![]() |
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: Hyperoperations and Related Studies (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=11) +--- Thread: Rational operators (a {t} b); a,b > e solved (/showthread.php?tid=653) |
Rational operators (a {t} b); a,b > e solved - JmsNxn - 06/06/2011 Well alas, logarithmic semi operators have paid off and have given a beautiful smooth curve over domain Which extends the ackerman function to domain real (given the restrictions provided). the upper superfunction of Logarithmic semi-operators contain infinite rings and infinite abelian groups. In so far as {t} and {t-1} always form a ring and {t-1} is always an abelian group (therefore any operator greater than {1} is not commutative and is not abelian). There is an identity function S(t), however its values occur below e and are therefore still unknown for operators less than {1} (except at negative integers where it is a variant of infinity (therefore difficult to play with) and at 0 where it is 0). Greater than {1} operators have identity 1. The difficulty is, if we use the lower superfunction of therefore rational roots, the inverse of rational exponentiation is defined so long as rational division and rational subtraction is possible if Here are some graphs, I'm sorry about their poor quality but I'm rather new to pari-gp so I don't know how to draw graphs using it. I'm stuck using python right now. Nonetheless here are the graphs. the window for these ones is xmin = -1, xmax = 2, ymin = 0, ymax = 100 ![]() ![]() ![]() If there's any transformation someone would like to see specifically, please just ask me. I wanted to do the transformation of Some numerical values: (I know I'm not supposed to be able to calculate the second one, but that's the power of recursion) I'm very excited by this, I wonder if anyone has any questions comments? for more on rational operators in general, see the identities they follow on this thread http://math.eretrandre.org/tetrationforum/showthread.php?tid=546 thanks, James PS: thanks go to Sheldon for the taylor series approximations of cheta and its inverse which allowed for the calculations. RE: Rational operators (a {t} b); a,b > e solved - sheldonison - 06/06/2011 (06/06/2011, 02:45 AM)JmsNxn Wrote: Well alas, logarithmic semi operators have paid off and have given a beautiful smooth curve over domainHey James, Sounds very exciting. fyi, I admit I don't yet understand your new functions, but I made some graphs, and it looks very promising. I thought I'd post the following snippet of pari-gp code, which implements the rational operator function you posted, which can be used with mylatest kneser.gp code, which includes Code: expeta(t,a) = { I've gotten as far as quickly verifying that for t=0, we have addition, t=1, we have multiplication, and t=2 is exponentiation. The existing code has problems for invcheta(z) or invsexpeta(z), where z is too close to e, and invcheta(z)<=-1000, or invsexpeta(z)>1000. Other than that, it seems to work great. For example, fatb(3,0,4)=7, which is 3+4 fatb(3,1,4)=12, which is 3x4 fatb(3,2,4)=81, which is 3^4 I wonder what it means that fatb(3,-1,4)=5.429897..? Also, is there a smooth continuation to a function for t=3, which would be tetration? - Sheldon RE: Rational operators (a {t} b); a,b > e solved - JmsNxn - 06/06/2011 (06/06/2011, 04:39 AM)sheldonison Wrote: mylatest kneser.gp code, which includes Oh that's great! I'd love to see what happens for complex operator values, I'll just modify your code slightly to say real(t) under the if conditions. (06/06/2011, 04:39 AM)sheldonison Wrote: I've gotten as far as quickly verifying that for t=0, we have addition, t=1, we have multiplication, and t=2 is exponentiation. fatb(3, -1, 4) = 5.429897, is the value of 3 delta 4; where deltation is the operator which satisfies the following conditions: Basically, the whole domain of operators from As of now, it's possible to compute The formula is given by: And lastly, I tried using sexp/slog base Edit: I edited your code and tested if fatb(3, i, 3) = fatb(3, i+1, 2), and it does. Ackerman function is entering the complex plane me thinks. Edit again: Quote: if a<e=2.718, then it appears that the non integer nodes have complex valuesI was wondering if you could explain what you mean by this? Is there a way to calculate a,b < e using the cheta function? RE: Rational operators (a {t} b); a,b > e solved - sheldonison - 06/06/2011 (06/06/2011, 05:34 AM)JmsNxn Wrote: ....Hey James, try my code snippet, which I updated while you were posting. It will work for values of a and b<e, seamlessly. - Shel RE: Rational operators (a {t} b); a,b > e solved - bo198214 - 06/06/2011 (06/06/2011, 02:45 AM)JmsNxn Wrote: But James, this is not analytic at We can say: where But then on the other hand there is a general problem with semioperators (note that your operator is 1 off the standard notation, i.e. {t}=[t+1]): (a [0] 0 = 1) a [1] 0 = a a [2] 0 = 0 a [3] 0 = 1 a [n] 0 = 1 for n>3 or (a [0] 1 = 2) a [1] 1 = a+1 a [2] 1 = a a [3] 1 = a for n>2 As soon as one defines a [t] 1 = a for t > 2, then the whole analytic function is already determined to be a, i.e. it must also be a for t=1 which is wrong. Hence an analytic t |-> a[t]1 will not be constant but somehow meandering between the a's, which is somehow really strange. But I see you gracefully avoided that problem by just defining it for a,b > e ![]() PS: 1. 2. RE: Rational operators (a {t} b); a,b > e solved - sheldonison - 06/06/2011 (06/06/2011, 05:34 AM)JmsNxn Wrote: Oh that's great! I'd love to see what happens for complex operator values, I'll just modify your code slightly to say real(t) under the if conditions....at t=1, which corresponds to multiplication, and is the boundary between the two definitions, it appears continuous in the complex plane (which is quite amazing), but it also looks like the function may not converge there. I tried generating a taylor series for fatb(2,t,3) around t=1, with different samping radii, and different numbers of samping points, and it looks like the derivatives grow without bounds, even though the function is continuous in the complex plane. So far, it appears analytic around t=0.5, or t=1.5. - Shel RE: Rational operators (a {t} b); a,b > e solved - JmsNxn - 06/06/2011 (06/06/2011, 06:53 AM)bo198214 Wrote: But James, this is not analytic at I like your definition better--it seems sleeker ![]() Quote:But I see you gracefully avoided that problem by just defining it for a,b > ewell hopefully I'll be having to tackle that problem soon. Quote:PS: Alright, from henceforth I shall refer to logarithmic semi operators with the following notation: And the inverse is given by: therefore: Quote:2. I knew there was something off about my equations. lol (06/06/2011, 06:02 AM)sheldonison Wrote: Hey James, try my code snippet, which I updated while you were posting. It will work for values of a and b<e, seamlessly. I'm wary about using fatb(e+0.0001, 2, pi*I) = -0.999999 - 0.00115551*I fatb(e+0.0001, 1.8, pi*I) = -1.883265702 - 0.00194696*I fatb(e+0.0001, 1.5, pi*I) = -5.707515375 - 0.011242371*I fatb(e+0.0001, 1.3, pi*I) = -4.091499848 - 8.531525563*I fatb(e+0.0001, 1.1, pi*I) = -1.002757644 - 8.536029475*I fatb(e+0.0001, 1, pi*I) = -8.53659263001*I Ignoring the drastic jumps in values, observe the hump that occurs in the real transformation. For no reason the values just spike to -5 randomly. This happens with all regular superfunctions of the logarithm. That's what makes the cheta function unique. But I think, is it possible to create an upper superfunction for RE: Rational operators (a {t} b); a,b > e solved - bo198214 - 06/06/2011 (06/06/2011, 08:47 AM)JmsNxn Wrote: I was sort of aware that there was no way I was gonna produce an analytic function over the whole complex domain, I'm happy with analytic in a few regions.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 Quote:Alright, from henceforth I shall refer to logarithmic semi operators with the following notation: oh I thought }t{ was kinda the inverse in t. But as I see that it is not ... there are already conventions that we use on the forum, even ASCII capable: a [1] b = a+ b a [2] b = a*b a [3] b = a^b so the operator is [t] (one off from your {t}). But then for the inverse functions b [t] x = y y /[t] x = b b [t]\ y = x for example x /[1] y = x - y x [1]\ y = -x + y x /[4] p is the p-th superroot (which is not equal to x [4] 1/p) b [4]\ x is the superlogarithm to base b. 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 \. RE: Rational operators (a {t} b); a,b > e solved - tommy1729 - 06/06/2011 what is the idea or intention behind working with base eta ? RE: Rational operators (a {t} b); a,b > e solved - nuninho1980 - 06/06/2011 (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 |