Wonderful new form of infinite series; easy solve tetration - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- 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 $\cdot, +$