While I've been trying to develop a linear operator that works well with hyper operators; I reinvestigated the iterated derivative I was working on and made a slight modification and got a new form of infinite series
Let
be referred to as the mega derivative. We define it as:
} \int_{-\infty}^1 \frac{f(t)}{(1-t)^{x+1}} dt)
We are referring to the exponential derivative; this implies the lower limit of the riemann liouville differintegral is negative infinity.
It is a linear operator; = \alpha \mathcal{M} f + \beta \mathcal{M}g)
We have
^x = (^{n-1} e)^x \cdot (^n e))
This is because = \ln(a)^x \cdot a)
for some constant C.
I'm sure everyone here sees the parallel to the power law. Using this we can make an infinite series.
 = \sum_{k=0}^{\infty} a_k (^k e)^x)
Easy to see that:
 = \sum_{k=0}^{\infty}a_k (^k e)^x \prod_{i=1}^n \,\,(^{k+i} e))
Which allows us to say that
}{\prod_{i=0}^{k} (^i e)})
We of course have the most powerful function: the fixpoint of the megaderivative:
 = \sum_{k=0}^{\infty} \frac{(^k e)^s}{\prod_{i=0}^{k} (^i e)})
This gives:
Or written more formally:
}{dt^s}_{t=1} = \lambda(s))
I'm currently putting aside research in hyper operators to investigate these series'. I think we can solve tetration with these.
In fact; we can deduce:
 = \,\,\,^n e)
And so tetration boils into iteration of the mega derivative at the fixpoint function.
This solution holds the recursive identity because it uses = (^{s-1} e))
This function also has the very cool result that:
 = \sum_{n=0}^{\infty} \frac{\lambda(n)}{n!}s^n)
or is its own generating function. So knowing it at integer arguments is enough.
I'm mostly thinking about representing tetration using these infinite series. I'm just wondering how.
Let
We are referring to the exponential derivative; this implies the lower limit of the riemann liouville differintegral is negative infinity.
It is a linear operator;
We have
This is because
I'm sure everyone here sees the parallel to the power law. Using this we can make an infinite series.
Easy to see that:
Which allows us to say that
We of course have the most powerful function: the fixpoint of the megaderivative:
This gives:
Or written more formally:
I'm currently putting aside research in hyper operators to investigate these series'. I think we can solve tetration with these.
In fact; we can deduce:
And so tetration boils into iteration of the mega derivative at the fixpoint function.
This solution holds the recursive identity because it uses
This function also has the very cool result that:
I'm mostly thinking about representing tetration using these infinite series. I'm just wondering how.
