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 \( \mathcal{M} \) be referred to as the mega derivative. We define it as:
\( \mathcal{M}f = \frac{d^x f}{dt^x}_{t = 1} = \frac{1}{\Gamma(-x)} \int_{-\infty}^1 \frac{f(t)}{(1-t)^{x+1}} dt \)
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; \( \mathcal{M} (\alpha f + \beta g ) = \alpha \mathcal{M} f + \beta \mathcal{M}g \)
We have
\( \mathcal{M}\,\,(^n e)^x = (^{n-1} e)^x \cdot (^n e) \)
This is because \( \mathcal{M} (a^x) = \ln(a)^x \cdot a \)
\( \mathcal{M}\,\,e^x = e \)
\( \mathcal{M} C = 0 \) for some constant C.
I'm sure everyone here sees the parallel to the power law. Using this we can make an infinite series.
\( f(x) = \sum_{k=0}^{\infty} a_k (^k e)^x \)
Easy to see that:
\( \mathcal{M}^n\,f(x) = \sum_{k=0}^{\infty}a_k (^k e)^x \prod_{i=1}^n \,\,(^{k+i} e) \)
Which allows us to say that
\( a_k = \lim_{x\to -\infty} \frac{\mathcal{M}^k\, f(x)}{\prod_{i=0}^{k} (^i e)} \)
We of course have the most powerful function: the fixpoint of the megaderivative:
\( \lambda(s) = \sum_{k=0}^{\infty} \frac{(^k e)^s}{\prod_{i=0}^{k} (^i e)} \)
This gives:
\( \mathcal{M} \lambda = \lambda \)
Or written more formally:
\( \frac{d^s \lambda(t)}{dt^s}_{t=1} = \lambda(s) \)
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:
\( \lim_{x \to - \infty} \mathcal{M}^n \lambda(x+1) = \,\,\,^n e \)
And so tetration boils into iteration of the mega derivative at the fixpoint function.
This solution holds the recursive identity because it uses \( \ln(^s e) = (^{s-1} e) \)
This function also has the very cool result that:
\( \lambda(s+1) = \sum_{n=0}^{\infty} \frac{\lambda(n)}{n!}s^n \) 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 \( \mathcal{M} \) be referred to as the mega derivative. We define it as:
\( \mathcal{M}f = \frac{d^x f}{dt^x}_{t = 1} = \frac{1}{\Gamma(-x)} \int_{-\infty}^1 \frac{f(t)}{(1-t)^{x+1}} dt \)
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; \( \mathcal{M} (\alpha f + \beta g ) = \alpha \mathcal{M} f + \beta \mathcal{M}g \)
We have
\( \mathcal{M}\,\,(^n e)^x = (^{n-1} e)^x \cdot (^n e) \)
This is because \( \mathcal{M} (a^x) = \ln(a)^x \cdot a \)
\( \mathcal{M}\,\,e^x = e \)
\( \mathcal{M} C = 0 \) for some constant C.
I'm sure everyone here sees the parallel to the power law. Using this we can make an infinite series.
\( f(x) = \sum_{k=0}^{\infty} a_k (^k e)^x \)
Easy to see that:
\( \mathcal{M}^n\,f(x) = \sum_{k=0}^{\infty}a_k (^k e)^x \prod_{i=1}^n \,\,(^{k+i} e) \)
Which allows us to say that
\( a_k = \lim_{x\to -\infty} \frac{\mathcal{M}^k\, f(x)}{\prod_{i=0}^{k} (^i e)} \)
We of course have the most powerful function: the fixpoint of the megaderivative:
\( \lambda(s) = \sum_{k=0}^{\infty} \frac{(^k e)^s}{\prod_{i=0}^{k} (^i e)} \)
This gives:
\( \mathcal{M} \lambda = \lambda \)
Or written more formally:
\( \frac{d^s \lambda(t)}{dt^s}_{t=1} = \lambda(s) \)
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:
\( \lim_{x \to - \infty} \mathcal{M}^n \lambda(x+1) = \,\,\,^n e \)
And so tetration boils into iteration of the mega derivative at the fixpoint function.
This solution holds the recursive identity because it uses \( \ln(^s e) = (^{s-1} e) \)
This function also has the very cool result that:
\( \lambda(s+1) = \sum_{n=0}^{\infty} \frac{\lambda(n)}{n!}s^n \) 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.
