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 Wonderful new form of infinite series; easy solve tetration JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 09/04/2012, 06:27 PM (This post was last modified: 09/04/2012, 08:49 PM by JmsNxn.) 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. JmsNxn Long Time Fellow Posts: 291 Threads: 67 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 »

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