Bounding error of Padé approximation

Question

I'm trying to understand how one would understand the error of a given Padé approximation for a function.

For instance, the $[2,1]$ approximant for $\log(1+x)$ is $\frac{x(6+x)}{6+4x}$. Is there a reasonable way to understand how good this approximation is for $|x|<1$ (without using the fact that we can actually go and calculate $\log(1+x)$ for any value we want)?

I understand there isn't a nice, closed form bound like what we have in Taylor's Theorem, but I suspect it is still possible to gain some understanding or bound to the theoretical error.

Answer 1

Yes, there are formulas that give you an error term depending on the regularity of the function. They are particularly useful for diagonal Pade approximants. This survey "Padé and Hermite-Padé approximation and orthogonality" by Walter Van Assche provides some useful information.

In practice, one can simply expand both functions, say $\log (1+x)$ and $\frac{x(6+x)}{6+4x}$ into the Taylor series in $|x|\le 1$ and after "cancelling" the corresponding equal terms, you will get an error term. Alternatively, you can set up the function $f(x)=\log{(x+1)}- \frac{x(6+x)}{6+4x}$ and study its behaviour using derivatives.

Answer 2

Maybe it defeats the purpose of using a rational approximation over a polynomial approximation, but perhaps you could compare to the Taylor polynomial. If we expect it to make sense, then write $$f(x) = f(x_0)+ f'(x_0)(x-x_0) +... + \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1},$$ for some $\xi$ between $x$ and $x_0$. Suppose $g(x)$ is your rational approximation and $f(x) = g(x) + R(x)$ where $R(x)$ is the remainder term. Then \begin{align*} R(x) &= f(x) - g(x)\\ & = f(x_0)+ f'(x_0)(x-x_0) +... + \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-x_0)^{n+1} - g(x). \end{align*} This is kind of unwieldy and I don't expect it will be very effective outside the set where the Taylor series converges, but there are cases where these error estimates are smaller than those on the Taylor polynomial.

Here is a Desmos graph that compares the error estimates of a [1,1] approximant with a second-order Taylor polynomial.

Answer 3

Just to get the error, I threw it into Wolfram Alpha which said the expansion of the error around $x=0$ is $-x^4/36 + (7 x^5)/135 - (11 x^6)/162 + (131 x^7)/1701 - (473 x^8)/5832 + O(x^9) $.