Rational operators (a {t} b); a,b > e solved - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Rational operators (a {t} b); a,b > e solved (/showthread.php?tid=653) Pages: 1 2 3 4 RE: Rational operators (a {t} b); a,b > e solved - JmsNxn - 06/06/2011 (06/06/2011, 11:59 AM)tommy1729 Wrote: what is the idea or intention behind working with base eta ? First I noticed that at base 2 humps would appear at around 0.5 and 1.5, and if we increased the base, the humps would get sharper and sharper and taller and taller. So I figured, there must be a base value where the humps disappear altogether. My first guess was eta, since everyone talks about how closely related it is to tetration, but when I made a graph of $f(t) = \eta \bigtriangleup_t 2$ there were still visible humps at 0.5 and 1.5. Disappointed, and intrigued by the cheta function, I tried using the upper superfunction of eta, and voila, the values converged miraculously. So it was more just a shot in the dark as opposed to a real mathematical deduction as to why we used base eta. (06/06/2011, 05:16 PM)nuninho1980 Wrote: (06/06/2011, 04:39 AM)sheldonison Wrote: I wonder what it means that fatb(3,-1,4)=5.429897..?no, it isn't correct, sorry. fatb(2,-1,2) = 2º2=4 -> 2+2=4 fatb(3,-1,3) = 3º3=5 -> 3+2=5 fatb(4,-1,3) = 4º3=5 - it's correct. because that's here down: aºb if a>b then a+1 if b>a then b+1 if a=b then a+2 or b+2 no, this isn't zeration, this a different operator designed to preserve the ring structure of operators [t-1] and [t]. (06/06/2011, 09:23 AM)bo198214 Wrote: Well not on the whole complex plane, but on the real axis, wouldnt that be nice? I anyway wonder whether thats possible at all. As $h_b$ is not even differentiable at 1, I wonder whether f is. Did you compare the derivations from left and right?Yes it would be nice to have it analytic on the real axis, I think it should be potentially analytic for at least (-oo, 1). I'm just not sure about the convergence radius. No I haven't compared the derivations, I'll do that though. It's hard for me to think up tests I can do with only highschool math under my belt (06/06/2011, 09:23 AM)bo198214 Wrote: This is also in sync with the convention for quasigroups, i.e. groups with non-associative operation but with left- and right-inverse. There the left- and right inverse are written as / and \. hmm, I really understand where you're coming from with using a standard notation instead of the triangles, it saves a lot of confusion, but I think that for the same reason the gamma function is written $\Gamma(n+1)$ instead of $n!$, logarithmic semi-operators should be written using their own notation--just to be clear this is only one extension of hyper operators. I know that simply stating this is the natural extension of hyper operators will step on a lot of people's toes. Also, I have something very interesting to report! My conjecture $\text{cheta}(t) = e\, \bigtriangleup_t \,e$ can be proved for $t \in (-\infty, 2], t \in Z$. $\text{cheta}(0) = 2e\\ \text{cheta}(-1) = \log_\eta(2e) = \log_\eta(2) + e$ and if anyone knows their deltation, like I just explained $a + \log_\eta(2) = a \,\,\bigtriangleup_{-1}\,\, a$ therefore: $\text{cheta}(-1) = e\,\, \bigtriangleup_{-1}\,\, e$ The proof can actually be made even simpler, consider, $\R(t),\R(p) \le 1; p,t \in C$: $log_\eta^{\circ -p}(a\,\, \bigtriangleup_t\,\, b) = log_\eta^{\circ -p}(a)\,\, \bigtriangleup_{t-p}\,\, \log_\eta^{\circ -p}(b)$ therefore since: $\text{cheta}(0) = e\,\, \bigtriangleup_0\,\, e$, $\log_\eta^{\circ -p}\text{cheta}(0) = \log_\eta^{\circ -p}(e\,\, \bigtriangleup_0\,\, e)$ $\text{cheta}(p) = \log_\eta^{\circ -p}(e)\,\, \bigtriangleup_p \,\,\log_\eta^{\circ -p}(e)$, and since there's a fixpoint at e, $\text{cheta}(p) = e\,\, \bigtriangleup_p\,\, e$ This proves a beautiful connection between logarithmic semi operators and the cheta function. And also gives me a new beautiful identity: $e^{\large{\frac{e\,\, \bigtriangleup_{t-1}\,\, e}{e}}} = e\, \bigtriangleup_t\, e$ RE: Rational operators (a {t} b); a,b > e solved - JmsNxn - 06/06/2011 Alright, testing the left hand right hand limit I get different values.... I'll refer to $\vartheta(a, b, \sigma) = a\,\,\bigtriangle_\sigma\,\,b$ $\{a,b, \sigma| \R(a), \R(b) > e\,;\,a,b,\sigma \in C\}$ from now on. So therefore: $\lim_{h\to 0^+} \frac{\large \vartheta(3, 4, 1 + h) - \vartheta(3, 4, 1)}{h} = 10.7633\\ \lim_{h\to 0^-} \frac{\large \vartheta(3, 4, 1 + h) - \vartheta(3, 4, 1)}{h} = 9.9206$ and $\lim_{h\to 0^+} \frac{\large \vartheta(3, 3, 1 + h) - \vartheta(3, 3, 1)}{h} = 5.4105\\ \lim_{h\to 0^-} \frac{\large \vartheta(3, 3, 1 + h) - \vartheta(3, 3, 1)}{h} = 5.3674$ And finally, I'd like to post a question for anyone more familiar with iteration dynamics than me. This is the open problem to extend $\R(\sigma) \in [2, 3]$ if $f(z) = a \,\,\bigtriangleup_{1+q}\,\, z = \vartheta(a, z, 1+q)$ then: $\vartheta(a, t, 2+q) = f^{\circ t}(1)$ and so solving for [2, 3] is solving for the iterate of f(1). f(z) is an analytic function since it's composed of analytic functions and q is restricted to [0,1]. hmmm, so the values come close to each other but don't quite make it. I wonder, does this disqualify it considering it's not analytic at t=1 and the derivative is undefined? The only case where it is analytic and it is defined is $\vartheta(e, e, \sigma) = \text{cheta}(\sigma)\,\,\,\, \R(\sigma) \le 2$ Which of course I tested using code over the exponential period [1, 2]. i.e.: $0 \le q \le 1$ $\lim_{h\to 0}\, \vartheta(e+h, e+h, 1 + q)\, =\, \text{cheta}(1+q)$ But I think that makes it even more beautiful, the fact that it's only analytic at 1 when a, b = e And finally, I'd like to post a question for anyone more familiar with iteration dynamics than me. This is the open problem to extend $\R(\sigma) \in [2, 3]$ if $f(z) = a \,\,\bigtriangleup_{1+q}\,\, z = \vartheta(a, z, 1+q)$ then: $\vartheta(a, t, 2+q) = f^{\circ t}(1)$ and so solving for [2, 3] is solving for the iterate of f(1). f(z) is an analytic function since it's composed of analytic functions. RE: Rational operators (a {t} b); a,b > e solved - sheldonison - 06/06/2011 (06/06/2011, 07:47 PM)JmsNxn Wrote: Alright, testing the left hand right hand limit I get different values.... I'll refer to $\vartheta(a, b, \sigma) = a\,\,\bigtriangle_\sigma\,\,b$ $\{a,b, \sigma| \R(a), \R(b) > e\,;\,a,b,\sigma \in C\}$ from now on. So therefore: $\lim_{h\to 0^+} \frac{\large \vartheta(3, 4, 1 + h) - \vartheta(3, 4, 1)}{h} = 10.7633\\ \lim_{h\to 0^-} \frac{\large \vartheta(3, 4, 1 + h) - \vartheta(3, 4, 1)}{h} = 9.9206$ and $\lim_{h\to 0^+} \frac{\large \vartheta(3, 3, 1 + h) - \vartheta(3, 3, 1)}{h} = 5.4105\\ \lim_{h\to 0^-} \frac{\large \vartheta(3, 3, 1 + h) - \vartheta(3, 3, 1)}{h} = 5.3674$This matches my results (exactly, actually). The 1st derivatives are close, but they don't exactly match, at the transition between the function that defines operators between addition...multiplication, as compared to the function that defines multiplication...exponentiation. I don't know why it works as well as it does, for base=eta. For other bases, which will also give the same results for integers, the resulting graphs are pretty ugly. (06/06/2011, 07:47 PM)JmsNxn Wrote: $\lim_{h\to 0}\, \vartheta(e+h, e+h, 1 + q)\, =\, \text{cheta}(1+q)$If you could get a definition about a complex circle around h=1, at a,b=e, that might be a big start. This would be 1+q, 1-q, 1+qi, 1-qi, also matching both of your initial definitions (which I haven't checked). If that were the case, you already have analytic functions defied for 0<=h<=1, and analytic functions defined for 1<=h<=2. Then, for one case, a=b=e, you might have a function defined for 0<=h<=2. Then the key is to morph this function, perhaps starting with the case a=b, as a=b becomes less than e, and greater than e, in such a way that it remains analytic. Of course, there is the small issue that the inverse superfunctions of eta have singularities at z=e, and the issue of the upper/lower superfunctions of eta, so there are many many challenges on this path. By the way, I agree with Henryk, that exponentiation should be rational operator three, and multiplication, rational operator 2, and addition rational operator 1. - Sheldon RE: Rational operators (a {t} b); a,b > e solved - bo198214 - 06/06/2011 (06/06/2011, 05:44 PM)JmsNxn Wrote: Yes it would be nice to have it analytic on the real axis, I think it should be potentially analytic for at least (-oo, 1).Yes. Quote:No I haven't compared the derivations, I'll do that though. It's hard for me to think up tests I can do with only highschool math under my belt You do very well Quote:but I think that for the same reason the gamma function is written $\Gamma(n+1)$ instead of $n!$ Oh dear thats another topic! This definition of the Gamma function (being off by one from factorial) is imho a history misdevelopment. There is no use at all of it, and it confuses the minds. Quote:, logarithmic semi-operators should be written using their own notation--just to be clear this is only one extension of hyper operators. I know that simply stating this is the natural extension of hyper operators will step on a lot of people's toes. Sure. I mean even for tetration there are different ways of extension. And long fights about which is the best/the most natural etc.... RE: Rational operators (a {t} b); a,b > e solved - JmsNxn - 06/07/2011 (06/06/2011, 08:43 PM)sheldonison Wrote: If you could get a definition about a complex circle around h=1, at a,b=e, that might be a big start. This would be 1+q, 1-q, 1+qi, 1-qi, also matching both of your initial definitions (which I haven't checked). If that were the case, you already have analytic functions defied for 0<=h<=1, and analytic functions defined for 1<=h<=2. Then, for one case, a=b=e, you might have a function defined for 0<=h<=2. Then the key is to morph this function, perhaps starting with the case a=b, as a=b becomes less than e, and greater than e, in such a way that it remains analytic. Of course, there is the small issue that the inverse superfunctions of eta have singularities at z=e, and the issue of the upper/lower superfunctions of eta, so there are many many challenges on this path. I don't understand what you mean about the first part, of making a definition about a complex circle, however I do understand what you mean by morphing the cheta function, and creating a generalization to $a\,\,\bigtriangle_\sigma\,\,a$. (06/06/2011, 08:43 PM)sheldonison Wrote: By the way, I agree with Henryk, that exponentiation should be rational operator three, and multiplication, rational operator 2, and addition rational operator 1. - SheldonHmm, this is just like the Gamma function predicament. Well, hear my arguments for centering addition at 0. Firstly I wanted to center the Identity function so that S(0) = 0, and S(1) = 1. Secondly, the function designed is naturally centered with addition as zero. Shifting it over by one is an offset. $\vartheta(a,b,\sigma) = a\,\,\bigtriangleup_\sigma\,\,b = \exp_\eta^{\circ \sigma}(\exp_\eta^{\circ -\sigma}(a) + h_b(\sigma))$ $h_b(\sigma)=\left{\begin{array}{c l} \exp_\eta^{\circ -\sigma}(b) & \sigma \le 1\\ \exp_\eta^{\circ -1}(b) & \sigma \in [1,2] \end{array}\right.$ Thirdly, the negative integers of operators were specifically designed to be negative because they aren't proper recursive. i.e: $p<=0$ $a\,\,\bigtriangleup_{p+1}\,\,\log_\eta^{\circ -p}(n) = a_1\,\,\bigtriangleup_{p}\,\,a_2\,\,\bigtriangleup_{p}\,\,a_3...\bigtriangleup_{p}\,\,a_n$, having 0 as the first non-proper recursive operator seems off. I feel it should be -1 (or technically -0.0000000001). Also, again you can see the series is centered with addition as 0. I don't really have any other arguments, it just seems simpler to center it at 0. It's just unfortunate that someone decided to call it "tetration" instead of "tertiation". RE: Rational operators (a {t} b); a,b > e solved - bo198214 - 06/07/2011 (06/07/2011, 02:45 AM)JmsNxn Wrote: It's just unfortunate that someone decided to call it "tetration" instead of "tertiation". You mean "triation". Its greek numbering not latin Actually I agree with you, in certain contexts I worked with it also seems more suitable to use addition as operation 0. (Even Ackermann did so.) On the other hand, on different contexts there it seems more preferable to start with 1. E.g. to have the zero-th operation a [0] x = x+1. And I dont think one can roll back the whole historic naming development, tetration, the forth operation is already ingrained in the minds. RE: Rational operators (a {t} b); a,b > e solved - JmsNxn - 06/08/2011 (06/07/2011, 06:59 AM)bo198214 Wrote: You mean "triation". Its greek numbering not latin Actually I agree with you, in certain contexts I worked with it also seems more suitable to use addition as operation 0. (Even Ackermann did so.) On the other hand, on different contexts there it seems more preferable to start with 1. E.g. to have the zero-th operation a [0] x = x+1. And I dont think one can roll back the whole historic naming development, tetration, the forth operation is already ingrained in the minds. I'm glad you agree with me about centering $\vartheta$ with addition as zero. I actually originally centered it at zero because Ackermann did. triation sounds nicer than tertiation. But besides this, I have something to report. $\lim_{y\to\infty}\vartheta(a, b, x\pm iy) = L$ which converges absolutely for all $a,b \in C$ and $x \in (-\infty, 2]$. Interestingly enough L does not change with a and b and seems to be only x dependent. L seems to always be greater than 2.6 but just around it. At first I thought it was a universal constant they were all converging to, but only $x \in [0, 0.5]$ showed convergence to this value. RE: Rational operators (a {t} b); a,b > e solved - JmsNxn - 06/08/2011 I can't believe I overlooked this! This is huge! $\vartheta(a, e, \sigma) = \exp_\eta^{\circ \sigma}(\exp_\eta^{\circ \sigma}(a) + h_e(\sigma))$ $h_e(\sigma)=\left{\begin{array}{c l} \exp_\eta^{\circ -\sigma}(e) & \sigma \le 1\\ \exp_\eta^{\circ -1}(e) & \sigma \in [1,2] \end{array}\right.$ but since $\exp_\eta(e) = e$, $h_e(\sigma) = e$ therefore: $\vartheta(a, e, \sigma) = \exp_\eta^{\circ \sigma}(\exp_\eta^{\circ -\sigma}(a) + e)$ And therefore $\vartheta$ is potentially analytic, and isn't piecewise, for all a, and b = e, $\sigma \in (-\infty, 2]$ This is crazy. RE: Rational operators (a {t} b); a,b > e solved - sheldonison - 06/08/2011 (06/08/2011, 07:31 PM)JmsNxn Wrote: I can't believe I overlooked this! This is huge! .... therefore: $\vartheta(a, e, \sigma) = \exp_\eta^{\circ \sigma}(\exp_\eta^{\circ -\sigma}(a) + e)$ And therefore $\vartheta$ is potentially analytic, and isn't piecewise, for all a, and b = e, $\sigma \in (-\infty, 2]$ This is crazy.So we have a definition for t=0..2 (addition, multiplication, exponentiation), for all values of a. Nice! $a\{t\}e = \exp_\eta^{\circ t}(\exp_\eta^{\circ -t}(a)+e)$ So, now, I'm going to define the function f(b), which returns the base which has the fixed point of "b". $f(e)=\eta$, since e is the fixed point of b=$\eta$ $f(2)=\sqrt{2}$, since 2 is the lower fixed point of b=sqrt(2) $f(3)=1.442249570$, since 3 is the upper fixed point of this base $f(4)=\sqrt{2}$, since 4 is the upper fixed point of b=sqrt(2) $f(5)=1.379729661$, since 5 is the upper fixed point of this base I don't know how to calculate the base from the fixed point, but that's the function we need, and we would like the function to be analytic! This also explains why the approximation of using base eta pretty well, since the base we're going to use isn't going to be much smaller than eta, as b gets bigger or smaller than e. Now, we use this new function in place of eta, in James's equation. Here, f=f(b). $a\{t\}b = \exp_f^{\circ t}(\exp_f^{\circ-t}(a)+b)$ - Sheldon RE: Rational operators (a {t} b); a,b > e solved - bo198214 - 06/08/2011 (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". $f(e)=\eta$, since e is the fixed point of b=$\eta$ $f(2)=\sqrt{2}$, since 2 is the lower fixed point of b=sqrt(2) $f(3)=1.442249570$, since 3 is the upper fixed point of this base $f(4)=\sqrt{2}$, since 4 is the upper fixed point of b=sqrt(2) $f(5)=1.379729661$, since 5 is the upper fixed point of this base I don't know how to calculate the base from the fixed point, but that's the function we need, and we would like the function to be analytic! Oh, Sheldon seems to be quite tired from all the calculation and discussion. Sheldon, give your self some time to rest! Your function is $f(z)=z^{1/z}$