Padé approximationi of polynomial

Question

Following wikipedia, I'm trying to compute the $[1/1]$-Padé approximation of $f(x)=x^2$. It should be of the form $\frac{a+bx}{1+cx}$, but this is either zero or a power series with a non-zero constant or linear term. In any case, it cannot approximate $x^2$ with an error of order $x^3$. Am I missing an assumption?

Answer 1

Let us seek a general expression for the $[1/1]$ Padé approximant in terms of derivatives. We have

$f(x)=\dfrac{a+bx}{1+cx}+O(x^3)$

and expand using the geometric series formula to get

$f(x)=a+(b-ac)x+(ac^2-bc)x^2+O(x^3)$

So matching with the Taylor series

$f(x)=f(0)+f'(0)x+\frac12f''(0)x^2+O(x^3)$

we get the system

$a=f(0)$

$b-ac=f'(0)\implies b=f'(0)+cf(0)$

$ac^2-bc=\frac12f''(0)\implies -f'(0)c=\frac12f''(0)$

$\therefore c=-\dfrac{f''(0)}{2f'(0)}, b=f'(0)-\dfrac{f(0)f''(0)}{2f'(0)},a=f(0).$

It becomes apparent that a singularity occurs when $f'(0)=0$. That is where your problem came with $f(x)=x^2$.

If your aim is to practice generating Padé approximants, choose a function that avoids this problem. For instance, with $f(x)=x^2+x$ whose derivative at $0$ is nonzero you should smoothly get $x/(1-x)$.

Answer 2

I give a necessary and sufficient criterion for the existence of a formal Padé $[m/n]$-approximation assuming that $f$ has a Taylor series, say $f=\sum_{k=0}^\infty a_kX^k$. Let $\beta=\sum b_kX^k$ and $\gamma=\sum c_kX^k$ be the polynomials of the approximation with degree $\le m$ and $\le n$ respectively and with $c_0=1$.
Then $f\gamma\equiv\beta\pmod{X^{m+n}}$ translates to the linear system $\sum_{i=0}^ka_{k-i}c_i=b_k$ for $k=0,\ldots,m+n$. Since $b_k=0$ for $k>m$ and $c_0=1$, we have $$\sum_{i=1}^{k}a_{k-i}c_i=-a_k$$ for $k=m+1,\ldots,m+n$. The Padé $[m/n]$-approximation exists if and only if this system has a solution (the $b_k$ are determined by $\sum_{i=0}^ka_{k-i}c_i=b_k$ for $k=0,\ldots,m$). For convenience let $a_k=0$ for $k<0$. Define $A:=(a_{m+i-j})_{i,j=1}^{n}$ and $b:=-(a_{m+i})_{i=1}^n$. Then $(c_i)_{i=1}^n$ is a solution of $Ax=b$. By linear algebra, the Padé $[m/n]$-approximation exists if and only if $A$ and the extended matrix $(A|b)$ have the same rank.

In the special case of the question we have $n=m=1$, $a_0=a_1=0$ and $A=(0)$, but $b=(-1)$. In general, $A$ is a Toeplitz matrix, which is invertible for instance, if $a_0=\ldots=a_{m-1}=0$ and $a_m\ne 0$.

Answer 3

Let us try to find a rational approximation (for $0 \leq x \leq t$) minimizing the norm $$\Phi=\int_0^t \left(x^2-\frac{a+b x}{1+c x}\right)^2\,dx$$ which is $$\Phi=\frac{2 (b c-1) (a c-b) \log (c t+1)}{c^4}+\frac{t (2 a c+b (b c-2))}{c^3}-\frac{(b-a c)^2}{c^3 (c t+1)}+$$ $$\frac{(b-a c)^2}{c^3}+\frac{t^2 (b-a c)}{c^2}-\frac{2 b t^3}{3 c}+\frac{t^5}{5}$$ Computing the partial derivatives $$c^3\frac{\partial \Phi}{\partial a}=\frac{c t \left(c \left(2 a c-2 b-c t^2+t\right)+2\right)}{c t+1}+2 (b c-1) \log (c t+1)\tag 1$$ $$3c^4\frac{\partial \Phi}{\partial b}=6 \left(a c^2-2 b c+1\right) \log (c t+1)+$$ $$\frac{c t (c (-6 a c+6 b (c t+2)+t (c t (1-2 c t)-3))-6)}{c t+1}\tag 2$$ So, for a given value of $t$,we have the expressions of $a(c)$ and $b(c)$ (two linear equations for two variables).

So, what is left is to solve for a given $t$ $$\frac{\partial \Phi}{\partial c}=0$$

This is a monster but using numerical methods, there is no major problem.

Trying for $t=2$ and making the numbers rational $$a \sim -\frac{91}{452} \qquad b \sim \frac{411}{439} \qquad c \sim -\frac{55}{184}\qquad \Phi=9.65\times 10^{-3}$$ which is not totally ridiculous.

At the midpoint $(x=1)$, the approximation gives $\frac{6707858}{6399303}=1.04822$.