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RE: Rational operators (a {t} b); a,b > e solved - JmsNxn - 06/08/2011
(06/08/2011, 08:32 PM)sheldonison Wrote: So we have a definition for t=0..2 (addition, multiplication, exponentiation), for all values of a. Nice! Woah, I wonder what consequences this will have on the algebra. I guess which I guess isn't too drastic. But ,the question is, of course, does the following still hold : , so no it doesn't. That's not good, we want operators to be recursive. And I'm unsure if the inverse is still well defined, so I think we lose: where S(q) is the identity function. We may even lose the identity function altogether, this is really bad. We also lose: These are all too many valuable qualities that are lost when redefining semi-operators the way that you do. Sure it's analytic over , but it loses all its traits which make it an operator in the first place. I'm going to have to stick with the original definition of that isn't fully analytic. However, I am willing to concede the idea of changing from base eta to base root 2. That is to say if we define: This will give the time honoured result, and aesthetic necessity in my point of view, of: for all I like this also because it makes and potentially analytic over since 2 and 4 are fix points. I also propose writing Sheldon's analytic function is then: that's still very pretty though, that isn't piecewise over and potentially analytic. RE: Rational operators (a {t} b); a,b > e solved - Gottfried - 06/11/2011
Hi James, I do not really know, whether the following matches your input here; but screening through older discussions I just found an older post of Mike (I'd saved it by copying from google.groups). He observed the following and asked Henryk had answered with some proof of convergence and rate of convergence. I had an idea to reformulate this in a way using somehow "lower degree operators" than addition but could not make it better computable, so I didn't involve then further. If I get your approach right this can be used for such "lower order" operators? Say and for the h-fold iterated log( 3) then Mike's limit can be expressed where the operator-precedence is lower the more negative the index at the plus is (so we evaluate it from the left). First question: is this in fact an application of your "rational operator"? And if it is so, then second question: does this help to evaluate this to higher depth of iteration than we can do it when we try it just by log and exp alone (we can do it to iteration 4 or 5 at max I think) ? Gottfried cite: Quote:In article RE: Rational operators (a {t} b); a,b > e solved - JmsNxn - 06/12/2011
(06/11/2011, 02:33 PM)Gottfried Wrote: Hi James, Yes this is right: I did a lot of investigation into this operator (well, to the best that I could). Quote:and for the h-fold iterated log( 3) then Mike's limit can be expressed Well it appears to be. I'm floored, I'm terrible at finding applications. Quote:And if it is so, then second question: does this help to evaluate this to higher depth of iteration than we can do it when we try it just by log and exp alone (we can do it to iteration 4 or 5 at max I think) ? Well, not so far since the calculations involved in lower order operators rely on iterations of exp. However, I investigated in seeing if was analytic (which should help in calculations). But I only made it to the sixth or seventh derivative before I realized I wasn't going to recognize the pattern. The thread's here http://www.mymathforum.com/viewtopic.php?f=23&t=20993 . I assume it would be analytic, (the function looks analytic when graphed, if that's any argument). Also, I think that if is analytic, then is probably analytic, since it's basically the same function with just a faster convergence to y=x and a higher starting point at negative infinity. RE: Rational operators (a {t} b); a,b > e solved - Xorter - 08/21/2016
(06/06/2011, 02:45 AM)JmsNxn Wrote: Well alas, logarithmic semi operators have paid off and have given a beautiful smooth curve over domain . This solution for rational operators is given by : Hello, James! Hello, Everyone! I am really interested in that how you could make that beautiful graph. Could you tell me, please? Thank you very much. RE: Rational operators (a {t} b); a,b > e solved - JmsNxn - 08/22/2016
(08/21/2016, 06:56 PM)Xorter Wrote: Hello, James! Hello, Everyone! Search for the cheta function on the forum and get its power series. Take define continuous solution which is analytic for t < 1 and t >1 with a singularity at t=1 oddly enough is analytic everywhere. RE: Rational operators (a {t} b); a,b > e solved - Xorter - 08/24/2016
(08/22/2016, 12:36 AM)JmsNxn Wrote:(08/21/2016, 06:56 PM)Xorter Wrote: Hello, James! Hello, Everyone! Thank you for the answer! Could you show me an example, too, please? E. g. How can I evaluate 3[0.5]3 or 3[1.5]3? RE: Rational operators (a {t} b); a,b > e solved - Xorter - 08/29/2016
According to James' formula I could not tetrate. For example: 2[3]3 should be 2^^3 = 16, but according to the formula: 2[3]3 = f(2,3f(-2,2)) = f(2,3*1.869...) = 18.125... But why? RE: Rational operators (a {t} b); a,b > e solved - JmsNxn - 09/01/2016
(08/29/2016, 02:06 PM)Xorter Wrote: According to James' formula I could not tetrate. For example: The formula only works for in x [t] y The formula isn't really that valuable. RE: Rational operators (a {t} b); a,b > e solved - tommy1729 - 09/02/2016
(06/08/2011, 09:14 PM)bo198214 Wrote:(06/08/2011, 08:32 PM)sheldonison Wrote: So, now, I'm going to define the function f(b), which returns the base which has the fixed point of "b". About those fixpoints and bases. Bo is correct ofcourse but i wanted to add how to find the other fixpoint. I end with a joke because i did NOT show which T is the correct one : it is the SMALLEST > 1 to be precise ... Analytic continuation is hard for a proof of that minimal property. The convex nature and fast growth of exp type functions is intuitive but also IMHO unconvincing / informal / weak. So no satisfying proof. Perhaps food for thought. ( certainly possible ! ) Anyway here it is ( and T = t ) X^1/x = y^1/y Ln(x)/ x = ln(y)/y Ln(x)/x = ln(T x)/ T x T ^ 1/T x^1/T = x T x = x^t X^(t - 1) = t Ofcourse new similar problem t1 ^ (1/(t1 - 1)) = t2 ^ (1/(t2 - 1)) Regards Tommy1729 The master RE: Rational operators (a {t} b); a,b > e solved - tommy1729 - 09/02/2016
Oh dear e ^(T - 1) = T for T <> 0 gave me the idea of " fake fixpoint theory " in analogue to " fake function theory ". Apart from that funny/annoying thing , another reason is that this innocent little thing apparantly f* Up Some proof strategies for the desired Nice proof mentioned b4. Regards Tommy1729 |