A quick inspection of the first Table suggests a rule to cover the case when <math>\texttt{u~=~v~=~1},</math> namely, <math>\texttt{du~=~dv~=~0}.</math> To put it another way, the Table characterizes Orbit 1 by means of the data: <math>(u, v, du, dv) = (1, 1, 0, 0).\!</math> Another way to convey the same information is by means of the extended proposition: <math>\texttt{u~v~(du)(dv)}.</math>
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A quick inspection of the first Table suggests a rule
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to cover the case when u = v = 1, namely, du = dv = 0.
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To put it another way, the Table characterizes Orbit 1
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by means of the data: <u, v, du, dv> = <1, 1, 0, 0>.
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Last but not least, yet another way to convey the same
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information is by means of the (first order) extended