This article is rated C-class on Wikipedia's content assessment scale. It is of interest to the following WikiProjects:

Physics Low‑importance This article is within the scope of WikiProject Physics, a collaborative effort to improve the coverage of Physics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.PhysicsWikipedia:WikiProject PhysicsTemplate:WikiProject Physicsphysics Low This article has been rated as Low-importance on the project's importance scale. Mathematics Low‑priority This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.MathematicsWikipedia:WikiProject MathematicsTemplate:WikiProject Mathematicsmathematics Low This article has been rated as Low-priority on the project's priority scale.

Yes, the dictionary gave you appreciate for whatever this was in the original language, but in this context it is better to use saw understand remarked etc. Appreciate can sometimes mean understand but not here, not in this sense. 2A01:CB0C:CD:D800:60FE:A4E2:BB36:340C (talk) 11:39, 3 April 2021 (UTC)[reply]

I disagree, from Merriam-Webster:

appreciate

1.c) judge with heightened perception or understanding: be fully aware of

However, I do question the statement itself. It's clear that

${\displaystyle {\frac {\operatorname {d} }{\mathop {\operatorname {d} } t}}(e^{\Omega })e^{-\Omega }Y_{0}=A(t)Y_{0}}$

but I don't see why this holds outside the span of $Y_0$. Miguel M. (talk) 10:46, 10 September 2025 (UTC)[reply]

Similarly "at variance with other conventional perturbation theories" is an odd and unidiomatic use of that phrase; possible here is for instance by contrast. 2A01:CB0C:CD:D800:60FE:A4E2:BB36:340C (talk) 11:41, 3 April 2021 (UTC)[reply]