02/12/2015, 10:27 AM
Hi.
I was wondering about this. I saw this:
http://arxiv.org/abs/1105.4735
made by two forum members here. It gives asymptotic formulae for the tetration and "upper super-exponential" to base \( e^{1/e] \), also called "\( \eta \)".
In particular, the formula is given for the tetration at base \( \eta \) with
\( F(z) = e \left(1 - \frac{2}{z} \left(1 + \sum_{m=1}^{\infty} \frac{P_m(-\log(z))}{(3z)^m}\right)\right) \)
so that \( ^z \eta \) is just a constant shift of this function. I'm not sure if this is a convergent formula or just an asymptotic, as the paper says asymptotic analysis determines the coefficients.
The first few polynomials \( P_n \) are given by
\( P_1(t) = t \)
\( P_2(t) = t^2 + t + 1/2 \)
\( P_3(t) = t^3 + \frac{5}{2}t^2 + \frac{5}{2}t + \frac{7}{10} \)
\( P_4(t) = t^4 + \frac{13}{3}t^3 + \frac{45}{6}t^2 + \frac{53}{10}t + \frac{67}{60} \)
\( P_5(t) = t^5 + \frac{77}{12}t^4 + \frac{101}{6}t^3 + \frac{83}{4}t^2 + \frac{653}{60}t + \frac{2701}{1680} \)
Now, is there some kind of explicit (at least a finite summation/product etc. is what I mean) formula for these polynomials?
Thanks to Richard Stanley on mathoverflow.net here:
http://mathoverflow.net/questions/57627/...lso-how-ca
an explicit formula was found for the polynomial coefficients for the regular iteration at the base \( e \), in particular if
\( F(z+1) = \exp(F(z)) \)
and
\( F(z) = \sum_{n=0}^{\infty} a_n L^{nz} \)
then
\( a_0 = L \)
\( a_1 = 1 \)
and for \( n > 1 \),
\( a_n = \frac{P_n(L)}{\prod_{j=2}^{n} j (L^{j-1} - 1)} \)
where
\( P_n(L) = \sum_{j=0}^{\frac{(n-1)(n-2)}{2}} \mathrm{mag}_{n,j} L^j \)
are polynomials, and the coefficients are given explicitly by
\( \mathrm{mag}_{n, {{n-1} \choose {2}} - j} = \sum_{S} \beta_n(S) \)
where \( S \) ranges over all subsets of \( \{1, 2, \cdots, n-2\} \) such that \( \sum_{s \in S} s = j \). And
\( \beta_n(S) = \sum_{T \subseteq S} (-1)^{|S - T|} \alpha_n(T) \)
where \( \alpha_n(T) \) is given for the given subset \( T \) by ordering its elements in the order \( t_1 < t_2 < \cdots < t_k \), as
\( \alpha_n(T) = \prod_{l=0}^{k-1} S(n - t_l, n - t_{l+1}) \).
where the \( S \)-numbers are just the familiar Stirling second-kind numbers and \( t_0 = 0 \).
So is there an explicit formula for the coefficients, or, perhaps more elegantly, the polynomials (e.g. as sums of multiplied simpler polynomials, for example) in the \( \eta \) tetration? Also, are these formulas of any interest? I don't know if anyone saw them on the MathOverflow site. I'm the original poster of the question linked there, btw.
I was wondering about this. I saw this:
http://arxiv.org/abs/1105.4735
made by two forum members here. It gives asymptotic formulae for the tetration and "upper super-exponential" to base \( e^{1/e] \), also called "\( \eta \)".
In particular, the formula is given for the tetration at base \( \eta \) with
\( F(z) = e \left(1 - \frac{2}{z} \left(1 + \sum_{m=1}^{\infty} \frac{P_m(-\log(z))}{(3z)^m}\right)\right) \)
so that \( ^z \eta \) is just a constant shift of this function. I'm not sure if this is a convergent formula or just an asymptotic, as the paper says asymptotic analysis determines the coefficients.
The first few polynomials \( P_n \) are given by
\( P_1(t) = t \)
\( P_2(t) = t^2 + t + 1/2 \)
\( P_3(t) = t^3 + \frac{5}{2}t^2 + \frac{5}{2}t + \frac{7}{10} \)
\( P_4(t) = t^4 + \frac{13}{3}t^3 + \frac{45}{6}t^2 + \frac{53}{10}t + \frac{67}{60} \)
\( P_5(t) = t^5 + \frac{77}{12}t^4 + \frac{101}{6}t^3 + \frac{83}{4}t^2 + \frac{653}{60}t + \frac{2701}{1680} \)
Now, is there some kind of explicit (at least a finite summation/product etc. is what I mean) formula for these polynomials?
Thanks to Richard Stanley on mathoverflow.net here:
http://mathoverflow.net/questions/57627/...lso-how-ca
an explicit formula was found for the polynomial coefficients for the regular iteration at the base \( e \), in particular if
\( F(z+1) = \exp(F(z)) \)
and
\( F(z) = \sum_{n=0}^{\infty} a_n L^{nz} \)
then
\( a_0 = L \)
\( a_1 = 1 \)
and for \( n > 1 \),
\( a_n = \frac{P_n(L)}{\prod_{j=2}^{n} j (L^{j-1} - 1)} \)
where
\( P_n(L) = \sum_{j=0}^{\frac{(n-1)(n-2)}{2}} \mathrm{mag}_{n,j} L^j \)
are polynomials, and the coefficients are given explicitly by
\( \mathrm{mag}_{n, {{n-1} \choose {2}} - j} = \sum_{S} \beta_n(S) \)
where \( S \) ranges over all subsets of \( \{1, 2, \cdots, n-2\} \) such that \( \sum_{s \in S} s = j \). And
\( \beta_n(S) = \sum_{T \subseteq S} (-1)^{|S - T|} \alpha_n(T) \)
where \( \alpha_n(T) \) is given for the given subset \( T \) by ordering its elements in the order \( t_1 < t_2 < \cdots < t_k \), as
\( \alpha_n(T) = \prod_{l=0}^{k-1} S(n - t_l, n - t_{l+1}) \).
where the \( S \)-numbers are just the familiar Stirling second-kind numbers and \( t_0 = 0 \).
So is there an explicit formula for the coefficients, or, perhaps more elegantly, the polynomials (e.g. as sums of multiplied simpler polynomials, for example) in the \( \eta \) tetration? Also, are these formulas of any interest? I don't know if anyone saw them on the MathOverflow site. I'm the original poster of the question linked there, btw.
