Sentence examples for methods with initial from inspiring English sources

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The system may be solved by numerical analysis methods with initial conditions, like the Runge-Kutta-Gear method.

Subject recruitment was by snowball methods with initial identification made by research contacts within Indigenous health.

Similar(58)

The dispersion relation is solved by a modified Newton Raphson method with initial values given by an asymptotic expansion or a winding integral method.

The existence of stable subharmonic motions found is confirmed by solving the differential equation of motion numerically by means of a time-difference method, with initial conditions being supplied by the harmonic balance approximation.

In [10], Park et al. proposed a FMD method with initial mode selection.

Then Newton's method with initial point x 0 ∈ G considered in [16] can be written in a coordinate-free form as follows.

Following [17], we define Newton's method with initial point x 0 for f on a Lie group as follows: x n + 1 = x n ⋅ exp ( − d f x n − 1 f ( x n ) ) for each  n = 0, 1, …. (3.1).

(28) In Fig. 3 we show a numerical simulation prescribed by the interface method, with initial data equivalent to that from the full space time simulation shown in Fig. 2.

The notations used in Theorem 3.3 are explained as follows: (Y_{k}) generated by the one-step method is an approximation to the exact solution (X(t_{k})) of (1.1) with (t_{k}=kh), (X_{t,x}(t+h)) denotes the exact solution of (1.1) with initial value x at time t and (bar{X}_{t,x}(t+h)) denotes a numerical solution generated by the one-step method with initial value x at time t.

Moreover, if β ≤ b, h has a unique zero respectively in [ 0, r 0 ] and [ r 0, R ], which are denoted by r 1 and r 2. Let { t n } denote the sequence generated by Newton's method with initial value t 0 = 0 for h, that is, t n + 1 = t n − h ′ ( t n ) − 1 h ( t n ) for each  n = 0, 1, …. (2.18).

Let β > 0 and L > 0. The quadratic majorizing function h is reduced to h ( t ) = L 2 t 2 − t + β for each  t ≥ 0. Let { t n } denote the sequence generated by Newton's method with initial value t 0 = 0 for h, that is, t n + 1 = t n − h ′ ( t n ) − 1 h ( t n ) for each  n = 0, 1, …. Assume that λ : = L β ≤ 1 2. Then h has two zeros r 1 and r 2 : r 1 = 1 − 1 − 2 λ L and r 2 = 1 + 1 − 2 λ L ; (5.1).

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