I just read Andrew Robbins' solution to the tetration problem, which I find very convincing, and want to use the opportunity to present and discuss it here.
The solution \( {}^y x \) satisfies the 2 natural conditions
1. \( {}^1b=b \) and \( {}^{x+1}b=b^{{}^xb} \)
2. \( x\mapsto b^x \) is infinitely differentiable.
However for avoiding difficulties with a later expansion, he instead solves for the tetration logarithm tlog (which he calls super logarithm but I find "tetration logarithm" somewhat more specific), which is the inverse of \( x\mapsto b^x \), i.e. \( {}^{\text{tlog}_b(x)}b=x \).
The first condition is then translated into
1'. \( \text{tlog}_b b=1 \) and \( \text{tlog}_b(x)=\text{tlog}_b(b^x)-1 \)
while the second condition is equivalent to that also
2'. \( x\mapsto \text{tlog}_b(x) \) is infinitely differentiable.
By 1' we merely need to consider the tlog on the interval \( (0,1\] \) and can then derive the values for the other intervals \( (1,b\] \), \( (b,b^b\] \), \( (b^b,b^{b^b}\] \) etc. above and \( (-\infty,0\] \) below.
The idea is now that we define a smooth (infinitely differentiable) function t on \( (0,1\] \) and ensure that the by 1' resulting function is also smooth at the joining points of the intervals. Obviously it suffices to ensure this for the joining point 0 (and by 1' this transposes to each other joining point).
We simply expand t into a power series at 0 and then try to determine the coefficients such that the resulting function is smooth at 0
\( t(x) = \sum_{i=0}^\infty \nu_i \frac{x^i}{i!} \)
\( \nu_i \) is the i-th derivative of t at 0.
The resulting function on \( (-\infty,0\] \) is
\( s(x)=t(b^x)-1 \)
We have to ensure that \( s^{(k)}(0)=t^{(k)}(0)=\nu_k \) for each \( k\ge 0 \). What is now the k-th derivative of s at 0?
For \( k=0 \) we get \( \nu_0=s(0)=t(b^0)-1=t(1)-1=\text{tlog}_b(b^1)-2=\text{tlog}_b(b)-2=1-2=-1 \).
For \( k\ge 1 \) the constant -1 vanishes and we make the following calculations:
\( s^{(k)}(x)=(t(b^x))^{(k)}=\left(\sum_{i=0}^\infty \nu_i \frac{b^{xi}}{i!}\right)^{(k)}=\sum_{i=0}^\infty\frac{\nu_i}{i!}(b^{xi})^{(k)} \)
The derivation of \( b^{xi} \) is easily determined to be
\( (b^{xi})'=b^{xi}\text{ln}(b) i \) and so the k-th derivative is \( (b^{xi})^{(k)} = b^{xi}(\text{ln}(b)i)^k \), which give us in turn
\( \nu_k=s^{(k)}(0)= \text{ln}(b)^k\sum_{i=0}^\infty\nu_i\frac{i^k}{i!} \) for \( k\ge 1 \).
This is an infinite linear equation system system in the variables \( \nu_1,\nu_2,\dots \).
The way of Andrew Robbins is now to approximate a solution by considering finite linear equation systems consisting of n equations and n variables \( \nu_{i,n} \) resulting from letting \( \nu_{k,n}=0 \) for \( k>n \).
First one can show that these equation systems have a unique solution for b>1 and numerical evidence then shows that \( \nu_k=\lim_{n\to\infty}\nu_{k,n} \) converges and that the resulting \( \nu_k \) are a solution of the infinite equation system.
Further numerical evidence shows, that the infinite sum in the definition of t converges for the so obtained \( \nu_i \).
However I would guess that the claimed uniqueness for a solution satisfying 1' and 2' is not guarantied. We can use different approximations, for example for a given constant we can consider the equation systems, resulting from letting \( \nu_k=c \) for \( k>n \). Because interestingly the sum \( \sum_{i=0}^\infty \frac{i^k}{i!} \) converges to \( eB_k \) where e is the Euler constant and B_k are the Bell numbers. So by setting \( \nu_k=c \) for \( k>n \) the remaining sum \( \sum_{i=n+1}^\infty c\frac{i^k}{i!} \) converges and merely introduces an additive constant in the linear equation system. The obtained solutions are different from the solution obtained by c=0.
However I didnt verify yet the convergence properties of these alternative solutions.
The solution \( {}^y x \) satisfies the 2 natural conditions
1. \( {}^1b=b \) and \( {}^{x+1}b=b^{{}^xb} \)
2. \( x\mapsto b^x \) is infinitely differentiable.
However for avoiding difficulties with a later expansion, he instead solves for the tetration logarithm tlog (which he calls super logarithm but I find "tetration logarithm" somewhat more specific), which is the inverse of \( x\mapsto b^x \), i.e. \( {}^{\text{tlog}_b(x)}b=x \).
The first condition is then translated into
1'. \( \text{tlog}_b b=1 \) and \( \text{tlog}_b(x)=\text{tlog}_b(b^x)-1 \)
while the second condition is equivalent to that also
2'. \( x\mapsto \text{tlog}_b(x) \) is infinitely differentiable.
By 1' we merely need to consider the tlog on the interval \( (0,1\] \) and can then derive the values for the other intervals \( (1,b\] \), \( (b,b^b\] \), \( (b^b,b^{b^b}\] \) etc. above and \( (-\infty,0\] \) below.
The idea is now that we define a smooth (infinitely differentiable) function t on \( (0,1\] \) and ensure that the by 1' resulting function is also smooth at the joining points of the intervals. Obviously it suffices to ensure this for the joining point 0 (and by 1' this transposes to each other joining point).
We simply expand t into a power series at 0 and then try to determine the coefficients such that the resulting function is smooth at 0
\( t(x) = \sum_{i=0}^\infty \nu_i \frac{x^i}{i!} \)
\( \nu_i \) is the i-th derivative of t at 0.
The resulting function on \( (-\infty,0\] \) is
\( s(x)=t(b^x)-1 \)
We have to ensure that \( s^{(k)}(0)=t^{(k)}(0)=\nu_k \) for each \( k\ge 0 \). What is now the k-th derivative of s at 0?
For \( k=0 \) we get \( \nu_0=s(0)=t(b^0)-1=t(1)-1=\text{tlog}_b(b^1)-2=\text{tlog}_b(b)-2=1-2=-1 \).
For \( k\ge 1 \) the constant -1 vanishes and we make the following calculations:
\( s^{(k)}(x)=(t(b^x))^{(k)}=\left(\sum_{i=0}^\infty \nu_i \frac{b^{xi}}{i!}\right)^{(k)}=\sum_{i=0}^\infty\frac{\nu_i}{i!}(b^{xi})^{(k)} \)
The derivation of \( b^{xi} \) is easily determined to be
\( (b^{xi})'=b^{xi}\text{ln}(b) i \) and so the k-th derivative is \( (b^{xi})^{(k)} = b^{xi}(\text{ln}(b)i)^k \), which give us in turn
\( \nu_k=s^{(k)}(0)= \text{ln}(b)^k\sum_{i=0}^\infty\nu_i\frac{i^k}{i!} \) for \( k\ge 1 \).
This is an infinite linear equation system system in the variables \( \nu_1,\nu_2,\dots \).
The way of Andrew Robbins is now to approximate a solution by considering finite linear equation systems consisting of n equations and n variables \( \nu_{i,n} \) resulting from letting \( \nu_{k,n}=0 \) for \( k>n \).
First one can show that these equation systems have a unique solution for b>1 and numerical evidence then shows that \( \nu_k=\lim_{n\to\infty}\nu_{k,n} \) converges and that the resulting \( \nu_k \) are a solution of the infinite equation system.
Further numerical evidence shows, that the infinite sum in the definition of t converges for the so obtained \( \nu_i \).
However I would guess that the claimed uniqueness for a solution satisfying 1' and 2' is not guarantied. We can use different approximations, for example for a given constant we can consider the equation systems, resulting from letting \( \nu_k=c \) for \( k>n \). Because interestingly the sum \( \sum_{i=0}^\infty \frac{i^k}{i!} \) converges to \( eB_k \) where e is the Euler constant and B_k are the Bell numbers. So by setting \( \nu_k=c \) for \( k>n \) the remaining sum \( \sum_{i=n+1}^\infty c\frac{i^k}{i!} \) converges and merely introduces an additive constant in the linear equation system. The obtained solutions are different from the solution obtained by c=0.
However I didnt verify yet the convergence properties of these alternative solutions.