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Exact(6)
Because of the Lipschitz continuity of ∇f and the fact that (Vert x_{i_{j}} -x_{j} Vert rightarrow0) the first and second terms converge to zero.
If 2 β − 1 > 0, then the term a n a n + 1 / b n → 1 as n → ∞, while the other terms converge to zero.
Then, for each (iin[n]), we have 1-frac{varepsilon^{q}}{np^{q}_{i}}=y^{q}_{i}-frac{p^{q} y^{q}}{p^{q} p^{q}}p_{i}^{q}, and since the right-hand side sequence is bounded, we can assume that the left-hand side terms converge to some (ainmathbb{R}).
Therefore, these reaction terms converge to zero as N→ ∞at least in the finite time interval.
Since these reaction terms have N γ −2= N−1 in their propensities when γ=1, which is smaller than the species number for S23 of order 1, these reaction terms converge to zero as N→ ∞.
Since the 2nd, 3rd, and 4th reaction terms have propensities with N=1 and the species number of S2 is of order 1, these reaction terms converge to nonzero limits in the limiting equation.
Similar(54)
Then, by letting ρ = x − a > 0 and multiplying both sides by ρ k + m), leads to a double series of positive terms converging to unity.
and for k → ∞ this term converges to zero.
If μ is sufficiently small, the first term converges to 0 and the second term converges to H R g X X R XX − 1.
The second term converges to zero by continuity of the function f in Θ and by the bounded convergence theorem, ▀ .
For ϵ - > ∞, the infection term converges to βTV.
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Justyna Jupowicz-Kozak
CEO of Professional Science Editing for Scientists @ prosciediting.com