Sentence examples for by multiplying the above from inspiring English sources

By multiplying the above inequalities, we get (2.10).

By multiplying the above inequalities, we obtain (2.4) and (2.5) (2.5).

By multiplying the above inequalities, we have the desired Corollary 2.14.

Now suppose that Θ is canonical, then by multiplying the above conditions by Θ, we find (C=operatorname{diag}_{frac{n_{y}}{2}}(J DB^{T}Theta) (and similarly, (C^{T}=Theta BD^{T}operatorname{diag}_{frac{n_{y}}{2}}(J))).

Therefore, an upper bound for E 5 is given by multiplying the above RHS by e − t 2 / 6 e H. Finally, the last term E 6 : = e − t 2 / 2 E [ 2 | H ˜ n ′ ( t ) | ⋅ [ | ( 2 A ˜ n − 1 ) t | + 3 | B ˜ n | t 2 ] e A ˜ n t 2 + | H ˜ n ( t ) | 1 T c ]. can be handled without further effort by resorting to (26), (29), (35), (38) and (53).

By multiplying the above equation with P 0 : V − 1 ( t ) T W, we notice that the channel estimation is decoupled from data detection so that the following expression is obtained P 0 : V − 1 ( t ) T W y n, k = ∑ k ̄ = 1 K P 0 : V − 1 ( t ) T H ̄ k, k ̄ ( n ) P 0 : V − 1 ( t ) t ~ k ̄ + P 0 : V − 1 ( t ) T W z n, k, (16).

end{aligned}Therefore, multiplying the above equation by (f_Omega ) and integrating by parts one gets begin{aligned} E(Omega )=-frac{1}{2}int _Omega |nabla f_Omega |^2,dx=-frac{1}{2}int _Omega f_Omega,dx.

Multiplying the above equation by u t, integrating (by parts) over Ω, and using (8 1, we obtain: 1 2 d d t ∫ ρ u t 2 d x + ∫ ( u + ζ ) ∇ u t 2 + div u t 2 d x ≤ ∫ ρ v ∇ u t u t d x + ∫ ρ t v ∇ u u t d x + ∫ ρ v t ∇ u u t d x + ∫ p t ∇ u t d x + 2 ζ ∫ w t ∇ u t d x = ∑ i = 1 5 I I i. (32).

Multiplying the above equation by and integrating in we obtain (4.60).

Multiplying the above inequality by, and integrating from to, we obtain (2.57).

Multiplying the above equation by and integrating over, we deduce that (2.31).