Exponential Quaternion
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Quaternion Exponential
This note is all about proving the statement for any quaternion $q = a + bi + cj + dk$ that $q = ||q||e^{\hat{n}\varphi} = ||q||(\cos(\varphi) + \hat{n}\sin({\varphi}))$. We can start with the geometrical interpretation of the quaternion and see that that it is made up of two parts: a real part and an imaginary part. Simply put, the quaternion can be written as…
\[\begin{equation} q = a + \mathbb{v} \end{equation}\]where $\mathbb{v} = bi + cj + dk$. If we look at this, it is a right triangle with angle $\varphi$ in between the hypotenuse and real part $a$. So, by geometry, one way we can write $a$ is $a = ||q||\cos(\varphi)$ and the length of the vertical (imaginary) part is $||\mathbb{v}|| = ||q||\sin(\varphi)$. Because the vertical part is pointing north always, all we have to do to get the vertical component of the triangle is scale by the normal vector which always points north too. The unit vector is
\[\begin{equation} \hat{n} = \frac{\mathbb{v}}{||\mathbb{v}||} \end{equation}\]So, we end up getting…
\[\begin{equation} \hat{n}||\mathbb{v}|| = \frac{\mathbb{v}}{||\mathbb{v}||}||\mathbb{v}|| = \mathbb{v} = ||q||\hat{n}\sin({\varphi}) \\ \implies \mathbb{v} = ||q|| \hat{n}\sin(\varphi) \end{equation}\]Putting all of this together gives…
\[\begin{equation} q = a + \mathbb{v} = ||q||\cos({\varphi}) + ||q||\hat{n}\sin(\varphi) \\ \implies q = ||q||(\cos(\varphi) + \hat{n}\sin({\varphi})) \end{equation}\]which is partially what we want. Now, we will compute the form for $e^q$. Using $(1)$, we arrive at…
\[\begin{equation} e^q = e^{a + \mathbb{v}} = e^ae^\mathbb{v} \end{equation}\]We know by Taylor series expansion that anytime we raise an exponential to an imaginary number, we end up with a rotation that puts us in the unit circle on the quaternion plane, $\mathbb{H}$. By vector rotation formula, we end up with…
\[\begin{equation} e^{\mathbb{v}} = \cos\left(||\mathbb{v}||\right) + \frac{\mathbb{v}}{||\mathbb{v}||}\sin(||\mathbb{v}||) \end{equation}\]So, when we raise and exponential by $q$, we end up with…
\[\begin{equation} e^q = e^a\left[\cos\left(||\mathbb{v}||\right) + \frac{\mathbb{v}}{||\mathbb{v}||}\sin(||\mathbb{v}||)\right] \end{equation}\]We want to show, $e^{\hat{n}\varphi} = \cos(\varphi) + \hat{n}\sin({\varphi})$ by intuition from looking at the equations we have so far. We already know $e^q$ from $(7)$ and all we are going to do now is input $\hat{n}\varphi$ instead of $q$ in the exponent. First we note that $\hat{n}\varphi$ has $a = 0$ and is purely imaginary due to $(2)$ and the fact that $\varphi \in \mathbb{R}$. In particular, we have…
\[\begin{equation} \hat{n}\varphi = \frac{\mathbb{v}}{||\mathbb{v}||}\varphi = \left[\frac{\varphi}{||\mathbb{v}||}\mathbb{v}\right] = \mathbb{v}^\ast \end{equation}\]Now, we can use $(7)$ and compute $e^{\hat{n}\varphi}$…
\[\begin{equation} e^{\hat{n}\varphi} = e^0\left[\cos\left(||\mathbb{v}^\ast||\right) + \frac{\mathbb{v}^\ast}{||\mathbb{v}^\ast||}\sin(||\mathbb{v}^\ast||)\right] \\ = \cos\left(\frac{\varphi}{||\mathbb{v}||}||\mathbb{v}||\right) + \hat{n}\sin\left(\frac{\varphi}{||\mathbb{v}||}||\mathbb{v}||\right) \\ = \cos(\varphi) + \hat{n}\sin(\varphi) \end{equation}\]So, $e^{\hat{n}\varphi} = \cos(\varphi) + \hat{n}\sin(\varphi)$. This is what we wanted to show for our intuition. Using $(4)$ and $(9)$, we end up with…
\[\begin{equation} q = ||q||e^{\hat{n}\varphi} = ||q||(\cos(\varphi) + \hat{n}\sin({\varphi})) \end{equation}\]for any quaternion, as desired. $\blacksquare$