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+--- Thread: Naturally solving the sexp (/showthread.php?tid=34)
Naturally solving the sexp - bo198214 - 08/21/2007
The Abel equation for the slog was \( \text{slog}_b(b^x)=\text{slog}_b(x)+1 \).
Our original recurring for the super exponential (which I will call similarly) sexp is \( \text{sexp}_b(x+1)=b^{\text{sexp}_b(x)} \).
So if we develop it at 0, say \( \text{sexp}_b(x)=\sum_{n=0}^\infty \rho_n x^n \) satisfying the inverted Abel equation:
\( \sum_{n=0}^\infty \rho_n (x+1)^n = \exp\left(\log(b)\sum_{n=0}^\infty \rho_n x^n\right) \)
we get
\( \sum_{n=0}^\infty \rho_n \sum_{k=0}^n \left(n\\k\right) x^k = 1 + \sum_{n=0}^\infty B(\log(b)\rho_1,\dots,\log(b)\rho_n) \frac{x^n}{n!} \)
where \( B(x_1,\dots,x_n) \) is the complete Bell polynomial.
The left side develops to \( \sum_{k=0}^\infty x^k \sum_{n=k}^\infty \rho_n \left(n\\k\right) \) and so we have the infinite equation system
\( \sum_{i=k}^\infty \rho_i \left(i\\k\right) = B(\log(b)\rho_1,\dots,\log(b)\rho_k) \frac{1}{k!} \)
And I wonder if we solve it the natural way whether we get exactly the inverse of the slog (which I assume). Unfortunately there is no complete Bell polynomial in Maple (at least I didnt find it) and I am too lazy in the moment to program it myself
And yes, it is not a linear equation system. Perhaps it is despite solvable, who knows ...
RE: Naturally solving the sexp - jaydfox - 08/22/2007
I've tried doing something similar. I used Faa di Bruno's formula, which, if I understand correctly, is similar to using a Bell polynomial (I can even see the reference in the article you linked to).
However, the system is of non-linear equations (as you point out, and Andrew points this out in his paper). The sytem can be solved iteratively by plugging approximations back into the system, and it converges exponentially for the first 11 terms. When I tried to solve the 12th term (with 11 unknowns, the first term being known), the system was unstable. I was able to manually pick the 11th unknown and solve the 10 unknown system, but this required two sets of iterations.
Solving beyond this would appear to require manually tweaking all the terms beyond the 10th and then iterating, and this becomes infeasible beyond about three or four additional terms, due to the non-linearity. If someone has a spiffier method that come overcome the non-linearity, it might be doable, but I don't see it being possible to solve for, say, a 100x100 system. To be honest, I was blown away that the slog function was a simple linear system, given how chaotic the sexp function is. Hats off to Andrew for his approach.
RE: Naturally solving the sexp - Gottfried - 08/22/2007
bo198214 Wrote:The Abel equation for the slog was \( \text{slog}_b(b^x)=\text{slog}_b(x)+1 \).
Our original recurring for the super exponential (which I will call similarly) sexp is \( \text{sexp}_b(x+1)=b^{\text{sexp}_b(x)} \).
So if we develop it at 0, say \( \text{sexp}_b(x)=\sum_{n=0}^\infty \rho_n x^n \) satisfying the inverted Abel equation:
\( \sum_{n=0}^\infty \rho_n (x+1)^n = \exp\left(\log(b)\sum_{n=0}^\infty \rho_n x^n\right) \)
we get
\( \sum_{n=0}^\infty \rho_n \sum_{k=0}^n \left(n\\k\right) x^k = 1 + \sum_{n=0}^\infty B(\log(b)\rho_1,\dots,\log(b)\rho_n) \frac{x^n}{n!} \)
where \( B(x_1,\dots,x_n) \) is the complete Bell polynomial.
The left side develops to \( \sum_{k=0}^\infty x^k \sum_{n=k}^\infty \rho_n \left(n\\k\right) \) and so we have the infinite equation system
\( \sum_{i=k}^\infty \rho_i \left(i\\k\right) = B(\log(b)\rho_1,\dots,\log(b)\rho_k) \frac{1}{k!} \)
And I wonder if we solve it the natural way whether we get exactly the inverse of the slog (which I assume). Unfortunately there is no complete Bell polynomial in Maple (at least I didnt find it) and I am too lazy in the moment to program it myself
And yes, it is not a linear equation system. Perhaps it is despite solvable, who knows ...
:-) Now I have a name for my matrix-computations. Expliciting the matrix-formulae into sum-notation for the multiplications I arrived exactly at these "complete Bell-polynomials". Thanks! I'll have to read and adapt my naming conventions.
Gottfried