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+--- Thread: Constructing the "analytical" formula for tetration. (/showthread.php?tid=576)
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RE: Constructing the "analytical" formula for tetration. - mike3 - 01/29/2011
(01/28/2011, 10:42 PM)Gottfried Wrote:(01/28/2011, 09:12 PM)mike3 Wrote: Well, here goes... this is LONG! I'm going to also give an explanation of the derivation, since you were asking for more "insight" into the problem.
Very interesting! and not soo long at all. I cannot chew it instantly but I'll take my time tomorrow and on sunday to interprete and relate it to the matrix-notation. Now since I see that 2,3,3,4,3,4,4,5,... pattern and the string-concatenation I get the impression that this might be one of the cases where the recursive description is replaced by an expansion back to the initial parameter(s), but where no obvious simplification is found - which would reflect my experience that I couldn't find a shorter description for the schröder-coefficients than by the explicite matrix-eigendecomposition (for the triangular case). But well - this is just an impression by the first reading, perhaps it is completely false.
Anyway - thanks so far. I'll try to understand everything next two days.
Gottfried
Upps: one question: the r's in this posts are the chi's of the earlier one?
Woo, that post was not actually finished, I had hit "post" too soon! There's more information there now.
And \( r_{n, m} \) would equal, in this case, \( r_{n, m} = \frac{u^{n-1}}{1 - u^{n-1}} \frac{m!}{n!} \left{{n \atop m}\right} \).
RE: Constructing the "analytical" formula for tetration. - mike3 - 02/10/2011
(01/28/2011, 09:12 PM)mike3 Wrote: Then, the not-quite-symbolic but still "explicit" and "non-recursive" formula is
\( a_n = \sum_{j=1}^{2^{n-2}} \prod_{k=1}^{(\mathrm{number\ of\ 1\ bits\ in\ }2^{n-1} + 2j - 1)} r_{(\mathrm{position\ of\ }k\mathrm{th\ 1\ bit\ in\ }2^{n-1} + 2j - 1\mathrm{,\ with\ the\ LSB\ counted\ as\ position\ 1}),(\mathrm{if\ }k-1 > 0\mathrm{,\ position\ of\ }k-1\mathrm{th\ 1\ bit\ in\ }2j - 1\mathrm{,\ with\ the\ LSB\ counted\ as\ position\ 1,\ otherwise\ 1})} \).
I just came up with a symbolic form for this formula. It may not be the best (the "nth one bit's index" translations may not be the most "elegant" possible), but at least it gives the result.
\( a_n = \sum_{j=1}^{2^{n-2}} \prod_{k=1}^{\left(2j -\ \sum_{i=1}^{\lfloor\log_2(2j - 1)\rfloor} \lfloor\frac{2j - 1}{2^i}\rfloor\right)} r_{\left(1 +\ \sum_{i=0}^{\lfloor\log_2(2^{n-1} + 2j - 1)\rfloor}\mathbf{1}_{\{\nu \in \mathbb{N}_0: \nu < k\}}\left(\sum_{s=0}^{i} \frac{1 - (-1)^{\lfloor\frac{2^{n-1} + 2j - 1}{2^s}\rfloor}}{2}\right)\right),\left(1 +\ \sum_{i=0}^{\lfloor\log_2(2j - 1)\rfloor}\mathbf{1}_{\{\nu \in \mathbb{N}_0: \nu < k-1\}}\left(\sum_{s=0}^{i} \frac{1 - (-1)^{\lfloor\frac{2j - 1}{2^s}\rfloor}}{2}\right)\right)} \).
Letting \( r_{n, m} = \frac{u^{n-1}}{1 - u^{n-1}} \frac{m!}{n!} \left{{n \atop m}\right} \), we now have \( \chi_n = a_n \), thus an explicit, non-recursive formula for the coefficients of the regular Schroder function of the decremented exponential.
However, as can be seen, this formula looks to be far too complicated and too general to be of any use in and of itself. However, it shows (or at least it will once the rigorous proof is complete!) that an explicit formula exists. The big question now is, is there a simpler one? Note that this can be used to derive an explicit formula for the Lagrange inversion formula, thus we can plug all that together to get the fully-explicit, non-recursive formula for the coefficients of the regular inverse Schroder function of the decremented exponential, and so a fully-explicit, non-recursive formula for the coefficients of the Fourier series regular superfunctions of the exponential. But plugging together all those substitutions in this form is just a horrid nightmare. Yet if this monster formula exists, then it would seem likely that a simpler one does, too, considering how general this is.
I got this by plugging in the following formulas:
\( \mathrm{number\ of\ 1\ bits\ in\ }N = N -\ \sum_{i=1}^{\lfloor\log_2(N)\rfloor} \lfloor\frac{N}{2^i}\rfloor \)
\( \mathrm{position\ of\ }k\mathrm{th\ 1\ bit\ in\ }N\mathrm{,\ with\ the\ LSB\ counted\ as\ position\ 1} =\ \sum_{i=0}^{\lfloor\log_2(N)\rfloor} \mathbf{1}_{\{\nu \in \mathbb{N}_0 : \nu < k}}\left(\sum_{s=0}^{i} \frac{1 - (-1)^{\lfloor\frac{N}{2^s}\rfloor}}{2}\right) \).
Also, note that the sth bit of N, with s = 0 being the LSB, is \( \frac{1 - (-1)^{\lfloor\frac{N}{2^s}\rfloor}}{2} \).
RE: Constructing the "analytical" formula for tetration. - sheldonison - 02/10/2011
(02/10/2011, 04:20 AM)mike3 Wrote: ....What is the decremented exponential? I'm guessing here, (I apologize for sometimes having trouble seeing the big picture behind the equations), but are these coefficients related to the superfunction of f(z)=exp(z)-1?
Letting \( r_{n, m} = \frac{u^{n-1}}{1 - u^{n-1}} \frac{m!}{n!} \left{{n \atop m}\right} \), we now have \( \chi_n = a_n \), thus an explicit, non-recursive formula for the coefficients of the regular Schroder function of the decremented exponential.
-Sheldon
RE: Constructing the "analytical" formula for tetration. - mike3 - 02/10/2011
(02/10/2011, 05:59 AM)sheldonison Wrote:(02/10/2011, 04:20 AM)mike3 Wrote: ....What is the decremented exponential? I'm guessing here, (I apologize for sometimes having trouble seeing the big picture behind the equations), but are these coefficients related to the superfunction of f(z)=exp(z)-1?
Letting \( r_{n, m} = \frac{u^{n-1}}{1 - u^{n-1}} \frac{m!}{n!} \left{{n \atop m}\right} \), we now have \( \chi_n = a_n \), thus an explicit, non-recursive formula for the coefficients of the regular Schroder function of the decremented exponential.
-Sheldon
Earlier here, I mention finding the regular Schroder function of the function \( e^{uz} - 1 \), about the fixed point of \( z = 0 \) of course. That function is the decremented exponential.