Sentence examples for such point exists from inspiring English sources

converges strongly to P Ω ( 0 ), whenever such point exists.

Starting from (y^{0}=x^{u}), the method, within a finite number of iterations, either yields an integer point in P or proves no such point exists.

Given any polytope, within a finite number of iterations, the method either yields an integer or mixed-integer point in the polytope or proves no such point exists.

Given a polytope, the method, within a finite number of iterations, either yields an integer or mixed-integer point in the polytope or proves no such point exists.

Given an integer point (yin D(P)) with (y_{1}>x_{1}^{l}), the method, within a finite number of iterations, either yields an integer point (x^in P) with (x^leq_{l} y) or proves no such point exists.

The method starts from an arbitrary integer point in the space and follows a simplicial path that either leads to an integer point in a polytope or proves no such point exists when the polytope is in a specific form.

Then F and G have a coupled coincidence point in X. Proof Let x 0, y 0 ∈ X be such that G ( x 0, y 0 ) ⪯ F ( x 0, y 0 ) and F ( y 0, x 0 ) ⪯ G ( y 0, x 0 ) (such points exist by hypothesis).

It can be mathematically characterized by finding points (xin{C}) and (yin{Q}) that satisfy the property begin{aligned} Ax=By, end{aligned} (1.2) if such points exist, where C and Q are nonempty closed convex subsets of Hilbert spaces (H_{1}) and (H_{2}), respectively, (H_{3}) is also a Hilbert space, (A:H_{1}rightarrow{H_{3}}) and (B H_{2}rightarrow{H_{3}}) are two bounded and linear operators.

Counting how many such points exist in the space is like trying to pack spheres of diameter D to fill the space as efficiently as possible.

Let x0 ∈ X such that gx0 ≼ fx0 (such a point exists by hypothesis).

Proof Let x 0 ∈ X be such that β ( x 0, T x 0, T x 0 ) ≥ 1 (such a point exists from the condition (ii) a ).