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 Is this entire expression for tetration legal JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 12/10/2012, 03:43 PM (This post was last modified: 12/11/2012, 02:09 PM by JmsNxn.) I'm going to prove to you an entire expression for tetration in one post for all b greater than eta. Is this valid? We start by defining: $\mathcal{E} f(s) = \frac{1}{\Gamma(-s)} \int_0^\infty f(-u) u^{-s-1}\, \partial u$ We call this the exponential derivative because: $\mathcal{E} f(s) = \frac{d^s f(t)}{dt^s} |_{t=0}$ Where this complex derivative is evaluated according to the Riemann-Liouville differ-integral where exponentiation is the fix point. By this definition we find: $\mathcal{E}(b^s) = \ln(b)^s$ Now we define a new multiplication across FUNCTIONS; i.e. you cannot plug in numbers into this multiplication. We refer to $\mathbb{E}$ as the complex linear span of $e^{\lambda s}\,\,;\,\,\lambda \in \mathbb{C}$. We now define: $\forall \alpha \in \mathbb{C}\,\,\,\,;\,\,\,\,\forall f,g,h \in \mathbb{E}$ $f \times g \in \mathbb{E}$ $\alpha \times f = \alpha f$ $f \times g = g \times f$ $(f \times g) \times h = f \times (g \times h)$ $f \times (g + h) = (f \times g) + (f \times h)$ With this we define $\forall \omega, \nu \in \mathbb{C}$: $(\Gamma(\omega + 1) (^\omega e)^s) \times(\Gamma(\nu + 1) (^\nu e)^s) = \Gamma(\omega + \nu + 1) (^{\omega+ \nu } e)^s$ This result gives us a beautiful isomorphism $\psi: \mathbb{E} \to \mathbb{C}[s]$: $\mathbb{U} = \{ 1, e^s, (^2 e)^s, (^3 e)^s, ..., (^N e)^s,...\}$ $\psi(\alpha f + \beta g) = \alpha \psi f + \beta \psi g$ $\psi(f \times g) = (\psi f) \cdot (\psi g)$ $\psi(\mathcal{E} f) = \frac{d \psi(f)}{ds}$ $\psi((^N e)^s) = \frac{s^N}{N!}$ Now thanks to Newton we have: $s^\omega = \sum_{N=0}^{\infty} \frac{\Gamma(\omega + 1)}{\Gamma(\omega - N + 1)N!} (s - 1)^N$ $s^{\omega} = \sum_{N=0}^{\infty} \frac{\Gamma(\omega + 1)}{\Gamma(\omega - N + 1)N!} \sum_{k=0}^N \frac{N!}{(N-k)! k!}(-1)^{N-k}s^k$ Now we apply the isomorphism to get the result $\exists \phi \in \mathbb{E}$: $\Gamma(\omega + 1) \phi_\omega(s) = \sum_{N=0}^{\infty} \frac{\Gamma(\omega + 1)}{\Gamma(\omega - N + 1)N!} \sum_{k=0}^N \frac{N!}{(N-k)! k!}(-1)^{N-k}k!(^k e)^s$ We can reduce this equation to: $\phi_\omega(s) = \sum_{N=0}^{\infty} \frac{\sum_{k=0}^{N} \frac{(-1)^{N-k}}{(N-k)!}(^k e)^s}{\Gamma(\omega - N +1)}$ We know these functions satisfy the equations: $\phi_\omega \times \phi_\nu = \frac{\Gamma(\omega + \nu + 1)}{\Gamma(\omega + 1)\Gamma(\nu + 1)} \phi_{\omega + \nu}(s)$ $\mathcal{E} \phi_\omega = \phi_{\omega - 1}$ There is only one function which satisfies this: $\phi_{\omega}(s) = (^\omega e)^s$ We can do the same procedure to arrive at the more general expression: $(^\omega b)^s = \sum_{N=0}^{\infty} \frac{\sum_{k=0}^{N} \frac{(-1)^{N-k}}{(N-k)!}(^k b)^s}{\Gamma(\omega - N +1)}$ This is probably the fastest way to derive tetration. It's analytic, entire for $b > \eta$ when we let $\Re(s) < 0$which is easy to show by the ratio test since tetration grows ridiculously faster than the Gamma function. Anyone see any mistakes I'm making? Or did I just solve tetration in fifteen minutes of work. LOL Being quick I'll write the glorious formula; $\forall b \in \mathbb{R}\,\,;\,\, b > e^{\frac{1}{e}}$: $\frac{1}{^\omega b} = \sum_{N=0}^{\infty} \frac{\sum_{k=0}^{N} \frac{(-1)^{N-k}}{(N-k)!(^k b)}}{ \Gamma(\omega - N +1)}$ I'm currently finishing a paper on $\mathcal{E}$ and this result just happened to fall in my lap. I'm wondering if it's valid so that I can keep it in the paper. « Next Oldest | Next Newest »

 Messages In This Thread Is this entire expression for tetration legal - by JmsNxn - 12/10/2012, 03:43 PM RE: Is this entire expression for tetration legal - by sheldonison - 12/13/2012, 02:30 AM RE: Is this entire expression for tetration legal - by andydude - 12/13/2012, 05:03 AM RE: Is this entire expression for tetration legal - by tommy1729 - 12/15/2012, 10:02 PM RE: Is this entire expression for tetration legal - by cosurgi - 01/20/2013, 07:32 PM

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