01/16/2008, 04:47 AM
Well, this is approximately what I was thinking. If \( f(x) = b^{b^x} \), then we are trying to solve \( f(x_0) = x_0 \) and \( f(x_1) = x_1 \) where \( x_0 \ne x_1 \)right? OK, so I started with a series expansion, and this is easier if you deal with a single exponential rather than an iterated exponential, so I found the series expansions:
\(
\begin{tabular}{rl}
b^x & =
x_1 + x_1\ln(b)(x-x_0) + \frac{1}{2}x_1\ln(b)^2(x-x_0)^2 + \cdots \\
\log_b(x) & =
x_1 + \frac{(x-x_0)}{x_0\ln(b)} - \frac{(x-x_0)^2}{2(x_0^2\ln(b))} + \cdots
\end{tabular}
\)
and with these expansions at hand, we notice that if \( x_0 = b^{b^{x_0}} \), then \( \log_b(x_0) = b^{x_0} = x_1 \), so these two expansions should be the same (this is not true, but for some reason I was thinking it was... It is true for \( x=x_0 \) but not for all x). If these two expansions are the same, then the first two coefficients should be equal, which means
\(
\begin{tabular}{rl}
x_1\ln(b) & = \frac{1}{x_0\ln(b)} \\
x_1\ln(b)x_0\ln(b) & = 1 \\
\ln(b^{x_1})\ln(b^{x_0}) & = 1 \\
\ln(x_0)\ln(b^{x_0}) & = 1 \\
\ln(x_0)\ln(x_1) & = 1
\end{tabular}
\)
Solving for \( b \) in the second-to-last equation gives \( b = \exp\left(\frac{1}{x_0 \ln(x_0)}\right) \), which I suppose is what I was thinking would be a general function, but I don't think it works that way. Also, solving for \( x_0 \) in the second-to-last equation gives \( x_0 = \frac{1}{\ln(b) W\left(\frac{1}{\ln(b)}\right)} \).
Andrew Robbins
\(
\begin{tabular}{rl}
b^x & =
x_1 + x_1\ln(b)(x-x_0) + \frac{1}{2}x_1\ln(b)^2(x-x_0)^2 + \cdots \\
\log_b(x) & =
x_1 + \frac{(x-x_0)}{x_0\ln(b)} - \frac{(x-x_0)^2}{2(x_0^2\ln(b))} + \cdots
\end{tabular}
\)
and with these expansions at hand, we notice that if \( x_0 = b^{b^{x_0}} \), then \( \log_b(x_0) = b^{x_0} = x_1 \), so these two expansions should be the same (this is not true, but for some reason I was thinking it was... It is true for \( x=x_0 \) but not for all x). If these two expansions are the same, then the first two coefficients should be equal, which means
\(
\begin{tabular}{rl}
x_1\ln(b) & = \frac{1}{x_0\ln(b)} \\
x_1\ln(b)x_0\ln(b) & = 1 \\
\ln(b^{x_1})\ln(b^{x_0}) & = 1 \\
\ln(x_0)\ln(b^{x_0}) & = 1 \\
\ln(x_0)\ln(x_1) & = 1
\end{tabular}
\)
Solving for \( b \) in the second-to-last equation gives \( b = \exp\left(\frac{1}{x_0 \ln(x_0)}\right) \), which I suppose is what I was thinking would be a general function, but I don't think it works that way. Also, solving for \( x_0 \) in the second-to-last equation gives \( x_0 = \frac{1}{\ln(b) W\left(\frac{1}{\ln(b)}\right)} \).
Andrew Robbins