Hello everyone. I've recently come across a way to perform continuum sums and I was wondering if anyone has any suggestions on how to formalize this or if they have any nice comments about how it works.
We start by defining an operator:
\( \mathcal{J} f(s) = \frac{1}{\Gamma(s)}\int_0^\infty f(-X)X^{s-1}dX \)
Thanks to Riemann and Liouville, if \( \mathcal{J} f \) and \( \mathcal{J} \frac{df}{ds} \) converge then:
\( \mathcal{J} \frac{df}{ds} = (\mathcal{J} f)(s-1) \)
And, neatly \( (\mathcal{J}\frac{df}{ds})(1) = f(0) \) therefore, by induction:
\( (\mathcal{J} f)(N) = \frac{d^{-N}}{ds^{-N}}f(s)|_{s=0} \)
It is noted that this operator can be inverted using Mellin inversion and Taylor series for certain analytic functions, and for some continuum Taylor transformation. So it makes sense to talk about \( \mathcal{J}^{-1} \)
However, we must use Riemann-Liouville differintegration.
Define:
\( \frac{d^{-s}f(t)}{dt^{-s}} = \frac{1}{\Gamma(s)}\int_{-\infty}^{t}f(u)(t-u)^{s-1}du \)
And we find when \( t=0 \) the operator reduces to \( \mathcal{J} \)
We then define our continuum sum operator which works beautifully:
\( \mathcal{Z}f(s) = \int_0^{\infty} e^{-t}\frac{d^{-s}f(t)}{dt^{-s}}dt \)
\( \mathcal{Z}f(s) = [-e^{-t}\frac{d^{-s}}{dt^{-s}}f(t)]_{t=0}^{\infty} + \int_0^{\infty} e^{-t}\frac{d^{1-s}f(t)}{dt^{1-s}}dt \)
Performing integration by parts, which is easy, and using the fact that \( \frac{d}{dt}\frac{d^{-s}f(t)}{dt^{-s}} = \frac{d^{1-s}f(t)}{dt^{1-s}} \)
We get:
\( \mathcal{Z}f(s) = \mathcal{J}f(s) + (\mathcal{Z}f)(s-1) \)
And if we take \( f = \mathcal{J}^{-1}g \) and say:
\( \phi(s) = \mathcal{Z}\mathcal{J}^{-1} g(s) \)
\( \phi(s) = g(s) + \phi(s-1) \)
Which is the glory of the continuum sum!
The real problem now is finding what kinds of functions g(s) does this work on
\( \mathcal{J}^{-1}g(s) = \int_{\sigma + -i\infty}^{\sigma + i\infty}\Gamma(t)g(t) e^{- \pi i t}s^{-t} dt \)
with \( \sigma \) chosen appropriately for g(s).
We also have:
\( \mathcal{J}^{-1}g(s) = \sum_{n=0}^{\infty} g(-n)\frac{s^n}{n!} \)
for some functions, this is remembering because \( g(-n) = \frac{d^n}{dt^n}f(t)|_{t=0} \) for some functions.
For example we can find a continuum sum, \( \lambda \in \mathbb{R} \) \( \mathcal{J}e^{\lambda s} = \lambda^{-s} \)
therefore for \( |\lambda| < 1 \) \( \Re \lambda < 1 \):
\( \phi(s) = \int_0^{\infty} e^{-t} \frac{d^{-s}e^{\lambda t}}{dt^{-s}}dt \)
\( \phi(s) = \lambda^{-s} \int_0^{\infty}e^{(\lambda-1)t}dt = \frac{\lambda^{-s}}{1-\lambda} = \sum_{n=0}^{\infty} \lambda^{n-s} \)
which satisfies the continuum sum rule.
We can do the same trick for \( e^{-s^2} \) we know that \( \mathcal{J} e^{-s^2} = \frac{ \Gamma(s/2)}{\Gamma(s)} \)
Therefore the continuum sum, or function that satisfies:
\( \phi(s) = \frac{\Gamma(s/2)}{\Gamma(s)} + \phi(s-1) \)
Is:
\( \phi(s) = \frac{1}{\Gamma(s)}\int_0^\infty e^{-t} \int_{-\infty}^t e^{-u^2}(t-u)^{s-1}dudt \)
I thought I would just write out some examples that I have worked out.
\( \phi(s) = \sin(\frac{\pi s}{2}) + \phi(s-1) \)
\( \phi(s) = \int_{0}^{\infty} e^{-t}\sin(t - \frac{\pi s}{2})dt \)
Same for cosine.
\( \psi(s) = \int_{0}^{\infty} e^{-t}\cos(t - \frac{\pi s}{2})dt \)
And miraculously: \( \frac{d\phi}{ds} = -\frac{\pi}{2}\psi(s) \)
Whats truly amazing is that this is a linear operator. Therefore we have an operator \( \mathcal{S}f(s) = \mathcal{Z}\mathcal{J}^{-1} f(s) \) which sends functions from themselves to their continuum sum.
What's truly even more amazing is that I have found an inverse expression for \( \mathcal{Z} \) This was an extra trick.
We start by defining an operator:
\( \mathcal{J} f(s) = \frac{1}{\Gamma(s)}\int_0^\infty f(-X)X^{s-1}dX \)
Thanks to Riemann and Liouville, if \( \mathcal{J} f \) and \( \mathcal{J} \frac{df}{ds} \) converge then:
\( \mathcal{J} \frac{df}{ds} = (\mathcal{J} f)(s-1) \)
And, neatly \( (\mathcal{J}\frac{df}{ds})(1) = f(0) \) therefore, by induction:
\( (\mathcal{J} f)(N) = \frac{d^{-N}}{ds^{-N}}f(s)|_{s=0} \)
It is noted that this operator can be inverted using Mellin inversion and Taylor series for certain analytic functions, and for some continuum Taylor transformation. So it makes sense to talk about \( \mathcal{J}^{-1} \)
However, we must use Riemann-Liouville differintegration.
Define:
\( \frac{d^{-s}f(t)}{dt^{-s}} = \frac{1}{\Gamma(s)}\int_{-\infty}^{t}f(u)(t-u)^{s-1}du \)
And we find when \( t=0 \) the operator reduces to \( \mathcal{J} \)
We then define our continuum sum operator which works beautifully:
\( \mathcal{Z}f(s) = \int_0^{\infty} e^{-t}\frac{d^{-s}f(t)}{dt^{-s}}dt \)
\( \mathcal{Z}f(s) = [-e^{-t}\frac{d^{-s}}{dt^{-s}}f(t)]_{t=0}^{\infty} + \int_0^{\infty} e^{-t}\frac{d^{1-s}f(t)}{dt^{1-s}}dt \)
Performing integration by parts, which is easy, and using the fact that \( \frac{d}{dt}\frac{d^{-s}f(t)}{dt^{-s}} = \frac{d^{1-s}f(t)}{dt^{1-s}} \)
We get:
\( \mathcal{Z}f(s) = \mathcal{J}f(s) + (\mathcal{Z}f)(s-1) \)
And if we take \( f = \mathcal{J}^{-1}g \) and say:
\( \phi(s) = \mathcal{Z}\mathcal{J}^{-1} g(s) \)
\( \phi(s) = g(s) + \phi(s-1) \)
Which is the glory of the continuum sum!
The real problem now is finding what kinds of functions g(s) does this work on
\( \mathcal{J}^{-1}g(s) = \int_{\sigma + -i\infty}^{\sigma + i\infty}\Gamma(t)g(t) e^{- \pi i t}s^{-t} dt \)
with \( \sigma \) chosen appropriately for g(s).
We also have:
\( \mathcal{J}^{-1}g(s) = \sum_{n=0}^{\infty} g(-n)\frac{s^n}{n!} \)
for some functions, this is remembering because \( g(-n) = \frac{d^n}{dt^n}f(t)|_{t=0} \) for some functions.
For example we can find a continuum sum, \( \lambda \in \mathbb{R} \) \( \mathcal{J}e^{\lambda s} = \lambda^{-s} \)
therefore for \( |\lambda| < 1 \) \( \Re \lambda < 1 \):
\( \phi(s) = \int_0^{\infty} e^{-t} \frac{d^{-s}e^{\lambda t}}{dt^{-s}}dt \)
\( \phi(s) = \lambda^{-s} \int_0^{\infty}e^{(\lambda-1)t}dt = \frac{\lambda^{-s}}{1-\lambda} = \sum_{n=0}^{\infty} \lambda^{n-s} \)
which satisfies the continuum sum rule.
We can do the same trick for \( e^{-s^2} \) we know that \( \mathcal{J} e^{-s^2} = \frac{ \Gamma(s/2)}{\Gamma(s)} \)
Therefore the continuum sum, or function that satisfies:
\( \phi(s) = \frac{\Gamma(s/2)}{\Gamma(s)} + \phi(s-1) \)
Is:
\( \phi(s) = \frac{1}{\Gamma(s)}\int_0^\infty e^{-t} \int_{-\infty}^t e^{-u^2}(t-u)^{s-1}dudt \)
I thought I would just write out some examples that I have worked out.
\( \phi(s) = \sin(\frac{\pi s}{2}) + \phi(s-1) \)
\( \phi(s) = \int_{0}^{\infty} e^{-t}\sin(t - \frac{\pi s}{2})dt \)
Same for cosine.
\( \psi(s) = \int_{0}^{\infty} e^{-t}\cos(t - \frac{\pi s}{2})dt \)
And miraculously: \( \frac{d\phi}{ds} = -\frac{\pi}{2}\psi(s) \)
Whats truly amazing is that this is a linear operator. Therefore we have an operator \( \mathcal{S}f(s) = \mathcal{Z}\mathcal{J}^{-1} f(s) \) which sends functions from themselves to their continuum sum.
What's truly even more amazing is that I have found an inverse expression for \( \mathcal{Z} \) This was an extra trick.