I was reading some work by Butcher and I came across Pade approximations and the correlation between them and stability functions for some Implicit Runge-Kutta methods. For example, in this Pade table for the exponential function, we see that the $(2,1)$ Pade approximation $$\frac{1+\frac{1}{3}z}{1-\frac{2}{3}z+\frac{1}{6}z^{2}}$$corresponds to the stability function for the RADAU IIA method. Similarly, the $(2,2)$ Pade approximation $$\frac{1+\frac{1}{2}z+\frac{1}{12}z^{2}}{1-\frac{1}{2}z+\frac{1}{12}z^{2}} $$corresponds to the stability function of the 2-stage Gauss Runge-Kutta method. However, I'm not sure if other Pade approximations correspond to stability functions; For example, I can't find the numerical method whos stability function corresponds to the $(2,0)$ Pade approximation$$\frac{1}{1-z+\frac{1}{2}z^{2}}$$So my question is two fold (I apologise for asking two questions in one post, it seems like a waste to create two new posts though): Is it always possible to generate an Implicit Runge-Kutta scheme whos stability function corresponds to a particular Pade approximation? And if soHow do we construct such an implicit scheme?
How to derive an Implicit Runge-Kutta method from Pade approximation
stability
There are a few people who have liked the question, so I'm posting an answer. The only way I could figure out how to do this is to just work backwards, starting with the stability function for implicit Runge-Kutta schemes. I'm not sure if there is a slicker way to do this, this is just the most obvious. Note that this only applies to implicit Runge-Kutta schemes, and I'm specifically using the Pade $(2,0)$ apprixmation as stated above as an example.Using the fact that the Pade approximation is given by\begin{align}R(z) &= \frac{P(z)}{Q(z)} \\&= \frac{1}{1-z+\frac{z^{2}}{2}}\end{align}and the stability function for an implicit Runge-Kutta is given by ($\vec e$ is a vector of ones)\begin{align}R(z) &= \frac{P(z)}{Q(z)} \\&= \frac{\det(I-zA+z \vec e b^{T})}{\det(I-zA)}\end{align}we just need to construct matrices such that\begin{align}\det(I-zA+z \vec e b^{T}) &= 1 \\\det(I-zA) &= 1-z+\frac{z^{2}}{2}\end{align}Setting $$A = \begin{pmatrix}a & b \\c & d\end{pmatrix}$$gives us\begin{align}\det(I-zA) &= 1 - z(a+d) + z^{2}(ad-cb) \\&= 1 - z + \frac{z^{2}}{2}\end{align}which gives us conditions on our coefficients\begin{align}a+d &= 1 \\ad-cb &= \frac{1}{2}\end{align}and now, we can arbitrarily choose values for $a,b,c,d$ such that they satisfy the above underdetermined system (for example, $a = d = \frac{1}{2}$, $b = \frac{1}{3}$, $c = -\frac{3}{4}$).We then solve $\det(I-zA+z \vec e b^{T})$ in a similar manner, by taking$$\vec e b^{T} = \begin{pmatrix}b_{1} & b_{2} \\b_{1} & b_{2}\end{pmatrix}$$constructing $I - zA + z \vec e b^{T}$ (using the previous matrix we constructed, $I - zA$, with the components of $A$ now known from $a,b,c,d$), taking the determinant and solving the corresponding system of equations in $b_{1}$ and $b_{2}$. The matrix $A$ and vector $b^{T}$ can now be written in a Butcher tableau.