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 Bounded Analytic Hyper operators fivexthethird Junior Fellow Posts: 9 Threads: 3 Joined: Nov 2013 03/28/2015, 06:46 AM (This post was last modified: 03/28/2015, 06:47 AM by fivexthethird.) It seems that this method is equivalent to the method of newton series. To see this, note that $e^x \sum_{k=0}^{\infty}f(k) \frac{(-1)^k x^k}{k!} = (\sum_{k=0}^{\infty}\frac{x^k}{k!})(\sum_{k=0}^{\infty}f(k) \frac{(-1)^k x^k}{k!}) =$ $\sum_{k=0}^{\infty}x^k \sum_{j=0}^k\frac{(-1)^{j}f(j)}{(k-j)!j!} = \sum_{k=0}^{\infty}\frac{x^k}{k!} \sum_{j=0}^k(-1)^{j}f(j){k \choose j} =$ $\sum_{k=0}^{\infty}{\Delta}^kf(0) \frac{(-1)^{k}x^k}{k!}$ where $\Delta f(x) = f(x+1)-f(x)$ So we can rewrite the integral as $\frac{1}{\Gamma(-z)}\int_0^{\infty} \sum_{k=0}^{\infty}{\Delta}^kf(0) \frac{(-1)^{k}x^{k-z-1}e^{-x}}{k! } dx$ Since the power series defines an entire function we can exchange the integral and sum so that we have $\sum_{k=0}^{\infty}{\Delta}^kf(0)\frac{(-1)^{k}}{k!\Gamma(-z)} \int_0^{\infty} x^{k-z-1}e^{-x} dx =$ $\sum_{k=0}^{\infty}{\Delta}^kf(0)\frac{(-1)^{k}\Gamma(k-z)}{k!\Gamma(-z)} =\sum_{k=0}^{\infty}{\Delta}^kf(0)\frac{(z)_{k}}{k!}$ where $(z)_{k} = z(z-1)(z-2)...(z-(k-1))$ is the falling factorial. This is just the newton series of f around 0. TPID 13 is thus solved, as $x^{\frac{1}{x}}$ satisfies the bounds required for Ramanujan's master theorem to apply. @tommy: I don't see how inverting around t helps us recover tetration from the interpolated super root... it gets us the slog, yes, but in either case we need to just invert around m to get tetration. Or am I misinterpreting your post? « Next Oldest | Next Newest »

 Messages In This Thread Bounded Analytic Hyper operators - by JmsNxn - 03/23/2015, 02:12 AM RE: Bounded Analytic Hyper operators - by MphLee - 03/24/2015, 08:34 PM RE: Bounded Analytic Hyper operators - by fivexthethird - 03/25/2015, 07:29 AM RE: Bounded Analytic Hyper operators - by marraco - 03/25/2015, 01:48 PM RE: Bounded Analytic Hyper operators - by MphLee - 03/25/2015, 07:43 PM RE: Bounded Analytic Hyper operators - by MphLee - 03/26/2015, 11:08 PM RE: Bounded Analytic Hyper operators - by tommy1729 - 03/27/2015, 09:02 AM RE: Bounded Analytic Hyper operators - by MphLee - 03/27/2015, 07:09 PM RE: Bounded Analytic Hyper operators - by tommy1729 - 03/27/2015, 11:20 PM RE: Bounded Analytic Hyper operators - by fivexthethird - 03/28/2015, 06:46 AM RE: Bounded Analytic Hyper operators - by tommy1729 - 03/28/2015, 08:59 PM RE: Bounded Analytic Hyper operators - by fivexthethird - 03/29/2015, 03:35 AM RE: Bounded Analytic Hyper operators - by tommy1729 - 03/29/2015, 04:36 PM RE: Bounded Analytic Hyper operators - by MphLee - 03/29/2015, 10:55 AM RE: Bounded Analytic Hyper operators - by tommy1729 - 03/29/2015, 04:22 PM RE: Bounded Analytic Hyper operators - by tommy1729 - 03/29/2015, 04:56 PM RE: Bounded Analytic Hyper operators - by MphLee - 03/29/2015, 07:08 PM RE: Bounded Analytic Hyper operators - by tommy1729 - 03/29/2015, 07:14 PM RE: Bounded Analytic Hyper operators - by tommy1729 - 03/29/2015, 09:37 PM RE: Bounded Analytic Hyper operators - by tommy1729 - 03/29/2015, 09:41 PM RE: Bounded Analytic Hyper operators - by JmsNxn - 03/29/2015, 11:25 PM RE: Bounded Analytic Hyper operators - by JmsNxn - 03/31/2015, 08:50 PM RE: Bounded Analytic Hyper operators - by MphLee - 03/31/2015, 09:16 PM RE: Bounded Analytic Hyper operators - by JmsNxn - 04/01/2015, 03:20 PM RE: Bounded Analytic Hyper operators - by MphLee - 04/01/2015, 06:09 PM

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