On a lark, I wanted to know if one can use Padé approximants to compute the exponential $\exp(z)$ of a quaternion $z=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k}$. Since Mathematica has a package meant for manipulating quaternions, as well as computing the exponential of a quaternion, I decided to try things out.
The numerator $N_{pq}(z)$ and denominator $D_{pq}(z)$ of the $(p,q)$ Padé approximant for the exponential function can be computed using the following formulae:
$\displaystyle N_{pq}(z)=\sum_{j=0}^p \frac{(p+q-j)!p!}{(p+q)!j!(p-j)!}z^j$
$\displaystyle D_{pq}(z)=\sum_{j=0}^q \frac{(p+q-j)!q!}{(p+q)!j!(q-j)!}(-z)^j$
after which the Padé approximant is $R_{pq}(z)=\frac{N_{pq}(z)}{D_{pq}(z)}$.
For simplicity (and since they are the ones most used in practice), I will only consider here the "diagonal" case, $p=q$.
The case where $z$ is a matrix is well-studied, and it is known that $R_{pq}(z)=N_{pq}(z)\cdot D_{pq}(z)^{-1}=D_{pq}(z)^{-1}\cdot N_{pq}(z)$. Banking on the isomorphism of quaternions to specially constructed matrices, I expected that the order of multiplication won't matter in the quaternion case, and experimentation seems to show that this holds.
What was mystifying in my experiments is that, although in general $|R_{pp}(z)-\exp(z)|$ can be made small enough by taking $p$ large enough, one pattern I saw seems to indicate that
$R_{pp}(a-b\mathbf{i}+c\mathbf{j}+d\mathbf{k})\approx \overline{\exp(a-b\mathbf{i}+c\mathbf{j}+d\mathbf{k})}$
for $b>0$, where the bar denotes conjugation.
I can't see an immediate equivalent of this phenomenon in the theory of the matrix exponential, so I'm wondering why the Padé approximants to the quaternion exponential behave this way.
For those who want to retry my Mathematica experiments:
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