Hey, Bo!
This is totally my domain. I can't work through all of this today. But I was planning on making a thread specifically about Borel summation of these parabolic iterations. I've just been absorbing this asymptotic expansion and trying to figure out.
But to begin;
If you have \(b_k = O(c^k k!)\); you don't just perform the standard way of Borel summing. So wikipedia won't help you. You have to be more clever. True "borel summing" is very sensitive to how you set up your integrals, and how you set up your sequences. There's no one size fits all manner. Borel sum, is more usually used to describe that they are all analytic continuations of each other. Not necessarily you just plug in the integral.
I'm not surprised your initial attempt failed. Because Euler's form (Gervery class 1) sequences do not converge in the manner you posted. When the coefficients are \(k!\). Again, we have to be much more clever. And largely, it equates to a bunch of esoteric changes of variables.
I will post a much better answer in the coming days. I don't want to tonight, because I'm still making sure I'm doing everything right. But, I will leave you with this tidbit.
If \(b_k = O(c^k k!)\)
Then:
\[
F(x) = \sum_{k=0}^\infty b_k\frac{x^{2k}}{2k!}
\]
Is an entire function; which is bounded by \(e^{b|x|}\)--or something close--for \(b<1\).
From here, you take the modified laplace transform:
\[
**\begin{align}
\int_0^\infty e^{-x} F(\sqrt{t}x)\,dx &= \sum_{k=0}^\infty b_k t^k\\
&= t^{-1/2}\int_0^\infty e^{-ut^{-1/2}} F(u)\,du
\end{align}
\]
Which is just a term wise evaluation of the integral***. This object converges for a specific domain in \(t\)--depending on where the asymptotic expansion works. Just be careful with \(t\). But this definitely, absolutely, no doubt about it, converges.
**Converges because \(|F(x)| < e^{b|x|}\) for \(b \approx 0\)--of which, we only need \(b<1\).
***
\[
\begin{align}
\int_0^\infty e^{-x} F(\sqrt{t}x)\,dx &= \int_0^\infty e^{-x}\sum_{k=0}^\infty b_k\frac{t^kx^{2k}}{2k!}\,dx\\
&=\sum_{k=0}^\infty b_k\frac{t^k}{2k!}\int_0^\infty e^{-x}x^{2k}\,dx\\
&= \sum_{k=0}^\infty b_k\frac{t^k}{2k!}2k!\\
&=\sum_{k=0}^\infty b_k t^k
\end{align}
\]
This stuff can get really tricky, bo. Especially when you want to expand in series. You have to remember all that Lebesgue stuff, all that interchange of series. And additionally, it doesn't hurt if you remember your Fourier analysis lessons, lol.
I want to make a long thread sort of detailing what I've learned so far. And I just need a bit of time
But I'll be able to make those graphs not diverge, just let me get my notes in order.
This is super interesting to me because "Fractional calculus" + "Parabolic Iteration" has always failed on all my attempts to crack it, and these past few weeks of results on parabolic iteration has opened my eyes as to how to approach it.
You guys might not notice it, but we've artfully used that \(2k!\) dominates \(k!^2\); and we've done two borel sums, by just modifying one variable. Which upon we just use the Mellin transform correspondence. \(\Gamma(z)^2\) looks a lot like \(\Gamma(2z)\), and you can relate these things exactly. That was a lot of Gauss' schtick--at least, anything to do with \(\Gamma\). #EulerDidItFirst
I have 2 weeks of free time no trouble; I am going to explain this much better. Trust me when I say, D. Zagier, and his paper solves all our problems
This is totally my domain. I can't work through all of this today. But I was planning on making a thread specifically about Borel summation of these parabolic iterations. I've just been absorbing this asymptotic expansion and trying to figure out.
But to begin;
If you have \(b_k = O(c^k k!)\); you don't just perform the standard way of Borel summing. So wikipedia won't help you. You have to be more clever. True "borel summing" is very sensitive to how you set up your integrals, and how you set up your sequences. There's no one size fits all manner. Borel sum, is more usually used to describe that they are all analytic continuations of each other. Not necessarily you just plug in the integral.
I'm not surprised your initial attempt failed. Because Euler's form (Gervery class 1) sequences do not converge in the manner you posted. When the coefficients are \(k!\). Again, we have to be much more clever. And largely, it equates to a bunch of esoteric changes of variables.
I will post a much better answer in the coming days. I don't want to tonight, because I'm still making sure I'm doing everything right. But, I will leave you with this tidbit.
If \(b_k = O(c^k k!)\)
Then:
\[
F(x) = \sum_{k=0}^\infty b_k\frac{x^{2k}}{2k!}
\]
Is an entire function; which is bounded by \(e^{b|x|}\)--or something close--for \(b<1\).
From here, you take the modified laplace transform:
\[
**\begin{align}
\int_0^\infty e^{-x} F(\sqrt{t}x)\,dx &= \sum_{k=0}^\infty b_k t^k\\
&= t^{-1/2}\int_0^\infty e^{-ut^{-1/2}} F(u)\,du
\end{align}
\]
Which is just a term wise evaluation of the integral***. This object converges for a specific domain in \(t\)--depending on where the asymptotic expansion works. Just be careful with \(t\). But this definitely, absolutely, no doubt about it, converges.
**Converges because \(|F(x)| < e^{b|x|}\) for \(b \approx 0\)--of which, we only need \(b<1\).
***
\[
\begin{align}
\int_0^\infty e^{-x} F(\sqrt{t}x)\,dx &= \int_0^\infty e^{-x}\sum_{k=0}^\infty b_k\frac{t^kx^{2k}}{2k!}\,dx\\
&=\sum_{k=0}^\infty b_k\frac{t^k}{2k!}\int_0^\infty e^{-x}x^{2k}\,dx\\
&= \sum_{k=0}^\infty b_k\frac{t^k}{2k!}2k!\\
&=\sum_{k=0}^\infty b_k t^k
\end{align}
\]
This stuff can get really tricky, bo. Especially when you want to expand in series. You have to remember all that Lebesgue stuff, all that interchange of series. And additionally, it doesn't hurt if you remember your Fourier analysis lessons, lol.
I want to make a long thread sort of detailing what I've learned so far. And I just need a bit of time
But I'll be able to make those graphs not diverge, just let me get my notes in order.This is super interesting to me because "Fractional calculus" + "Parabolic Iteration" has always failed on all my attempts to crack it, and these past few weeks of results on parabolic iteration has opened my eyes as to how to approach it.
You guys might not notice it, but we've artfully used that \(2k!\) dominates \(k!^2\); and we've done two borel sums, by just modifying one variable. Which upon we just use the Mellin transform correspondence. \(\Gamma(z)^2\) looks a lot like \(\Gamma(2z)\), and you can relate these things exactly. That was a lot of Gauss' schtick--at least, anything to do with \(\Gamma\). #EulerDidItFirst
I have 2 weeks of free time no trouble; I am going to explain this much better. Trust me when I say, D. Zagier, and his paper solves all our problems