The equation for the exponential function of a quaternion $q = a + b i + c j + dk$ is supposed to be $$e^{q} = e^a (\cos(\sqrt{b^2+c^2+d^2})+\frac{(b i + c j + dk)}{\sqrt{b^2+c^2+d^2}} \sin(\sqrt{b^2+c^2+d^2}))$$

I'm having a difficult time finding a derivation of this formula. I keep trying to derive it, but I end up getting different results. Would someone be able to point me to a proof of this formula or do the derivation here?

Note: I also don't understand why some people say $e^q = e^a e^{b i + c j + d k}$. Can you please explain this, too?

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    $\begingroup$ That exponential function is from the quaternions... to where ? And how do you define, say $\;e^j\;,\;\;e^k\;$ , etc.? $\endgroup$ – Timbuc Nov 20 '14 at 12:53
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    $\begingroup$ I think a natural way to extend the exponential function to quaternions would be to use the Taylor series of the exponential over the complex and just extend domain and range to include quaternions. $\endgroup$ – Raskolnikov Nov 20 '14 at 12:56
  • $\begingroup$ @Raskolnikov, perhaps. In the meantime the OP hasn't yet addressed my doubts and I'm not in the mood for guessing posters' intentions. I agree with you, though. $\endgroup$ – Timbuc Nov 20 '14 at 13:10
  • $\begingroup$ The exponential function would map the quaternions to the quaternions. $\endgroup$ – Jade196 Nov 20 '14 at 13:25
  • $\begingroup$ $e^j=\sum_{n=0}^\infty \frac{j^n}{n!} = \sum_{n=0}^\infty (-1)^n \frac{1}{(2n)!} + j \sum_{n=0}^\infty (-1)^n \frac{1}{(2n+1)!} = cos(1) + j sin(1)$ $\endgroup$ – Jade196 Nov 20 '14 at 13:32

The definition of quaternionic exponential is given by the absolutely convergent series $$ e^z=\sum_{k=0}^\infty\dfrac{z^k}{k!} $$ It is well known that, from this definition, if $x, y$ commute we have $e^xe^y=e^ye^x=e^{x+y}$. Since real quaternions commute with all other quaternions, for $a \in \mathbb{R}$ we have $e^{a+z}=e^ae^z \; \forall z\in \mathbb{H}$ so, if $z=a+b\mathbf{i}+c\mathbf{j}+d\mathbf{k} = a+\mathbf{v}$, we have $e^z=e^ae^\mathbf{v}$, where $\mathbf{v}$ is an imaginary (or vector) quaternion. Now we have:


If $ \mathbf{v} \in \mathbb{H}_P$ is an imaginary quaternion, putting $\theta=|\mathbf{v}|$ we have: $$ e^\mathbf{v}= \cos\theta + \mathbf{v}\;\dfrac{\sin \theta}{\theta} $$ proof

We note that: $$ \mathbf{v}^2= (b \mathbf{i}+c \mathbf{j} +d \mathbf{k})(b \mathbf{i}+c \mathbf{j} +d \mathbf{k})= -b^2-c^2-d^2=-|\mathbf{v}|^2 $$ so: $$ \mathbf{v}^2= -\theta^2 \quad,\quad \mathbf{v}^3= -\theta^2\mathbf{v} \quad,\quad \mathbf{v}^4= \theta^4 \quad,\quad \mathbf{v}^5= \theta^4 \mathbf{v} \quad,\quad \mathbf{v}^6= -\theta^6 \quad,\quad \cdots $$ and the series become. $$ \begin{split} e^\mathbf{v}&=\sum_{k=0}^\infty\dfrac{\mathbf{v}^k}{k!}=\\ % &=1+\dfrac{\mathbf{v}}{1!}-\dfrac{\theta^2}{2!}-\dfrac{\theta^2\mathbf{v}}{3!}+\dfrac{\theta^4}{4!}+\dfrac{\theta^4\mathbf{v}}{5!}-\dfrac{\theta^6}{6!}+\cdots=\\ % &=1+\dfrac{\theta\mathbf{v}}{1!\,\theta}-\dfrac{\theta^2}{2!}-\dfrac{\theta^3\mathbf{v}}{3!\,\theta}+\dfrac{\theta^4}{4!}+\dfrac{\theta^5\mathbf{v}}{5!\,\theta}-\dfrac{\theta^6}{6!}+\cdots=\\ % &=\left(1-\dfrac{\theta^2}{2!}+\dfrac{\theta^4}{4!}-\dfrac{\theta^6}{6!}\cdots\right)+\dfrac{\mathbf{v}}{\theta}\left( \dfrac{\theta}{1!}-\dfrac{\theta^3}{3!}+\dfrac{\theta^5}{5!}\cdots\right)=\\ % &=\cos\theta +\dfrac{\mathbf{v}}{\theta}\sin\theta \end{split} $$

So the exponential of a quaternion is: $$ e^z = e^{a+\mathbf{v}}=e^a \left( \cos |\mathbf{v}| +\dfrac{\mathbf{v}}{|\mathbf{v}|} \,\sin |\mathbf{v}| \right) $$

  • $\begingroup$ Does this have a link with the matrix exponential and quaternion representation in the form of a 2x2 complex matrix ? $\endgroup$ – Psylex Oct 20 at 13:20
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    $\begingroup$ Yes. It is special case because a pure imaginary quaternion is represented by a $2\times 2$ matrix $M$ with null trace, and for such a matrix it is easy to show that $M^2=-|M|I$ $\endgroup$ – Emilio Novati Oct 21 at 15:30

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