# Exponential Function of Quaternion - Derivation

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

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?

• That exponential function is from the quaternions... to where ? And how do you define, say $\;e^j\;,\;\;e^k\;$ , etc.? Commented Nov 20, 2014 at 12:53
• 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. Commented Nov 20, 2014 at 12:56
• @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. Commented Nov 20, 2014 at 13:10
• The exponential function would map the quaternions to the quaternions. Commented Nov 20, 2014 at 13:25
• @Jade196 I would first check that if $e^x:=\sum_{n=0}^{\infty}x^n/n!$ for $x$ in a normed algebra then $e^{x+y}=e^xe^y$. Then $e^{q}=e^ae^{bi}e^{cj}e^{dk}$, which is maybe more comfortable to compute. You have essentially computed already $e^{xr}$, for $x$ real and $r=i,j,k$. $e^{xr}=\sum_{n=0}^{\infty}(xr)^n/n!=\sum_{n=0}^{\infty}(-1)^nx^{2n}/(2n)!+r\sum_{n=0}^{\infty}(-1)^nx^{2n+1}/(2n+1)!=\cos(x)+r\sin(x)$.
– user192614
Commented Nov 20, 2014 at 15:57

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:

claim

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: $$e^z = e^{a+\mathbf{v}}=e^a \left( \cos |\mathbf{v}| +\dfrac{\mathbf{v}}{|\mathbf{v}|} \,\sin |\mathbf{v}| \right)$$

• Does this have a link with the matrix exponential and quaternion representation in the form of a 2x2 complex matrix ? Commented Oct 20, 2019 at 13:20
• 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 $M^2=-|M|I$ Commented Oct 21, 2019 at 15:30
• Hi @EmilioNovati, in case $a = \cos \phi, \mathbf{v} = (\sin \phi) \mathbf{v_{\sin}}$ ($z$ is a unit quaternion, $\|\mathbf{v_{\sin}}\| = 1$), does $\sin |\mathbf{v}|$ means $\sin (\text{abs}(\sin \phi))$ or $\sin (\sin \phi)$? As far as I know $\phi \in [-\pi, \pi]$, that $\sin (\sin \phi)$ can be negative. Commented Aug 12, 2020 at 7:58
• Since you had to multiply top and bottom by theta, sinc(x) is probably more accurate than sin(x)/x here just for clarity because it avoids the singularly of dividing by something with norm of zero. Commented Feb 28, 2021 at 17:52
• @PineappleFish $\mathbf{v}$ goes to zero just as fast as $\sin|\mathbf{v}|$, so $\frac{\mathbf{v}}{|\mathbf{v}|}$ handles the approach to the singularity just as well as $\mathrm{sinc}$ would. Since the actual numerator is quadratic, there isn't any numerical concern anyway. This leaves the singularity itself to take care of, and in a computer implementation you might want to call $\mathrm{sinc}$ (if provided) to avoid having to write if (v == 0) yourself. It's not very common though, as far as standard libraries neither C++ nor Python provide it (numpy has it though). Commented Aug 31, 2022 at 2:09

If $$q=\alpha+i\beta+j\gamma+k\delta$$, it is always possible to represent $$q$$ as $$q=a+Ib$$, where $$a=\alpha$$, $$b=\sqrt{\beta^2+\gamma^2+\delta^2}$$ and $$I=\dfrac{i\beta+j\gamma+k\delta}{b}$$. Note that $$I^2=-1$$. Since for complex numbers it holds $$e^{a+ib}=e^a(\cos(b)+i\sin(b))$$, the same holds for $$e^q=e^{a+Ib}=e^a(\cos(b)+I\sin(b))=e^a\left(cos\left(\sqrt{\beta^2+\gamma^2+\delta^2}\right)+\dfrac{i\beta+j\gamma+k\delta}{\sqrt{\beta^2+\gamma^2+\delta^2}}\sin\left(\sqrt{\beta^2+\gamma^2+\delta^2}\right)\right).$$