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That is to ask, is $e^{\ln(q_0)}$ = $\ln(e^{q_0})$?

May Oakes
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  • This condition does not imply that the two maps are inverse, but yes, this is true for sufficiently small q_0. In fact the inverse relationship between the logarithm and the exponential is formal in the sense that it is an equality of power series so it holds whenever the series converge. – Qiaochu Yuan Feb 24 '11 at 18:52

4 Answers4

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$e^{\ln(q_0)} = q_0$ always holds for all branches of the logarithm and even for square matrices (because this is essentially the definition of the log).

$\ln(e^{q_0}) = q_0$ is already for complex numbers not always true (because the logarithm is multivalued)

Fabian
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I wrote software to play with quaternions numerically on the command line (http://sourceforge.net/projects/quaternions/).

q_ln 1 2 3 4 | q_exp 1.0000000000000011 2.0000000000000000 3.0000000000000004 4.0000000000000000
This confirms $e^{ln(q_0)}=q_0$. Flip the order: q_exp 1 2 3 4 | q_ln 1.0000000000000000 -0.3335164408915938 -0.5002746613373907 -0.6670328817831875 So $\ln(e^{q_0}) \ne q_0$.

sweetser
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Same here. I have code that seems correct, but I get the following results:

log(exp({2,4,3,3})) = {2,-0.31023,-0.232672,-0.232672}
exp(log({2,4,3,3})) = {2,4,3,3}
Frank
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Within a limited domain exp(ln(q0)) = ln(exp(q0)) is true; where q0 is originally a valid unit quaternion for rotation. for some other lnQ0, exp(lnQ0) is sometimes = q0; but there are many lnQ0 which also equal q0; but ln(q0) will only get a certain subset of log-quaternions.

I made this interactive demo to explore log-quaternions as the native or natural operating space. The link to the repository is at the bottom of the page. There's documentation about what the demo does(sort of buried) in the 'Curvature.md' file. It hints at improved accuracy and simplification of calculations.

J Decker
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