Ai Feedback
Exact(6)
So Eq. (4.8) has no root with modulus one for all (tau>0).
So (2.10) has no root with modulus one for all (tau>0).
(i) If (cu_{ast}<1), from Lemmas 1 and 2, we know that (2.10) has no root with modulus one for all (tau>0).
For any step-size h, if (cu_{ast}<1), then (2.10) has no root with modulus one for all (tau>0).
For any step-size h, if (frac{1-alpha}{alpha c}< bu_{ast}<2+frac{1-alpha}{alpha c}), then Eq. (4.8) has no root with modulus one for all (tau>0).
(1) If (frac{1-alpha}{alpha c}< bu_{ast}<2+frac{1-alpha}{alpha c}), from Lemmas 2 and 3 we know that Eq. (4.8) has no root with modulus one for all (tau>0).
Similar(54)
The authors found that there were two types of cells which differed in Young's modulus: one type had a Young's modulus of 5.6 ± 3.5 kPa, while others had one of 1.5 ± 0.76 kPa.
All roots of (2.10) have modulus less than one for sufficiently small (tau>0).
Applying Corollary 2.4 in [20], all roots of (2.10) have modulus less than one for all (tau>0).
Applying Corollary 2.4 in [20], all roots of Eq. (4.8) have modulus less than one for all (tau>0).
If (bu_{ast}>frac{1-alpha}{alpha c}), then all roots of Eq. (4.8) have modulus less than one for sufficiently small (tau>0).
Write better and faster with AI suggestions while staying true to your unique style.
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