Continued fractions I

We can use Padé approximants to build a sequence of truncated continued fractions representing a given function. For example using $P(1,1)$ :

$P(1,1) = \frac{A_0 + A_1 x}{1 + B_1 x} = C_0 + C_1 x + C_2 x^2$

Solving the linear systems presented in post Computing Padé approximants sequentially leads to:

$C_1 B_1 = - C_2 \quad \Rightarrow \quad B_1 = -\frac{C_2}{C_1}$

$\begin{pmatrix} C_0 & 0 \\ C_1 & C_0 \end{pmatrix} \begin{pmatrix} 1 \\ B_1 \end{pmatrix} = \begin{pmatrix} A_0 \\ A_1 \end{pmatrix}$

$A_0 = C_0, \quad A_1 = C_1 + C_0 B_1$

Setting $C_0 = 1$ :

$P(1,1) = \frac{1 + (C_1 + B_1) x}{1 + B_1 x}$

$P(1,1) = \frac{1 + (C_1 - \frac{C_2}{C_1}) x}{1 - \frac{C_2}{C_1} x}$

$P(1,1) = \frac{1 + (C_1 - \frac{C_2}{C_1}) x}{1 + (C_1 - \frac{C_2}{C_1}) x - C_1 x}$

$P(1,1) = \frac{1}{1 - \frac{C_1 x}{1 + (C_1 - \frac{C_2}{C_1}) x}}$

$P(1,1) = \frac{1}{1 - \frac{C_1 x}{1 - (\frac{C_2}{C_1} - C_1) x}}$

Now define:

$\frac{b_0}{1 - \frac{b_1 x}{1 - b_2 x}} = \frac{1}{1 - \frac{C_1 x}{1 - \left(\frac{C_2}{C_1} - C_1\right) x}}$

We have:

$b_0 = 1, \quad b_1 = C_1, \quad b_2 = \frac{C_2}{C_1} - C_1, \quad \ldots$

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