_{n=0}^\infty e^{2\pi i \overline{x}n^2}}\\
&= \sum_{n=0}^\infty e^{-2\pi i x n^2}\\
\end{align}
\]
And this finds \(F(z)\) for \(|z| >1\). But each are not holomorphic on the unit circle \(|z| = 1\). Where upon we have two different functions; one holomorphic for \(|z| < 1\)--the other for \(|z| > 1\). Where each complements the other; and both blow up as \(|z|\to 1\).
In conclusion; we have shown asymptotics on \(\lambda(n)\) and \(\Omega(n)\)... Please state them!
Regards, James
Let's say we write:
\[
\lambda(n) =(-1)^{\Omega(n)}\\
\]
Where \(\Omega(n)\)--counts how many prime factors \(n\) has. When we write:
\[
\Omega(s)\\
\]
For \(\Re(s) > 1\); where \(\Omega(s) \Big{|}_{\mathbb{N}} = \Omega(n)\). We have to make an additional distinction claim on \(\Omega(s)\). Let's say that \(\Omega(s)\) is in our Fractional Calculus space. So let's say:
\[
\Omega(s+1) = \frac{d^{s}}{dw^s}\Big{|}_{w=0} \sum_{n=0}^\infty \Omega(n+1) \frac{(-w)^n}{n!}\\
\]
In order to call this an interpolation, you have to show that:
\[
\begin{align}
f(w) &= \sum_{n=0}^\infty \Omega(n+1) \frac{(-w)^n}{n!}\\
\Omega(s+1) &= \frac{1}{\Gamma(1-s)} \int_0^\infty f(w) w^{-s}\,dw\\
\int_0^\infty |f(w)|\,dw &< \infty\\
\end{align}
\]
Now this would perfectly solve your problem. BUT, AND THIS IS A HUGE FUCKING BUT!
It's on you to show that:
\[
\int_0^\infty |f(w)|\,dw < \infty
\]
In order to say that:
\[
\lambda(s)\\
\]
Is holomorphic. Or it's on you to say that:
\[
\lambda(s+1) = \frac{d^s}{dw^s}\Big{|}_{w=0} \sum_{n=0}^\infty \lambda(n+1) \frac{(-w)^n}{n!}\\
\]
And test that:
\[
\int_0^\infty \left| \sum_{n=0}^\infty \lambda(n+1) \frac{(-w)^n}{n!}\right| \,dw < \infty\\
\]
BUT YOU HAVE TO PROVE THIS. AND IT IS SUPER NONTRIVIAL!!!!!!!
Let's instead look at your problem at face value:
\[
\sum_{n=0}^\infty \frac{\lambda(n) z^n}{1-z^n} = \sum_{n=0}^\infty z^{n^{2}}\\
\]
Setting \(\lambda(0) = 1\) and \(\Omega(0) = 0\).
The question, and area of research you are broaching on, Caleb. Is an old avenue of research. I am not that well versed in it; but it's referred to as: THE CIRCLE METHOD. Or, Ramanujan's and Hardy's CIRCLE METHOD. What happens is close to what you are writing. But it gets very harder.
\[
\sum_{n=0}^\infty z^{n^2} = \sum_{n=0}^M \frac{\lambda(n)z^n}{1-z^n} + \sum_{j=0}^M H_j(z)\\
\]
Where as \(M\to \infty\), the sum \(H_j(z)\) turns into asymptotic series. Which allow for estimates on growth. END OF STORY. Your quest to analytically continue this sum is futile.
The function:
\[
F(z) = \sum_{n=0}^\infty z^{n^2}\\
\]
Has a dense amount of singularities on \(|z| =1\). This function is reducible into:
\[
F(e^{2\pi i x}) = \sum_{n=0}^\infty e^{2\pi i x n^2}\\
\]
Which is holomorphic for \(\Im(x) > 0\). Where a natural boundary appears along \(x \in \mathbb{R}\). There is no analytic continuation. But for; \(\Im(x) < 0\) we have:
\[
\begin{align}
F(e^{2\pi i x}) &= \overline{\sum_{n=0}^\infty e^{2\pi i \overline{x}n^2}}\\
&= \sum_{n=0}^\infty e^{-2\pi i x n^2}\\
\end{align}
\]
And this finds \(F(z)\) for \(|z| >1\). But each are not holomorphic on the unit circle \(|z| = 1\). Where upon we have two different functions; one holomorphic for \(|z| < 1\)--the other for \(|z| > 1\). Where each complements the other; and both blow up as \(|z|\to 1\).
In conclusion; we have shown asymptotics on \(\lambda(n)\) and \(\Omega(n)\)... Please state them!
Regards, James