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A tiny base-dependent formula for tetration (change-of-base?) - Gottfried - 03/14/2009
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/PascalMatrixTetrated.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:
ยด
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 2003RE: A tiny base-dependent formula for tetration (change-of-base?) - Ivars - 03/15/2009
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
RE: A tiny base-dependent formula for tetration (change-of-base?) - sheldonison - 03/16/2009
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).
\( \text{sexp}_a(x + \theta(x)) =
\text{ } \lim_{n \to \infty}
\text{log}_a^{\circ n}(\text{sexp}_b (x + \text{slog}_b(\text{sexp}_a(n))) \)
RE: A tiny base-dependent formula for tetration (change-of-base?) - Gottfried - 03/16/2009
Hi sheldonison (??? How can I adress you more personally?)
sheldonison Wrote:?? 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???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)?
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
RE: A tiny base-dependent formula for tetration (change-of-base?) - Gottfried - 03/16/2009
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
RE: A tiny base-dependent formula for tetration (change-of-base?) - sheldonison - 03/16/2009
Gottfried Wrote:Hi sheldonison (??? How can I adress you more personally?)Sheldon
Quote:sheldonison Wrote:?? 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???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)?
typo, I meant b, p, the two bases in your equation.
Quote: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.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
RE: A tiny base-dependent formula for tetration (change-of-base?) - Gottfried - 03/17/2009
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
RE: A tiny base-dependent formula for tetration (change-of-base?) - Gottfried - 03/17/2009
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
RE: A tiny base-dependent formula for tetration (change-of-base?) - Gottfried - 03/18/2009
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|<exp(-1)
[<...>]The 2.nd msg:
Code:
> > (Galidakis replies) :
> > However, the recursive expression for the coefficients
> > given in (6) [<in mathworld, G.H.>] 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...)