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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?

LudvigH
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Jade196
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    That exponential function is from the quaternions... *to where* ? And how do you **define**, say $\;e^j\;,\;\;e^k\;$ , etc.? – Timbuc Nov 20 '14 at 12:53
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    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. – Raskolnikov Nov 20 '14 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. – Timbuc Nov 20 '14 at 13:10
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    The exponential function would map the quaternions to the quaternions. – Jade196 Nov 20 '14 at 13:25
  • $e^j=\sum_{n=0}^\infty \frac{j^n}{n!} = \sum_{n=0}^\infty (-1)^n \frac{1}{(2n)!} + j \sum_{n=0}^\infty (-1)^n \frac{1}{(2n+1)!} = cos(1) + j sin(1)$ – Jade196 Nov 20 '14 at 13:32
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    @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 $r=i,j,k$. $e^{xr}=\sum_{n=0}^{\infty}(xr)^n/n!=\sum_{n=0}^{\infty}(-1)^nx^{2n}/(2n)!+r\sum_{n=0}^{\infty}(-1)^nx^{2n+1}/(2n+1)!=\cos(x)+r\sin(x)$. –  Nov 20 '14 at 15:57
  • @Jade196 We get $e^q=e^a(\cos(b)+i\sin(b))(\cos(c)+j\sin(c))(\cos(d)+k\sin(d))$. Now we open parentheses, use the product table of $i,j,k$, and do some trigonometry. –  Nov 20 '14 at 15:59
  • @Jade196 Alternatively, we could expand the right-hand side in your formula and check it is equal to the product in the previous comment. Maybe this way is more comfortable because we don't have to synthesize a trigonometric expression, instead it is all about opening brackets of a product of series. –  Nov 20 '14 at 16:03

1 Answers1

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

Ryan Pavlik
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Emilio Novati
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  • Does this have a link with the matrix exponential and quaternion representation in the form of a 2x2 complex matrix ? – Psylex Oct 20 '19 at 13:20
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    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 $M^2=-|M|I$ – Emilio Novati Oct 21 '19 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. – Yuki.F Aug 12 '20 at 07:58
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    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. – Pineapple Fish Feb 28 '21 at 17:52