Why bases 0<a<1 don't get love on the forum?

[marraco]

What is the justification to demand that the derivative at zero must be zero?

Every time I tried to develop a Taylor series of tetration, I found that there is ever a degree of liberty, in choosing the derivative at 0.

This was part of one of my failed attempts.

Let suppose that there exist a series for \( \\[15pt]

{^xa} \) convergent in a radius equal or larger than 1.

Since °a=1, the first coefficient must be a₀=1:

\( ^xa \,=\, 1 \,+\, \sum _{n=1}^{\infty } a_n
. x^n \)


Let's call p to the sum of even terms, and q to the sum of odd ones
\( p \,=\, \sum _{n=even}^{\infty }
a_n \\
\\
\\
q \,=\, \sum _{n=odd}^{\infty }
a_n \)


Since we know the values of \( \\[15pt]

{^xa} \) at -1 and 1:
\(
\left.\begin{matrix}\\
\\[15pt]

{^1a=a}
\,\,\,&\Rightarrow\,\,\,
\sum _{n=0}^{\infty } a_n=a

\,\,\,&\Rightarrow\,\,\,
p+q=a
\\
\\
\\[15pt]

{^{-1}a=0}
\,\,\,&\Rightarrow\,\,\,
1+\sum _{n=1}^{\infty } (-1)^n a_n=0
\,\,\,&\Rightarrow\,\,\,
p-q=0
\end{matrix}\right\}
\Rightarrow
\left\{\begin{matrix}
p=\frac{a}{2}
\\
\\
q=\frac{a}{2}
\end{matrix}\right.
\)

If we need to make a linear systems of equations to solve the coefficients, this would be the first one.


Looking at the series, is evident that the functions generated by the coefficients p, P(x) and q, Q(x) would be calculable from \( \\[15pt]

{^xa} \) this way:

\( {\\[15pt]

{\color{Red} P}} \)\(
(x) \,=\,
1+\sum _{n=1}^{\infty } a_{2n} \,.\, x^{2n}\,=\,
\frac{^xa \,+\,{^{-x}a}}{2}
\)

\( {\\[15pt]

{\color{Green} Q}} \)\(
(x) \,=\,
\sum _{n=0}^{\infty } a_{1+2n} \,.\, x^{1+2n}\,=\,
\frac{^xa\,-\,{^{-x}a}}{2}
\)

Those functions may be related to an analogous decomposition of e^x into sines and cosines.




We also know this differential equation:

\( \frac{\mathrm{d} ^xa}{\mathrm{d} x} \,=\, ln(a)\,.\,^xa\,.\,\frac{\mathrm{d} (^{x-1}a)}{\mathrm{d} x} \)

The derivative at 0 is \( \\[15pt]

{D_0^1=a_1} \)

Then:

\( \left.\begin{matrix}\\
\\[15pt]

{D_0^1\,=\,ln(a).\,^0a.D_{-1}^1}
\\
D_{x}^1 \,=\,
\sum _{n=1}^{\infty } n.a_n.
x^{n-1}
\,=\,\dot{P}+\dot{Q}

\end{matrix}\right\}

\,\,a_1\,=\,
ln(a)\,.\, \sum _{n=1}^{\infty } n.a_n. (-1)^{n-1}
\,=\,
ln(a)\,.\, ( \dot{P}_{-1}+\dot{Q}_{-1})
\)

The derivative at x=1 is


\(
\\[15pt]

{\dot{P}_1+\dot{Q}_1 \,=\,
ln(a).\,^1a.D_{0}^1 \,=\,
ln(a).\,a.\,a_1
}
\)

But, P series only has even powers of x, and Q only odd, so:
\( \\[15pt]

{
\dot{P}_{-1}\,=\,-\dot{P}_{1}
\\
\dot{Q}_{-1}\,=\,\dot{Q}_{1}
} \)

Then

\( \\[15pt]


{ \frac{1}{ln(a)} \,.\, a_1 \,+\,
( \dot{P}_{1}-\dot{Q}_{1})
\,=\, 0
\\
ln(a^{-a}) .\,a_1 \,+\, \dot{P}_1+\dot{Q}_1
\,=\, 0
}
\)

Which is a system of 2 equations with 3 unknown. If I try to add anything else, I ever get one extra unknown.