# Why do exponential functions of quaternions not work as expected?

The exponential function for the imaginary part of the quaternion is defined like this: $\exp(q) = \sin(|q|)+\cos(|q|)\times q|q|$

When I set $q=q_1+q_2$, then $\exp(q)=\exp(q_1+q_2)=\exp(q_1)\exp(q_2)$. Is this already a mistake here?

When I then multiply it out with some quaternions $q_1=i$ and $q_2=j$, then $\exp(i+j)=\sin(\sqrt2)+\cos(\sqrt2)(i+j)/\sqrt2$, but\begin{align*}\exp(i+j)&=\exp(i)\exp(j)\\&=(\sin(1)+\cos(1)i)+(\sin(1)+\cos(1)j)\\&=(\sin(1)^2+\sin(1)\cos(1)(i+j)-\cos(1)^2k,\end{align*}which aren't the same.

Why are these results different? Where is the mistake and why does it occur?

This question also applies to geometric algebra, but since this seems to be less understood, I ask about quaternions, which have the same problems for the exponential function.

• I'm not familiar with the definition you provide, and I'd check the placement of parentheses there as a precaution. In general the sum of exponents giving a product of exponentials does depend on commutativity of the underlying algebra, so I'd wonder with you whether the next statement in your post is "already a mistake". – hardmath Aug 9 '17 at 22:41
• – Count Iblis Aug 9 '17 at 22:50
• Link doesn't work correctly. When clicking it, it will direct me to the Article about "Baker" ;) – porky11 Aug 9 '17 at 22:52

A basic problem is that it is not true, in general, that $$\exp(q_1+q_2)=\exp(q_1)\exp(q_2)$$. This holds, however, if $$q_1$$ and $$q_2$$ commute. But, since $$i.j=-j.i$$, you have no reason to believe that $$\exp(i+j)=\exp(i).\exp(j)$$.
By the way, the usual definition of the exponentional function within the quaternions (and elsewhere) is$$\exp(q)=\sum_{n=0}^\infty\frac{q^n}{n!}.$$
• @porky11 Suppose that $\exp(q_1)$ and $\exp(q_2)$ don't commute. Then $\exp(q_1)\exp(q_2)\neq\exp(q_2)\exp(q_1)$. But addition is commutative and so $\exp(q_1+q_2)=\exp(q_2+q_1)$ and therefore $\exp(q_1+q_2)$ cannot be equal to both numbers $\exp(q_1)\exp(q_2)$ and $\exp(q_2)\exp(q_1)$. – José Carlos Santos Aug 9 '17 at 23:03
• Even if multiplication is not commutative in general, it may be possible, that in some cases some numbers commute. So this could also mean, that when $\exp(q_1+q_2)=exp(q_1)exp(q_2)$, then all possible results of the exponential function will form a commutative subalgebra. – porky11 Aug 9 '17 at 23:16