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(iii) The equality is clear for n = 0. .
The assertion is clear for n = 0. Assume that for a given integer n ≥ 1, the function ( x, t ) → x 2 − α G n − 1 ( x, t ) is continuous on [ 0, 1 ] × [ 0, 1 ].
Proof (i) The assertion is clear for n = 0. Assume that inequality in (i) holds for some n ≥ 0, then by using (3.2) and (1.11), we obtain G n + 1 ( x, t ) ≤ α q n ∫ 0 1 G ( x, r ) G ( r, t ) q ( r ) d r ≤ α q n + 1 G ( x, t ).
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It is clear that for n = 2 and 3 (but not 4), constructed polymorphisms have allele frequencies that are more even than our random expectation.
F3′5′H enzymes from Catharanthus roseus and Petunia x hybrida have achieved highest activities with naringenin and apigenin [ 37], and N and DHK are equally hydroxylated by Osteospermum hybrida F3′5′H, whereas F3′H from Gerbera hybrida exhibits a clear substrate preference for N [ 36].
Originally it was clear for each of n water droplets whether they were members of the cloud to degree 1 or degree 0. So there were 2n candidate clouds, and the Problem of the Many is finding out how to preserve the intuition when faced with all these objects.
The quantities analyzed in Figure 2 are not defined well and it is not clear how the numbers for N and n correspond to each other.
It is clear that Tx n ≥ x n for all n ≥ 1.
By the definition of μ, it is clear that (2.3) is true for n = 1, 2, …, k.
Lemma 3 The inequality f ( ( x 1 ∗ ⋯ ∗ x n ) ( 1 / n ) ∗ ) ≤ ( f ( x 1 ) ∘ ⋯ ∘ f ( x n ) ) ( 1 / n ) ∘. holds for all n ∈ N and x 1, …, x n ∈ I. Proof It is clear that the lemma holds for n = 1.
It is clear that β n ≤ β n ∗ for all n ≥ 1.
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Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com