Wonderful new form of infinite series; easy solve tetration
#2
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 \( \cdot, + \)
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RE: Wonderful new form of infinite series; easy solve tetration - by JmsNxn - 09/06/2012, 02:01 AM

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