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Found this problem on a blackboard today. It looks like some form of quaternion analysis, something to do with the MacLaurin series of $e^x$ My question is, what all is going on here, and how might I go about solving it? Here’s the problem:

Let $\text{pow}(q, n) = q^n$ for all $n \in \mathbb{Z}, q \in \mathbb{H}$. It follows that: $$ D_{\text{pow}}(n) \circ \delta_q = \sum_{k=1}^n{q^{n-k}\delta_{q}q^{k-1}} \quad\forall \;n \in \mathbb{Z}^+$$ Also, let $\exp(q) = \sum_{j=0}^{\infty}{\frac{1}{j!}q^j}$ for all $q \in \mathbb{H}$. It follows that: $$ D_{\exp} \circ \delta_q = \sum_{j=1}^{\infty}{\frac{1}{j!}\sum_{k=1}^j{q^{j-k}\delta_{q}q^{k-1}}}$$ Given the differential equation: $$D_{\exp} \circ \delta_q = \exp(a)\delta_{q}\exp(b)\quad |\; a, b \in \mathbb{H} $$ Find the values of $a$ and $b$, in terms of $q$.

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  • $\begingroup$ Could you specify what $\delta_q$, $D_{pow}$ and $D_{exp}$ are ? $\endgroup$ Commented Oct 22 at 5:46

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This looks like directional derivatives in some kind of linear-algebra formalism. $\delta_q$ would then be the direction of derivation or the operator that extracts the tangent direction from a point in the tangent space (as fibre bundle) over $q$.

In another formalism one could write $$ \frac{d}{dt}q^n = \sum_{k=1}^nq^{n-k}\dot qq^{k-1} $$ where $q(t)$ is some differentiable path, $q=q(0)$, $\dot q=\dot q(0)=\frac{d}{dt}q(t)|_{t=0}$ etc.

Now we want to know if in the derivative of the exponential $e^{q(t)}$ one can factor out the linearly occurring $\dot q$. This is usually done working not from the series of the exponential, but from its power limit definition $$ e^q=\lim_{n\to\infty}e_n(q)=\lim_{n\to\infty}\left(1+\frac qn\right)^n $$ so that $$ \frac{d}{dt}e_n(q)=\frac1n\sum_{k=1}^ne_n(q)^{1-k/n}\,\dot q \,e_n(q)^{k/n-1/n}. $$ In the limit, claiming that everything is tame and exists, and reading the sum as a Riemann sum, thus replacing that with the corresponding integral, $$ \frac{d}{dt}e^q=\int_0^1e^{(1-s)q}\dot qe^{sq}\,ds. $$ This does not look very amenable to factoring $\dot q$ out. In matrices and Lie groups one continues with the adjoint operator and its exponential I do not see off-hand how that translates to quaternions in some organic way.

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  • $\begingroup$ Thank you so much! I think this integral form is what the problem was getting at. Sorry for the strange notation ^^` $\endgroup$ Commented Oct 22 at 17:24
  • $\begingroup$ As announced, for matrix valued paths the derivative of the exponential is since long explored: en.wikipedia.org/wiki/Derivative_of_the_exponential_map $\endgroup$ Commented Oct 22 at 18:49

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