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07/20/2010, 04:13 AM
(This post was last modified: 07/24/2010, 02:23 AM by bo198214.)
Hello all you tetration brainies out there,
looking somewhat deeper into the intuitive Abel function of f(x)=b*x (which is supposed to be log_b(x) however unproven until now), I found a somewhat direct expression of the coefficients, which boils down to the following challenging question:
Let the sequence _{n\in\mathbb{N}}) be defined recursively in the following way for  :
 and  (1-b)^{n-m} b^m) for 
Is }) ?
Does it converge? The following graph of the sequence for  , \approx 1.442695) leaves the question open:
(The messed up numbers on the left side are due to a bug in sage *sigh*)
An equivalent slightly nicer formulation of the problem
Let the sequence be defined recursively in the following way for :
 and  (1-b)^{n-m} b^m) for
Is }) ?
edit: this can be found now as TPID 9 in the open problems thread.
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the abel function is the functional inverse of the superfunction , thus
abel function of b*x = funct inverse of b^x = log_b(x).
i guess what you did is trying to solve an abel equation with some unmentioned method ... thus what is unproven ?
i noticed you added "intuitive" abel function.
i dont know what you used to compute that ... a fixpoint ? matrix ?
log_b(x) is not analytic at x = 0 ... but 0 is the fixpoint of bx ...
maybe you first computed a superfunction and later inverted it ?
plz clarify.
regards
tommy1729
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(07/20/2010, 09:11 PM)tommy1729 Wrote: i noticed you added "intuitive" abel function.
i dont know what you used to compute that ... a fixpoint ? matrix ?
log_b(x) is not analytic at x = 0 ... but 0 is the fixpoint of bx ...
maybe you first computed a superfunction and later inverted it ?
This thread is not about the intuitive Abel function (which you can find described in the overview paper) but about the stated formula. Actually its too cumbersome for me (and I surely guess to the reader too!) to put down all steps that led me to the given limit expression.
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07/24/2010, 02:19 AM
(This post was last modified: 07/24/2010, 11:10 PM by bo198214.)
News, news!
I managed to obtain a direct formula as sum:
 \frac{k(-1)^{k}}{1-b^{k}})
But it still doesnt remind me of something useful.
Why does this simple sequence tend to ) ???
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(07/24/2010, 02:19 AM)bo198214 Wrote: News, news!
I managed to obtain a direct formula as sum:
 \frac{k(-1)^{k}}{1-c^{k}})
But it still doesnt remind me of something useful.
Why does this simple sequence tend to ???
i assume you mean 'b' instead of 'c' in your formula.
considering all that is written about hypergeometric series , this must be provable if true.
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(07/24/2010, 09:33 PM)tommy1729 Wrote: i assume you mean 'b' instead of 'c' in your formula.
Ya, changed it accordingly.
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07/25/2010, 03:39 PM
(This post was last modified: 07/25/2010, 03:56 PM by bo198214.)
Slowly I doubt that this sequence converges.
Have a look again at the picture:
![[Image: attachment.php?aid=719]](http://math.eretrandre.org/tetrationforum/attachment.php?aid=719)
You see, that it oscillates around 1/log(2). Now I considered only the local maxima.
If there is maxima at m then the next maxima is roughly located at 2*m+1.
I computed the values at the maxima:
Code: 00004 1.44761904761905
00013 1.44294012536393
00028 1.44275389908130
00057 1.44272403814808
00115 1.44271536072053
00231 1.44271205805011
00464 1.44271061914438
00929 1.44270994540048
01859 1.44270961966486
03719 1.44270945950856
07439 1.44270938009939
14879 1.44270934056096
29759 1.44270932083314
oo 1.44269504088896?
If you compare with the actual value of 1/log(2.0) (at the end of the code section), it looks as if it would never be reached, because the changing digits just wander too quickly to the right. What do you think?
If this is true, then the whole intuitive method questionable, because it does not converge in the simplest case of a linear function.
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07/26/2010, 12:14 PM
(This post was last modified: 07/26/2010, 12:25 PM by Gottfried.)
(07/25/2010, 03:39 PM)bo198214 Wrote: Code: 14879 1.44270934056096
29759 1.44270932083314
oo 1.44269504088896?
Hmm, I got no better result yet, but maybe you see something further.
I reformulated this into
 \sum_{k=1}^{\infty} \( 1-\frac1{2^k} \)^n*\frac1{2^k} =^{?}_{ } \frac1{\ln\(2\)} )
or
^n*\frac1{2^k} =^{?}_{ } \frac1{\ln\(2^{n+1}\)} <br />
)
to remove the binomials first. So far I get numerically the same results as in your examples (first version of my equation).
Though I don't see much more clearer here.
Perhaps we can relate this to Euler's "false logarithmic series" which provides a polynomial approximation to the log which is only valid at the integers (but not at zero(!)) and has a sinusoidal deviation from the true log-function - but I didn't go deeper into this yet...
Gottfried
Gottfried Helms, Kassel
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(07/26/2010, 12:14 PM)Gottfried Wrote: I reformulated this into
 \sum_{k=1}^{\infty} \( 1-\frac1{2^k} \)^n*\frac1{2^k} =^{?}_{ } \frac1{\ln\(2\)} )
Yes, expanding  and then applying the binomial formula. But I was still unable to make something nice out of this formula.
I was now going one step further and multiplied the sequence  with ^k}{k}) , i.e. the logarithmic series which we know must tend to ) (for  ), so if converges then  should converge to 1.
The graph for b=2 (which is on the boundary of convergence) behaves very nice, the period is now exactly 2.
And this is the graph for argumetns 2n:
Quote:Perhaps we can relate this to Euler's "false logarithmic series"
which is that formula?
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07/26/2010, 05:08 PM
(This post was last modified: 07/26/2010, 05:13 PM by Gottfried.)
(07/26/2010, 03:56 PM)bo198214 Wrote: Quote:Perhaps we can relate this to Euler's "false logarithmic series"
which is that formula? Hmm, maybe best is to read the article. The formula is on page 2 and 3.
Euler's "false logarithmic series" (Ed Sandifer)
I've found the taylor-expression of the truncations of this series when I interpolated 2^0,2^1,2^2,2^3,... polynomially (*1) ... 
But anyway - I still don't know whether this can be related to your problem here, so ...
Gottfried
(*1) or was it this "exponential polynomial interpolation"? don't have it exactly in mind - I had a certain question and tried with a lot of options recently without archiving all the partial successes and failings. I'll see whether I can reproduce it
Gottfried Helms, Kassel
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