Comments (3)
Yes, you would get (I might have put the signs wrong)
Jl(a) = I + [a]x / 2 + [a]x^2 / 6 + ...
and you may decide to truncate before the quadratic term -- this is up to you.
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To find small-angle approximations, substitute sin(theta)
and cos(theta)
by their Taylor expansions, cancel the terms with those in the denominator, and truncate the remaining terms at the desired length.
For example the following term appearing in the right Jacobian:
(1 - cos (a) ) / a^2 ~ (1 - (1 - a^2 / 2 + a^4 / 24)) / a^2 = (a^2 / 2 - a^4 / 24) / a^2 = 1/2 - a^2 / 24 ~ 1/2
You proceed akin with the other term (a - sin(a) ) / a^3
from manif.
Aha, that makes sense, thanks for the quick answer!
The term (a - sin(a)) / a^3
gives me 1/6
, but I guess that's ignored because it multiplies a quadratic term, [theta]_x^2
, right?
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