Padé approximants are a particular type of rational approximation. The $L, M$ Padé approximant is denoted by
$$[L/M] = P_L(x)/Q_M(x)$$
where $P_L(x)$ is a polynomial of degree less than or equal to $L$, and $Q_M(x)$ is a polynomial of degree less than or equal to $M$. The formal power series
$$f(x) = \sum _{j=0}^\infty f_jx^j,$$
determines the coefficients by the equation
$$f(x) -P_L(x)/Q_M(x)=O(x^{L+M+1}).$$
There is also an extra condition added by Baker which states that
$$Q_M(0)= 1.$$
A proof at the beginning of Essentials of Padé approximants shows that the $[L/M]$ Padé approximant is unique when it exists (Frobenius, $1881$).
Now, it is also well known that Jacobi made a breakthrough in $1846$. In particular, he applied Cramer's rule to get a determinant solution to the Padé approximant equations.
For $P_L(x)$ and $Q_M(x)$, a simplification of Cauchy's solution to the problem of rational interpolation produces two determinant solution formulas:
The numerator, $P_L(x)$, can be represented by
$$\label{ten}
P_L(x)=\mathrm{det}\,\begin{vmatrix} f_{L-M+1} & f_{L-M+2} & \cdots & f_{L+1}\\ \vdots & \vdots & \ddots & \vdots \\f_L & f_{L+1} & \cdots &f_{L+M} \\ \sum_{j=M}^L f_{j-M}x^j & \sum_{j=M-1}^L f_{j-M+1}x^j & \cdots & \sum_{j=0}^L f_jx^j \end{vmatrix}\ $$
Similarly, the denominator can be represented by
$$\label{nine}
Q_M(x)=\mathrm{det}\,\begin{vmatrix} f_{L-M+1} & f_{L-M+2} & \cdots & f_{L+1}\\ \vdots & \vdots & \ddots & \vdots\\f_L & f_{L+1} & \cdots &f_{L+M} \\ x^M & x^{M-1} & \cdots & 1 \end{vmatrix}\ $$
However, it isn't clear to me how Jacobi found these determinant solutions. If I go and review the source:
Jacobi, C. G. J. $(1846)$ Uber die Darstellung einer Reihe Gegebner Werthe durch eine Gebrochne Rationale Function. J. Reine Angew. Math. (Crelle) $30, 127-156.$
I find that Crelle's journal contains the necessary information on pages $127-156$ of the $1846$ issue. However, the paper is in German and I cannot read it.
Does anyone know if this paper has been translated into English? Is there some way I could find this out on my own? I am looking to do the derivation myself and wasn't able to find an equivalent article in English.
Best Answer
If we add the expression of $Q_{n,m}$, $Q_{n,m}(z)=b_{0}+b_{1}z+\dots+b_{m}z^{m}$, as a new equation in the original system, we have \begin{equation*} \begin{matrix} f_{n+1}b_{0}&+&f_{n}b_{1}&+&\cdots&+&f_{n-m+1}b_{m}&=&0 \\ \vdots&&\vdots &&&&\vdots &\;\;\;\;\;&\vdots\\ f_{n+m}b_{0}&+&f_{n+m-1}b_{m-1}&+&\cdots&+&f_{n}b_{m}&=&0\\ b_{0}&+&b_{1}z&+&\cdots&+&b_{m}z^{m}&=&Q_{n,m}(z), \end{matrix} \end{equation*} considering the $z\in\mathbb{C}$ such that \begin{equation}\label{a} \Delta_{m}(z)=\begin{vmatrix} f_{n+1}&f_{n} &\cdots &f_{n-m+1}\\ \vdots &\vdots & &\vdots \\ f_{n+m}&f_{n+m-1}&\cdots&f_{n}\\ 1 & z &\cdots & z^{m} \end{vmatrix}\neq0, \end{equation} by obtaining the expression of $ b_{0}$ using the Cramer rule, we conclude $$b_{0}=(-1)^{m}Q_{n,m}(z)\frac{H_{n,m}}{\begin{vmatrix} f_{n+1}&f_{n}&\cdots&f_{n-m+1} \\ \vdots&\vdots& &\vdots \\ f_{n+m}&f_{n+m-1} &\cdots &f_{n}\\ 1&z&\dots&z^{m} \end{vmatrix}},$$ for $z$ that verify the formula of $P_{n,m}$. Therefore, we have shown that is true for all $z\in\mathbb{C}$ such that $\Delta_{m}(z)\neq0 $. It follows that is true $\forall z\in\mathbb{C}$ since $Q_{n, m}(z)$ and $\frac{(-1)^{m}b_{0}}{H{n,m}}\Delta_{m}(z)$ are polynomials at the most $m$ degree and it is enough that they coincide in $m+1$ points so that both polynomials are equal.