Taylor polynomial. System of equations for the coefficients.

Taylor polynomial. System of equations for the coefficients. - Printable Version

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RE: Taylor polinomial. System of equations for the coefficients. - marraco - 05/06/2015

(05/05/2015, 07:40 AM)Gottfried Wrote: P*A = A*Bb

I think that we are speaking of different things.

Obviously, there should be a way to demonstrate the equivalence of both, because they are trying to solve the same problem; looking for the same solution.

But as I understand, the Carleman matrix A only contains powers of a_i coefficients, yet if you look at the red side, it cannot be written as a matrix product A*Bb, because it needs to have products of a_i coefficients (like $ a_1^3.a_3^2.a_5^8.a_... $). Maybe it is a power of A.Bb, or something like A^Bb?

The Pascal matrix on the blue side is the exponential of a much simpler matrix

$
\exp
\left (
\left [
\begin{matrix}
. & 1 & . & . & . & . & . \\
. & . & 2 & . & . & . & . \\
. & . & . & 3 & . & . & . \\
. & . & . & . & 4 & . & . \\
. & . & . & . & . & 5 & . \\
. & . & . & . & . & . & 6 \\
. & . & . & . & . & . & .
\end{matrix}
\right ]
\right )
=
\left [
\begin{matrix}
1 & 1 & 1 & 1 & 1 & 1 & 1 \\
. & 1 & 2 & 3 & 4 & 5 & 6 \\
. & . & 1 & 3 & 6 & 10 & 15 \\
. & . & . & 1 & 4 & 10 & 20 \\
. & . & . & . & 1 & 5 & 15 \\
. & . & . & . & . & 1 & 6 \\
. & . & . & . & . & . & 1
\end{matrix}
\right ] $

Maybe the equation can be greatly simplified by taking a logarithm of both sides.

RE: Taylor polinomial. System of equations for the coefficients. - Gottfried - 05/06/2015

(05/06/2015, 02:42 PM)marraco Wrote:
(05/05/2015, 07:40 AM)Gottfried Wrote: P*A = A*Bb

I think that we are speaking of different things.

Obviously, there should be a way to demonstrate the equivalence of both, because they are trying to solve the same problem; looking for the same solution.

But as I understand, the Carleman matrix A only contains powers of a_i coefficients, yet if you look at the red side, it cannot be written as a matrix product A*Bb, because it needs to have products of a_i coefficients (like $ a_1^3.a_3^2.a_5^8.a_... $). Maybe it is a power of A.Bb, or something like A^Bb?

No, no ... In your convolution-formula you have in the inner of the double sum powers of powerseries (the red-colored formula $ a^{ \;^x a} $ in your first posting ) with the coefficients of the a()-function (not of its single coefficients), and if I decode this correctly, then this matches perfectly the composition of

V(x)*A * Bb = (V(x)*A) * Bb = [1,a(x), a(x)^2, a(x)^3),...] * Bb = V(a(x))*Bb

Only, that after removing of the left V(x)-vector we do things in different order:

V(x)*A * Bb = V(x)*(A * Bb )

and I discuss that remaining matrix in the parenthese of the rhs. That V(x) can be removed on the rhs and on the lhs of the matrix-equation must be justified; if anywhere occur divergent series, this becomes difficult, but as far as we have nonzero intervals of convergence for all dot-products, this exploitation of associativity can be done /should be possible to be done (as far as I think). (The goal of this all is of course to improve computability of A, for instance by diagonalization of P or Bb and algebraic manipulations of the occuring matrix-factors).

Anyway - I hope I didn't actually misread you (which is always possible given the lot of coefficients... )

Gottfried

RE: Taylor polynomial. System of equations for the coefficients. - marraco - 05/07/2015

I misinterpreted what the Carleman matrix was. I tough that it contained the powers of the derivatives of a function (valued at zero), but it contains the derivatives of the powers of a function, so it actually haves the products of the aᵢ coefficients (of bᵢ in your notation).

________________

I tried to use this method to find the coefficients for exponentiation: bˣ=Σbᵢ.xⁿ

The condition is
b.(x+1)=b.Σbᵢ.xⁿ

which translates into
P.[bᵢ]=b.[bᵢ]

or
[P-b.I].[bᵢ]=0

The solution should be bᵢ=ln(b)ⁱ / i!

I found bᵢ=c. (ln(b)ⁱ/i!), where c is an arbitrary constant, because, obviously c.b⁽ˣ⁺¹⁾=b.(c.bˣ)

I was bugged for the fact that any equation for solving tetration I tried seems to have at least one degree of liberty. I think now that it should be explained by one (at least) arbitrary constant in the solution.

This looks analogous to constants found in the solution of differential equations, so I wonder if the evolvent of the curves generated by the constant is also a solution, and what is his meaning.

RE: Taylor polynomial. System of equations for the coefficients. - marraco - 01/13/2016

So, we want the vector $ \\[15pt]

{[a_i]} $, from the matrix equation:

$ \\[15pt]

{\left [ {{i} \choose {r}} \right ]\cdot \left [a_i \right ]
=
\left [ \sum_{n=1}^{P(i)} \frac{a.ln(a)^{\sum_{j=1}^{i}c_{n,j}}}{ \prod_{j=1}^{i} c_{n,j}!}\prod_{j=1}^{i}a_j^{c_{n,j}} \right ]} $

where "r" is the row index of the first matrix at left, and "i" his column index.
Note that in the last equation, both r and i start counting from zero for the first row and column.

______________________________________
P(i) is the partition function

The first few values of the partition function are (starting with p(0)=1):

1, 1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231, 297, 385, 490, 627, 792, 1002, 1255, 1575, 1958, 2436, 3010, 3718, 4565, 5604, … (sequence A000041 in OEIS; the link has valuable information about the partition function).

______________________________________
$ \\[15pt]

{c_{n,j}} $ is the number of repetitions of the integer j in the $ \\[15pt]

{n^{th}} $ partition of the number i

______________________________________
Solving the equation

If we do the substitution $ \\[25pt]

{a_i=\frac {b_i} {ln(a)} } $, we simplify the first equation to:

$ \\[15pt]

{\left [ {{i} \choose {r}} \right ]\cdot \left [b_i \right ]
= ln(^2a) \,
\left [ \sum_{n=1}^{P(i)} \frac{1}{ \prod_{j=1}^{i} c_{n,j}!}\prod_{j=1}^{i}b_j^{c_{n,j}} \right ]} $

______________________________________
Special base.

This equation suggest a special number, which is m=1.7632228343518967102252017769517070804...
m is defined by $ \\[15pt]

{^2m=e} $

For the base a=m, the equation gets simplified to:

$ \\[35pt]

{\left [ {{i} \choose {r}} \right ]\cdot \left [b_i \right ]
=
\left [ \sum_{n=1}^{P(i)} \frac{1}{ \prod_{j=1}^{i} c_{n,j}!}\prod_{j=1}^{i}b_j^{c_{n,j}} \right ]} $

But let's forget about m for now.

______________________________________

We are now very close to the solution. The only obstacle remaining is the product:
$ \\[15pt]

{ \frac{1}{ \prod_{j=1}^{i} c_{n,j}!} } $

If we can do a substitution that get us rid of him, we have the solution:

$ \\[15pt]

{\left [ {{i} \choose {r}} \right ]\cdot \left [b_i \right ]
= ln(^2a) \,
\left [ \sum_{n=1}^{P(i)}\prod_{j=1}^{i}b_j^{c_{n,j}} \right ]} $

At this point we only need to substitute $ \\[25pt]

{b_i=f^i} $, where f is arbitrary, to get:

$ \\[15pt]

{\left [ {{i} \choose {r}} \right ]\cdot \left [f^i \right ]
= ln(^2a) \,
\left [ \sum_{n=1}^{P(i)} f^i \right ] \,=\, ln(^2a) . [P(i) . f^i]} $

... and we get:

$ \\[15pt]

{ \left [f^i \right ]
= \left [ {{i} \choose {r}} \right ]^{-1} \cdot [ ln(^2a) . P(i) . f^i]} $

The choice of f, very probably, determines the value for °a, and the branch of tetration.

(01/03/2016, 11:24 PM)marraco Wrote:

RE: Taylor polynomial. System of equations for the coefficients. - marraco - 01/14/2016

^^ Sorry. I made a big mistake. We cannot substitute $ \\[25pt]

{b_i=f^i} $ of course.

Maybe $ \\[15pt]

{b_i=f^{k.i}} $ would work as an approximation, because we know that $ \\[15pt]

{b_i} $ tends very rapidly to a line on logarithmic scale. Anyways, it would be of little use.

We know that the $ \\[15pt]

{a_i} $ are the derivatives of $ \\[15pt]

{^xa|_0} $ , so a Fourier or Laplace transform would turn the derivatives into products. But that would mess with the rest of the equation.

RE: Taylor polinomial. System of equations for the coefficients. - marraco - 01/14/2016

Here I make an expansion of a row, in hope that it helps somebody to digest the equation.

(01/13/2016, 04:32 AM)marraco Wrote: We are now very close to the solution. The only obstacle remaining is the product:
$ \\[15pt]

{ \frac{1}{ \prod_{j=1}^{i} c_{n,j}!} } $

The product is what I called "the integer divisor"

(05/03/2015, 04:35 AM)marraco Wrote: $ \mathbf{
-The \,\ integer \,\ divisor \,\ is \,\ the \,\ product \,\ of \,\ the \,\ factorials \,\ of \,\ the \,\ exponents \,\ of \,\ a_i
} $

$

+a.(lna.a_9 + lna^2.a_4 .a_5 + lna^2 .a_3 .a_6 + \frac{lna^3}{6} .a_3^3 + lna^2.a_2 .a_7 + lna^3.a_2 .a_3 .a_4 + \frac{lna^3}{2}.a_2^2 .a_5 + \frac{lna^4}{6} .a_2^3 .a_3 + lna^2.a_1 .a_8 + \frac{lna^3}{2} .a_1 .a_4^2 + \\
lna^3 .a_1 .a_3 .a_5 + lna^3 .a_1 .a_2 .a_6 + \frac{lna^4}{2} .a_1 .a_2 .a_3^2 + \frac{lna^4}{2} .a_1 .a_2^2 .a_4 + \frac{lna^5}{24}.a_1 .a_2^4 + \frac{lna^3}{2}.a_1^2 .a_7 + \frac{lna^4}{2}.a_1^2 .a_3 .a_4 + \frac{lna^4}{2} .a_1^2 .a_2 .a_5 + \frac{lna^5}{4}.a_1^2 .a_2^2 .a_3 + \frac{lna^4}{6} .a_1^3 .a_6 +\\
\frac{lna^5}{12} .a_1^3 .a_3^2 + \frac{lna^5}{6} .a_1^3 .a_2 .a_4 + \frac{lna^6}{36}.a_1^3 .a_2^3 + \frac{lna^5}{24}.a_1^4 .a_5 + \frac{lna^6}{24}.a_1^4 .a_2 .a_3 + \frac{lna^6}{120}.a_1^5 .a_4 + \frac{lna^7}{240} .a_1^5 .a_2^2 + \frac{lna^7}{720} .a_1^6 .a_3 + \frac{lna^8}{5040}.a_1^7 .a_2 + \frac{lna^9}{362880}.a_1^9 ).x^9
$

\( \small {
9\,\,\right\,\, a_9^1\,\,\right\,\, 1! \,=\, 1 \\
1,\, 8\,\,\right\,\,a_1^1\,.\,a_8^1\,\,\right\,\,1!.1! \,=\, 1 \\
2,\, 7\,\,\right\,\, a_2^1\,.\,a_7^1\,\,\right\,\, 1!.1! \,=\, 1 \\
3,\, 6\,\,\right\,\, a_3^1\,.\,a_6^1\,\,\right\,\, 1!.1! \,=\, 1 \\
4,\, 5\,\,\right\,\, a_4^1\,.\,a_5^1\,\,\right\,\, 1!.1! \,=\, 1 \\
1,\, 1,\, 7\,\,\right\,\,a_1^2\,.\,a_7^1\,\,\right\,\,2!.1! \,=\, 2 \\
1,\, 2,\, 6\,\,\right\,\,a_1^1\,.\,a_2^1\,.\,a_6^1\,\,\right\,\,1!.1!.1! \,=\, 1 \\
1,\, 3,\, 5\,\,\right\,\,a_1^1\,.\,a_3^1\,.\,a_5^1\,\,\right\,\,1!.1!.1! \,=\, 1 \\
1,\, 4,\, 4\,\,\right\,\,a_1^1\,.\,a_4^2\,\,\right\,\,1!.2! \,=\, 2 \\
2,\, 2,\, 5\,\,\right\,\, a_2^2\,.\,a_5^1\,\,\right\,\, 2!.1! \,=\, 2 \\
2,\, 3,\, 4\,\,\right\,\, a_2^1\,.\,a_3^1\,.\,a_4^1\,\,\right\,\, 1!.1!.1! \,=\, 1 \\
3,\, 3,\, 3\,\,\right\,\, a_3^3\,\,\right\,\, 3! \,=\, 6 \\
1,\, 1,\, 1,\, 6\,\,\right\,\,a_1^3\,.\,a_6^1\,\,\right\,\,3!.1! \,=\, 6 \\
1,\, 1,\, 2,\, 5\,\,\right\,\,a_1^2\,.\,a_2^1\,.\,a_5^1\,\,\right\,\,2!.1!.1! \,=\, 2 \\
1,\, 1,\, 3,\, 4\,\,\right\,\,a_1^2\,.\,a_3^1\,.\,a_4^1\,\,\right\,\,2!.1!.1! \,=\, 2 \\
1,\, 2,\, 2,\, 4\,\,\right\,\,a_1^1\,.\,a_2^2\,.\,a_4^1\,\,\right\,\,1!.2!.1! \,=\, 2 \\
1,\, 2,\, 3,\, 3\,\,\right\,\,a_1^1\,.\,a_2^1\,.\,a_3^2\,\,\right\,\,1!.1!.2! \,=\, 2 \\
2,\, 2,\, 2,\, 3\,\,\right\,\, a_2^3\,.\,a_3^1\,\,\right\,\, 3!.1! \,=\, 6 \\
1,\, 1,\,1,\, 1,\, 5\,\,\right\,\,a_1^4\,.\,a_5^1\,\,\right\,\,4!.1! \,=\, 24 \\
1,\, 1,\, 1,\, 2,\, 4\,\,\right\,\,a_1^3\,.\,a_2^1\,.\,a_4^1\,\,\right\,\,3!.1!.1! \,=\, 6 \\
1,\, 1,\, 1,\, 3,\, 3\,\,\right\,\,a_1^3\,.\,a_3^2\,\,\right\,\,3!.2! \,=\, 12 \\
1,\, 1,\, 2,\, 2,\, 3\,\,\right\,\,a_1^2\,.\,a_2^2\,.\,a_3^1\,\,\right\,\,2!.2!.1! \,=\, 4 \\
1,\, 2,\, 2,\, 2,\, 2\,\,\right\,\,a_1^1\,.\,a_2^4\,\,\right\,\,1!.4! \,=\, 24 \\
1,\, 1,\, 1,\, 1,\, 1,\, 4\,\,\right\,\,a_1^5\,.\,a_4^1\,\,\right\,\,5!.1! \,=\, 120 \\
1,\, 1,\, 1,\, 1,\, 2,\, 3\,\,\right\,\,a_1^4\,.\,a_2^1\,.\,a_3^1\,\,\right\,\,4!.1!.1! \,=\, 24 \\
1,\, 1,\, 1,\, 2,\, 2,\, 2\,\,\right\,\,a_1^3\,.\,a_2^3\,\,\right\,\,3!.3! \,=\, 36 \\
1,\, 1,\, 1,\, 1,\, 1,\, 1,\, 3\,\,\right\,\,a_1^6\,.\,a_3^1\,\,\right\,\,6!.1! \,=\, 720 \\
1,\, 1,\, 1,\, 1,\, 1,\, 2,\, 2\,\,\right\,\,a_1^5\,.\,a_2^2\,\,\right\,\,5!.2! \,=\, 240 \\
1,\, 1,\, 1,\, 1,\, 1,\, 1,\, 1,\, 2\,\,\right\,\,a_1^7\,.\,a_2^1\,\,\right\,\,7!.1! \,=\, 5040 \\
1,\, 1,\, 1,\, 1,\, 1,\, 1,\, 1,\, 1,\, 1\,\,\right\,\,a_1^9\,\,\right\,\,9! \,=\, 362880 \\
} \)

^^ Here I expanded the row for i=9 of the equation:

$ \\[15pt]

{
\left [ \sum_{n=1}^{P(9)} \frac{a.ln(a)^{\sum_{j=1}^{9}c_{n,j}}}{ \prod_{j=1}^{9} c_{n,j}!}\prod_{j=1}^{9}a_j^{c_{n,j}} \right ]} $

after the substitution $ \\[25pt]

{a_i=\frac {b_i} {ln(a)} } $:

\(
= ln(^2a) \,
\left [ \sum_{n=1}^{P(9)} \prod_{j=1}^{9}{\frac{b_j^{c_{n,j}}} { c_{n,j}!} \right ] \,=\,

ln(^2a) \, \left [\frac {b_9^1} {1!}+\frac {b_1^1} {1!}\frac {b_8^1} {1!}+\frac {b_2^1} {1!}\frac {b_7^1} {1!}+\frac {b_3^1} {1!}\frac {b_6^1} {1!}+\frac {b_4^1} {1!}\frac {b_5^1} {1!}+\frac {b_1^2} {2!}\frac {b_7^1} {1!}+\frac {b_1^1} {1!}\frac {b_2^1} {1!}\frac {b_6^1} {1!}+\frac {b_1^1} {1!}\frac {b_3^1} {1!}\frac {b_5^1} {1!}+\frac {b_1^1} {1!}\frac {b_4^2} {2!}+\frac {b_2^2} {2!}\frac {b_5^1} {1!}+\frac {b_2^1} {1!}\frac {b_3^1} {1!}\frac {b_4^1} {1!}+\frac {b_3^3} {3!}+\frac {b_1^3} {3!}\frac {b_6^1} {1!}+\frac {b_1^2} {2!}\frac {b_2^1} {1!}\frac {b_5^1} {1!}+\frac {b_1^2} {2!}\frac {b_3^1} {1!}\frac {b_4^1} {1!}+\frac {b_1^1} {1!}\frac {b_2^2} {2!}\frac {b_4^1} {1!}+\frac {b_1^1} {1!}\frac {b_2^1} {1!}\frac {b_3^2} {2!}+\frac {b_2^3} {3!}\frac {b_3^1} {1!}+\frac {b_1^4} {4!}\frac {b_5^1} {1!}+\frac {b_1^3} {3!}\frac {b_2^1} {1!}\frac {b_4^1} {1!}+\frac {b_1^3} {3!}\frac {b_3^2} {2!}+\frac {b_1^2} {2!}\frac {b_2^2} {2!}\frac {b_3^1} {1!}+\frac {b_1^1} {1!}\frac {b_2^4} {4!}+\frac {b_1^5} {5!}\frac {b_4^1} {1!}+\frac {b_1^4} {4!}\frac {b_2^1} {1!}\frac {b_3^1} {1!}+\frac {b_1^3} {3!}\frac {b_2^3} {3!}+\frac {b_1^6} {6!}\frac {b_3^1} {1!}+\frac {b_1^5} {5!}\frac {b_2^2} {2!}+\frac {b_1^7} {7!}\frac {b_2^1} {1!}+\frac {b_1^9} {9!}

\right ] \)
The problematic terms come from the factors $ \\[25pt]

{\frac {b_i^q}{q!} $. The q! divisors may emerge not from the term raised to q. q! could emerge from the absence of the other terms: $ \\[25pt]

{b_i^0} $.
For example, the term $ \\[25pt]

{\frac {b_1^2} {2!} .\frac {b_2^2} {2!} .\frac {b_3^1} {1!} } $ is actually $ \\[25pt]

{\frac {b_1^2} {2!} .\frac {b_2^2} {2!} .\frac {b_3^1} {1!} \,.\, \frac {b_4^0} {0!}.\frac {b_5^0} {0!}.\frac {b_6^0} {0!}.\frac {b_7^0} {0!}.\frac {b_8^0} {0!}.\frac {b_9^0} {0!}} $

RE: Taylor polynomial. System of equations for the coefficients. - marraco - 03/13/2016

(01/13/2016, 04:32 AM)marraco Wrote: So, we want the vector $ \\[15pt]

{[a_i]} $, from the matrix equation:

$ \color{Blue}\\[15pt]

{\left [ {{i} \choose {r}} \right ]\cdot \left [a_i \right ] $$ \,=\, $$ {\color{Red}
\left [ \sum_{n=1}^{P(i)} \frac{a.ln(a)^{\sum_{j=1}^{i}c_{n,j}}}{ \prod_{j=1}^{i} c_{n,j}!}\prod_{j=1}^{i}a_j^{c_{n,j}} \right ]} $

Thanks to Daniel advice, is easy to see that the red side can be derived as a direct application of Faà di Bruno's formula

$ {d^i \over dx^i} f(g(x))
=\sum \frac{i!}{m_1!\,m_2!\,\cdots\,m_i!} \cdot
f^{(m_1+\cdots+m_i)}(g(x)) \cdot
\prod_{j=1}^i\left(\frac{g^{(j)}(x)}{j!}\right)^{m_j} $

in the blue red equation,

(04/30/2015, 03:24 AM)marraco Wrote: We want the coefficients aᵢ of this Taylor expansion:

$ {^xa}=\sum_{n=0}^{\infty}{a_n .x^n} $

They should match this equation:
$ {\color{Blue} {^{x+1}a} $$ \,=\, $$ {\color{Red} a^{^xa}} $

is easy to see on the red side, that

$ {m_j=c_{n,j}} $

$ {f(x)=a^x} $
$ {g(x)={^xa}=\sum_{n=0}^{\infty}{a_n .x^n}} $
$ f^{(m_1+\cdots+m_i)}\,_{(g(0))}=a.ln(a)^{\sum_{j=1}^{i}c_{n,j}} $

$ {\frac{g^{(j)}(0)}{j!}=a_j} $

RE: Taylor polinomial. System of equations for the coefficients. - Gottfried - 08/23/2016

(05/01/2015, 01:57 AM)marraco Wrote: This is a numerical example for base a=e
(...)

So you get this systems of equations (blue to the left, and red to the right):

Code:
[1 1 1 1 1 1 1 1 1] [ 1] [e] [0 1 2 3 4 5 6 7 8] [a₁] [e.a₁] [0 0 1 3 6 10 15 21 28] [a₂] [e.a₂+e/2.a₁²] [0 0 0 1 4 10 20 35 56] [a₃] [e.a₃+e.a₁.a₂+e/6.a₁³] [0 0 0 0 1 5 15 35 70] * [a₄] = [e.a₄+e/2.a₂²+e.a₃.a₁+e/2.a₂.a1²+e/24.a₁⁴] [0 0 0 0 0 1 6 21 56] [a₅] [...] [0 0 0 0 0 0 1 7 28] [a₆] [...] [0 0 0 0 0 0 0 1 8] [a₇] [...] [0 0 0 0 0 0 0 0 1] [a₈] [...]

Quote:It is a non linear system of equations, and the solution for this particular case is:

Code:
a₀= 1,00000000000000000 a₁= 1,09975111049169000 a₂= 0,24752638354178700 a₃= 0,15046151104294100 a₄= 0,12170896032120000 a₅= 0,16084324512292400 a₆= -0,02254254634348470 a₇= -0,10318144159688800 a₈= 0,06371479195361670
(...)

This is perhaps a good starting point to explain the use of Carleman-matrices in my (Pari/GP-supported) matrix-toolbox, because you've just applied things analoguously to how I do this, only you didn' express it in matrix-formulae.
To explain the basic idea of a Carlemanmatrix:
consider a powerseries
$ \hspace{100} \small f(x) = a_0 + a_1 x + a_2 x^2 + ... $

We express this in terms of the dot-product of two infinite-sized vectors
$ \hspace{100} \small V(x) \cdot A_1 = f(x) $
where the column-vector A_1 contains the coefficients $ \small A_1=[a_0,a_1,a_2,...] $ and the row-vector $ \small V(x)=[1,x,x^2,x^3,x^4,...] $
Now to make that idea valuable for function-composition /- iteration it would be good, if the output of such an operation were not simple a scalar, but of the same type ("vandermonde vector") as the input ($ \small V(x) $ ).

This leads to the idea of Carlemanmatrices: we just generate the vectors $ \small A_0,A_1,A_2,A_3,A_4,... $ where the vector $ \small A_k $ contains the coefficients for powers of f(x), such that $ \small V(x) \cdot A_k = f(x)^k $
$ \hspace{400} $ ... in a matrix $ \small A $ getting the operation:
$ \hspace{100} \small V(x) \cdot A = [ 1, f(x), f(x)^2, f(x)^3 ,... ] $ or
$ \hspace{100} \small V(x) \cdot A = V(f(x)) $

Having this general idea we can fill our toolbox with Carlemanmatrices for the composition of functions for a fairly wide range of algebra.
For instance the operation INC
$ \hspace{100} \text{INC := } \hspace{50} \small V(x) \cdot P = V(x+1) $
and its h'th iteration ADD
$ \hspace{100} \text{ADD(h) := INC ^h:= } \hspace{50} \small V(x) \cdot P^h = V(x+h) $
is then only a problem of powers of P

The operation MUL needs a diagonal vandermonde vector:
$ \hspace{100} \text{MUL(w) := } \hspace{50} \small V(x) \cdot ^dV(w) = V(x*w) $

The operation DEXP (= exp(x)-1) needs the matrix of Stirlingnumbers 2nd kind, similarity-scaled by factorials:
$ \hspace{100} \text{DEXP := } \hspace{50} \small V(x) \cdot S2 = V(e^x -1) $

and as an exercise, we see, that if we right-compose this with the INC -operation, we get the ordinary EXP operator, for which I give the matrix-name B:
$ \hspace{100} \text{EXP := } \hspace{50} \small V(x) \cdot S2 \cdot P = V( e^x -1) \cdot P = V(( e^x -1) +1) = V(e^x) $
$ \hspace{100} \text{EXP := } \hspace{50} \small V(x) \cdot B = V( e^x) $

Of course, iterations of the EXP require then only powers of the matrix B.

To see, that this is really useful, we need a lemma on the uniqueness of power-series. That is, in the new matrix-notation:
If a function $ \small V(x) \cdot A_1 = f(x)$ $ is continuous for a (even small) continuous range of the argument x, then the coefficients in A_1 are uniquely determined.
That uniqueness of the coefficients in A_1 is the key, that we can look at the compositions of Carleman-matrices alone without respect of the notation with the dotproduct by V(x) and for instance, we can make use of the analysis of Carlemanmatrix-decompositions like
$ \hspace{100} \small V(x) \cdot S2 \cdot P = V(x) \cdot B $
and can analyze
$ \hspace{100} \small S2 \cdot P = B $
directly, for instance to arrive at the operation LOGP : log(1+x)
$ \hspace{100} \small S2 \cdot P = B \\
\hspace{100} \small P = S2^{-1} \cdot B \qquad \text{ where } \qquad S2^{-1} = S1 \\
\hspace{100} \small S2 = B \cdot P^{-1} \\
$

<hr>
Now I relate this to that derivation which I've quoted from marraco's post.

(05/01/2015, 01:57 AM)marraco Wrote: This is a numerical example for base a=e
(...)
So you get this systems of equations (blue to the left, and red to the right):

Code:
[1 1 1 1 1 1 1 1 1] [ 1] [e] [0 1 2 3 4 5 6 7 8] [a₁] [e.a₁] [0 0 1 3 6 10 15 21 28] [a₂] [e.a₂+e/2.a₁²] [0 0 0 1 4 10 20 35 56] [a₃] [e.a₃+e.a₁.a₂+e/6.a₁³] [0 0 0 0 1 5 15 35 70] * [a₄] = [e.a₄+e/2.a₂²+e.a₃.a₁+e/2.a₂.a1²+e/24.a₁⁴] [0 0 0 0 0 1 6 21 56] [a₅] [...] [0 0 0 0 0 0 1 7 28] [a₆] [...] [0 0 0 0 0 0 0 1 8] [a₇] [...] [0 0 0 0 0 0 0 0 1] [a₈] [...]

First we see the Pascalmatrix P on the lhs in action, then the coefficients $ \small [1,a_1,a_2,...] $ of the Abel-function $ \small \alpha(x) $ in the vector, say, A_1 . So the left hand is
$ \hspace{100} \small P \cdot A_1 $

To make things smoother first, we assume A as complete Carlemanmatrix, expanded from A_1. If we "complete" that left hand side to discuss this in power series we have
$ \hspace{100} \small V(x) \cdot P \cdot A = V(x+1) \cdot A = V( \alpha (x+1)) $

It is very likely, that the author wanted to derive the solution for the equation $ \small \alpha(x+1) = e^{\alpha(x)} $ ; so we would have for the right hand side
$ \hspace{100} \small V(x) \cdot A \cdot B = V(\alpha(x)) \cdot B = V( e^{\alpha (x)}) $
and indeed, expanding the terms using the matrixes as created in Pari/GP with, let's say size of 32x32 or 64x64 we get very nice approximations to that descriptions in the rhs of the quoted matrix-formula.

What we can now do, depends on the above uniqueness-lemma: we can discard the V(x)-reference, just writing $ \small A \cdot B $ and looking at the second column of B only $ \small A \cdot B[,1] = Y $ we get $ \small y_1 = e \cdot a_1 , y_2 =e \cdot (... ) $ as shown in the quoted post.

So indeed, that system of equations of the initial post is expressible by
$ \hspace{100} \small P \cdot A = A \cdot B $
and the OP searches a solution for A.

<hr>
While I've -at the moment- not yet a solution for A this way, we can, for instance, note that if A is invertible, then the equation can be made in a Jordan-form:
$ \hspace{100} \small P = A \cdot B \cdot A^{-1} $
which means, that B can be decomposed by similarity transformations into a triangular Jordan block, namely the Pascalmatrix - and having a Jordan-solver for finite matrix-sizes, one could try, whether increasing the matrix-sizes the Jordan-solutions converge to some limit-matrix A.

For the alternative: looking at the "regular tetration" and the Schröder-function (including recentering the powerseries around the fixpoint) one gets a simple solution just by the diagonalization-formulae for triangular Carlemanmatrices which follow the same formal analysis using the "matrix-toolbox" which can, for finite size and numerical approximations, nicely be constructed using the matrix-features of Pari/GP.