Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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:

<<Algebra`Quaternions`

z=Quaternion[1.,-1.,1.,1.](*Quaternion[2 Random[] - 1, 2 Random[] - 1, 2 Random[] - 1, 2 Random[] - 1]*)

p=q=8;

num=Sum[(((p+q-j)!p!)/((p+q)!j!(p-j)!)) z^j, {j, 0, p}];

den=Sum[(((p+q-j)!q!)/((p+q)!j!(q-j)!)) (-z)^j, {j, 0, q}];

{num**(den^-1),(den^-1)**num,Exp[z],Conjugate[Exp[z]]}
share|improve this question
2  
Hmm. Can this be explained starting from the observation that if $z=a+bi+cj+dk$, then $izi^{-1}=a+bi-cj-dk$? IOW different kind of conjugations working somehow together? Anyway, you seem to be mostly working inside the ring $\mathbf{R}[z]$, and that is just a copy of $\mathbf{C}$: write $z=a+ru$, with $r=\sqrt{b^2+c^2+d^2}$ and $u$ a suitable unit vector. Therefore $u^2=-1$, and you may as well equate $z$ with the complex number $a+ri$. Looks like your Padé approximants never leave this copy of $\mathbf{C}$, so the results from the complex numbers should carry over to quaternions. –  Jyrki Lahtonen Oct 30 '11 at 20:16
    
Wow, I haven't looked at this in a while! Interesting thoughts, @Jyrki. –  J. M. Oct 30 '11 at 22:46

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.