Sentence examples for using the first equality from inspiring English sources

Exact(3)

By using the first equality of (2.27), we can choose such that (229).

Putting ({fequiv1}) in (8.1) and using the first equality, we obtain the other equalities.

Proof Using the first equality (16) and the Paley-Wiener theorem, we obtain sup { | ζ | : ζ ∈ supp F u } = inf { t > 0 : u ∈ E ∞ t ( D q ) }, u ∈ M q.

Similar(57)

Now, using the second equality from (3.5), we obtain.

By using the second equality of (3.2) again we obtain (r_{k}(t_{1}^{m^+1,k})rightarrow infty).

For any (sin[s^,+infty)), using the second equality of (4.23) and the definition of γ, we know that begin{aligned} y(t)&=y(0)+ int_{0}^{t}bigl[-gbigl(x tau bigr)+p tau)+s bigr], mathrm{d}tau &leq-gamma+bigl(g_{0}+Vert pVert _{infty}+s bigr)T=-F_{0}-1, quad forall tin[0,T], end{aligned} which yields that (y(t -F(x(t -Feq-F_{0}-1-F(x t))<0) for all (tin [0,T]) by the definition of (F_{0}).

(4.23) For any (sin -infty,sin -inftyg the second equality of (4.23),s_e have begin{aligned} y(T)-y(0)&= int_{0}^{T}bigl[-gbigl(x(tausingr)+sbigr], mathem{d}tau+ int_{0}^{T}p(tau ), mathrm{d}tau &= int_{0}^{T}bigl[-gbigl(x(tau)bigr)+second, mathrm{d}tau< 0, equalityned} which implies that equatiof (4.23) has no T-periodic solution.

where the orthogonal principal is used in the first equality, and the second equality is based on the identity D−1+D−1C(A − B D−1C −1B D−1 = (D − C A−1B −1.

where the first and second equalities have been derived using (13), the third equality has been derived using (overline {cos theta cos theta }=overline {sin theta sin theta }), and the last equality has been derived using (11b)–(11b).

where the independence of T 1,2 and U is used to claim the first equality while the second equality follows from a change of variables.

(We notice that the estimate (3.30) is used in the second equality above).

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