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+--- Thread: Wonderful new form of infinite series; easy solve tetration (/showthread.php?tid=749)
Wonderful new form of infinite series; easy solve tetration - JmsNxn - 09/04/2012
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.
RE: Wonderful new form of infinite series; easy solve tetration - JmsNxn - 09/06/2012
At the moment I'm trying reconstruct the polynomial ring over the field of complex numbers with a product across the mega derivative that satisfies the product law and is commutative/associative and dist. across addition. It's very beautiful in terms of a relationship with tetration. Very much so as if it were exponentiation; and regular polynomials. Using the mega differential operator we create a multiplication across functions belonging to the vector space \( \mathbb{V} \) such that the basis elements are functions \( (^k e)^s: \mathbb{C} \to \mathbb{C} \) where \( k \) is some integer greater to or equal to zero. Call these tetranomials.
The product can bedefined as follows:
if \( A,B \) are tetranomials:
\( \mathcal{M} (A \times B) = \mathcal{M}A \times B + A \times \mathcal{M}B \)
then using mega integration; which distributes across addition and is easy for tetranomials by the power law of mega differentiation we can write the product law for tetranomials. It's a rather cumbersome sum that you get; but nonetheless; it's commutative and associative and distributes across addition; is compatible with scalar multiplication; and is destroyed by zero; however; the only assumption I've made is \( 1 \times A = A \). Which ends the recurrence relation in the multiplication since a finite number of mega differentiations on a tetranomial reduces it to a constant.
I've found a lot of rich discoveries,. Particularly: if \( \lambda(\beta) = 0 \) which exists because of picard's theorem.
then:
\( e^{-\pi i z}\int_{-\infty}^{\beta} \lambda(s) \times (^{z-1}\,e)^s \mathcal{M}s = \cdot \gamma(z) \)
where here:
\( ^ze \cdot \gamma(z) = \gamma(z+1) \)
\( \gamma(k+1) = \prod_{i=0}^{k} ^i e \)
where this is a definite mega integral. This is quite incredible. There is no geometric interpretation of the definite mega integral; however; it acts as a limit involving the anti mega derivative and is a convenient notation.
This formula is easily verified by the product law and the power law of tetranomials and the fact that lambda is a fix point of the mega derivative.
Miraculously; \( (^k e)^s \times (^j e)^s = C (^{k+j} e)^s \) for some constant C depending on k and j. We must remember we are performing a differential operator on the tetranomial not on the tetrated number or the tetration function. A tetranomial has natural tetration values but complex exponentiation values.
I'm working on finding a product representation of \( \gamma \). I'm finding a lot of parallels here between tetranomials and \( \times, + \) and polynomials an \( \cdot, + \)