Sentence examples for using the implicit function from inspiring English sources

This result is proved by using the implicit function theorem.

By using the implicit function theorem and Taylor series expansion, the observer-based control law and the weight update law of the HFNN adaptive controller are derived.

(2.20) Integrating the above formula after separating the variable and using the implicit function theorem, one can claim that the system (2.20) has a solution.

Therefore, solving (F-B=0) using the implicit function theorem with respect to the rescaled parameter A (recall (a=sqrt{varepsilon }A)) allows us to prove the presence of saddle-node homoclinics of ('jump-back') canard type.

Therefore, using the implicit function theorem, the map u = u is infinitely differentiable at any ℓ in the interior of A. We now compute the first and second derivatives of the coefficient-to-solution map.

Similar to the proof in Theorem 1, using the implicit function theorem, we can prove that the power series R ( z ) = z + ∑ n = 2 ∞ C n z n is convergent in a neighborhood of the origin.

Proof of Theorem 4.1 By using (4.2), Lemma 4.2 and Lemma 4.3, we can use the implicit function theorem ([25], p.61) that permits us to say that there exist a neighborhood of 0 in E 2 ( R ), a neighborhood of 0 in P and a C 1 -mapping p ↦ x ̲ [ p ] from into, which satisfies the following conditions: (a) x ̲ [ 0 ] = 0, that is, the condition (i) of Theorem 4.1.

We follow the idea of [8, 9] and use the implicit function theorem to solve (2.1 - 2.2 2.1 - 2.2

The bifurcation argument uses the implicit function theorem and relies on the linearization.

Hence, if (alpha ) is large enough, we can use the implicit function theorem to construct a smooth map (G: D_alpha (x) rightarrow N) such that (bar{F} circ G = mathrm{id}_{D_alpha (x)}) and (G(F x)) = x).

To fulfill this, we use the Implicit Function Theorem to show that there exists a δ > 0 such that there are functions a 0 , a 0 ~ , a 1 and a 1 ~ which are continuous for 0 ≤ α - 1 < δ and a 0 ( 1 ) = a 0 ~ ( 1 ) = 0, a 1 ( 1 ) = a 1 ~ ( 1 ) = a.