I fiddled a bit around with Gottfried's suggestion of LU decomposition of the Abel matrix (though in the end the formula is independent of the LU decomposition).

The annoying thing about calculating the intuitive Abel function (by solving the equation Ax=b where A is the Abel matrix, x the powerseries development of the Abel function and b=(1,0,...)) that if you want to increas the matrix size you have to solve the complete equation again without being able to use your previous solution.

Now I found a way how you can compute the inverse of the \( A_n \) matrix by using the \( A_{n-1} \) Abel matrix. I dissect the matrix as follows, for brevity I set \( A=A_{n-1} \):

\(

A_n=\left(\begin{array}{ccc|c}

\phantom{1}& &\phantom{1} & \\

&A& &\acute{a}\\

& & & \\\hline

&\grave{a}& &a_n

\end{array}\right)

\)

\( \acute{a} \) means column vector and \( \grave{a} \) means row vector.

The final incremental formula is then:

\( {A_n}^{-1} =(A^{-1})_{+0} + \frac{(-A^{-1}\acute{a}\oplus

1)(-\grave{a}A^{-1}\oplus 1)}{a_n-\grave{a}A^{-1}\acute{a}} \)

Where \( \oplus 1 \) means adding the entry 1 to the vector and \( (A^{-1})_{+0} \) is \( A^{-1} \) extended to a nxn matrix by filling with 0's.

The deriviation is perhaps too uninteresting and cumbersome to put, but I can post it if inquired.

The annoying thing about calculating the intuitive Abel function (by solving the equation Ax=b where A is the Abel matrix, x the powerseries development of the Abel function and b=(1,0,...)) that if you want to increas the matrix size you have to solve the complete equation again without being able to use your previous solution.

Now I found a way how you can compute the inverse of the \( A_n \) matrix by using the \( A_{n-1} \) Abel matrix. I dissect the matrix as follows, for brevity I set \( A=A_{n-1} \):

\(

A_n=\left(\begin{array}{ccc|c}

\phantom{1}& &\phantom{1} & \\

&A& &\acute{a}\\

& & & \\\hline

&\grave{a}& &a_n

\end{array}\right)

\)

\( \acute{a} \) means column vector and \( \grave{a} \) means row vector.

The final incremental formula is then:

\( {A_n}^{-1} =(A^{-1})_{+0} + \frac{(-A^{-1}\acute{a}\oplus

1)(-\grave{a}A^{-1}\oplus 1)}{a_n-\grave{a}A^{-1}\acute{a}} \)

Where \( \oplus 1 \) means adding the entry 1 to the vector and \( (A^{-1})_{+0} \) is \( A^{-1} \) extended to a nxn matrix by filling with 0's.

The deriviation is perhaps too uninteresting and cumbersome to put, but I can post it if inquired.