11/12/2007, 09:14 AM
Base 1 and base infinity? I'm not quite sure I follow...
~ Jay Daniel Fox
Andrew Robbins' Tetration Extension

11/12/2007, 09:56 AM
The best way to illustrate it is through this graph:
where the thin dotted line is tetration base1, the solid line is tetration basee, and the thick dotted line is tetration baseinfinity. Neither of these are very interesting (because they are straight lines), and baseinfinity isn't really solvable, but it doesn't matter because you can take limits. In the limit as the base goes to 1 the curve gets closer and closer to the thin dotted line, whereas in the limit as the base goes to infinity, the curve gets closer and closer to the thick dotted line, so these aren't really functions but asymptotes of tetration! Some of these trends can be seen on one of my graphs on my website. However, I didn't come to this conclusion by looking at graphs; I have proof. So the superlogarithms of the previously mentioned bases would be the reflection of those asymptotes across the y=x line. Andrew Robbins
11/12/2007, 08:05 PM
Quote:I have proof. So what do you actually prove? bo198214 Wrote:So what do you actually prove? That: (expanded about z=0) and: (expanded about z=0), which means in the limit: and: One of the nice things about having an exact form for all approximations is that the approximations are invertible (so we can find tetration as well) but the limit to infinity makes it discontinuous, which is not invertible. Andrew Robbins
11/13/2007, 10:21 AM
andydude Wrote:That: Ah, ok. But you still assume that the coefficients of converge. I asked because it sounded in the beginning as if you had a proof for the convergence of the coefficients for base and which would be in itself somehow strange statement.
11/13/2007, 05:45 PM
Indeed, that would be strange. I'm not assuming convergence, but I am assuming that:
What I can prove is that for the nth approximation of the superlog (the finite series):
From this it is easy to show Andrew Robbins
Hi 
I reread this thread yesterday. Now I tried a matrixversion of the slog  I'd now like to see, if the two methods agree. In short: I suppose most matrices as known, as a reminder I only recall that the matrix Ut performs the decremented iterated exponentiation Ut(x) to base t here: Code: ´  Now to ask for the superlog is to ask for the height h, given y and x=1. This can surprisingly easy be solved using the known eigenmatrices Z of Ut let u=log(t) then Code: ´ Ut = Z * dV(u) * Z^1 then also Code: ´ Ut^h = Z * dV(u^h) * Z^1 Then the equation Code: ´ V(x)~ * Ut^h = V(y)~ can be rewritten as Code: ´ Code: ´ (V(x)~ * Z) * dV(u^h) = (V(y)~ * Z)  Since we need only the second column of the evaluated parentheses, let's denote the entries of the r'th row of the second column of Z as z_r Then define a function Code: ´ g(x) = z0 + z1*x + z2*x^2 + z3*x^3 + ... Code: ´ g(x)*u^h = g(y) Code: ´ If g(x) diverges conditionally, it may still be Eulersummable (see **) For base t=2 the sequence of z_k diverges with a small rate, and seem all to be positive. If x is negative, then this is surely Eulersummable, if x<1/2 then it is even convergent. Exampleterms z_r for t=2 Code: ´ By the definition b=t^(1/t) we may use this also for the tetrationfunction (or better "iterated exponential" in Andrew's wording) of base b, since Code: ´ Tb(x) = Ut(x')" where x'=x/t1 and x"=(x+1)*t (but remember, that this use of fixpointshift gives varying results dependent of the choice of the fixpoint  however small the differences may be, according to our current state of discussion) So Code: ´ for Ut°h(x) = y the heightfunction hghU() gives Code: ´ h = hghU_t(y) = log( (g(y)/g(1)) / log(log(t)) and for Code: ´ Tb°h(x) = y the heightfunction hghT() gives Code: ´ Gottfried  (**) the Eulersummation adds coefficients e_k of weight to each term in g(x), so the Eulersummed variant eg(x) of the dimtruncated powerseries is then eg(x) = e0*z0 + e1*z1*x + e2*z2*x^2 + e3*z3*x^3 + ... + e_dim*z_dim*x^dim where the e_k have to be determined by a given size of matrixtruncation and a given appropriate Eulerorder.
Gottfried Helms, Kassel
03/17/2008, 06:09 PM
Gottfried Wrote:and forActually this means, the hghfunctions compute the heightdifferences; call lgg(x)= log(g(x))/log(u) then more generally hghT_b(x1,x0) = lgg(x1')  lgg(x0') hghU_t(x1,x0) = lgg(x1)  lgg(x0) and hghT_b(x) = lgg(x')  lgg(1') and hghU_t(x) = lgg(x)  lgg(1) may be taken as short notations for a default case. (Remember, that the function g and lg are dependent on the baseparameters b and/or t) Gottfried
Gottfried Helms, Kassel
06/26/2009, 10:51 PM
(08/19/2007, 09:50 AM)bo198214 Wrote: I just verified numerically that the superlog critical function (originally defined on ) for base satisfies as for the radius of convergence : let A be the smallest fixpoint => b^A = A then ( andrew's ! ) slog(z) with base b should satisfy : slog(z) = slog(b^z)  1 => slog(A) = slog(b^A)  1 => slog(A) = slog(A)  1 => abs ( slog(A) ) = oo so the radius should be smaller or equal to abs(A) maybe i missed it , but i didnt see that mentioned. also this makes me doubt  especially considering that for every base b slog should 'also' ( together with the oo value at the fixed point A mentioned above ) have a period ( thus abs ( slog(A + period) ) = oo too ! )  , however thats just an emotion and no math of course ... ( btw the video link mentioned in this thread doesnt work for me bo , maybe it isnt online anymore ? )
06/27/2009, 09:39 AM
(06/26/2009, 10:51 PM)tommy1729 Wrote: as for the radius of convergence :Its not only valid for Andrew's slog but for every slog and also not only for the smallest but for every fixed point. However not completely: One can not expect the slog to satisfy slog(e^z)=slog(z)+1 *everywhere*. Its a bit like with the logarithm, it does not satisfy log(ab)=log(a)+log(b) *everywhere*. What we however can say is that log(ab)=log(a)+log(b) *up to branches*. I.e. for every occuring log in the equation there is a suitable branch such that the equation holds. The same can be said about the slog equation. So if we can show that Andrew's slog satisfies slog(e^z)=slog(z)+1 e.g. for then it must have a singularity at A. Quote:also this makes me doubt  especially considering that for every base b slog should 'also' ( together with the oo value at the fixed point A mentioned above ) have a period ( thus abs ( slog(A + period) ) = oo too ! )  , however thats just an emotion and no math of course ...I just showed in some post before but dont remember which one: again up to branches. Quote:( btw the video link mentioned in this thread doesnt work for me bo , maybe it isnt online anymore ? ) well it seems to not exist anymore. It didnt give concrete solutions to our problems, it was just interesting that others also deal with the solution of infinite equation systems and approximation via finite equation systems. 
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