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Determine the Padé approximation of degree $5$ with $ n =2 $ and $ m= 3$ for $f(x) = e^{-x}$.


Suppose $r$ is a rational function of degree $N$.$$ r(x) = \frac{p(x)}{q(x)} = \frac{p_0 +p_1x + \cdots + p_nx^n}{q_0 +q_1x+ \cdots + q_mx^m}. $$ The Padé approximation technique is$$ f(x)-r(x) = f(x)-\frac{p(x)}{q(x)} = \frac{f(x)q(x)-p(x)}{q(x)} = \frac{\left(\sum\limits_{i=0}^{\infty} a_ix^i \right)\left(\sum\limits^m_{i=0}q_{i}x^i \right)-\sum\limits^n_{i=0} p_ix^i}{q(x)}. $$ To find the Padé approximation we need to choose $p_0,p_1,p_2,q_1$ and $q_2,q_3$. The Maclaurin expansion of $e^{-x}$ is $\sum\limits_{i = 0}^{\infty} \dfrac{(-1)^i}{i!} x^i$. Since $q_0 \neq 0$, thus $q_0= 1$ if $x= 0$, and $p_0 =1$. How does one find $p_1,p_2,q_1,q_2,q_3$ ao that$$ r_{2,3}(x) = \frac{p_0+p_1x+p_2x^2}{q_0+q_1x+q_2x^2+q_{3}x^3}? $$ Does one have to transform this into a linear equation to solve?$$ \begin{bmatrix} a_0 & & & &\\ a_1 & a_0 & & &\\ a_2 & a_1 & a_0&& \\ a_3 & a_2 & a_1& a_0&\\ a_4 & a_3 & a_2 & a_1&a_0\\ \end{bmatrix} \begin{bmatrix} q_1\\ q_2\\ q_3\\ 0\\ 0 \end{bmatrix} - \begin{bmatrix} p_1\\ p_2\\ 0\\ 0\\ 0 \end{bmatrix} = -\begin{bmatrix} a_1\\ a_2\\ a_3\\ a_4\\ a_5 \end{bmatrix} $$ How do I solve this linear system so I may obtain the solution? Can someone demonstrate a concise method of Gaussian elimination to find $r_{2,3}$ and that can also work for $r_{1,4}$ or any $r_{n,m}$ for Padé approximation?

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  • $\begingroup$ @keith I found a website that shows a method using Gaussian Elimination with backward substitution to approximate the answer. Unfortunately it does not state how to use this method to find other r(x) such as $n= 2$ and $m=3$. Website $\endgroup$
    – Jon
    Commented Apr 20, 2018 at 0:40
  • $\begingroup$ They do $n=3$ and $m=2$. I think the method is similar in your case. $\endgroup$ Commented Apr 20, 2018 at 4:54

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$\newenvironment{gmatrix}{\left\lgroup\begin{matrix}}{\end{matrix}\right\rgroup}$In general, to find the Padé approximation $r(x) = \dfrac{\sum\limits_{k = 0}^n p_k x^k}{\sum\limits_{k = 0}^m q_k x^k}$ for $f(x) = \sum\limits_{k = 1}^\infty a_k x^k$, it reduces to find the solution to$$ \begin{gmatrix} a_0 &&&\\ a_1 & a_0 &&\\ \vdots & \ddots & \ddots &\\ a_{m + n} & \cdots & a_1 & a_0 \end{gmatrix} \begin{gmatrix} q_0 \\ q_1 \\ \vdots \\ q_{m + n} \end{gmatrix} = \begin{gmatrix} p_0 \\ p_1 \\ \vdots \\ p_{m + n} \end{gmatrix}, \tag{1} $$ where $p_{n + 1} = \cdots = p_{m + n} = 0$, $q_{m + 1} = \cdots = q_{m + n} = 0$. In order to solve this system of equations, usually it is assumed that $q_0 = 1$ and (1) becomes$$ \begin{gmatrix} a_0 &&\\ \vdots & \ddots &\\ a_{m + n - 1} & \cdots & a_0 \end{gmatrix} \begin{gmatrix} q_1 \\ \vdots \\ q_{m + n} \end{gmatrix} - \begin{gmatrix} p_1 \\ \vdots \\ p_{m + n} \end{gmatrix} = \begin{gmatrix} a_1 \\ \vdots \\ a_{m + n} \end{gmatrix},\ p_0 = a_0. $$ Since\begin{align*} &\mathrel{\phantom{=}}{} \begin{gmatrix} a_0 &&\\ \vdots & \ddots &\\ a_{m + n - 1} & \cdots & a_0 \end{gmatrix} \begin{gmatrix} q_1 \\ \vdots \\ q_{m + n} \end{gmatrix} = \begin{gmatrix} a_0 &&\\ \vdots & \ddots &\\ a_{m + n - 1} & \cdots & a_0 \end{gmatrix} \begin{gmatrix} q_1 \\ \vdots \\ q_m \\ 0 \\ \vdots \\ 0 \end{gmatrix}\\ &= \begin{gmatrix} a_0\\ \vdots & \ddots\\ a_{m - 1} & \cdots & a_0\\ a_m & \cdots & a_1 & 0\\ \vdots & \ddots & \vdots & \vdots & \ddots\\ a_{m + n - 1} & \cdots & a_n & 0 & \cdots & 0 \end{gmatrix} \begin{gmatrix} q_1 \\ \vdots \\ q_m \\ 0 \\ \vdots \\ 0 \end{gmatrix} = \begin{gmatrix} A & B\\ C & O \end{gmatrix} \begin{gmatrix} q_1 \\ \vdots \\ q_m \\ 0 \\ \vdots \\ 0 \end{gmatrix}, \end{align*} where$$ A = \begin{gmatrix} a_0\\ \vdots & \ddots\\ a_{m - 1} & \cdots & a_0 \end{gmatrix},\ B = -I_n,\ C = \begin{gmatrix} a_m & \cdots & a_1\\ \vdots & \ddots & \vdots\\ a_{m + n - 1} & \cdots & a_n \end{gmatrix}, $$ and $O$ is an $m × n$ zero matrix, and$$ -\begin{gmatrix} p_1 \\ \vdots \\ p_{m + n} \end{gmatrix} = -\begin{gmatrix} p_1 \\ \vdots \\ p_n \\ 0 \\ \vdots \\ 0 \end{gmatrix} = \begin{gmatrix} A & B\\ C & O \end{gmatrix} \begin{gmatrix} 0 \\ \vdots \\ 0 \\ p_1 \\ \vdots \\ p_n \end{gmatrix}, $$ then (1) is equivalent to$$ \begin{gmatrix} A & B\\ C & O \end{gmatrix} \begin{gmatrix} q_1 \\ \vdots \\ q_m \\ p_1 \\ \vdots \\ p_n \end{gmatrix} = \begin{gmatrix} a_1 \\ \vdots \\ a_{m + n} \end{gmatrix},\ p_0 = a_0. \tag{2} $$

Now, to find the Padé approximation $r(x) = \dfrac{\sum\limits_{k = 0}^2 p_k x^k}{\sum\limits_{k = 0}^3 q_k x^k}$ for $f(x) = \mathrm{e}^{-x} = \sum\limits_{k = 1}^\infty \dfrac{(-1)^k}{k!} x^k$, from (2) there is $q_0 = 1$, $p_0 = 1$, and$$ \begin{gmatrix} 1 &&& -1\\ -1 & 1 &&& -1\\ \dfrac{1}{2} & -1 & 1\\ -\dfrac{1}{6} & \dfrac{1}{2} & -1\\ \dfrac{1}{24} & -\dfrac{1}{6} & \dfrac{1}{2} \end{gmatrix} \begin{gmatrix} q_1 \\ q_2 \\q_3 \\ p_1 \\ p_2 \end{gmatrix} = \begin{gmatrix} -1 \\ \dfrac{1}{2} \\ -\dfrac{1}{6} \\ \dfrac{1}{24} \\ -\dfrac{1}{120} \end{gmatrix}, $$ thus $q_1 = -\dfrac{3}{5}$, $q_2 = -\dfrac{3}{20}$, $q_3 = -\dfrac{1}{60}$, $p_1 = \dfrac{2}{5}$, $p_2 = -\dfrac{1}{20}$. Therefore,$$ r_{2, 3}(x) = \frac{1 + \dfrac{2}{5}x - \dfrac{1}{20} x^2}{1 - \dfrac{3}{5} x - \dfrac{3}{20} x^2 - \dfrac{1}{60} x^3}. $$

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  • $\begingroup$ Are the blank spaces in your matrix zeros? $\endgroup$
    – Jon
    Commented Apr 22, 2018 at 15:39
  • $\begingroup$ One last question I know you might have answered it in the answer but why does the -1 columns go at the 2nd to last column and the last column of A? $\endgroup$
    – Jon
    Commented Apr 22, 2018 at 15:52
  • $\begingroup$ One last thing if you add the negative on the matrix $b$ that should give you $\frac{1-2/5x+1/20x^2}{1+3/5x+3/20x^2+1/60x^3}$ which gives you the final answer. $\endgroup$
    – Jon
    Commented Apr 23, 2018 at 2:41
  • $\begingroup$ @Jon Why reversing the signs? They're not the solution to (2) if signs are reversed. $\endgroup$ Commented Apr 23, 2018 at 2:49
  • $\begingroup$ I just do that to match the answer given in the solution that's all. $\endgroup$
    – Jon
    Commented Apr 23, 2018 at 13:25

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