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Therefore, if L + ϵ < 1 L + ϵ, eventually γ + ( P ) will intersect the line x − x 1 L + ϵ − x 1 f ( x 1 ) at some x ¯ < x 1, and hence also Δ at some x ¯ < x < x 1.
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No one should be surprised, therefore, if one morning we wake up to read that, in this ongoing drama, "Client 9" recovers a repressed memory of his own.
Therefore, if A = L − W, then ⦀ e − A ⦀ ≤ e α ∗ − l.
With the feature notation, we define: (6) Assuming strong complementarity slackness, we obtain the following result (see Supplementary Material for detailed proof): T heorem 1 (7) (7) Therefore, if after the corrector step, (and its associated parameter ) must be added to the active set.
2. If w ∈ L l, then (w ) ∈ ℒ l (bar-bracket representation of a stacked pair).
4. If w ∈ L l, then + (w ) + ⊂ L l (bar-bracket representations of interior loops).
If l = 0, the proof is complete.
Therefore, if γ ( z ) ≥ z for every z > 1, then (7) follows.
Consequently, p0 is approximately equal to 1, and therefore, (7) P 4 / P 3 ≈ 10 8 p 12 (Q − 1 ).
Therefore, if pi is random it contains accidental order.
Therefore, if you're thinking about, Should we do x?
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Since I tried Ludwig back in 2017, I have been constantly using it in both editing and translation. Ever since, I suggest it to my translators at ProSciEditing.

Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com