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 Theorem in fractional calculus needed for hyperoperators JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 07/03/2014, 02:17 PM Hey everybody! Well I've boiled down my requirements for solving tetration, pentation, semi operators, and a whole list of recursive relationships using fractional calculus into a single theorem. I am pretty certain this theorem will be true. Well I'll start by saying, if $f(x) = \sum_{n=0}^\infty a_n x^n/n!$ where $F(a_n) = a_{n+1}$ then under certain conditions $F(\frac{d^{z}}{dx^z}|_{x=0}f(x)) = \frac{d^{z+1}}{dx^{z+1}}|_{x=0}f(x)$ Now of course, the problem is that when $a_n$ is something like tetration, or pentation, or whatever, this doesn't converge and we are stuck in the mud. So I've boiled a way to fix this. Now I don't have this theorem yet, but if its solved, all that's required is a bunch of lemmas I know how to prove and we will have tetration, pentation, hexation, semi operators, and some more. So without further ado, here is the theorem we need. Assume $a_n$ is a sequence of complex numbers such that $f(x) = \sum_{n=0}^\infty a_n \frac{x^n}{n!}$ is entire. Then, there always exists $b_n$ such that, $g(x) = \sum_{n=0}^\infty b_n \frac{x^n}{n!}$ is entire and Weyl differintegrable on all of $\mathbb{C}$ and $h(x) = \sum_{n=0}^\infty a_n b_n \frac{x^n}{n!}$ is such that $\frac{d^z}{dx^z}{|}_{x=0} h(x)$ exists for all z. If this theorem is shown, then... define $G(z) = \frac{\frac{d^z}{dx^z}|_{x=0} h(x)}{\frac{d^z}{dx^z}|_{x=0} g(x)}$ and $F(G(z)) = G(z+1)$ and we are done. Any one have any advice on how I can show this theorem? this is quite a stump. MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 07/03/2014, 04:23 PM (This post was last modified: 07/03/2014, 04:28 PM by MphLee.) Hoc can you find the sequence $b_n$ from the sequence $a_n$? Quote:Assume $a_n$ is a sequence of complex numbers such that $f(x) = \sum_{n=0}^\infty a_n \frac{x^n}{n!}$ is entire. Then, there always exists $b_n$ such that, $g(x) = \sum_{n=0}^\infty b_n \frac{x^n}{n!}$ is entire and Weyl differintegrable on all of $\mathbb{C}$ and[...] And what happens if the sequence $a_n$ gives you a non-entire $f$? Looking at the definition of entire function I saw that $(|a_n|/n!)^{(1/n)}$ should converge to zero or $(ln|a_n/n!|)/n$ to $- \infty$(for n that goeas to infinity), but for some sequences seems it does have a uggly behaviour...I guess that more the sequence grows fast and bigger are the problem...but sequences that grows very fast are exactly the ones defined using the Hyperops. I apologize if I made some errors. MathStackExchange account:MphLee JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 07/03/2014, 04:52 PM (This post was last modified: 07/03/2014, 09:01 PM by JmsNxn.) (07/03/2014, 04:23 PM)MphLee Wrote: Hoc can you find the sequence $b_n$ from the sequence $a_n$? Quote:Assume $a_n$ is a sequence of complex numbers such that $f(x) = \sum_{n=0}^\infty a_n \frac{x^n}{n!}$ is entire. Then, there always exists $b_n$ such that, $g(x) = \sum_{n=0}^\infty b_n \frac{x^n}{n!}$ is entire and Weyl differintegrable on all of $\mathbb{C}$ and[...] And what happens if the sequence $a_n$ gives you a non-entire $f$? Looking at the definition of entire function I saw that $(|a_n|/n!)^{(1/n)}$ should converge to zero or $(ln|a_n/n!|)/n$ to $- \infty$(for n that goeas to infinity), but for some sequences seems it does have a uggly behaviour...I guess that more the sequence grows fast and bigger are the problem...but sequences that grows very fast are exactly the ones defined using the Hyperops. I apologize if I made some errors. ^_^ $a_n = 1/(^nb)$ not $a_n = (^n b)$ MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 07/03/2014, 05:29 PM xD looks like you are hacking in this way...jk Btw you can ingore my last observation(was just a doubt) but I'm curious about the other two questions. I'd like to understand more your idea. MathStackExchange account:MphLee JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 07/03/2014, 06:13 PM If its not entire there's no point in doing anything. It doesn't matter, it will be entire for inverse hyp operators. Constructing b_n from a_n is the goal, I'm trying to think of how to do this. I've made some progress but its very taxing. I've come up with some conditions on how the function g(x) should behave, however, its hard to boil this into how b_n behave. The problem is rather tricky. I have a lot of confidence we can do this though, it will require lots of tricks though. MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 07/07/2014, 06:47 PM (This post was last modified: 07/08/2014, 02:59 PM by MphLee.) For the first time I think I reached a general understanding of your method. Ignoring the fine details, and by focusing on the skeleton of your method I think I have a schematic image of how your method works. Sure... to know WHY it works I should know all those fine details (and understand the proof you gave in your paper) but the fact that I can follow your idea is enough exciting at the moment, even with a tiny knowledge of the analysis. Let me try to use a high-order funtion notation (it's easier to me) and tell me if my understanding is wrong. Given a function $f$, lets call $F_{\beta}$ its $\beta$-based superfunction over the naturals $F_{\beta}(n):=f^{\circ n}(\beta)$ your method gives an extension $F'_{\beta}$ of $F_{\beta}$ to the complex numbers. $F_{\beta}\subset F'_{\beta}$ --------------------------------- Let me define 1-$\mathcal{S}_{\beta}\{f\}(n) := f^{\circ n}(\beta)$ 2-$\mathcal{T}\{f\}(x) := \sum_{n=0}^\infty f(n) x^n/n!$ 3-$\mathcal{F}\{f\}(z) :=\frac{d^{z}}{dx^z}|_{x=0}f(x)$ 4- $f \cdot g(x):=\cdot \{f,g\}(x):=f(x)g(x)$; $\frac{f} { g}(x):=/ \{f,g\}(x):=\frac{f(x)}{g(x)}$ ------------------------------------ You say that if $\mathcal{S}_{\beta}\{f\}$ has some nice properties then your method give an extension of it to the complex: if $\mathcal{S}_{\beta}\{f\}$ has some properties then $\mathcal{S}_{\beta}\{f\} \subset \mathcal{F}\{ \mathcal{T}\{ \mathcal{S}_{\beta}\{f\} \} \}$ and $ f\circ \mathcal{F}\{ \mathcal{T}\{ \mathcal{S}_{\beta}\{f\} \} \} = \mathcal{F}\{ \mathcal{T}\{ \mathcal{S}_{\beta}\{f\} \} \} \circ S$ to be more precise we have that $ \mathcal{F}\{ \mathcal{T}\{ \mathcal{S}_{\beta}\{f\} \} \}(n)=\mathcal{S}_{\beta}\{f\}(n)$ (07/03/2014, 02:17 PM)JmsNxn Wrote: Now of course, the problem is that when $a_n$ is something like tetration, or pentation, or whatever, this doesn't converge and we are stuck in the mud. [...] So without further ado, here is the theorem we need. Assume $a_n$ is a sequence of complex numbers such that $f(x) = \sum_{n=0}^\infty a_n \frac{x^n}{n!}$ is entire. Then, there always exists $b_n$ such that, $g(x) = \sum_{n=0}^\infty b_n \frac{x^n}{n!}$ is entire and Weyl differintegrable on all of $\mathbb{C}$ and $h(x) = \sum_{n=0}^\infty a_n b_n \frac{x^n}{n!}$ is such that $\frac{d^z}{dx^z}{|}_{x=0} h(x)$ exists for all z. If this theorem is shown, then... define $G(z) = \frac{\frac{d^z}{dx^z}|_{x=0} h(x)}{\frac{d^z}{dx^z}|_{x=0} g(x)}$ and $F(G(z)) = G(z+1)$ and we are done. Any one have any advice on how I can show this theorem? this is quite a stump. So you say that from every $f$ we can find an unique function that we call $f^{\psi}$ such that i) - $\mathcal{T}\{f^{\psi}\}$ " is entire and Weyl differintegrable on all of $\mathbb{C}$" ii) - $\mathcal{F}\{ \mathcal{T} \{f \cdot f^{\psi}\} \}$ is defined for all the complex numbers ................................. from the existence and the uniqueness (it has to be bijective?) of such map ${-}^{\psi}:f\mapsto f^{\psi}$ and if it satiefies i) and ii) then $\frac{ \mathcal{F}\{ \mathcal{T} \{\mathcal{S}_{\beta}\{f\} \cdot \mathcal{S}_{\beta}\{f\}^{\psi}\} \} } {\mathcal{F}\{ \mathcal{T} \{ \mathcal{S}_{\beta}\{f\}^{\psi}\} \}} (n)=\mathcal{S}_{\beta}\{f\}(n)$ and $ f(\frac{ \mathcal{F}\{ \mathcal{T} \{\mathcal{S}_{\beta}\{f\} \cdot \mathcal{S}_{\beta}\{f\}^{\psi}\} \} } {\mathcal{F}\{ \mathcal{T} \{ \mathcal{S}_{\beta}\{f\}^{\psi}\} \}}(z) )= \frac{ \mathcal{F}\{ \mathcal{T} \{\mathcal{S}_{\beta}\{f\} \cdot \mathcal{S}_{\beta}\{f\}^{\psi}\} \} } {\mathcal{F}\{ \mathcal{T} \{ \mathcal{S}_{\beta}\{f\}^{\psi}\} \}}(z+1)$ ------------------------------- I hope that this is fine...then you have to work on the map ${-}^{\psi}:f\mapsto f^{\psi}$ Questions A- I'm sure that I can't understand how you got there (with my knowledge) but where is the link betwen the fractional iteration and the fractional calculus and how you discovered this link? B- The way you use $\mathcal{T}\{f\}(x) := \sum_{n=0}^\infty f(n) x^n/n!$ ... it reminds me the taylor series (when is centred at a=0 but with the derivative index replaced by the iteration...) I'm sure it is not a case... where is the link? C- the results depends alot on the choice of $\beta$ ? D-What is the relation betwen $\frac{ \mathcal{F}\{ \mathcal{T} \{\mathcal{S}_{\beta}\{f\} \cdot \mathcal{S}_{\beta}\{f\}^{\psi}\} \} } {\mathcal{F}\{ \mathcal{T} \{ \mathcal{S}_{\beta}\{f\}^{\psi}\} \}}$ and $ \mathcal{F}\{ \mathcal{T}\{ \mathcal{S}_{\beta}\{f\} \} \}$? When the two expression coincides (if they do)? E -About the hyperoperations...how you can get them using this method? If I have to think about it..the only way to reach the fractional hyperoperations i see is the fractional iteration of the operator $\mathcal{J}_{\beta}^{\circ z}$ where $\mathcal{J}_{\beta}\{f\}:= \frac{ \mathcal{F}\{ \mathcal{T} \{\mathcal{S}_{\beta}\{f\} \cdot \mathcal{S}_{\beta}\{f\}^{\psi}\} \} } {\mathcal{F}\{ \mathcal{T} \{ \mathcal{S}_{\beta}\{f\}^{\psi}\} \}}$ ...where is the trick to avoid that? ps: i did a mistake in the tex code... ill fix it soon (is important) pps: fixed MathStackExchange account:MphLee « Next Oldest | Next Newest »

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