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martin Wrote:bo198214 Wrote:Thats interesting! Can you give a proof for n=0, i.e that
?
Erm... no. Hasn't this already been proven somewhere?
Yes, but knowing an believing are two different things
When I considered that formula I realized that I never learned the logarithm formula during studying:
This is just what you get when you invert the famous Euler formula .
Knowing that formula, we can do much more with the limit
This is just a particular case with , and in the formula .
with the above base idea it is then no more difficult to prove the general case.
This formula is really amazing
Quote:It should merely interpolate 2^^x between x=0 and x=1  the graph seemed smooth enough (together with the other iterated values beyond) to assume it represented the actual values to at least four decimal digits, but in other tables I've seen, only three decimal digits match.
Aha and there is the answer why you considered your formula to be tetration. Because you compared it with some table values!
Now that you are catched, you have to tell where you found this table!
However, to all my knowledge, there is no *the* tetration. There are different approaches which result in different tetrations (though some only differ at the 10th or higher significant digit) that all satisfy those properties like being analytic strictly increasing and so on.
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07/19/2008, 07:04 AM
(This post was last modified: 07/19/2008, 09:36 AM by Ivars.)
martin Wrote:By the way, I achieved slightly better results when I changed that n=0.345627(2x) into n=0.6910,3450832*x. Gosh, I think this is getting a bit out of hand here...
Well this looks very much like ln(2)  ln(2)/2*x ....
May be it is worth trying ( just out of pure interpolation) without really following what has been done in this thread) something like
ln(2)ln(2)/2*x+(ln(2)/3!)^x^2  (ln(2)/4!)*x^3 +
or similar (may be not factorial, but just 1/2, 1/3, 1/4, etc) alternating series with ln(2) in coefficient.
The inversed Euler limit it very interesting because it involves ROOTs and roots can have many values these in the end should correspond to the infinitely many values of complex logarithm, the thing I tried to understand here:
Cardinality of infinite number of values of complex logarithm via ROOTS of I
As You take the limit of inversed formula, the number of roots with n> infinity grows (since roots are nvalued) as does the possible combinations of the values. However , there are infinitely many ways n can reach infinity via subsets of N ( like 2^n, n^2, primes, n, etc etc) each way generating a different tree structure, and each of them corresponding to convergent or divergent series over 1/n , 1/(2^n), 1/(n^2) so root series are such may converge or diverge ( if they are placed in exponents via product of roots of I in complex logarithm, as I did in the tread I mentioned above, these sums make certain sense, but they do not hold all information neither about TREE structures underlaying them). If we turn this around, it may be that this underlaying Tree structure is the cause and reason for divergence/convergence if certain sums.
In the end infinite values of complex logarithm obtained via roots should EXACTLY match the infinite number of values complex Ln(x) can have, but that means the infinite number of values of complex logarithm has rather clear internal relations via TREE structure related to the combinations of multivalued roots I do not know if that is made explicit somewhere.
My opinion is that complex logarithm ( as imaginary unit itself) is far from being understood.
Ivars
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07/19/2008, 09:03 AM
(This post was last modified: 07/19/2008, 04:47 PM by martin.)
bo198214 Wrote:Aha and there is the answer why you considered your formula to be tetration. Because you compared it with some table values!
Now that you are catched, you have to tell where you found this table!
One of them is here: http://tetration.itgo.com/txt/tabletetbin.txt
(I think that's the only one I found for 2^^x)
bo198214 Wrote:However, to all my knowledge, there is no *the* tetration. There are different approaches which result in different tetrations that all satisfy those properties like being analytic strictly increasing and so on.
This is exactly my state of knowledge, too. And it's still hard to believe (aren't there any bounds or something?)
Ivars Wrote:May be it is worth trying ( just out of pure interpolation) without really following what has been done in this thread) something like
ln(2)ln(2)/2*x+(ln(2)/3!)^x^2  (ln(2)/4!)*x^3 +
or similar (may be not factorial, but just 1/2, 1/3, 1/4, etc) alternating series with ln(2) in coefficient.
I shall try that next week during the lunch breaks. I recognized I was close to ln(2) there, but didn't think that far.
bo198214 Wrote:Can you give a proof for n=0, i.e that
?
Come think of it, I could actually try that...
Recall that Newton approximated his square roots by taking the arithmeticharmonic mean:
(2+3)/2=2.5 or 5/2 and 2/(1/2+1/3)=2.4 or 12/5
(5/2+12/5)/2=49/20 and 2/(2/5+5/12)=120/49 etc., eventually obtaining the square root of 6:
(sqr(6)+sqr(6))/2=sqr(6) and 2/(1/sqr(6)+1/sqr(6))=sqr(6)
That lies between n=1 and n=1. Is the result really in the middle at n=0? I try that with n=1/2 and 1/2:
((sqr(2)+sqr(3))/2)^2=5/4+sqr(6)/2 and (2/(1/sqr(2)+1/sqr(3)))^2=12048*sqr(6)
etc., I end up with the same solution. Moreover, the geometric mean of the two values at each step is exactly sqr(6), in both examples.
I hope this was somewhat understandable.
Does this count as a proof?
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If in series there is 1/n!, then formula becomes:
n= ln(2) * ((1exp(x))/x)
if just 1/n, then:
n= ln(2) * (ln(1+x)/x)
Ivars
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Also, I noticed you gave a link to my website, but there is actually a newer version of that table, that I have yet to incorporate into the rest of my site:
http://tetration.itgo.com/newtxt/tabletetbin.txt
Andrew Robbins
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07/21/2008, 06:31 PM
(This post was last modified: 07/21/2008, 06:32 PM by martin.)
andydude Wrote:http://tetration.itgo.com/newtxt/tabletetbin.txt
Wow, that's an extensive and interesting table. I especially like the values below 2.
Ivars Wrote:If in series there is 1/n!, then formula becomes:
n= ln(2) * ((1exp(x))/x)
if just 1/n, then:
n= ln(2) * (ln(1+x)/x)
Nope, neither of them worked. But hey, it was worth trying.
Meanwhile, I figured out linear parameters for 3^^x.
With n = 0.08130.328823x, for 0 < x < 1: 3^^x ~~ [x(3^n1)+1]^(1/n)
Again, quite close, but still with minimal irregularities.
Either I didn't try long enough to find exact parameters, or I'm really on the wrong track. (Most probably the latter. (Then again, thinking about n to be a polynomial of infinite degree or some other equation ... nah, I leave that to someone else who's interested.))
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martin Wrote:Ivars Wrote:If in series there is 1/n!, then formula becomes:
n= ln(2) * ((1exp(x))/x)
if just 1/n, then:
n= ln(2) * (ln(1+x)/x)
Nope, neither of them worked. But hey, it was worth trying.
I may have been mistaken with formulas which I derived from series, the series of type ln(2)*(1 a*x +b*x^2  c*x^3+..) itself looked more promising
May be it is just ln(2)ln(2)/2*x.
Ivars
