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Breaking New Ground In The Quest For The "Analytical" Formula For Tetration.
#1
Hi.

I just managed to put together the hard proof of the explicit, non-recursive formula for the solution of general recurrences of the form




that I mentioned here:

http://math.eretrandre.org/tetrationforu...42#pid5542

The note with the proof is attached to this post. Comments, critique, etc. would be welcomed.

This proves the existence of the explicit non-recursive expression for the coefficients of the regular Schroder function of , and indeed, of the coefficients of regular Schroder functions in general.

So the question now becomes: is there a more "elegant" expression? I'll give some attempts at trying to take a whack at it here.

1. The regular Schroder function (at the natural fixed point 0) coefficients are given by the above recurrence formula, or the general explicit formula, with

.
(where is a Stirling number of the 2nd kind.)
Let the resulting coefficients "" be denoted . Then the Schroder function is
.

2. Now, we go and expand these out. The first few coefficients look like:




.
...

Factoring the denominators and pattern recognition suggests



where are the so-called "Schroder function magic numbers".

3. We want to turn this into a recurrence on the numerators. Let denote the numerator of the nth term . Then we have

.

Now insert the recurrent expression for :

.

But beyond here we run out of luck. The next step would be to set the sum for in this and try to solve for a recurrence for the and then try to come up with an explicit non-recursive formula for that, but it gets hairy and we don't have an expression for the expanded-out product there. So the first order of business here would be to find the explicit, non-recursive formula for the coefficients of

.

Any ideas? The degree of the resulting polynomial is .

Also, . We don't set it to 1, since if you look at the denominator formula, you'll notice it is equal to -1 there and we have .


Attached Files
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#2
(03/20/2011, 03:45 AM)mike3 Wrote: But beyond here we run out of luck. The next step would be to set the sum for in this and try to solve for a recurrence for the and then try to come up with an explicit non-recursive formula for that, but it gets hairy and we don't have an expression for the expanded-out product there. So the first order of business here would be to find the explicit, non-recursive formula for the coefficients of

.

Any ideas? The degree of the resulting polynomial is .

Also, . We don't set it to 1, since if you look at the denominator formula, you'll notice it is equal to -1 there and we have .

Add:
For this product, we can write



where is another sequence of magic numbers defined for . To find the equations for , we use



then

.

Now, letting , we have



The last formula allows for equating coefficients, yielding a three-piece recurrence:

, for
, for
, for

with the appropriate initial conditions, e.g.

,
, for ,
.

Not sure if this is of any use or helps in getting explicit coefficient formulas, though.
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#3
Going further, we can get recurrence formulas for . We have

.

.

Now consider all occurrences of for a given j. That will be the recurrence for generating the . Let's see where that goes. We multiply the two sums together to get

.

Then we have

.

To get the coefficient for a given j, we note that the sum over j will only include that j when , which means . So we get



as the atrocious recurrence for the magic numbers. What to do with that?

(P.S. when should be 1)
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#4
Breaking news:

On this thread on sci.math:
http://groups.google.com/group/sci.math/...e91f?hl=en

I discussed what amounted to a special case of finding the coefficients for the finite product in the formulas. Apparently, this does not have a known solution, and the solution for a related problem (partition numbers) looked to be extremely non trivial, involving sophisticated mathematics.

So it seems that for now, the binary-counting based formula will just have to do...
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#5
(04/10/2011, 08:33 PM)mike3 Wrote: Breaking news:

On this thread on sci.math:
http://groups.google.com/group/sci.math/...e91f?hl=en

I discussed what amounted to a special case of finding the coefficients for the finite product in the formulas. Apparently, this does not have a known solution, and the solution for a related problem (partition numbers) looked to be extremely non trivial, involving sophisticated mathematics.

So it seems that for now, the binary-counting based formula will just have to do...

what about contour integrals ? euler expressed it as an n'th derivative and those can be given as contour integrals.

its not a sum though nor simple.

( btw as for p(n) being a fractal , i said that before. )

tommy1729
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#6
Meh, it looks like the whole approach of trying to directly find the "magic" numbers is insoluble. So it seems we're left with trying to simplify/"elegantize" the "ugly" binary formula in some other way.

Anyway, I noticed that that "binary" formula can be written simply as a sum over subsets of an interval of naturals (note that binary = indicator functions), giving a neater, more compact form:


.

Then,

.

I didn't try a form like this before since I didn't consider it "explicit" enough for my liking Smile (I'm a fan of sums over straight indices, and I was hoping the straight-index form would help, but alas, no dice.) But this lets us write the Schroder coefficients as

.

We can eliminate the factorial quotient from the product since the product multiplies it to , giving

.

A "simpler" formula for this would then mean one that sums over fewer terms and also has straight indices and no binary/subset/etc. stuff. This sums over terms. In theory, it should be possible to get it to . How can we do this? The above is very general -- we can express Bernoulli numbers and what not in a similar form. But in that case, and in a lot of other cases, there exists a form with fewer terms, straight indices and no binary/subset/etc. stuff -- just some nested sums and products. Is that possible here as well? Is this formula already known somewhere?

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