I am officially submitting this function before tommy.
\[
F_\epsilon(c,z) = \sum_{k=-\infty}^\infty c^k e^{-\epsilon k^2}f^{\circ k}(z)
\]
Which is actually the integral:
\[
\int_{-\infty}^\infty c^ke^{-\epsilon k^2}f^{\circ k}(z)\,d\mu(k)
\]
Where \(\mu\) is a measure of moles. Where it has dirac deltas at the integers.
We can write \(F_\epsilon \to F\) as \(\epsilon \to 0\)--We can do this better than I said before. But there's another trick I'm hiding; which looks like:
\[
G(s,z) = \sum_{k=-\infty}^\infty e^{-\epsilon(s-k)^2}f^{\circ k}(z)\\
\]
Which satisfies \(G(s+1,z) = G(s,f(z))\).
We can do this a lot of ways. I'm happy to give you 80% of the credit for some of the things I'm about to do, Tommy. Raes method, would be how I'd say it. I think it aligns very well with everything you've been doing lately.
Not trying to steal your thunder, just trying to get in on it
TO BEGIN:
\[
G(s,z) = \int_{-\infty}^\infty e^{-\pi(s-k)^2}f^{\circ k}(z)\,d\mu(k)
\]
Then, this is actually a fourier transform:
\[
e^{-\pi(s-k)^2} = e^{-\pi s^2} e^{2\pi s k}e^{-\pi k^2}
\]
If we move the variable \(s\) to \(-is\); then we get a Fourier looking expansion.
\[
G(-is,z) = e^{\pi s^2}\int_{-\infty}^\infty e^{-2\pi i s k}e^{-\pi k^2}f^{\circ k}(z)\,d\mu(k)
\]
Now, a little rearrangement:
\[
e^{-\pi s^2} G(-is,z) = \int_{-\infty}^\infty e^{-2 \pi i s k} e^{-\pi k^2}f^{\circ k}(z)\,d\mu(k)
\]
Right hand side is a fucking fourier transform!!!!!!
Fecking ya!
The inverse then, is that:
\[
e^{-\pi k^2}f^{\circ k}(z) = \int_{-\infty}^\infty e^{2 \pi i s k} e^{-\pi s^2} G(-is,z)\,d\widehat{\mu}(s)
\]
I'm a little lazy at the moment to find \(\widehat{\mu}\); but a rough memory is that in the above instance \(\widehat{\mu} = \mu\) (EDIT: THIS IS WRONG! \(\widehat{\mu}\) is the indicator function on \([0,1]\)). When we choose the integers as our moles/indicators/dirac shit, then it's idempotent. But if not, and I'm misremembering. For any measure \(\mu\), there is a Fourier transform dual \(\widehat{\mu}\)--and that's what this notation means.
Fuck
So Essentially. Tommy, you have sent me down a path where I believe I can finally show the following result.
Let:
\[
H(s,z) = \sum_{k=-\infty}^{\infty} e^{-\pi(is+k)^2} f^{\circ k}(z)
\]
Then:
\[
e^{-k^2}f^{\circ k}(z) = \sum_{j=-\infty}^\infty e^{-\pi j^2} e^{-2\pi i k j} H(j,z)
\]
EDIT: THIS IS WRONG. IT SHOULD BE:
\[
e^{-k^2}f^{\circ k}(z) = \int_0^1e^{-\pi j^2} e^{2\pi i k j} H(j,z)\,dj
\]
Again, I might have screwed up some things; but fuck this is it:
Fuck
Raes Method
\[
F_\epsilon(c,z) = \sum_{k=-\infty}^\infty c^k e^{-\epsilon k^2}f^{\circ k}(z)
\]
Which is actually the integral:
\[
\int_{-\infty}^\infty c^ke^{-\epsilon k^2}f^{\circ k}(z)\,d\mu(k)
\]
Where \(\mu\) is a measure of moles. Where it has dirac deltas at the integers.
We can write \(F_\epsilon \to F\) as \(\epsilon \to 0\)--We can do this better than I said before. But there's another trick I'm hiding; which looks like:
\[
G(s,z) = \sum_{k=-\infty}^\infty e^{-\epsilon(s-k)^2}f^{\circ k}(z)\\
\]
Which satisfies \(G(s+1,z) = G(s,f(z))\).
We can do this a lot of ways. I'm happy to give you 80% of the credit for some of the things I'm about to do, Tommy. Raes method, would be how I'd say it. I think it aligns very well with everything you've been doing lately.
Not trying to steal your thunder, just trying to get in on it
TO BEGIN:
\[
G(s,z) = \int_{-\infty}^\infty e^{-\pi(s-k)^2}f^{\circ k}(z)\,d\mu(k)
\]
Then, this is actually a fourier transform:
\[
e^{-\pi(s-k)^2} = e^{-\pi s^2} e^{2\pi s k}e^{-\pi k^2}
\]
If we move the variable \(s\) to \(-is\); then we get a Fourier looking expansion.
\[
G(-is,z) = e^{\pi s^2}\int_{-\infty}^\infty e^{-2\pi i s k}e^{-\pi k^2}f^{\circ k}(z)\,d\mu(k)
\]
Now, a little rearrangement:
\[
e^{-\pi s^2} G(-is,z) = \int_{-\infty}^\infty e^{-2 \pi i s k} e^{-\pi k^2}f^{\circ k}(z)\,d\mu(k)
\]
Right hand side is a fucking fourier transform!!!!!!
Fecking ya!
The inverse then, is that:
\[
e^{-\pi k^2}f^{\circ k}(z) = \int_{-\infty}^\infty e^{2 \pi i s k} e^{-\pi s^2} G(-is,z)\,d\widehat{\mu}(s)
\]
I'm a little lazy at the moment to find \(\widehat{\mu}\); but a rough memory is that in the above instance \(\widehat{\mu} = \mu\) (EDIT: THIS IS WRONG! \(\widehat{\mu}\) is the indicator function on \([0,1]\)). When we choose the integers as our moles/indicators/dirac shit, then it's idempotent. But if not, and I'm misremembering. For any measure \(\mu\), there is a Fourier transform dual \(\widehat{\mu}\)--and that's what this notation means.
Fuck
So Essentially. Tommy, you have sent me down a path where I believe I can finally show the following result.
Let:
\[
H(s,z) = \sum_{k=-\infty}^{\infty} e^{-\pi(is+k)^2} f^{\circ k}(z)
\]
Then:
\[
e^{-k^2}f^{\circ k}(z) = \sum_{j=-\infty}^\infty e^{-\pi j^2} e^{-2\pi i k j} H(j,z)
\]
EDIT: THIS IS WRONG. IT SHOULD BE:
\[
e^{-k^2}f^{\circ k}(z) = \int_0^1e^{-\pi j^2} e^{2\pi i k j} H(j,z)\,dj
\]
Again, I might have screwed up some things; but fuck this is it:
Fuck
Raes Method