To recap, we know a simple parameterization of \( x = b(x)^{b(x)^x} \) on the middle branch is just \( b(x) = x^{1/x} \), and the other branches can be described by the differential equation given in my previous post, along with a given initial condition. The initial condition used for the previous series expansion was the point of intersection: \( (x=\frac{1}{e}, b=e^{-e}) \). I just found another point (probably found before) on the outer branches, one that has many advantages over the point of intersection. This point is \( (x=\frac{1}{2}, b=\frac{1}{16}) \) which is 2-periodic in the sense:
The problem with the point \( (x=\frac{1}{e}, b=e^{-e}) \) is that its really not a periodic point at all, its a fixed point, so it doesn't qualify as a 2-periodic point. Another problem with the this is that when it is used as an expansion point, the corresponding series has a radius of convergence of 1/e. The reason why is because there are singularities at (x=0,b=0) and (x=1,b=0), if you don't believe me, then do the math, you'll see the singularities. This means that we can't evaluate the series near x=1, because this is outside the radius of convergence.
The advantage of the point \( (x=\frac{1}{2}, b=\frac{1}{16}) \) is that not only is it a true 2-periodic point, but it is half way between x=0 and x=1 (obviously), which means the radius of convergence would be 1/2. Another advantage of this point is that the derivative of b is non-zero, which means the series is invertible. The problem here is that since it is so close to the other point (where the series is not invertible) the inverse series would have a radius of convergence of \( |(e^{-e})-\frac{1}{16}| \approx 0.00349 \) which is pretty useless.
Despite these short-commings, it still converges between x=0 and x=1, so it is still useful for this. Here is the series expansion that I found about the second point:
\(
\begin{tabular}{rl}
b(x)
& = \frac{1}{16}
\\ & +\ (-2 + 4 \ln(2)^2) \frac{(x-\frac{1}{2})}{4(2\ln(2)-1)}
\\ & +\ (1-8\ln(2)+28\ln(2)^2-20\ln(2)^3-32\ln(2)^4+32\ln(2)^5) \frac{(x-\frac{1}{2})^2}{2(2\ln(2)-1)^3}
\\ & +\ (24-192\ln(2)+2176\ln(2)^3-5568\ln(2)^4+4544\ln(2)^5+2048\ln(2)^6-5120\ln(2)^7+2048\ln(2)^
\frac{(x-\frac{1}{2})^3}{12(2\ln(2)-1)^5}
\\ & +\ \cdots
\end{tabular}
\)
It might be interesting to note that the highest power of \( \ln(2) \) displays a pattern in the numerator (3n-1) and the denominator (2n-1). Which means a closed form might be possible some day.
Andrew Robbins
- \( (1/16)^{(1/2)} = (1/4) \)
- \( (1/16)^{(1/4)} = (1/2) \)
The problem with the point \( (x=\frac{1}{e}, b=e^{-e}) \) is that its really not a periodic point at all, its a fixed point, so it doesn't qualify as a 2-periodic point. Another problem with the this is that when it is used as an expansion point, the corresponding series has a radius of convergence of 1/e. The reason why is because there are singularities at (x=0,b=0) and (x=1,b=0), if you don't believe me, then do the math, you'll see the singularities. This means that we can't evaluate the series near x=1, because this is outside the radius of convergence.
The advantage of the point \( (x=\frac{1}{2}, b=\frac{1}{16}) \) is that not only is it a true 2-periodic point, but it is half way between x=0 and x=1 (obviously), which means the radius of convergence would be 1/2. Another advantage of this point is that the derivative of b is non-zero, which means the series is invertible. The problem here is that since it is so close to the other point (where the series is not invertible) the inverse series would have a radius of convergence of \( |(e^{-e})-\frac{1}{16}| \approx 0.00349 \) which is pretty useless.
Despite these short-commings, it still converges between x=0 and x=1, so it is still useful for this. Here is the series expansion that I found about the second point:
\(
\begin{tabular}{rl}
b(x)
& = \frac{1}{16}
\\ & +\ (-2 + 4 \ln(2)^2) \frac{(x-\frac{1}{2})}{4(2\ln(2)-1)}
\\ & +\ (1-8\ln(2)+28\ln(2)^2-20\ln(2)^3-32\ln(2)^4+32\ln(2)^5) \frac{(x-\frac{1}{2})^2}{2(2\ln(2)-1)^3}
\\ & +\ (24-192\ln(2)+2176\ln(2)^3-5568\ln(2)^4+4544\ln(2)^5+2048\ln(2)^6-5120\ln(2)^7+2048\ln(2)^
\frac{(x-\frac{1}{2})^3}{12(2\ln(2)-1)^5}\\ & +\ \cdots
\end{tabular}
\)
It might be interesting to note that the highest power of \( \ln(2) \) displays a pattern in the numerator (3n-1) and the denominator (2n-1). Which means a closed form might be possible some day.
Andrew Robbins