Sentence examples for using the above equality from inspiring English sources

Using the above equality, we show that the equalities (3.23).

Multiplying the weak form of (4.3) by z λ ′ and using the above equality, we have (4.4).

Using the above equality and (2), equation (5) can be rewritten as follows: mathcal{J}_{a+}^{alpha}mathcal{D}_{a+}^{alpha}Y t)=Y t) ominus _{g}frac{ t-a)^{alpha-1}}{Gamma(alpha)}{B}.

On the other hand, we use the above equality to get lim_{n toinfty} frac{1}{a^{4n+2}} Phi bigl( a^{2n+1}x bigr) = frac{1}{a^{2}} lim_{n toinfty} frac{1}{a^{4n}} Phi bigl( a^{2n}ax bigr) = 0 for all (x in V backslash{ 0 }).

Recalling that I λ, δ ′ ( u, v ) = 0, we can use the above equality and (2.8) to obtain lim m → ∞ ( ũ m, ṽ m ) E p = k = lim m → ∞ ∫ Ω F ( x, ũ m, ṽ m ) d x, c ≥ 1 p - 1 p * k = 1 N k, where k is a nonnegative number.

Finally, since equalities, hold, we can apply inequality (4.2), using the above relations to get the required inequality.

Using the inequalities (3.31)–(3.35) and the above equality, we can define the mappings by (3.37).

Letting (klongrightarrowinfty) in the above equality, using the continuity of θ and Lemma 14, we get theta bigl( d ( x_{ast},Tx_{ast} ) bigr) leq bigl[ theta bigl( d ( x_{ast},Tx_{ast} ) bigr) bigr] ^{k}< theta bigl( d ( x_{ast},Tx_{ast} ) bigr), which is a contradiction.

end{aligned} Taking the limit as (mtoinfty) in the above equality and using condition (15), one obtains lim_{mtoinfty}sigma_{mr}=frac{1}{1-z} quad mbox{uniformly in }r if and only if (zin R).

On the other hand, since by (3.15), we have V ( q u ) ≤ V ( q ω ) ≤ ω < ∞, then by applying the operator ( I + V ( q ⋅ ) ) on both sides of the above equality and using (3.11) and (3.12), we conclude that u satisfies u = ω − V ( u φ ( ⋅, u ) ). (3.17).

Using point (i) from Theorem 1 in the above equality we obtain the inequalities of the statement.