Okay, so the Euler Formula I was able to get is not as nice as I'd hoped, but still pretty good.
I still think there is a better one out there. We start by writing:
\[
g(m) = (-1)^{m+1}\sum_{n\le m} \frac{\log(n)^{\frac{\log(m)}{\log(n)}}}{\frac{\log(m)}{\log(n)}!}\chi_m(n)\\
\]
This is naively bounded as:
\[
|g(m)| \le \log(m) |\sqrt{\mathcal{I}_m}|\\
\]
If you don't trust my math on that, as it is a little tricky--this entire discussion can be done with the bound \(|g(m)| \le \log(m)^{\log(m)} |\sqrt{\mathcal{I}_m}|\), and everything still applies
.
For Mphlee's radical. Now we can bound:
\[
|\sqrt{\mathcal{I}_m}| \le \Pi(m)\\
\]
We can also bound:
\[
\log(m) \le m\\
\]
So we can bound:
\[
|g(m)| \le m \Pi(m)\\
\]
This is not the best bound (a lot of this is using pretty weak bounds). We can actually strengthen this by finding \(\tau : \mathbb{N}\to\mathbb{N}\) such that \(\tau(m) \ge \log(m)\) but \(\tau(ab) = \tau(a) \tau(b)\) when \((a,b) = 1\) are coprime. I'm just using \(\tau(m) =m\) at the moment. So now, let's take:
\[
T(x,2) = \sum_{n=1}^\infty n^{n^{-x}} = \sum_{m=1}^\infty g(m)m^{-x}\\
\]
We can bound, for \(\Re(x) = \sigma\):
\[
|T(x,2)| \le \sum_{m=1}^\infty m\Pi(m)m^{-\sigma} = |T|(\sigma)\\
\]
Now I suggest to anyone who hasn't seen this Master Class result by Euler, to go look at it instantly! https://en.wikipedia.org/wiki/Proof_of_t...a_function This is a great result to learn the history of, because it exemplifies just how bananas Euler was. This is largely considered the first Analytic Number Theory result. He used it to create the product expansion of \(\frac{\pi^2}{6}\) using primes!!!!!
I'm going to go quick, but we begin by writing:
\[
|T|(\sigma) = \prod_{p \,\text{prime}} \left(1+\sum_{j=1}^\infty p^j \sigma(j) p^{-j\sigma}\right)\\
\]
Where the product is over all prime numbers.... To explain this, think of how the product:
\[
\left(1 + a^1 + a^2 +....\right)\left(1 + b^1 + b^2 +....\right) = \sum_{j=1}^\infty \sum_{k=1}^\infty a^j b^k\\
\]
We are doing the same thing, but \(a\) and \(b\) are primes. And by the fundamental theorem of arithmetic, every number is represented. The function \(\Pi(mn) = \Pi(m) \Pi(n)\) when \((m,n)=1\), and additionally the prime powers are just the divisor function...
Also, I would like to note, as I haven't at this point, that this is an analytic continuation of \(T(x,2)\), as I've constructed it. As it necessarily converges in a half plane. I never actually proved these things converged, until this bound I just gave. I do believe we can analytically continue this even further. I am optimistic there exists some kind of reflection formula. I'm not upto date on most advanced zeta function stuff; but I believe we can create at least an analytic continuation of \(T(x,2)\) do a bigger domain than \(\Re(x)>1\). I'd put my money on at least \(\Re(x) > 1/2\), we may have trouble in the critical strip, simply because of the way \(g(m)\) looks a lot like a product of divisor functions--and this encodes too much info about primes....
But this is really important to note, because the function \(T(x,2)\) is holomorphic in weird domains as \(\Re(x) < 0\). Where, again, I did the variable change \(x \mapsto -x\) at the beginning of this discussion. So the function Gottfried is detailing in his work, is the left half plane of the function I am detailing. So we're almost there with an analytic continuation....... You would call Gottfried's work the "zeta regularization"--which is just a fancy way of saying the analytic continuation to the left of the half- plane the zeta sum converges at.
But this successfully analytically continues Gottfried's function, to at least \(\Re(x) > 1\), where for \(\Re(x) > 2\) we have a cool prime bound. Gottfried's concerns were with \(\Re(x) < 0\) as far as I can tell. This leaves us, low and behold, with the troublesome "critical strip" of \( 0 < \Re(x) < 1\). This is solvable. I'll post the analytic continuation soon. This is a result using Abel's interpretation of Zeta functions, and Abel sums and integrals acting on it. It shouldn't be hard at all because we have such nice Abel sums with plain bounds...
So mostly so far, I have shown that Gottfried's sum is formally equal to the sum I wrote. I have shown my sum converges, and additionally Gottfried's sum does converge in some of the areas my sum converges. So it is a viable analytic continuation... I'm very interested to see if we can extend this function cleaner into one function...
I still think there is a better one out there. We start by writing:
\[
g(m) = (-1)^{m+1}\sum_{n\le m} \frac{\log(n)^{\frac{\log(m)}{\log(n)}}}{\frac{\log(m)}{\log(n)}!}\chi_m(n)\\
\]
This is naively bounded as:
\[
|g(m)| \le \log(m) |\sqrt{\mathcal{I}_m}|\\
\]
If you don't trust my math on that, as it is a little tricky--this entire discussion can be done with the bound \(|g(m)| \le \log(m)^{\log(m)} |\sqrt{\mathcal{I}_m}|\), and everything still applies
.For Mphlee's radical. Now we can bound:
\[
|\sqrt{\mathcal{I}_m}| \le \Pi(m)\\
\]
We can also bound:
\[
\log(m) \le m\\
\]
So we can bound:
\[
|g(m)| \le m \Pi(m)\\
\]
This is not the best bound (a lot of this is using pretty weak bounds). We can actually strengthen this by finding \(\tau : \mathbb{N}\to\mathbb{N}\) such that \(\tau(m) \ge \log(m)\) but \(\tau(ab) = \tau(a) \tau(b)\) when \((a,b) = 1\) are coprime. I'm just using \(\tau(m) =m\) at the moment. So now, let's take:
\[
T(x,2) = \sum_{n=1}^\infty n^{n^{-x}} = \sum_{m=1}^\infty g(m)m^{-x}\\
\]
We can bound, for \(\Re(x) = \sigma\):
\[
|T(x,2)| \le \sum_{m=1}^\infty m\Pi(m)m^{-\sigma} = |T|(\sigma)\\
\]
Now I suggest to anyone who hasn't seen this Master Class result by Euler, to go look at it instantly! https://en.wikipedia.org/wiki/Proof_of_t...a_function This is a great result to learn the history of, because it exemplifies just how bananas Euler was. This is largely considered the first Analytic Number Theory result. He used it to create the product expansion of \(\frac{\pi^2}{6}\) using primes!!!!!
I'm going to go quick, but we begin by writing:
\[
|T|(\sigma) = \prod_{p \,\text{prime}} \left(1+\sum_{j=1}^\infty p^j \sigma(j) p^{-j\sigma}\right)\\
\]
Where the product is over all prime numbers.... To explain this, think of how the product:
\[
\left(1 + a^1 + a^2 +....\right)\left(1 + b^1 + b^2 +....\right) = \sum_{j=1}^\infty \sum_{k=1}^\infty a^j b^k\\
\]
We are doing the same thing, but \(a\) and \(b\) are primes. And by the fundamental theorem of arithmetic, every number is represented. The function \(\Pi(mn) = \Pi(m) \Pi(n)\) when \((m,n)=1\), and additionally the prime powers are just the divisor function...
Also, I would like to note, as I haven't at this point, that this is an analytic continuation of \(T(x,2)\), as I've constructed it. As it necessarily converges in a half plane. I never actually proved these things converged, until this bound I just gave. I do believe we can analytically continue this even further. I am optimistic there exists some kind of reflection formula. I'm not upto date on most advanced zeta function stuff; but I believe we can create at least an analytic continuation of \(T(x,2)\) do a bigger domain than \(\Re(x)>1\). I'd put my money on at least \(\Re(x) > 1/2\), we may have trouble in the critical strip, simply because of the way \(g(m)\) looks a lot like a product of divisor functions--and this encodes too much info about primes....
But this is really important to note, because the function \(T(x,2)\) is holomorphic in weird domains as \(\Re(x) < 0\). Where, again, I did the variable change \(x \mapsto -x\) at the beginning of this discussion. So the function Gottfried is detailing in his work, is the left half plane of the function I am detailing. So we're almost there with an analytic continuation....... You would call Gottfried's work the "zeta regularization"--which is just a fancy way of saying the analytic continuation to the left of the half- plane the zeta sum converges at.
But this successfully analytically continues Gottfried's function, to at least \(\Re(x) > 1\), where for \(\Re(x) > 2\) we have a cool prime bound. Gottfried's concerns were with \(\Re(x) < 0\) as far as I can tell. This leaves us, low and behold, with the troublesome "critical strip" of \( 0 < \Re(x) < 1\). This is solvable. I'll post the analytic continuation soon. This is a result using Abel's interpretation of Zeta functions, and Abel sums and integrals acting on it. It shouldn't be hard at all because we have such nice Abel sums with plain bounds...
So mostly so far, I have shown that Gottfried's sum is formally equal to the sum I wrote. I have shown my sum converges, and additionally Gottfried's sum does converge in some of the areas my sum converges. So it is a viable analytic continuation... I'm very interested to see if we can extend this function cleaner into one function...
