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comment Applying Collatz function iterations to large integers
The given number (with the term $\small 3^{666669}$) can be handled in Pari/GP, it has about 318095 decimal digits - and is internal in binary representation. From this we need (in principle) only remove the trailing binary zeros to compute the next iterate. The first and last digits of $\small {375... 685\cdot 3^{666669}-1 \over 2}$ are $\small 167321202936474 .... 4706927$
11h
comment An obvious pattern to $i\uparrow\uparrow n$ that is eluding us all?
I tried a similar picture for the fixpoint with which you work in your answer ($^\infty i$) - unfortunately there are regions of complex starting values where the iteration escapes (at least temporarily) to gigantic values and I don't know whether from that regions they even might diverge to infinity. The blue colored area in the picture go.helms-net.de/math/tetdocs/bilder/BATBaseI_3Exp.png indicate initial values where the iteration goes towards the fixpoint and from the grey area it is unknown to me so far where iterations go. Some examples looked at individually manually converged.
11h
comment An obvious pattern to $i\uparrow\uparrow n$ that is eluding us all?
Just to avoid misunderstanding: the three additional periodic points do not occur because of using a different branch in the logarithm. While the above fixpoint $ ^\infty i$ is approached by iterated exponentiation, the other fixpoint (resp the periodic points) are approached by repeated logarithmizing (always the principal value was taken!) So in short $ ^{- \infty} i $ arrives at $-1.86-0.41$ when started from a value in the green area and one of the periodic points when started from the red,magenta or seagreen area in the picture go.helms-net.de/math/tetdocs/bilder/BATBaseI_3Log.png
1d
comment Applying Collatz function iterations to large integers
You might find it interesting that not only numbers of the form $2^k \cdot p -1$ (with sufficient high $k$) have this feature to increase to the form $3^k \cdot p -1$ but also numbers of the form $2^k \cdot p -5$ and/or $2^k \cdot p -7$ go to $3^k \cdot p -5$ and/or $3^k \cdot p -7$ and similarly one more of these nice patterns can be described with the initial residue $-17$ : the reason is, that in the negative integers, $-1$ forms a one-step-loop, $-5,-7$ a two-step-loop and $-17,...$ a 7-step loop (the latter is from the top of my head, no guarantee for the number of steps until cycling)
1d
comment An obvious pattern to $i\uparrow\uparrow n$ that is eluding us all?
We have two simple fixpoints: 0.4383+0.3606i (attracting for i^z) , -1.86-0.41i (attracting for $\log _i(z)$) , and three periodic points : -1.14 + 0.71i, 1.65-0.19i, -0.07 - 0.32i (attracting for $\log _i(z)$)
1d
comment An obvious pattern to $i\uparrow\uparrow n$ that is eluding us all?
Hi Sheldon - thanks, that is an interesting information for me (possibly I'll start understanding that method once :-) ) . As the question of fixpoints are of relevance now I've tried to look systematically at the fixpoints to base (i) and observe, that we do not only have on simple attracting fixpoint for the log but also a set of 3 periodic points attracting for the log. Is such a property in any way relevant for the (generalized) Kneser-ansatz? (See a picture in the tetration-forum math.eretrandre.org/tetrationforum/… )
2d
comment How to do a regression which includes reciprocals?
Hi Claude! Very nice - I had given up because of an insufficient test-environment for simultanously adapt four parameters... I'll see what this gives to me, thanks so far!
2d
comment What the minimum of infinite tetration divided by $\sqrt{x}$?
The key, why $x=\sqrt{2}$ works is not because it is a root, but an exponential of the form $x=b^{1\over b}$ at $b=2$. Thus a better generalization is for instance $ ^3 \sqrt{3}$ etc. After that try to find the number $b$ where $x=2 = b^{1 \over b}$ and look what happens when you evaluate x^b,x^x^b,x^x^x^b,... resp 2^b,2^2^b, 2^2^2^b,...
Feb
5
comment How to do a regression which includes reciprocals?
Step 1: Yes, I've done the same regression and got the same/a similar result. Step 2: But after that I looked at the residuals and noticed, that they have a trend to become absolutely smaller. So I considered to do a further improvement to find a characteristic of that. Looked at the residuals and at their reciprocals - observing/guessing that there is another linear trend. Thus the residuals of Step 1(as of your finding) are in the denominator of my formula and I tried -as a higher order approximation- to get something useful by removing that trend. That's the reason for my complete formula.
Feb
5
comment How to do a regression which includes reciprocals?
I think now, that a simple answer is not possible since it is a non-linear regression-problem and I've to finetune any final approximation manually. So I just want to say thank you very much for all your help!
Feb
5
comment How to do a regression which includes reciprocals?
I added a link to a *.csv-file in my answer which contains the sample of data
Feb
5
comment How to do a regression which includes reciprocals?
@Yves - well actually I used that linear interpolation as rough approximation (as well as a more sophisticated formula, but that's complete different path, see the new link in my updated answer). I just saw room for improvement and tried to make sense of the above regression-model. My problem is simply to derive the solution for the parameters for a regression, where the data-to-be-fit (resp. the intended residuals) are in the denominator of the regression-formula which is quite unusual for me. I thought there might be some (simple) canonical answer for such a question.
Feb
5
comment How to do a regression which includes reciprocals?
Hmm, the rationale of the "regression" is the minimization of the error of prognosis in the sense of the least (sum-of) squares of the residuals. Now I'm missing the "squares" and the "sum" and the "minimum" in your first equation where you equal the lhs to zero. Speaking of "minimization" the derivative of a function should become zero (and the second derivative should become positive), not the function itself. Did I misread something in your ansatz?
Feb
5
comment How to do a regression which includes reciprocals?
... was an instrument to uncover the inappropriateness of the polynomial interpolation at all. I think the same applies to any method - even with this regression-model.
Feb
5
comment How to do a regression which includes reciprocals?
@Yves: the splitting is done to have some sanity check of the procedure. For how many current data-points I'll do any optimizaton, there will be infinitely many more for which that optimization must be sensical. The problem is that an estimation will always face the problem that I have to extrapolate to hiher indexes. So this splitting shall give some indication, whether the found optimization can be meaningful at all. Consider the polynomial interpolation. I did it with first forty points and got exact fit. But extrapolated to further points showed catastrophic error, so this splitting...
Feb
5
comment How to do a regression which includes reciprocals?
@Yves : I've put the data into an answer-box to not to clutter the question too much.
Feb
2
comment Can all series in the elementary Ramanujan class R = 2 be shifted?
Hmm, so we get $-1/2+5/16=-3/16$?
Feb
2
comment Can all series in the elementary Ramanujan class R = 2 be shifted?
I'm still surprised about your result for the second series ($1-2+4-6+...$ which in an earlier question of mine was constructed by the partial sums of $1-3+6-10+15-...$) and which has from there a generating polynomial as $ {1\over(1+x)^3 (1-x) } $ (You gave already there that answer of $1/4$). The Cauchy principal value is ${ 3\over 16 }$ and as I understood, the Cauchy principal value is in the examples where I was confronted with it the "best"/the "logical" representation for the divergent expressions (for instance with $\zeta(1)$) . So why should this be/is this different here?
Feb
1
comment Can we assign a value to the sum of the reciprocals of the natural numbers?
@Jaume , (at) Lucian: I just had a look in my working paper about the "Eulerian summation" on nonalternating divergent series and the $\ln(2)$ - bit was not the "result" but rather somehow a residue. If you like to read into this it is in go.helms-net.de/math/binomial_new/EulerianSumsV2.pdf at page 15 and 16. (Unfortunately I don't have much mathematical conversation so the writing style is (as said by some correspondent) at many places still difficult to read - but perhaps interesting enough to give it a try anyway...)
Feb
1
comment Can we assign a value to the sum of the reciprocals of the natural numbers?
@Jaume : wow, that's a curious formula. But like Lucian I thought I'd seen a direct relation of the $\ln(2)$ -substitution anywhere (but without concrete idea where - I think I should reconsider Konrad Knopp's monography on series). I'm experimenting with the matrix of Eulerian numbers and created a summation-procedure from this (partially described in a question at math.stackexchange.com/questions/310123). It works well for convergent and many divergent series (even for the infamous 0!-1!+2!-3!+...) and arrives for the nonalternating harmonic series at the $\ln(2)$