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

[tommy1729]

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=-1
\end{matrix}\right\}
\Rightarrow
\left\{\begin{matrix}
p=\frac{a-1}{2}
\\
\\
q=\frac{a+1}{2}
\end{matrix}\right.
\)

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

I also find it interesting because \( \\[15pt]

{p=q-1 \,\,\Rightarrow\,\, ^qa\,=\,a^{^pa}} \)



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.

I think \( a_1=ln(a) a \) makes sense.

Regards

Tommy1729