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 Wonderful new form of infinite series; easy solve tetration JmsNxn Long Time Fellow Posts: 571 Threads: 95 Joined: Dec 2010 09/06/2012, 02:01 AM (This post was last modified: 09/06/2012, 02:50 AM by JmsNxn.) 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 $\cdot, +$ « Next Oldest | Next Newest »

 Messages In This Thread Wonderful new form of infinite series; easy solve tetration - by JmsNxn - 09/04/2012, 06:27 PM RE: Wonderful new form of infinite series; easy solve tetration - by JmsNxn - 09/06/2012, 02:01 AM

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