Exact(9)
On the other hand, we choose N 1 even larger so that K ( t ) is equivalent to E ( t ), i.e., K ( t ) ∼ E ( t ).
Again, employing that G is equivalent to E, we get E t)leq K e^{-nuint_{t_{0}}^{t} zeta(s),ds}, quad forall tgeq t_{0}, where K is a positive constant.
Employing that G is equivalent to E, we get E t)leq C e^{- nuint^{t}_{t_{0}} zeta(s),ds} quadhbox{for } t geq t_{0}, where C and ν are positive constants.
It is easy to see that E φ ( A, X ) = L φ ( A, X ) is equivalent to E φ ( A ) = L φ ( A ), this means that the equality depends only on φ.
It is to be noted that the equation (3.3) is equivalent to e i α z f ′ ( z ) f ( z ) = β + i p sin α - ( β - p cos α ) q ( z ).
Using (7) and (4), the condition of unbiased estimates is equivalent to: E log E i ̂ = log E i + ∑ j = 1 P w ij σ − 2 E i, Δ t j ν E i, Δ t j ∑ j = 1 P w ij σ − 2 E i, Δ t j. (23).
Similar(50)
A solution of 50 nm vesicles was prepared, with a concentration equivalent to E-DNO based on rhodamine absorbance.
For the case where (omegain B^{c}), the inequality is equivalent to bigl 3-2g(2)bigr)e^{2lambda}+bigl 3-5g(2)bigr)e^{lambda}+ bigl 3-2g(2)bigl 3-2g.
(hat{b}_1 = 0) is equivalent to (frac{partial E(Pi ^mathrm{C})}{partial b_1} < 0).
Since (Vert cdot Vert ) is equivalent to the E norm on (E^), there exists a constant (C>0) such that (Vert (v^{j})^ Vert _{E}leqslant C).
Defining s ≜ | r | T a, the above expression is equivalent to P e = ∫ L Δ / 2 L Δ p S ( s | b ) d s. (8).
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Justyna Jupowicz-Kozak
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