Bounded Analytic Hyper operators JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 Joined: Dec 2010 03/29/2015, 11:25 PM (This post was last modified: 03/29/2015, 11:27 PM by JmsNxn.) Hey everyone. Sorry I've taken so long to comment. Been very busy working on the super factorial I think I've found a way to solve it. This does work on the continuum sum Tommy. It takes a lot of extra fiddling but look here: http://arxiv.org/abs/1503.07555 This one is a little more rough but it explains how continuum sums can be created. I made a few modifications (had some typos and etc) but heres the paper in full force: http://arxiv.org/abs/1503.07555 I'm currently trying to make this work on half operators, I do believe that $\frac{d^{z-1}}{dw^{z-1}} |_{w=0} \sum_{n=0}^\infty \alpha \uparrow^{n+1} x \frac{w^{n}}{n!}$ produces a holomorphic function. This will interpolate the hyper operators and satisfy the recursive identity. Sadly it will only work for operators greater than or equal to multiplication. It fails for addition by simple means. Thanks for the replies, all your questions have been things I'm interested in and investigating. This will probably work on the super root and more complicated patterns. The goal is to encourage using only integer valued iterations to produce complex iterations. Any questions I'll be glad to answer. This is really a simple idea that has been pushed to its extreme. If this works on bases greater than eta Id be blown away. I think we can modify it somehow to produce this, but it would require understanding iteration when the multiplier is complex and sends to complex domains rather than when it is only real. This is something I've been working on for a while and I have a lot more built up knowledge than just what's in this paper. If you have qualms or something look at the arxiv paper, it is slightly touched up, again, I fixed some typos and a few brazen quick techniques that weren't right. (03/29/2015, 03:35 AM)fivexthethird Wrote: (03/28/2015, 08:59 PM)tommy1729 Wrote: I repeat : Does this method agree with koenigs ?What is koenigs? Are you talking about the regular iteration? If so, in the case of JmsNxn's tetration, yes, any two tetrations for bases $1 are equal if they have the same period. See proposition 10 and corollary 8 in this paper I'm petty sure that this can also be proved using the techniques in JmsNxn's paper As for the super root thing... it's not even clear if it even results in a tetration, since the super root has no nice recurrence relation we can exploit to prove it. Quote:Im wondering if this method also simplifies towards the continuum sum. ( that might have affect on methods based on continuum sums that are in an " unfinished state " ) With this we can easily define a continuum sum: $\sum_{x=0}^{z-1}f(x) = \frac{d^z}{dx^z}|_{k=0} \sum_{k=1}^{\infty}\sum_{i=0}^{k-1}f(i) \frac{x^k}{k!}$ That this works can be easily verified with the newton series identity. Using that, it can also be seen that it extends faulhaber's formula (i.e. it maps polynomials to polynomials) Now the continuum sum formula is $\log_b(\frac{\mathrm{tet}_b'(z)}{\mathrm{tet}'_b(0)\ln(b)^z}) = \sum_{k=0}^{z-1}\mathrm{tet}_b(k)$ for all positive integers z. If the LHS satisfies the bounds for Ramanujan's master theorem to apply, then this equality applies for all z in the right half-plane and is just a restatement of that fact. I'm almost certain that the equality applies to JmsNxn's tetration, but I have no idea how to prove that. It would also imply that the iterational form probably converges to it, too. you're quite right, see my second paper on the indefinite sum. It's a little rough but true. I actually develop an analytic formula involving a triple integral that justifies the expression you put there. I do believe this will work on many many recursive relationships. Again, thanks to everyone for reading this. If you want your mind blown away just wait for how I apply this to $\alpha \uparrow^{1/2} x$ I think it's almost completely solved. I just need to prove a few lemmas. This method will generalize to many more operators. Including the super iterations. (iterations of iteration.) MphLee Long Time Fellow Posts: 373 Threads: 29 Joined: May 2013 03/30/2015, 07:41 PM (This post was last modified: 03/30/2015, 09:16 PM by MphLee.) (03/29/2015, 11:25 PM)JmsNxn Wrote: This one is a little more rough but it explains how continuum sums can be created. http://arxiv.org/abs/1503.07555Did you mean this link? http://arxiv.org/abs/1503.06211 (I guess you pasted the wrong link) Quote:I'm currently trying to make this work on half operators, I do believe that $\frac{d^{z-1}}{dw^{z-1}} |_{w=0} \sum_{n=0}^\infty \alpha \uparrow^{n+1} x \frac{w^{n}}{n!}$ produces a holomorphic function. This will interpolate the hyper operators and satisfy the recursive identity. That is exatly what I was thinking you were hoping for! We hope that the sequence $\phi:n\mapsto\phi_n:=\alpha\uparrow^n x$ for a fixed pair of suitable values $(\alpha,x)$ and where the uparrow is computed using your method. The main problem is that if we believe that $\frac{d^{z-1}}{dw^{z-1}} |_{w=0} \sum_{n=0}^\infty \alpha \uparrow^{n+1} x \frac{w^{n}}{n!}=\phi_z$ and $\phi_z=\alpha\uparrow^z x$... how can we check the recursive identity for the sequence $\phi_z$? I mean... ackermann function is defined using recursion at the same time on two different variables (rank and exponent) so that the only recurrence that we have should be of this form Let $\phi_{z,x}:=\alpha\uparrow^z x$ then $\phi_{z+1,x}=\phi_{z,\phi_{z+1,x-1}}$ In other words, the sequence $\phi_{z,x}$ doens't occours on the right side of the recurrence... so where is the recurrence if $\phi_{z+1,x}$ is evaluated whithout using another element of the sequence $\phi_{z,x}$ (rememebr that x is fixed)? Anyways I see that if we watch the problem from the point of view of creating a new auxiliary functions $\vartheta_E$ relative to another Operator $E:f\mapsto E[f]$ (and its natural iterations) the problem seems to disappear: you outlined this at page 4 of your paper. The concept seems really mysterious an at the same time amazing... I'm asking to my self (mhh... asking to you) if there are ... "hidden simmetries" or something like a big picture behind it. In fact I can see alot of analogies: $\vartheta_E(w,\xi,f):= \sum_{n=0}^\infty E^{\circ n}[f](\xi) \frac{w^{n}}{n!}$ An interesting pattern appears: For E=D $\vartheta_D(w,0,f)$ is the Maclaurin series of f at zero. If $E={\mathcal C}_f$ then $\vartheta_{{\mathcal C}_f}(w,\xi,f)$ is the auxiliary function that you are using in your paper. So if we define the "Superfunction operator" ${\mathcal S}_\beta:f\mapsto {\mathcal S}_\beta[f]=F$ and $f^z(\beta)=F(z)$ then iteration of iteration should be reachable working on this auxiliary function $\vartheta_{{\mathcal S}_\beta}(w,\xi,f):= \sum_{n=0}^\infty {\mathcal S}_\beta^{\circ n}[f](\xi) \frac{w^{n}}{n!}$ Since you say Quote:Sadly it will only work for operators greater than or equal to multiplication. It fails for addition by simple means. Let ${\rm knu}_0(x)=\alpha\uparrow^0 x=\alpha\cdot x$ for a suitable fixed alpha and for beta=1 then ${\mathcal S}_{\beta=1}^{\circ n}[{\rm knu}_0]={\rm knu}_n$ so that $\vartheta_{{\mathcal S}_1}(w,\xi,{\rm knu}_0):= \sum_{n=0}^\infty {\mathcal S}_1^{\circ n}[{\rm knu}_0](\xi) \frac{w^{n}}{n!}= \sum_{n=0}^\infty {\rm knu}_n(\xi) \frac{w^{n}}{n!}= \sum_{n=0}^\infty \alpha\uparrow^n\xi \frac{w^{n}}{n!}$ There seems to be some indexes that look weird/wrong (the sequence inside the summation should start with n=one?) but nothing that you can't fix. Anyways I think we could use the antirecursion/subfunction operator as well: $\vartheta_{{\mathcal S}_\beta}(w,t,f):= \sum_{n=0}^\infty {\mathcal S}^{\circ -n}[f](t) \frac{w^{n}}{n!}$ And that is equivalent to using my tau-sequence or theta-sequence. Maybe those sequences are more easy to handle... if the the rank of f is finite then after some n the tau sequence becomes t+1 (t-1 if the theta sequence) and also all the information about its tau-sequence is already inside f. And there exist surely functions with a tau sequence that converges to t+1... because the successor is a fixed point of the subfunction operator. Quote:Any questions I'll be glad to answer. This is really a simple idea that has been pushed to its extreme. If this works on bases greater than eta Id be blown away. I think we can modify it somehow to produce this, but it would require understanding iteration when the multiplier is complex and sends to complex domains rather than when it is only real. This is something I've been working on for a while and I have a lot more built up knowledge than just what's in this paper. I really wish you good luck! MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 Joined: Dec 2010 03/31/2015, 08:50 PM (This post was last modified: 03/31/2015, 08:52 PM by JmsNxn.) MPHLEE You're taking the words out of my mouth. I use the super function operator which makes the whole thing come together. The trick is constructing an iterate of this operator and then interpolating it in the manner I posted. This requires generalizing Koenigs linearization theorem considerably. It does come out to this series and checking the convergence is a lot easier, which upon the recursive identity DOES come out using Ramanujan's master theorem, but the trick is hidden. The goal is to do it in a much more general setting. I am working very hard on cleaning it up and making sure all the proofs pop out like clock work from the more general schema. There are tons of operators that can be iterated through my methods, which is quite beguiling, but each requires a special treatment on their own. The key word, contraction operators will appear as objects that are easy to iterate. Just having a knowledge of how these operators iterate proves to be very valuable towards the super function operator. I'll get it out soon. Everything you're saying appears but there is much trickery to the solution. I'm just trying to rigorize everything now. The skeleton is there, I just need to prop it up with some muscular rigor. $\mathcal{C}_\phi$ is a contraction operator, this proves invaluable to the generalization. But we must be guaranteed a form of koenigs linearization for the interpolation method to work, and then we can scrap the construction and evaluation of the quasi schroder functions (built for arbitrary operators) and just use the fractional calculus interpolation. The recursion is the simple part, its proving there exists an exponentially bounded function, "factorable," that interpolates hyper operators. That's the difficult part. MphLee Long Time Fellow Posts: 373 Threads: 29 Joined: May 2013 03/31/2015, 09:16 PM (This post was last modified: 03/31/2015, 09:19 PM by MphLee.) (03/31/2015, 08:50 PM)JmsNxn Wrote: The goal is to do it in a much more general setting. I am working very hard on cleaning it up and making sure all the proofs pop out like clock work from the more general schema. [...] I'm just trying to rigorize everything now. The skeleton is there, I just need to prop it up with some muscular rigor.I can't really follow some of the step but I think I got what your going for. The complex analysis that youre using is really crazy for me atm. Anyways I always believed that one of the ways to reach the solution was inside operator theory (aka application of the analysis to higher order functions). Quote:$\mathcal{C}_\phi$ is a contraction operator, this proves invaluable to the generalization. About the contraction operator, the concept is new for me (Operator theory is still new for me) but maybe I'm missing something...Did you present a normed space of functions in your paper? Because the contraction op. need the operator norm in order to be defined, and operator norm is defined using the norm(s) of the space(s) (if I recall correctly it is the operator norm is the smallest number that bounds the operator on its domain). Wich norm is used usually in these kind of frameworks? (sorry if the question is stupid). MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ JmsNxn Ultimate Fellow Posts: 1,176 Threads: 123 Joined: Dec 2010 04/01/2015, 03:20 PM (03/31/2015, 09:16 PM)MphLee Wrote: (03/31/2015, 08:50 PM)JmsNxn Wrote: The goal is to do it in a much more general setting. I am working very hard on cleaning it up and making sure all the proofs pop out like clock work from the more general schema. [...] I'm just trying to rigorize everything now. The skeleton is there, I just need to prop it up with some muscular rigor.I can't really follow some of the step but I think I got what your going for. The complex analysis that youre using is really crazy for me atm. Anyways I always believed that one of the ways to reach the solution was inside operator theory (aka application of the analysis to higher order functions). Quote:$\mathcal{C}_\phi$ is a contraction operator, this proves invaluable to the generalization. About the contraction operator, the concept is new for me (Operator theory is still new for me) but maybe I'm missing something...Did you present a normed space of functions in your paper? Because the contraction op. need the operator norm in order to be defined, and operator norm is defined using the norm(s) of the space(s) (if I recall correctly it is the operator norm is the smallest number that bounds the operator on its domain). Wich norm is used usually in these kind of frameworks? (sorry if the question is stupid). To answer your question: Quote:Wich norm is used usually in these kind of frameworks? It is usually defined formally on any metric space. This can be generalized to many different types of mathematical objects. Sometimes they're functions, sometimes they're points on a plane, sometimes they're operators themselves acting on more operators. The normed space part can be cleverly avoided. We do not need a complicated metric for the functions we are investigating. It's a certain type of contraction operator--namely one that works on points on a plane, not on the functions themselves. For example in the paper we note that $\phi^{\circ n} (\xi) \to \xi_0$ as $n\to\infty$ for all $\xi$ in the immediate basin of attraction. This is equivalent to the statement that $|\phi(\xi) - \phi(\xi') |< q|\xi - \xi'|$ in the basin for some $0. We are contracting the points, based off the operator $\mathcal{C}_\phi$. It's a little tricky to wrap your head around but I'll explain it much more clearly in the next paper. We do not have to create a normed space that the super function operator acts on, but we do have to talk about values converging to a fixed point under the super function operator. This will be equivalent to the values "contracting" under the point wise norm. Namely the family of functions we use are those such that $\uparrow^{n} f \to Constant$ as $n\to\infty$, with some additional conditions attached to $f$. The $\uparrow$ notation is the notation for the super function operator that I am using. This also implies under the pointwise norm it is a type of contraction mapping. We don't need to induce a metric anymore than the traditional one, but again, the trick is hidden. I'm close to getting a well rounded rough copy out. I don't want to give away all the details at the moment but it's simpler than it seems. I just have to make sure I don't have any holes in my ideas. I'm going to talk to some of my professors to make sure it all makes sense and the advanced theorems I need to reference are not used incorrectly. This is going to be more complicated than my last two papers, and at a few parts I feel like I'm out of my depth, but I'll manage. It will probably take a month or two to get this out well enough that I'm willing to post it. MphLee Long Time Fellow Posts: 373 Threads: 29 Joined: May 2013 04/01/2015, 06:09 PM (This post was last modified: 04/01/2015, 07:58 PM by MphLee.) (04/01/2015, 03:20 PM)JmsNxn Wrote: We do not have to create a normed space that the super function operator acts on, but we do have to talk about values converging to a fixed point under the super function operator. This will be equivalent to the values "contracting" under the point wise norm. Namely the family of functions we use are those such that $\uparrow^{n} f \to Constant$ as $n\to\infty$, with some additional conditions attached to $f$. The $\uparrow$ notation is the notation for the super function operator that I am using. This also implies under the pointwise norm it is a type of contraction mapping.Yeah! That sounds awesome! Reading about dynamics I was just guessing about the basin of attraction on the space of functions while studyng the dynamics of the superfunction/subfunction operators. For example studing the dynamics of the antirecursion(subfunction) we see that the set ${\rm FR}:=\{f:{\rm rank}(f)<\infty\}$ of finite-rank functions is a proper subset of the set of functions that tends to the successor (wich is a fixed point). ${\rm FR}\subseteq \{\phi:\lim_{n\to \infty}{\mathcal S}^{-n}[\phi]=S\}$ Imho there is alot of interesting stuff about the classifications of functions... I know you are involved in an analytic theory of "iterated iteration" but I'm still working on some synthetic framework (to be precise to an abstract-"algebraic" translation of the problems) that maybe could give us some existence/non-existence proofs inside some simpler models (some well behaved toy models) and then go for the generalization ... maybe not very soon but I hope for it. BTW I'm sure that you have noticed that the sequence of hyperexponentiations seems to converge to the successor function in the interval [0,1] as the rank grows ... and probably its on "omegation" too (I rememeber Romerio/Rubtsov and also an old post of you here on the TF ) Quote:. This is going to be more complicated than my last two papers, and at a few parts I feel like I'm out of my depth, but I'll manage. It will probably take a month or two to get this out well enough that I'm willing to post it. Ok, good luck! MSE MphLee Mother Law $$(\sigma+1)0=\sigma (\sigma+1)$$ S Law $$\bigcirc_f^{\lambda}\square_f^{\lambda^+}(g)=\square_g^{\lambda}\bigcirc_g^{\lambda^+}(f)$$ « Next Oldest | Next Newest »

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