# Tetration Forum

Full Version: Infinite products
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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
$
\text{tet}(x) = \sum_{k=0}^{\infty}c_k(x+1)^k
$
and using the definition, we can rewrite this as
$
\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 $a_k = e^{c_k}$

I wonder if this simplifies the coefficients or just makes things more complicated?

Andrew Robbins
andydude Wrote: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
$
\text{tet}(x) = \sum_{k=0}^{\infty}c_k(x+1)^k
$
and using the definition, we can rewrite this as
$
\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 $a_k = e^{c_k}$

I wonder if this simplifies the coefficients or just makes things more complicated?

Andrew Robbins

that could have been my own post

i wondered about it too.
andydude Wrote:
$
\begin{array}{rl}
\text{tet}(x) =
\dots & = \prod_{k=0}^{\infty} a_k^{x^k} \\
\end{array}
$
where $a_k = e^{c_k}$

I wonder if this simplifies the coefficients or just makes things more complicated?

To calculate the coefficients you either need on both sides a product or on both sides a sum. So I dont see how the product representation can be useful.