• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 A tiny base-dependent formula for tetration (change-of-base?) Gottfried Ultimate Fellow Posts: 757 Threads: 116 Joined: Aug 2007 03/14/2009, 09:17 PM (This post was last modified: 03/14/2009, 10:48 PM by Gottfried.) A tiny base-dependent formula for tetration (related to "change-of-base"?) Well, the following is not yet deeply considered; it is more that I stumbled across this observation, and did not yet really analyze all its implication but want to announce, so someone else may crosscheck and look at it (I've my head elsewhere currently) When rereading my article about "pascalmatrix tetrated" (*1) and playing/checking some computations, it came to my attention, that I already have a formula for sexp with fixed (integer) height parameter, but variable with the base. So feeding log(2) as parameter to the series may give 2^^3, and feeding log(3) gives then 3^^3 . In all my recent consideration the base-parameter was deeply calculated in the coefficients and the height-parameter was isolated and thus subject to change; here the height-parameter is hidden in the coefficients and the base-parameter is explicit and can be subject to change. We need simply the first column of the tetrated pascalmatrix, scale it by reciprocal factorials and use it as coefficients for the powerseries: ${b\^\^}^h = \sum_{k=0}^{\infty} \frac{{P\^\^}^h_{k,0}}{k!}\log(b)^k$ where P^^h is the tetrated pascalmatrix as described in the article with a given integer height h, and the indexes in the formula denote the k'th rows in first column. (Clearly this has range of convergence and such, just to be determined soon.) Gottfried (*1) http://go.helms-net.de/math/tetdocs/Pasc...trated.pdf The fact of representing an exponential generating-function was also given in the OEIS but it didn't appear to me, that this is (thus) a base-dependent formula; for instance see for h=2: (*2) http://www.research.att.com/~njas/sequences/A000248 Code:´ FORMULA   E.g.f.: exp(x*exp(x)).           G.f.: Sum_{k>=0} x^k/(1-k*x)^(k+1). - Vladeta Jovovic (vladeta(AT)Eunet.yu), Oct 25 2003 Gottfried Helms, Kassel Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/15/2009, 08:50 AM (This post was last modified: 03/15/2009, 09:22 AM by Ivars.) Sorry for being little out of touch, to me "upping " things one step looks very interesting in general ( because of log-Poisson distribution): $e^{\lambda^k /k!*e^{-\lambda}$ Can we not replace $\lambda^k/k!$ by any function $f(\lambda)$ which has a powerseries expansion, as that expansion would usually contain something like $c* \lambda^k/k!$ ? Of course, coefficients at powers of $\lambda$ can be also anything else. I can only see so far that terms of powerseries expansion has to be put into "double exponentiation operator" one by one: $e^{f_k (\lambda)*e^{-\lambda}}$ Then the summation of terms of powerseries has to happen in exponent, and than we end up with : $e^{f(\lambda)*e^{-\lambda}}$ Perhaps than we can replace $f(\lambda) = f (T(z))$ and see what happens. In case $f( \lambda) = \lambda =T(z)$ we get $e^{T(z)*e^{-T(z)}=e^z$ $T(z)$ is Euler tree function. if now $z=\ln w$ then $e^{T(\ln w) * e^{-T(\ln w)} = e^{\ln w}= w$ $T(\ln w) = h ( w) * \ln w$ so $e^{h(w)*\ln w*e^{-(h(w)*\ln w) }} = w$ Just an idea. Ivars sheldonison Long Time Fellow Posts: 626 Threads: 22 Joined: Oct 2008 03/16/2009, 01:36 PM (This post was last modified: 03/16/2009, 02:13 PM by sheldonison.) Gottfried Wrote:....When rereading my article about "pascalmatrix tetrated" (*1) and playing/checking some computations, it came to my attention, that I already have a formula for sexp with fixed (integer) height parameter, but variable with the base. So feeding log(2) as parameter to the series may give 2^^3, and feeding log(3) gives then 3^^3 . In all my recent consideration the base-parameter was deeply calculated in the coefficients and the height-parameter was isolated and thus subject to change; here the height-parameter is hidden in the coefficients and the base-parameter is explicit and can be subject to change. We need simply the first column of the tetrated pascalmatrix, scale it by reciprocal factorials and use it as coefficients for the powerseries: ${b\^\^}^h = \sum_{k=0}^{\infty} \frac{{P\^\^}^h_{k,0}}{k!}\log(b)^k$ .... Gottfried, Does your equation work for values of a,p > e^(1/e)? Assuming it does, my next question would be does your equation need a $b\^\^(h+\theta(h))$ term? Below is my base conversion equation, (which is more or less the same as Jay's earlier results), where n is an integer, so sexp(n) is always well defined and $\theta(x)$ is a small 1-cyclic sinusoid transfer function, equal to zero for integers. $\theta(x)$ is important if the desired sexp function is required to have all odd derivatives have positive values for all x>-2, otherwise the 5th order derivatives showed negative values. In my post, I tried to characterize $\theta(x)$, where one base=e and the other base is a little larger than e^(1/e). $\text{sexp}_a(x + \theta(x)) = \text{ } \lim_{n \to \infty} \text{log}_a^{\circ n}(\text{sexp}_b (x + \text{slog}_b(\text{sexp}_a(n)))$ Gottfried Ultimate Fellow Posts: 757 Threads: 116 Joined: Aug 2007 03/16/2009, 02:59 PM (This post was last modified: 03/16/2009, 03:05 PM by Gottfried.) Hi sheldonison (??? How can I adress you more personally?) sheldonison Wrote:Gottfried Wrote:We need simply the first column of the tetrated pascalmatrix, scale it by reciprocal factorials and use it as coefficients for the powerseries: ${b\^\^}^h = \sum_{k=0}^{\infty} \frac{{P\^\^}^h_{k,0}}{k!}\log(b)^k$ .... Gottfried, Does your equation work for values of a,p > e^(1/e)??? No a,p in my formula ... What do you mean? Since you refer to e^(1/e) I assume you mean the base, but then you have "a,p" - how can this be a base??? Quote:Assuming it does, my next question would be does your equation need a $b\^\^(h+\theta(h))$ term? Hmm. Again ??? "need your formula..." - for what? Well, also I've the problem that I cannot go very deeply into this currently; what I've done the past days was to make some things straight (from my sketchy notes-pads) to put them out to you folks for consideration, because next days I'm absent and possibly take a complete rest for some weeks from mathematics at all (let's see). I think the description of this formula in my article is pretty clear: what it gives..., what it is for... - only I did not yet discuss range of convergence (which I also use to extend by Euler-summation ... ) There is no other term needed and the base-parameter is just involved in the form of powers of log(base) , and the formula can be used for a certain range for log(base), whose bounds should be determined. The "base-conversion" by this formula is surely not useful in general. This is, because it is only defined for integer iterates and it is much easier to compute with the integer iterates by the common function than by a powerseries... whose coefficients stem from a tetrated pascalmatrix. I could, for instance, think, that your's (and Jay's) may easily come out to be the reference-formula for this type of problem, why not. Sorry I can't write more at the moment - Regards - Gottfried Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 757 Threads: 116 Joined: Aug 2007 03/16/2009, 03:20 PM Ivars Wrote:Sorry for being little out of touch, to me "upping " things one step looks very interesting in general ( because of log-Poisson distribution): $e^{\lambda^k /k!*e^{-\lambda}$ Can we not replace $\lambda^k/k!$ by any function $f(\lambda)$ which has a powerseries expansion, as that expansion would usually contain something like $c* \lambda^k/k!$ ? Of course, coefficients at powers of $\lambda$ can be also anything else. I can only see so far that terms of powerseries expansion has to be put into "double exponentiation operator" one by one: $e^{f_k (\lambda)*e^{-\lambda}}$ Then the summation of terms of powerseries has to happen in exponent, and than we end up with : $e^{f(\lambda)*e^{-\lambda}}$ Perhaps than we can replace $f(\lambda) = f (T(z))$ and see what happens. In case $f( \lambda) = \lambda =T(z)$ we get $e^{T(z)*e^{-T(z)}=e^z$ $T(z)$ is Euler tree function. if now $z=\ln w$ then $e^{T(\ln w) * e^{-T(\ln w)} = e^{\ln w}= w$ $T(\ln w) = h ( w) * \ln w$ so $e^{h(w)*\ln w*e^{-(h(w)*\ln w) }} = w$ Just an idea. Ivars Hi Ivars - yes, "upping up" is always interesting: even if it doesn't work, it sharpens the contour of the subject. We know some thing not only by the "what it can" but also (and sometimes even more in depth) by the "what it cannot". So, let's see to what you can proceed... Cordially - Gottfried Gottfried Helms, Kassel sheldonison Long Time Fellow Posts: 626 Threads: 22 Joined: Oct 2008 03/16/2009, 11:01 PM Gottfried Wrote:Hi sheldonison (??? How can I adress you more personally?)Sheldon Quote:sheldonison Wrote:Gottfried Wrote:We need simply the first column of the tetrated pascalmatrix, scale it by reciprocal factorials and use it as coefficients for the powerseries: ${b\^\^}^h = \sum_{k=0}^{\infty} \frac{{P\^\^}^h_{k,0}}{k!}\log(b)^k$ .... Gottfried, Does your equation work for values of a,p > e^(1/e)??? No a,p in my formula ... What do you mean? Since you refer to e^(1/e) I assume you mean the base, but then you have "a,p" - how can this be a base??? typo, I meant b, p, the two bases in your equation. Quote:Quote:Assuming it does, my next question would be does your equation need a $b\^\^(h+\theta(h))$ term? Hmm. Again ??? "need your formula..." - for what? .... The "base-conversion" by this formula is surely not useful in general. This is, because it is only defined for integer iterates ... Regards - Gottfried The $b\^\^(h+\theta(h))$ term is only for real values, and only to meet the uniqueness criterion that the odd derivatives are all positive. Gottfried Ultimate Fellow Posts: 757 Threads: 116 Joined: Aug 2007 03/17/2009, 12:12 PM (This post was last modified: 03/17/2009, 12:32 PM by Gottfried.) Hi Sheldon - I'm very sorry... I think I've expressed myself misleading. sheldonison Wrote:Gottfried Wrote:${b\^\^}^h = \sum_{k=0}^{\infty} \frac{{P\^\^}^h_{k,0}}{k!}\log(b)^k$ .... The meaning of this formula,..., it does not relate two different bases. Only it gives a series for an iterate at a certain height and has the base-parameter b explicit (or: isolated) so we can work with it: extract it(possibly), build sums/series of powertowers of different bases, replace one base-parameter by another one. So - although this is working with the base parameter, it is surely not what we are discussing with the focus on "change-of-base"-formula. Sorry for mixing this up. (So also the question about the theta is inapplicable here.) Btw, the capital P is simply the name for the Pascalmatrix, and P^^h with indexes means the entries of the first column of the h'th tetrated pascalmatrix, so the series gives just a(nother) definition of b^^h Regards - Gottfried Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 757 Threads: 116 Joined: Aug 2007 03/17/2009, 06:36 PM (This post was last modified: 12/30/2017, 09:20 PM by Gottfried.) Hmm, curious... curious... , things evolve a bit more. From the tetrated pascalmatrices we can also derive a fractional iteration if we apply the binomial-formula to the sequence of powerseries for different heights. Since the powerseries have this stunning smooth behave for increasing heights, the binomially interpolated series are accordingly smooth. This seems all to be too crazy.... very smooth for the convergence bases and possibly continuable for the divergent cases, don't see it yet. See the new chapter 2 in PascalmatrixTetrated Gottfried Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 757 Threads: 116 Joined: Aug 2007 03/18/2009, 07:26 PM (This post was last modified: 03/18/2009, 08:23 PM by Gottfried.) Here I present three postings in sci.math. It seems, that the method is not well suited for the interpolation to fractional heights (as I hoped it would be). But - perhaps we can find a workaround. On the other hand: it is not needed that many different methods exist, so... Also Ioannis (Galidakis) reminded me of the entry in mathworld,"powertower", where he already characterized this type of series. (http://mathworld.wolfram.com/PowerTower.html) Here the current msgs aus sci.math: ( some edits in double-brackets [< >]) . subject: tetration: another family of powerseries for fractional iteration Code:Maybe this is all known; I didn't see it so far. The idea was triggered by the comments of V Jovovic in the OEIS concerning the below generating functions. Consider the sequence of functions T0(x) = 1, T1(x) = exp(x*1), T2(x) = exp(x*exp(x)), T_h(x) = exp(x*T_{h-1}(x)),... They are also the generation-functions for the following sequence of powerseries: T0:  1 + 0 +   0   + .... T1:  1 + x + 1/2*x^2 +  1/6*x^3 +   1/24*x^4 +    1/120*x^5 +     1/720*x^6 +      1/5040*x^7 ... T2:  1 + x + 3/2*x^2 + 10/6*x^3 +  41/24*x^4 +  196/120*x^5 +  1057/720*x^6 +   6322/5040*x^7 +... T3:  1 + x + 3/2*x^2 + 16/6*x^3 + 101/24*x^4 +  756/120*x^5 +  6607/720*x^6 + 160504/5040*x^7 + ... T4:  1 + x + 3/2*x^2 + 16/6*x^3 + 125/24*x^4 + 1176/120*x^5 + 12847/720*x^6 + 229384/5040*x^7 + ... T5:  1 + x + 3/2*x^2 + 16/6*x^3 + 125/24*x^4 + 1296/120*x^5 + 16087/720*x^6 + 257104/5040*x^7 + ... T6:  1 + x + 3/2*x^2 + 16/6*x^3 + 125/24*x^4 + 1296/120*x^5 + 16807/720*x^6 + 262144/5040*x^7 + ... T7:  1 + x + 3/2*x^2 + 16/6*x^3 + 125/24*x^4 + 1296/120*x^5 + 16807/720*x^6 + 262144/5040*x^7 + ... T8:  1 + x + 3/2*x^2 + 16/6*x^3 + 125/24*x^4 + 1296/120*x^5 + 16807/720*x^6 + 262144/5040*x^7 + ... T9:  1 + x + 3/2*x^2 + 16/6*x^3 + 125/24*x^4 + 1296/120*x^5 + 16807/720*x^6 + 262144/5040*x^7 + ... ... Too: 1 + x + 3/2*x^2 + 4^2/3!*x^3 + 5^3/4!*x^4 + 6^4/5!*x^5 +    7^5/6!*x^6 +      8^6/7!*x^7 + ... //limit h->inf That means, if x = log(b), we have by this T0(x) = 1 T1(x) = b      = b^^1 T2(x) = b^b    = b^^2 T3(x) = b^b^b  = b^^3 ... Too(x) = ...^b^b   = b^^oo and for the limit h->inf we have with Too(x) the series for the h-function of b: Too(x) = h(b) which is convergent for |x|] The 2.nd msg: Code:> > (Galidakis replies) : > >                          However, the recursive expression for the coefficients > > given in (6) [] does not seem to allow that. > > > > If you can find a way to interpolate between those coefficients for non-natural > > heights using your matrix method AND at the same time you manage to preserve the > > functional equation F(x + 1) = e^{x*F(x)}, then, by Jove, you've got a nice > > analytic solution to tetration :-) Ok, let's give a start. Recall: T0:  1 + 0 +   0   + .... T1:  1 + x + 1/2*x^2 +  1/6*x^3 +   1/24*x^4 +    1/120*x^5 +     1/720*x^6 +      1/5040*x^7 ... T2:  1 + x + 3/2*x^2 + 10/6*x^3 +  41/24*x^4 +  196/120*x^5 +  1057/720*x^6 +   6322/5040*x^7 +... T3:  1 + x + 3/2*x^2 + 16/6*x^3 + 101/24*x^4 +  756/120*x^5 +  6607/720*x^6 + 160504/5040*x^7 + ... T4:  1 + x + 3/2*x^2 + 16/6*x^3 + 125/24*x^4 + 1176/120*x^5 + 12847/720*x^6 + 229384/5040*x^7 + ... T5:  1 + x + 3/2*x^2 + 16/6*x^3 + 125/24*x^4 + 1296/120*x^5 + 16087/720*x^6 + 257104/5040*x^7 + ... T6:  1 + x + 3/2*x^2 + 16/6*x^3 + 125/24*x^4 + 1296/120*x^5 + 16807/720*x^6 + 262144/5040*x^7 + ... T7:  1 + x + 3/2*x^2 + 16/6*x^3 + 125/24*x^4 + 1296/120*x^5 + 16807/720*x^6 + 262144/5040*x^7 + ... T8:  1 + x + 3/2*x^2 + 16/6*x^3 + 125/24*x^4 + 1296/120*x^5 + 16807/720*x^6 + 262144/5040*x^7 + ... T9:  1 + x + 3/2*x^2 + 16/6*x^3 + 125/24*x^4 + 1296/120*x^5 + 16807/720*x^6 + 262144/5040*x^7 + ... ... Too: 1 + x + 3/2*x^2 + 4^2/3!*x^3 + 5^3/4!*x^4 + 6^4/5!*x^5 +    7^5/6!*x^6 +      8^6/7!*x^7 + ... //limit h->inf We want to interpolate for coefficients of T0.5, means between rows T0 and T1. I'll rewrite the powerseries without the powers of x. And since we do the binomial composition of coefficients at like powers of x, we compose the coefficients down a column; so the common denominator(the factorial) of a column can be omitted for the scheme. Thus I get for the original coefficients, only rescaled   T0:   1  0  0   0    0     0      0       0   ...   T1:   1  1  1   1    1     1      1       1   ...   T2:   1  1  3  10   41   196   1057    6322   T3:   1  1  3  16  101   756   6607   65794   T4:   1  1  3  16  125  1176  12847  160504   T5:   1  1  3  16  125  1296  16087  229384   T6:   1  1  3  16  125  1296  16807  257104   T7:   1  1  3  16  125  1296  16807  262144   T8:   1  1  3  16  125  1296  16807  262144   T9:   1  1  3  16  125  1296  16807  262144     ... The first binomial-composition along the columns gives   X0:   1   0    0    0     0      0       0        0 ...   X1:   0   1    1    1     1      1       1        1 ...   X2:   0  -1    1    8    39    194    1055     6320   X3:   0   1   -3  -11   -19    171    3439    46831   X4:   0  -1    5    8   -37   -676   -7243   -64744   X5:   0   1   -7    1   105   1021    7357    21589   X6:   0  -1    9  -16  -161  -1026   -3301    67304   X7:   0   1  -11   37   181    631   -3605  -168125   X8:   0  -1   13  -64  -141    104   10961   246224   X9:   0   1  -15   97    17   -999  -16007  -278711   ...       ... The second binomial-composition (using h=0.5) [< Table 5: this will be the reference-table for the composition of coefficients of T05 >]   Y0:   1          0             0            0            0              0                0  ...   Y1:   0        1/2           1/2          1/2          1/2            1/2              1/2  ...   Y2:   0        1/8          -1/8           -1        -39/8          -97/4          -1055/8   Y3:   0       1/16         -3/16       -11/16       -19/16         171/16          3439/16   Y4:   0      5/128       -25/128        -5/16      185/128         845/32        36215/128   Y5:   0      7/256       -49/256        7/256      735/256       7147/256        51499/256   Y6:   0    21/1024     -189/1024        21/64    3381/1024      10773/512       69321/1024   Y7:   0    33/2048     -363/2048    1221/2048    5973/2048     20823/2048     -118965/2048   Y8:   0  429/32768   -5577/32768      429/512  60489/32768     -5577/4096   -4702269/32768   Y9:   0  715/65536  -10725/65536  69355/65536  12155/65536  -714285/65536  -11445005/65536  ...      ...      ... ---------------------------------------------------------------------------------------------------- sum.    s0     s1             s2           s3     ... ==================================================================================================== T0.5:   c0     c1             c2           c3     ... and                T0.5(x) = c0 + c1*x + c2*x^2/2! + c3*x^/3! + ... the interpolated coefficients c0,c1,c2,... for h=0.5 should then be computed by the column-sums (and finally the rescaling by the omitted factorials). The partial sums in the columns converge only badly if at all, so let's look, whether we can find some analytic solution. The denominators in the rows can be majorized by powers of 4, and all can then be divided by 2, so let's rewrite this                                                                                                        common scaling   Y0:   1/2     0       0       0      0         0          0           0            0             0   *2 /4^0   Y1:     0     1       1       1      1         1          1           1            1             1   *2 /4^1   Y2:     0     1      -1      -8    -39      -194      -1055       -6320       -41391       -293606   *2 /4^2   Y3:     0     2      -6     -22    -38       342       6878       93662      1219314      16331654   *2 /4^3   Y4:     0     5     -25     -40    185      3380      36215      323720      2128445      -5199340   *2 /4^4   Y5:     0    14     -98      14   1470     14294     102998      302246     -9722034    -332756410   *2 /4^5   Y6:     0    42    -378     672   6762     43092     138642    -2826768    -93176118   -1954258068   *2 /4^6   Y7:     0   132   -1452    4884  23892     83292    -475860   -22192500   -463551132   -7659247332   *2 /4^7   Y8:     0   429   -5577   27456  60489    -44616   -4702269  -105630096  -1778712507  -23047084632   *2 /4^8   Y9:     0  1430  -21450  138710  24310  -1428570  -22890010  -398556730  -5760084330  -51266562490   *2 /4^9      ...      ... ---------------------------------------------------------------------------------------------------- sum.    s0     s1     s2    s3     ... ==================================================================================================== T0.5:   c0     c1     c2    c3     ... and                T0.5(x) = c0 + c1*x + c2*x^2/2! + c3*x^/3! + ... Let's look at the columnsums of the table; that sums, divided by the factorial, give the coefficients c_k for the T0.5(x)-powerseries. First, s0 = 1, (remember the scaling extracted to the rhs) ,     so c0 = 1 Next, s1. Here we recognize, that the numbers are the catalan-numbers, and, with the current scaling have the generation-function  1- sqrt(1-z). Since we want to know simply the sum, we set z=1 and get for the sum         s1 = 1- sqrt(1-1) = 1 so     c1 =1 Next, s2. It becomes more difficult. We can add columns s2 and s1 to get a sequence, which can formally be expressed as the derivative of the sqrt(1 - z)-function, where possibly we need also a scaling at z, so likely something like   1 - sqrt(1 - a z)' It looks, as if the series is divergent, too, so we'll have to see, whether this operation (and the following, which surely are similar) can be justified/make sense at all. ----------------- I proceeded for the first few terms s2,s3,s4,s5... Things seem to come out uneasy... Code:(msg 3) (...) Formally composed by derivatives of sqrt(1-z) I get for the series s1,s2,s3,... the following generating functions s0:    1 s1:    1 -    1*sqrt(1-z) s2:    3 -    3*sqrt(1-z) +    2*z*(sqrt(1-z)') s3:   16 -   16*sqrt(1-z) +   15*z*(sqrt(1-z)') -   3*z^2*(sqrt(1-z)'') s4:  125 -  125*sqrt(1-z) +  124*z*(sqrt(1-z)') -  42*z^2*(sqrt(1-z)'') +  4*z^3*(sqrt(1-z)''') s5: 1296 - 1296*sqrt(1-z) + 1295*z*(sqrt(1-z)') - 550*z^2*(sqrt(1-z)'') + 90*z^3*(sqrt(1-z)''')  - 5*z^4*(sqrt(1-z)'''') ... which have to be evaluated at z=1 to give the value for the sums. Now the derivatives have a vertical asymptote at z=1, so here are infinities everywhere... Even more obvious, if I expand the derivatives into terms of sqrt(1-z) I get the following explicite generating functions for the series of s0,s1,s2,...: s0: 1 s1: 1     - sqrt(1-z)        * (   1 ) s2: 3     - sqrt(1-z)/(1-z)^1* (   3 -    4/2*z) s3: 16    - sqrt(1-z)/(1-z)^2* (  16 -   49/2*z +    31/4*z^2 ) s4: 125   - sqrt(1-z)/(1-z)^3* ( 125 -  626/2*z +   962/4*z^2 -   408/8*z^3 ) s5: 1296  - sqrt(1-z)/(1-z)^4* (1296 - 9073/2*z + 22784/4*z^2 - 23462/8*z^3 +  7561/16*z^4) where all except the first two grow unboundedly, if z->1 So for that approach: it looks as if we cannot express a half-iterate based on this type of powerseries. Pity.... Maybe we can find a workaround - change order of summation or something else, don't have an idea. Another idea around? Or: can we work differently with the binomial-compositions (I've only basic understanding/knowledge of the generation-function-concept, for instance...) Gottfried Helms, Kassel « Next Oldest | Next Newest »

 Possibly Related Threads... Thread Author Replies Views Last Post Approximation to half-iterate by high indexed natural iterates (base on ShlThrb) Gottfried 1 151 09/09/2019, 10:50 PM Last Post: tommy1729 [MSE] Shape of orbit of iterations with base b on Shell-Thron-region Gottfried 3 284 08/18/2019, 02:22 PM Last Post: sheldonison Complex Tetration, to base exp(1/e) Ember Edison 7 936 08/14/2019, 09:15 AM Last Post: sheldonison b^b^x with base 0

Users browsing this thread: 1 Guest(s)