Approximaths

Padé approximants of sqrt(1+x) and 1/sqrt(1+x)

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In this post we will have a look at the Padé approximants of \sqrt{1+x} and \frac{1}{\sqrt{1+x}}. The Maclaurin series of \sqrt{1+x} is:

\displaystyle 1+ \frac{1}{2}x - \frac{1}{8} x^2 + \frac{1}{16}x^3 - \frac{5}{128} x^4 + \frac{7}{256}x^5 + \dots

The MacLaurin series of \frac{1}{\sqrt{1+x}} is:

\displaystyle 1 - \frac{1}{2}x + \frac{3}{8}x^2 - \frac{5}{16}x^3 + \frac{35}{128}x^4 + \dots

According to the section concerning the calculation of Padé approximants using a matrix notation (see post Computing Padé approximants, we have to solve two linear systems sequentially. For \sqrt{1+x} we therefore have to first solve:

\displaystyle \begin{pmatrix} C_1 & C_2 \\ C_2 & C_3 \end{pmatrix} \begin{pmatrix} B_2 \\ B_1 \end{pmatrix} = -\begin{pmatrix} C_3 \\ C_4 \end{pmatrix}
\displaystyle \begin{pmatrix} \frac{1}{2} & -\frac{1}{8} \\ -\frac{1}{8} & \frac{1}{16} \end{pmatrix} \begin{pmatrix} B_2 \\ B_1 \end{pmatrix} = \begin{pmatrix} -\frac{1}{16} \\ \frac{5}{128} \end{pmatrix}
\displaystyle \begin{pmatrix} B_2 \\ B_1 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & -\frac{1}{8} \\ -\frac{1}{8} & \frac{1}{16} \end{pmatrix}^{-1} \begin{pmatrix} -\frac{1}{16} \\ \frac{5}{128} \end{pmatrix}
\displaystyle \begin{pmatrix} B_2 \\ B_1 \end{pmatrix} = \begin{pmatrix} 4 & 8 \\ 8 & 32 \end{pmatrix} \begin{pmatrix} -\frac{1}{16} \\ \frac{5}{128} \end{pmatrix}
\displaystyle \begin{pmatrix} B_2 \\ B_1 \end{pmatrix} = \begin{pmatrix} \frac{1}{16} \\ \frac{3}{4} \end{pmatrix}

Injecting the B_n coefficients calculated above in the second linear system we have:

\displaystyle \begin{pmatrix} C_0 & 0 & 0 \\ C_1 & C_0 & 0 \\ C_2 & C_1 & C_0 \end{pmatrix} \begin{pmatrix} B_0 \\ B_1 \\ B_2 \end{pmatrix} = \begin{pmatrix} A_0 \\ A_1 \\ A_2 \end{pmatrix}
\displaystyle \begin{pmatrix} 1 & 0 & 0 \\ \frac{1}{2} & 1 & 0 \\ -\frac{1}{8} & \frac{1}{2} & 1 \end{pmatrix} \begin{pmatrix} 1 \\ \frac{3}{4} \\ \frac{1}{16} \end{pmatrix} = \begin{pmatrix} A_0 \\ A_1 \\ A_2 \end{pmatrix}

From this system we obtain A_0 = 1, A_1 = \frac{5}{4}, A_2 = \frac{5}{16}. Therefore:

\displaystyle P(2,2) = \frac{A_0 + A_1 x + A_2 x^2}{1 + B_1 x + B_2 x^2}
\displaystyle = \frac{1 +\frac{5}{4} x + \frac{5}{16} x^2}{1 + \frac{3}{4} x + \frac{1}{16} x^2}

For \frac{1}{\sqrt{1+x}} we have to first solve:

\displaystyle \begin{pmatrix} C_1 & C_2 \\ C_2 & C_3 \end{pmatrix} \begin{pmatrix} B_2 \\ B_1 \end{pmatrix} = -\begin{pmatrix} C_3 \\ C_4 \end{pmatrix}
\displaystyle \begin{pmatrix} -\frac{1}{2} & \frac{3}{8} \\ \frac{3}{8} & -\frac{5}{16} \end{pmatrix} \begin{pmatrix} B_2 \\ B_1 \end{pmatrix} = \begin{pmatrix} \frac{5}{16} \\ -\frac{35}{128} \end{pmatrix}
\displaystyle \begin{pmatrix} B_2 \\ B_1 \end{pmatrix} = \begin{pmatrix} -\frac{1}{2} & \frac{3}{8} \\ \frac{3}{8} & -\frac{5}{16} \end{pmatrix}^{-1} \begin{pmatrix} \frac{5}{16} \\ -\frac{35}{128} \end{pmatrix}
\displaystyle \begin{pmatrix} B_2 \\ B_1 \end{pmatrix} = \begin{pmatrix} -20 & -24 \\ -24 & -32 \end{pmatrix} \begin{pmatrix} \frac{5}{16} \\ -\frac{35}{128} \end{pmatrix}
\displaystyle \begin{pmatrix} B_2 \\ B_1 \end{pmatrix} = \begin{pmatrix} \frac{5}{16} \\ \frac{5}{4} \end{pmatrix}

Injecting the B_n coefficients calculated above in the second linear system we have:

\displaystyle \begin{pmatrix} C_0 & 0 & 0 \\ C_1 & C_0 & 0 \\ C_2 & C_1 & C_0 \end{pmatrix} \begin{pmatrix} B_0 \\ B_1 \\ B_2 \end{pmatrix} = \begin{pmatrix} A_0 \\ A_1 \\ A_2 \end{pmatrix}
\displaystyle \begin{pmatrix} 1 & 0 & 0 \\ -\frac{1}{2} & 1 & 0 \\ \frac{3}{8} & -\frac{1}{2} & 1 \end{pmatrix} \begin{pmatrix} 1 \\ \frac{5}{4} \\ \frac{5}{16} \end{pmatrix} = \begin{pmatrix} A_0 \\ A_1 \\ A_2 \end{pmatrix}

From this system we obtain A_0 = 1, A_1 = \frac{3}{4}, A_2 = \frac{1}{16}. Therefore:

\displaystyle P(2,2) = \frac{A_0 + A_1 x + A_2 x^2}{1 + B_1 x + B_2 x^2}
\displaystyle = \frac{1 + \frac{3}{4} x + \frac{1}{16} x^2}{1 + \frac{5}{4} x + \frac{5}{16} x^2}

We observe that:

\displaystyle P(2,2)_{\frac{1}{\sqrt{1+x}}} = \frac{1}{P(2,2)_{\sqrt{1+x}}}

These calculations suggest that, if g = \frac{1}{f} then:

P(n,m)_{g} = \frac{1}{P(n,m)_{f}}

Where P(n,m)_{g} and P(n,m)_{f} are the Padé approximants of g and f respectively. This proposition is actually true and can be proved formally.

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