# Tetration Forum

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Gottfried Wrote:[quote=Ivars]
[quote=Gottfried]
Hmm, for 1+I-1-I+1+I-1-I+.... I get 1/(1-I) = 1/2 + 1/2I by the formula just invoking Pari/GP on that, and also 1/2 + 1/2I by conventional Euler-summation.
0 + Ix +0x^2 - Ix^3 + 0 x^4 + I x^5 .... = I-I+I-I+..= Ix *(1/(1+x^2)) -> 1/2 I
or
I - Ix + Ix^2 - Ix^3 .... = I*(1/(1+x)) -> 1/2 I

Gottfried

That seems logical. Thanks, I forgot the trick with generating series where x->1. Instead, and BTW, why can't it go to I as well (x->I) ? Certainly putting x=I will also lead to some sort of summation, albeit slightly different generating series. E.g.

0 + x -0x^2 + x^3 - 0 x^4 + x^5 .... = I-I+I-I+.. =x* (1/(1-x^2)) ->x=I-> I*(1/(1+1))= 1/2* I

But also I/2= sin(I*ln(GoldenMean)) = sin(I*ln(1+0.618033..)= sin (I*( 1+0.618+0.618^2/2+0.618^3/3+0.618^4/4+...+0.618^n/n) = sin (I+I*0.618+I/2*0.618^2+I/3*0.618^3+..+I/n*0.618^n+..) -> plus extend sinus in series.

See here Formula 26.

while I try to understand increasing height series, what would be Your take on harmonic imaginary series Sum : I/n? Or easier case, I/n^2?

I was similarly(?) to You wandering that perhaps we need infinite product power series to evaluate tetration ( because of its speed)?

so b[4]n= b^b^b^b... =(a1*b)^1*(a2*b)^2*(a3* b)^3*(a4* b)^4*...*(an*b)^n = ?= c1*b^1+c^2*b^b+c^3*b^b^b+..?

You probably have a more clear view on that.

Ivars
Ivars Wrote:while I try to understand increasing height series...

I think, that is not too difficult. Just take a base, where the trajectory converges (for instance b=sqrt(2)) and alternate sum the points of the trajectory. (In pari/gp there is the sumalt()-function).
The problem is then only, if b>e^(1/e). Then the trajectory begins to strongly diverge - but for a certain range of b you may still be able to apply one common summation-method, for instance the Shanks-method.

But while even this method allows bases a little bit greater, the diagonalization method seems to be able to sum that to arbitrary bases...
:cool:

Best is, you try with convergent example - just add the points of your trajectory in the list-matrix with alternating signs (even Cesaro-sum should be sufficient for this)
Quote:I was similarly(?) to You wandering that perhaps we need infinite product power series to evaluate tetration ( because of its speed)?

Yes, sometimes I was speculating about this. But I doubt it is of any use... Infinite product-series can be evaluated by the additive series of their logarithms. In the context of tetration this means, if we get an improvement at all (by this reformulation) then it is only occuring for height 1. For greater heights we are still lost. So everytime this idea pops up in my head, I just dismiss it...

Gottfried
Gottfried Wrote:Infinite product-series can be evaluated by the additive series of their logarithms. In the context of tetration this means, if we get an improvement at all (by this reformulation) then it is only occuring for height 1. For greater heights we are still lost. So everytime this idea pops up in my head, I just dismiss it...

Gottfried

Hmm. But if we have tetration of $h(e^z)$ , then we can continue taking logarithms forever, reducing all heights? No.
Then my beloved $h(e^{\pi/2}) = h(I^{1/I})$ just ends where everything else:

$\ln(\ln(\ln(....h(e^{\pi/2})) = \ln(\ln(\ln(....\pm..I))=0.318131505... \pm 1.337235701...I=-W(-1)$(both branches), where $W(z)$ is Lambert W function.

$-W(-1)= \sum_{n=1}^{\infty} \frac{n^{(n-1)}} {n!}$

So would than divergent products (and even some sums) similarly to divergent tetration end up in complex plane?

Ivars
If we continiue like this, then we can check:
tetration of $h(e^{-1})=\Omega$

Here we have $\ln(\Omega)=- \Omega.$ : Interestingly, the same logarithms appeared here where I changed the issue by always using module of ln(x):

Iterations of ln(mod(x))

Anyway, if we take logarithms of negative numbers via complex numbers, we again end at $-W(-1)$:

Everything ends at that point. But also infinitely iterated logarithm of any number AT ALL if allowed to go into complex number when it first gets negative seems to end at this point(s). Is it true, or just oscillations die out so fast that all numbers give the same result... Anyway, why exactly close to -W(-1) which reminds me of Eulers Lambert function -W(-z)? His work about it is probably only in Latin.

Jaydfox reffered to these 2 fixed points here:Imaginary Iterates of exponentiation

This number $=0.318131505... \pm 1.337235701...I) = -W(-1)$ and $\Omega=0.567143 =W(1)$ are kind of symmetric since they are solutions of equations:

$e^z=z$ and
$e^{(-z)} = z$ respectively.

Also,
$-W(-1)^{1/-W(-1)}=e$

$W(1)^{1/W(1)}={1/e}$

Ivars
Gottfried Wrote:Refering to the header of this thread and your obvious love to iteration and trajectories, I challenge you to compute the alternating sum of all the intermediate steps of the trajectory
I, 2^I, 2^(2^I), 2^(2^(2^I)...

AS(2,I) = I - 2^I + 2^2^I - 2^2^2^I + ... -... = ??

My proposal is

AS(2,I) = -0.440033096027+0.928380628227*I

What do you think?

Gottfried

I can only say it diverges, and I understand You want me to find a method to sum it and a result. Unfortunately, I can not do it yet, but I am interested in 2^I in general and divergent series as well as they might be doing the same as tetration-somehow sending real numbers to complex domain , though the example I made with infinite logarithms ended in a kind of disaster as every base b ( except h(0) and h(1)) atracted to single value -W(-1). So let us keep Your result in mind while learn the basics in my strange ways-backwards

Ivars
Ivars Wrote:
Gottfried Wrote:AS(2,I) = I - 2^I + 2^2^I - 2^2^2^I + ... -... = ??

My proposal is

AS(2,I) = -0.440033096027+0.928380628227*I

What do you think?

Gottfried

I can only say it diverges, and I understand You want me to find a method to sum it and a result. (...)
Ivars

Well, diagonalization seems to give an answer here.

Remember, in my matrix-notation, we have

$\hspace{24} V(x)\sim * B_b = V(b^x)\sim$

or more general

$\hspace{24} V(T_b^{{o}h}(x)) )\sim * B_b = V(T_b^{{o}h+1}(x)) )\sim$

and using digonalization

$\hspace{24} B_b^h = W *^dV(u^h) * W^{-1}$

Now the alternating sum is (using the small letter "i" for the imaginary unit to prevent confusion with "I" as the identity matrix)

$\hspace{24} AS(2,i) = V(i)\sim * B_b^0 - V(I)\sim *B_b^1 + V(I)\sim*B_b^2 - ... + ...$

$\hspace{24} = V(i)\sim*(B_b^0 - B_b^1 + B_b^2 - ... +...)$

$\hspace{24} = V(i)\sim*W *(^dV(u^0)-^dV(u^1)+^dV(u^2)-...+...)*W^{-1}
$

$\hspace{24} = V(i)\sim*W *(X)*W^{-1}
$

where we have to determine the entries of the diagonal-matrix X. But they can all be determined by the rule of geometric series:

$\hspace{24} X[0] = (u^0)^0 - (u^1)^0+(u^2)^0 ... = 1/(1+u^0) = 1/(1+1)=1/2$

$\hspace{24} X[1] = (u^0)^1 - (u^1)^1+(u^2)^1 ... = 1/(1+u^1)$

$\hspace{24} X[2] = (u^0)^2 - (u^1)^2+(u^2)^2 ... = 1/(1+u^2)$
...
(Note, that we need no fractional powers, so the multiplications in the exponents are commutative and can be reordered)
Then
$\hspace{24} X = diag([1/(1+u^0),1/(1+u^1),1/(1+u^2),...])$

which can be described by

$\hspace{24} X = (I +^dV(u))^{-1}$ // "I" means here the identity-matrix

so the eigenvalues of

$\hspace{24} AS_b = W * X * W^{-1}$

are not a vandermonde-vector, btw.

In most short form we may simply write, again invoking the identities of diagonalization

$\hspace{24} AS(2,i) = V(i)\sim *\(I + B_b)^{-1} [,1]$ // [,1] means: use column 1 (the second) only for coefficients

The nice aspect now is, that even for b>e^(1/e), or abs(u)>1, which leads to divergent trajectories, the eigenvalues of the matrix (I+Bb)^-1 form a convergent sequence (well, I've to check for the precise range) and AS() gives a reasonable result.

I've given a plot for some bases, where x=1 (instead of I as in the case here), which compares the (cesaro-and) Euler-summable bases and the Shanks-summable bases with values of the diagonalization-summable bases, where the diagonalization extends the summablity to arbitrary high bases. (Note: this is an old picture, for instance I used "s" for base and "matrix-method" for "diagonalization")
[attachment=77]
The result looks pretty smooth...

Because of the nice eigenvalue-configuration, you may arrive at these results even without fixpoint-shifts and expression by the Ut-matrix; simply invert -using an empirical Bb-matrix- I+Bb of reasonable size, say 32x32 or 64x64, to get good approximations.

Gottfried
I corrected the mistakes of taking logarithms from h(z) in previous posts.

But one could still ask if there exists transformation H such that :

$H(h(z))= z*z*z*z*z*z*z..........$

Then

$ln(H(z)) = z+z+z+z+z+z.........$ would always be a divergent sum, related to h(z). Instead of all z being similar, one can have e.g z=2^n or other general term, then these sums will coincide with usually known divergent sums like 1+2+4+8+16+32........ or 1+1+1+1+1+1....

Similarly 1/h(z) could be transformed in z/z/z/z/z/z/z.......... and sums would be -z-z-z-z-z-z-...?

Ivars

regards

tommy1729
Tommy1729, thank you for bringing this thread back to our/to my attention!

With my current knowledge I'm pretty sure, that my challenge for the alternating sum at the time of the original post was most likely false; a (still only "most likely") more meaningful version uses now the "polynomial method" for the tetration for bases outside the Shell-Thron-region, thus for instance of all real bases $b \gt \eta \approx 1.444$.

The results using this method gives the following table for some bases b and the argument x=I :
Code:
_   base           log(base)           AS(base,I)                   1.1000000  0.095310180   -0.45069485+0.91385251*I   1.2000000   0.18232156   -0.40503707+0.85093158*I   1.3000000   0.26236426   -0.36504196+0.80486309*I   1.4000000   0.33647224   -0.33093412+0.77025573*I   1.5000000   0.40546511   -0.30187962+0.74327224*I   1.6000000   0.47000363   -0.27685754+0.72148441*I   1.7000000   0.53062825   -0.25501392+0.70337461*I   1.8000000   0.58778666   -0.23569667+0.68796239*I   1.9000000   0.64185389   -0.21841589+0.67458888*I   2.0000000   0.69314718   -0.20280135+0.66279493*I   2.1000000   0.74193734   -0.18856952+0.65225041*I   2.2000000   0.78845736   -0.17550012+0.64271115*I   2.3000000   0.83290912   -0.16341960+0.63399285*I   2.4000000   0.87546874   -0.15218883+0.62595235*I   2.5000000   0.91629073   -0.14169584+0.61847841*I   2.6000000   0.95551145   -0.13184894+0.61147999*I   2.7000000   0.99325177   -0.12257089+0.60488549*I   2.8000000    1.0296194   -0.11380069+0.59863552*I   2.9000000    1.0647107   -0.10548489+0.59267760*I   3.0000000    1.0986123  -0.097576064+0.58697104*I   3.1000000    1.1314021  -0.090038424+0.58148334*I   3.2000000    1.1631508  -0.082843993+0.57618212*I   3.3000000    1.1939225  -0.075964723+0.57103646*I   3.4000000    1.2237754  -0.069372706+0.56602343*I
from where my challenge should have been:
Code:
AS(2,I) =  -0.20280135+0.66279493*I

I haven't (in the year of the above posts) been aware of the problem of the attracting and repelling fixpoints and the sensitivity of the "Alternating Series with increasing heights" (which I call now "alternating iteration series") around which fixpoint the series is developed. Assuming that the "polynomial method" for tetration approximates (and generalizes) the Kneser-solution and also assuming that the Kneser-solution is the best for real tetration the AS()-function should be based on that "polynomial method".

The Pari/GP-procedures are even simpler than that of the original posting;
Code:
n=32  \\ constant gives dimension for matrices default(realprecision,200)  \\ internal real-arithmetic computation uses 200 digits default(format,"g0.12") \\ show 12 digits in user-interface {ASinit(b)=local(a);   \\ define the matrix and coefficients for powerseries   a=log(b);   B = matrix(n,n,r,c, (a*(c-1))^(r-1)/(r-1)!);  \\ the carlemanmatrix   C = matsolve(matid(n)+B,matid(n)[,2]);    \\ vector of coefficients   return("ok");} AS(x) = sum(0,n-1, x^k*C[1+k])  \\ the value for the (truncated) and approximate powerseries AS(base,x)

Note, that the coefficients in C form the (head of a) powerseries, which seems to have nonzero (and surprisingly interesting) range of convergence even if base b is outside of the Shell/Thron (resp Euler)-range!

After that, do the computation:
Code:
base=2 ASinit(base) %486 = "ok" AS( 1 )      \\ consider only x for which the truncated powerseries is convergent %487 = 0.28740870 AS( I )   \\ reproduce the new result (if n is at least 32  ) %488 = -0.20280135 + 0.66279493*I

Note finally that for bases inside the Euler-interval one can use the builtin Pari/GP-function "sumalt" to simply do a summation for the (then conditionally converging) alternating iteration series and use this to crosscheck the AS(x)-evaluation taken by the "polynomial method" for tetration.

Gottfried
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