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 Bounded Analytic Hyper operators tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 03/28/2015, 08:59 PM @ 5x 3rd : You are correct ! I wonder what you guys will call it I repeat : Does this method agree with koenigs ? 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 " ) regards tommy1729 fivexthethird Junior Fellow Posts: 9 Threads: 3 Joined: Nov 2013 03/29/2015, 03:35 AM (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. MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 03/29/2015, 10:55 AM Off Topic: @Fivexthethird About continuum sum (is the inverse of the difference operator right?) if yes have you checked the other paper of JmsNxn? The paper on the bounded Hos. was uploaded on arxiv paired with one about indefinite sum operator. James Nixon - On indefinite sum in fractional calculus MathStackExchange account:MphLee tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 03/29/2015, 04:22 PM Very intresting. Im having a bad week it seems , im forgetting things ... Why was it again that interpolation of 1/( b^^a ) thus D^z 1/( b^^0) + 1/( b^^1) x + 1/( b^^2) x^2 + ... is not a good way for getting at tetration ? Is it because 1/exp(1/x) = exp(-1/x) is not analytic at 0 ? Or is it the problems with negative or imaginary iterations ? Maybe this is C^oo but not analytic ? Maybe I should read a few things again.... regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 03/29/2015, 04:36 PM (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. What about the 1-periodic wave then ? Maybe I misunderstood you. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 03/29/2015, 04:56 PM Hmm after some consideration the continuum sum methods remain problematic ... It seems all conditions of everything we have so far are to strong for the continuum sum methods. This requires additional thinking. Its on my to-do list. regards tommy1729 MphLee Fellow Posts: 95 Threads: 7 Joined: May 2013 03/29/2015, 07:08 PM Since on the other thread about Zeration I learnt about the base change formula... what happens if we apply it to JmsNxn's tetration? Could we reach every base? MathStackExchange account:MphLee tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 03/29/2015, 07:14 PM (03/29/2015, 07:08 PM)MphLee Wrote: Since on the other thread about Zeration I learnt about the base change formula... what happens if we apply it to JmsNxn's tetration? Could we reach every base? I think "We" believe the base change is not analytic. Going from an analytic function to another analytic function with a nonanalytic function is usually not possible. With "We" I mean the top posters at this forum. Sheldon has expressed his strong disbelief in base change being analytic and probably others too. This does not mean that idea will never be usefull. But probably not in the way you think right now. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 03/29/2015, 09:37 PM Hmm the superroots probably do no iterate analytically from ssqrt_n to ssqrt_(n+1) so that gives us a problem. Maybe I was to fast in assuming we can have tetration for any base from these. Or maybe we do get tetration but with many singularities. Need to consider this. regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,370 Threads: 335 Joined: Feb 2009 03/29/2015, 09:41 PM As for the continuum sum , I fear that we do not have the condition of carlson's theorem so this might fail for fast growing functions or equations involving continuum sums and fast growing functions ? regards tommy1729 « Next Oldest | Next Newest »

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