This is actually reducible to a zeta function, in case you guys didn't know. I'm going to switch to \(-x\) rather than \(x\).
\[
n^{n^{-x}} = \sum_{k=0}^\infty \log(n)^k \frac{n^{-kx}}{k!}\\
\]
Then interchanging the sum:
\[
T(x,2) = 1 + \sum_{n=2}^\infty \sum_{k=0}^\infty (-1)^{n+1}\log(n)^k \frac{n^{-kx}}{k!}\\
\]
This becomes a counting function with \(g(m)\) such that:
\[
T(x,2) = \sum_{m=1}^\infty g(m) m^{-x}\\
\]
The counting function \(g(m)\) is definitely "power related", where it has a weighted count of some log series at each point. So let's write:
\[
g(m) = \sum_{k,n = 1\,\,m = n^k} (-1)^{n+1}\frac{\log(n)^k}{k!}
\]
Then there's your zeta function. I forget the name of this type of zeta function, but similar ones exist, a famous one, is actually really close to this:
\[
h(m) = \sum_{k,n = 1\,\,m = n^k} (-1)^{n+1} = (-1)^{m+1} \sum_{k,n = 1\,\,m = n^k} 1
\]
This is actually counting how many even/odd powers exist of a number--and finding out if there are more even/odd power functions. And I believe it's open whether this function is analytically continuable to \(1/2 < \Re(x)\)... Gottfried's function is basically just a weighted version of this function.
Pick up your Analytic number theory book and get cracking Gottfried
It's probably equivalent to the Riemann Hypothesis to prove some specific thing about this function, lmao.
Is there anything specific you would like to know? I believe these types of zeta functions appear often in "smooth number theory". https://en.wikipedia.org/wiki/Smooth_number
I'm a little bored, so I thought I'd apply some basic analysis; when we write:
\[
q(m) = \sum_{k,n=1\,\,m=n^k} 1\\
\]
A cleaner way to write this, and something more common in analytic number theory is:
\[
q(m) = \sum_{n \le m\,\,m=n^k} 1\\
\]
I was mistaken with smooth number theory; smooth number theory would restrict the primes available. So I screwed that up, lmao. This is much more standard than smooth numbers. This is just guessing the powers. I believe \(q(m)\) has a name, I just can't remember it off the top of my head. Your weighted version is easily discovered from this form though (though it requires a bit more finesse). Essentially it becomes:
\[
q(m) = \sum_{n\le m} \chi_m(n)\\
\]
Where \(\chi_m(n) = 1\) if there exists a \(k\) such that \(n^k = m\) and \(0\) other wise. From this we are weighting \(\chi_m(n)\) with \(\log\) weight. Getting a bound is easy from this...
The function \(q(m)\) asks how many natural number roots \(m\) has. Gottfried's functions are weighing the solutions, which complicates the matter, but not too much.
This is a very common function, and adding weights is nothing new. I don't even think Gottfried's weights are anything too complicated...
We have to guess how fast \(q(m)\) grows. I think a modest estimate is \(q(m) = m^{1/2+\delta}\), whereby, the estimate of:
\[
g(m) = (-1)^{m+1} \sum_{k,n=1\,\,m=n^k} \frac{\log(n)^k}{k!}\\
\]
is about \(O(m^{3/2}\log(m))\)... Which is accomplished through Abel's summation technique. It's probably a good amount smaller, I think a decent estimate would be \(g(m) = O(m\log(m))\). If you could prove \(g(m) = O(\sqrt{m}\log(m))\) you'd probably have proven the Riemann Hypothesis.
I double checked my analytic number theory library of papers, and I believe I can enlighten you on some good "guesstimates" on the behaviour of your function Gottfried. So again, anything specific you are interested in? The function \(q(m)\) is fairly well studied, and I believe there is an expression involving it and the zeta function. In which your function would only be a weighted version--which means we can relate the two through convolution and Abel's summation technique.
Oh my god. I forgot how easy it is:
\[
g(m) = \sum_{n\le m} (-1)^{n+1} \frac{\log(n)^{\frac{\log m}{\log n}}}{\frac{\log m}{\log n}!}\chi_m(n)
\]
Where \(\chi_m\) was defined as above; if \(n^k = m\) for some \(k\), then \(\chi_m(n)= 1\)--otherwise zero. .... The value \(\log(m)/\log(n) = k\), and we are just reorganizing a sum...
We can fucking pull a lazy cauchy bound which is
\[
|g(m)| \le \log(m) \sum_{n\le m} \chi_m(n) = \log(m) q(m)\\
\]
This means:
\[
T(x,2) = \sum_{m=1}^\infty g(m) m^{-x} = \sum_{m=1}^\infty O\left(\log(m) q(m)\right)m^{-x}\\
\]
The function \(\chi_m(n)\) is called something. I can't remember right now. It's on the tip of my tongue...
Regards, James
\[
n^{n^{-x}} = \sum_{k=0}^\infty \log(n)^k \frac{n^{-kx}}{k!}\\
\]
Then interchanging the sum:
\[
T(x,2) = 1 + \sum_{n=2}^\infty \sum_{k=0}^\infty (-1)^{n+1}\log(n)^k \frac{n^{-kx}}{k!}\\
\]
This becomes a counting function with \(g(m)\) such that:
\[
T(x,2) = \sum_{m=1}^\infty g(m) m^{-x}\\
\]
The counting function \(g(m)\) is definitely "power related", where it has a weighted count of some log series at each point. So let's write:
\[
g(m) = \sum_{k,n = 1\,\,m = n^k} (-1)^{n+1}\frac{\log(n)^k}{k!}
\]
Then there's your zeta function. I forget the name of this type of zeta function, but similar ones exist, a famous one, is actually really close to this:
\[
h(m) = \sum_{k,n = 1\,\,m = n^k} (-1)^{n+1} = (-1)^{m+1} \sum_{k,n = 1\,\,m = n^k} 1
\]
This is actually counting how many even/odd powers exist of a number--and finding out if there are more even/odd power functions. And I believe it's open whether this function is analytically continuable to \(1/2 < \Re(x)\)... Gottfried's function is basically just a weighted version of this function.
Pick up your Analytic number theory book and get cracking Gottfried
It's probably equivalent to the Riemann Hypothesis to prove some specific thing about this function, lmao.Is there anything specific you would like to know? I believe these types of zeta functions appear often in "smooth number theory". https://en.wikipedia.org/wiki/Smooth_number
I'm a little bored, so I thought I'd apply some basic analysis; when we write:
\[
q(m) = \sum_{k,n=1\,\,m=n^k} 1\\
\]
A cleaner way to write this, and something more common in analytic number theory is:
\[
q(m) = \sum_{n \le m\,\,m=n^k} 1\\
\]
I was mistaken with smooth number theory; smooth number theory would restrict the primes available. So I screwed that up, lmao. This is much more standard than smooth numbers. This is just guessing the powers. I believe \(q(m)\) has a name, I just can't remember it off the top of my head. Your weighted version is easily discovered from this form though (though it requires a bit more finesse). Essentially it becomes:
\[
q(m) = \sum_{n\le m} \chi_m(n)\\
\]
Where \(\chi_m(n) = 1\) if there exists a \(k\) such that \(n^k = m\) and \(0\) other wise. From this we are weighting \(\chi_m(n)\) with \(\log\) weight. Getting a bound is easy from this...
The function \(q(m)\) asks how many natural number roots \(m\) has. Gottfried's functions are weighing the solutions, which complicates the matter, but not too much.
This is a very common function, and adding weights is nothing new. I don't even think Gottfried's weights are anything too complicated...
We have to guess how fast \(q(m)\) grows. I think a modest estimate is \(q(m) = m^{1/2+\delta}\), whereby, the estimate of:
\[
g(m) = (-1)^{m+1} \sum_{k,n=1\,\,m=n^k} \frac{\log(n)^k}{k!}\\
\]
is about \(O(m^{3/2}\log(m))\)... Which is accomplished through Abel's summation technique. It's probably a good amount smaller, I think a decent estimate would be \(g(m) = O(m\log(m))\). If you could prove \(g(m) = O(\sqrt{m}\log(m))\) you'd probably have proven the Riemann Hypothesis.
I double checked my analytic number theory library of papers, and I believe I can enlighten you on some good "guesstimates" on the behaviour of your function Gottfried. So again, anything specific you are interested in? The function \(q(m)\) is fairly well studied, and I believe there is an expression involving it and the zeta function. In which your function would only be a weighted version--which means we can relate the two through convolution and Abel's summation technique.
Oh my god. I forgot how easy it is:
\[
g(m) = \sum_{n\le m} (-1)^{n+1} \frac{\log(n)^{\frac{\log m}{\log n}}}{\frac{\log m}{\log n}!}\chi_m(n)
\]
Where \(\chi_m\) was defined as above; if \(n^k = m\) for some \(k\), then \(\chi_m(n)= 1\)--otherwise zero. .... The value \(\log(m)/\log(n) = k\), and we are just reorganizing a sum...
We can fucking pull a lazy cauchy bound which is
\[
|g(m)| \le \log(m) \sum_{n\le m} \chi_m(n) = \log(m) q(m)\\
\]
This means:
\[
T(x,2) = \sum_{m=1}^\infty g(m) m^{-x} = \sum_{m=1}^\infty O\left(\log(m) q(m)\right)m^{-x}\\
\]
The function \(\chi_m(n)\) is called something. I can't remember right now. It's on the tip of my tongue...
Regards, James
