jaydfox Wrote:I've come up with these forumulae as well, though in a different form. They follow from basic principles and are "ignorant" of the underlying solution (i.e., the function for the critical interval doesn't matter).Oops. I wrote a subscript b, as though the base were arbitrary, but then I went ahead and assumed base e when I was solving. Amateur mistake.
To start with, let's bear in mind the following identity:
\( T(b, x) = \exp_b(T(b, x-1)) \)
From this starting point, let's compute the first derivative using the chain rule:
\(
\begin{eqnarray}
T'(b, x)
& = & \text{D}_x \left(T(b, x)\right) \\
& = & \text{D}_x \left(\exp_b(T(b, x-1))\right) \\
& = & \exp_b(T(b, x-1))\text{D}_x\left(T(b, x-1)\right) \\
& = & T(b, x)\text{D}_x\left(T(b, x-1)\right) \\
& = & T(b, x)T'(b, x-1)
\end{eqnarray}
\)
From here, we simply rearrange to get \( T(b, x) = \frac{T'(b, x)}{T'(b, x-1)} \).
As you can see, this is an identity, so there's no need to compute numerically to verify. If for some reason a particular solution doesn't satisfy this condition, then we can be sure it either doesn't satisfy the iterative exponential property, or it some other problem such as not being at least twice differentiable.
I haven't tried with the W function, but I would assume that the process is similar.
Let's try again:
\(
\begin{eqnarray}
T'(b, x)
& = & \text{D}_x \left(T(b, x)\right) \\
& = & \text{D}_x \left(\exp_b(T(b, x-1))\right) \\
& = & \ln(b)\exp_b(T(b, x-1))\text{D}_x\left(T(b, x-1)\right) \\
& = & \ln(b)T(b, x)\text{D}_x\left(T(b, x-1)\right) \\
& = & \ln(b)T(b, x)T'(b, x-1)
\end{eqnarray}
\)
As before, we simply rearrange to get \( T(b, x) = \frac{T'(b, x)}{T'(b, x-1)\ln(b)} \).
Again, as before, we have a recurrence relation we can leverage:
\(
\begin{eqnarray}
T'(b, x)
& = & \ln(b)T(b, x)T'(b, x-1) \\
& = & \left(\ln(b)\right)^{2}T(b, x)T(b, x-1)T'(b, x-2) \\
& = & \left(\ln(b)\right)^{3}T(b, x)T(b, x-1)T(b, x-2)T'(b, x-3)
\end{eqnarray}
\)
Et cetera.
~ Jay Daniel Fox
