Sentence examples for continuing this argument from inspiring English sources

Continuing this argument, we have (varphi ^{n} tau ) > varepsilon ) for any (n in mathbb {N}).

Continuing this argument, we can prove (varphi^{n} (t) leqtheta_^{-1} circpsi^{n} circtheta(t) ) by induction.

Since φ is nondecreasing, we have (varphi ^{2} tau ) geq varphi (tau ) ). Continuing this argument, we can show that ({ varphi ^{n} tau ) }) is nondecreasing.

So by continuing this argument we obtain two sequences { x n } n = 0 ∞ and { y n } n = 0 ∞ such that x 0 ≤ x 1 ≤ ⋯ ≤ x n ≤ ⋯ and y 0 ≥ y 1 ≥ ⋯ ≥ y n ≥ ⋯  .

Hence, there exists n 2 > m 2 such that τ + δ 4 < t n 2, q n 2 < τ + δ 2. Continuing this argument, we can define a subsequence { n i } of { n } satisfying.

Continuing this argument, it is easily seen how the contrasting TACS-3 pattern found in the literature, i.e. collagen fibers oriented perpendicular to the tumor boundary, may allow metastasizing cells to travel outward along these collagen "tracks" to more readily invade the host [ 5, 10].

A week ago, Orbán continued this argument at a press conference with the rightwing Polish politician Jarosław Kaczyński, in which he called for Europe and its institutions to be reworked in favour of a rightwing vision.

Continuing this line of argument, councils may use their legal voice less often than was presented in the results.

He continues this line of argument in "God's Name in Vain," claiming that our culture marginalizes religion, making it "easy to paint people who put God first as dangerous fanatics".

Continuing this "house that Watts built" argument, Mr. Genser then told a tale of how Mr. Watts was playing tennis one day, became frustrated with his serve and took his anger out on the automatic ball feeder.

By continuing this process and following an argument similar to that in Theorem 4 in [2] (see also [1]), it is easy to see that d ( T m + 1 x 0, T m x 0 ) ≤ l v, where m = 2 l or m = 2 l + 1 for l ≥ 1 or d ( T m + 1 x 0, T m x 0 ) ≤ l − 1 v, where m = 2 l or m = 2 l − 1 for l ≥ 2. Thus, d ( T m + 1 x 0, T m x 0 ) → 0. By using Lemma 2.1, T has an approximate fixed point.