@Ivars: First of all those are not my definitions, they are Knoebel's. His paper, called "Exponentials Reiterated" is an amazing exposition of the infinitely iterated exponential. It covers almost all aspects of it, and mentions many aspects of tetration as well.
@Gottfried: Your graph confuses me, why is the 1-tick for b=... and the 1-tick for f=... not aligned? and why do you have X's for points, do you have any options for boxes? diamonds? circles? no dots? That would make the graph much nicer to look at. But anyway, I got the general idea of the graph, I just think it could use some work. I'm not sure what the benefit of such a change-of-variables would be, unless it would simplify the series for b^b^x.
@Gottfried: As for your series expansion, I really, really like your presentation of the series for (b^x-1), it shows patterns in the coefficients, where I never saw patterns before
Good job.
About the role of x1 and x2, the role is as follows. The whole point of \( b^{b^x} = x \) is to parameterize the two branches of odd-H, and even-H as described by Knoebel. Individually, both x0 and x1 satisfy this equation, but when viewed together, they satisfy:
Anyways, I found a way to construct a power series for this function that expresses the coefficients. Lets call it \( b(x) \) since its a function that gives b as an output, and it takes x as an input, such that \( x = b(x)^{b(x)^x} \) for all x in [0, 1].
Starting with the definition of this function, and differentiating:
\(
\begin{tabular}{rl}
b^{b^x} & = x \\
(b^{b^x}) b^x (b \ln(b)^2 + (1 + x \ln(b)) b')/b & = 1 \\
(b^{b^x}) b^x (b \ln(b)^2 + (1 + x \ln(b)) b') & = b \\
x b^x (b \ln(b)^2 + (1 + x \ln(b)) b') & = b \\
x b^x b \ln(b)^2 + x b^x (1 + x \ln(b)) b' & = b \\
x b^x b \ln(b) \ln(b) + x b^x (1 + x \ln(b)) b' & = b \\
x b \ln(b^{b^x}) \ln(b) + x b^x (1 + x \ln(b)) b' & = b \\
x b \ln(x) \ln(b) + x b^x (1 + x \ln(b)) b' & = b \\
x b \ln(x) \ln(b) + x b^x b' + x^2 b^x b' \ln(b) & = b \\
x b \ln(x) \ln(b) + x b^x b' + x^2 b' \ln(b^{b^x}) & = b \\
x b \ln(x) \ln(b) + x b^x b' + x^2 b' \ln(x) & = b \\
x b^x b' + x^2 b' \ln(x) & = b - x b \ln(x) \ln(b)\\
x b^x b' + x^2 b' \ln(x) & = b - x b \ln(x) \ln(b)\\
x(b^x + x\ln(x))b' & = b(1 - x \ln(x) \ln(b)) \\
b'(x) & = \frac{b(1 - x \ln(x) \ln(b))}{x(b^x + x\ln(x))}
\end{tabular}
\)
And we now have a differential equation, or in other words, a closed form for the derivative. Many cases in which we have a closed form for the derivative, we can find a series expansion around (almost) any point. So for example, we know the closed form of the derivative of the Lambert W function, which is one way its coefficients can be found (this is the hard way). Anyways, to get back to my point. Using the second-to-last equation (which allows either side to be 0, the last equation prevents the denominator from being 0), We can start with an "unknown" series expansion for \( b(x) \) and solve for its coefficients to satisfy this differential equation. Doing this at \( x=\frac{1}{e} \), we get (factoring out \( e^{-e} \) to simplify things):
\(
\begin{tabular}{rl}
\frac{b(x)}{e^{-e}}
& = 1
\\ & +\ \frac{1}{6} e^3 (-1) (x-\frac{1}{e})^2
\\ & +\ \frac{1}{18} e^4 (x-\frac{1}{e})^3
\\ & +\ \frac{1}{1080} e^5 (-82 + 15e)(x-\frac{1}{e})^4
\\ & +\ \frac{1}{1620} e^6 (79 - 15e)(x-\frac{1}{e})^5
\\ & +\ \frac{1}{45360} e^7 (-2236 + 644e - 35e^2)(x-\frac{1}{e})^6
\\ & +\ \frac{1}{226800} e^8 (9004 - 2800e + 175e^2)(x-\frac{1}{e})^7
\\ & +\ \cdots
\end{tabular}
\)
With this series expansion, we can then test to see if we have a "correct" function, by testing its derivatives, rather than continually guessing. Maybe all we have to do to find the right function is look it up on OEIS, maybe its already there...
Andrew Robbins
PS. I love finding new series
@Gottfried: Your graph confuses me, why is the 1-tick for b=... and the 1-tick for f=... not aligned? and why do you have X's for points, do you have any options for boxes? diamonds? circles? no dots? That would make the graph much nicer to look at. But anyway, I got the general idea of the graph, I just think it could use some work. I'm not sure what the benefit of such a change-of-variables would be, unless it would simplify the series for b^b^x.
@Gottfried: As for your series expansion, I really, really like your presentation of the series for (b^x-1), it shows patterns in the coefficients, where I never saw patterns before
Good job.About the role of x1 and x2, the role is as follows. The whole point of \( b^{b^x} = x \) is to parameterize the two branches of odd-H, and even-H as described by Knoebel. Individually, both x0 and x1 satisfy this equation, but when viewed together, they satisfy:
- \( b^{x_0} = x_1 \) and
- \( b^{x_1} = x_0 \),
Anyways, I found a way to construct a power series for this function that expresses the coefficients. Lets call it \( b(x) \) since its a function that gives b as an output, and it takes x as an input, such that \( x = b(x)^{b(x)^x} \) for all x in [0, 1].
Starting with the definition of this function, and differentiating:
\(
\begin{tabular}{rl}
b^{b^x} & = x \\
(b^{b^x}) b^x (b \ln(b)^2 + (1 + x \ln(b)) b')/b & = 1 \\
(b^{b^x}) b^x (b \ln(b)^2 + (1 + x \ln(b)) b') & = b \\
x b^x (b \ln(b)^2 + (1 + x \ln(b)) b') & = b \\
x b^x b \ln(b)^2 + x b^x (1 + x \ln(b)) b' & = b \\
x b^x b \ln(b) \ln(b) + x b^x (1 + x \ln(b)) b' & = b \\
x b \ln(b^{b^x}) \ln(b) + x b^x (1 + x \ln(b)) b' & = b \\
x b \ln(x) \ln(b) + x b^x (1 + x \ln(b)) b' & = b \\
x b \ln(x) \ln(b) + x b^x b' + x^2 b^x b' \ln(b) & = b \\
x b \ln(x) \ln(b) + x b^x b' + x^2 b' \ln(b^{b^x}) & = b \\
x b \ln(x) \ln(b) + x b^x b' + x^2 b' \ln(x) & = b \\
x b^x b' + x^2 b' \ln(x) & = b - x b \ln(x) \ln(b)\\
x b^x b' + x^2 b' \ln(x) & = b - x b \ln(x) \ln(b)\\
x(b^x + x\ln(x))b' & = b(1 - x \ln(x) \ln(b)) \\
b'(x) & = \frac{b(1 - x \ln(x) \ln(b))}{x(b^x + x\ln(x))}
\end{tabular}
\)
And we now have a differential equation, or in other words, a closed form for the derivative. Many cases in which we have a closed form for the derivative, we can find a series expansion around (almost) any point. So for example, we know the closed form of the derivative of the Lambert W function, which is one way its coefficients can be found (this is the hard way). Anyways, to get back to my point. Using the second-to-last equation (which allows either side to be 0, the last equation prevents the denominator from being 0), We can start with an "unknown" series expansion for \( b(x) \) and solve for its coefficients to satisfy this differential equation. Doing this at \( x=\frac{1}{e} \), we get (factoring out \( e^{-e} \) to simplify things):
\(
\begin{tabular}{rl}
\frac{b(x)}{e^{-e}}
& = 1
\\ & +\ \frac{1}{6} e^3 (-1) (x-\frac{1}{e})^2
\\ & +\ \frac{1}{18} e^4 (x-\frac{1}{e})^3
\\ & +\ \frac{1}{1080} e^5 (-82 + 15e)(x-\frac{1}{e})^4
\\ & +\ \frac{1}{1620} e^6 (79 - 15e)(x-\frac{1}{e})^5
\\ & +\ \frac{1}{45360} e^7 (-2236 + 644e - 35e^2)(x-\frac{1}{e})^6
\\ & +\ \frac{1}{226800} e^8 (9004 - 2800e + 175e^2)(x-\frac{1}{e})^7
\\ & +\ \cdots
\end{tabular}
\)
With this series expansion, we can then test to see if we have a "correct" function, by testing its derivatives, rather than continually guessing. Maybe all we have to do to find the right function is look it up on OEIS, maybe its already there...
Andrew Robbins
PS. I love finding new series