04/06/2009, 09:20 AM
I just remembered a method I tried like 5 years ago, but haven't visited in a while.
Now that we have better power series expansions of tetration, for example, at \( x=(-1) \) we have
I wonder if this simplifies the coefficients or just makes things more complicated?
Andrew Robbins
Now that we have better power series expansions of tetration, for example, at \( x=(-1) \) we have
\(
\text{tet}(x) = \sum_{k=0}^{\infty}c_k(x+1)^k
\)
and using the definition, we can rewrite this as\text{tet}(x) = \sum_{k=0}^{\infty}c_k(x+1)^k
\)
\(
\begin{array}{rl}
\text{tet}(x)
& = \exp\left(\text{tet}(x-1)\right) \\
& = \exp\left(\sum_{k=0}^{\infty}c_kx^k\right) \\
& = \prod_{k=0}^{\infty} \exp(c_k{x^k}) \\
& = \prod_{k=0}^{\infty} a_k^{x^k} \\
\end{array}
\)
where \( a_k = e^{c_k} \)\begin{array}{rl}
\text{tet}(x)
& = \exp\left(\text{tet}(x-1)\right) \\
& = \exp\left(\sum_{k=0}^{\infty}c_kx^k\right) \\
& = \prod_{k=0}^{\infty} \exp(c_k{x^k}) \\
& = \prod_{k=0}^{\infty} a_k^{x^k} \\
\end{array}
\)
I wonder if this simplifies the coefficients or just makes things more complicated?
Andrew Robbins