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which ends the proof of this case.
The proof of this case is completed.
(iv) The proof of this case is parallel to (iii).
The proof of this case is parallel to (iii).
The proof of this case is completed by application of Proposition 2.1 ii).
Next, take g ∗=g+ℓ≠g, which has the same moment sequence as g. ii) The proof of this case is similar to that of case (i). iii) If one ϕ j is not odd, then it is done as in (i) (by taking (ell (x)=frac {1}{2}[!phi _{j}(x)+phi _{j} -x ]) and g ∗=g+ℓ).
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Also, the rest of the proof for this case is similar to that of Theorem 2.5 in (Jahanshahloo et al. 2004) and is omitted. .
This completes the proof of Case 3. Case 4. Let (1.9) ((r=-1)) and (1.11) hold true.
This completes the proof of the case (a).
This completes the proof of the case n = 2. Now, suppose that lim i → ∞ sup j ≥ 1 ∥ ( T 1 k S n − 1, 1 x i, 1 − S n − 1, 1 T 1 k x i, 1, …, T j k S n − 1, j x i, j − S n − 1, j T j k x i, j ) ∥ j = 0. uniformly for each k ≥ 1.
Proof of the case (k=4).
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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