Approximaths

Padé approximants of sin(x)

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We are interested in the P(2,2) Padé approximant of the sinus function. We first have to consider de Taylor series of sin(x) up to terms 2+2=4:

\displaystyle sin(x) = x - \frac{x^3}{3!} + \mathcal{O}(x^5)

For m=2 and n=2, the Hankel determinant (see previous post) is:

\displaystyle H_{2,2}(f) = \begin{vmatrix} C_{1} & C_{2} \\ C_{2} & C_{3} \\ \end{vmatrix}

The Hankel determinant corresponding to the sinus series above is therefore:

\displaystyle H_{2,2}(sin) = \begin{vmatrix} 1 & 0 \\ 0 & -\frac{1}{6} \\ \end{vmatrix} = -\frac{1}{6}

According to the previous section we have to first solve:

\displaystyle \begin{pmatrix} 1 & 0 \\ 0 & -\frac{1}{6} \end{pmatrix} \begin{pmatrix} B_2 \\ B_1 \end{pmatrix} = - \begin{pmatrix} C_{3} \\ C_{4} \\ \end{pmatrix}

Since C_3 = -\frac{1}{6} and C_4 = 0 we have to solve:

\displaystyle \begin{pmatrix} 1 & 0 \\ 0 & -\frac{1}{6} \end{pmatrix} \begin{pmatrix} B_2 \\ B_1 \end{pmatrix} = \begin{pmatrix} \frac{1}{6} \\ 0 \\ \end{pmatrix}

So B_2 = \frac{1}{6} and B_1 = 0.

Finally we have to solve this system (B_0 being set to one):

\displaystyle \begin{pmatrix} C_0 & 0 & 0 \\ C_1 & C_0 & 0 \\ C_2 & C_1 & C_0 \end{pmatrix} \begin{pmatrix} 1 \\ B_1 \\ B_2 \end{pmatrix} = \begin{pmatrix} A_0 \\ A_1 \\ A_2 \end{pmatrix}
\displaystyle \begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ \frac{1}{6} \end{pmatrix} = \begin{pmatrix} A_0 \\ A_1 \\ A_2 \end{pmatrix}

This implies that A_0 = 0, A_1 = 1 and A_2 = 0. The P(2,2) Padé approximant for sin(x) is therefore:

\displaystyle P(2,2) = \frac{x}{1 + \frac{1}{6}x^2}

The graph of sin(x) and its corresponding P(2,2) approximant are shown in figure 1.

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