Sentence examples for term can be rewritten from inspiring English sources

The first term can be rewritten as h 2 ∑ i = 1 N ( σ x i σ x i + 1 + σ y i σ y i + 1 ).

The first term on the right hand side of Eq. (8) is then ∂N/∂z⋅u PV = ∂N/∂z⋅w⋅sin2(I), with the help of Eq. (6), and the second term can be rewritten as N⋅∂u PV/∂z = N⋅∂w/∂z⋅sin2 (I) = i⋅N⋅k⋅w⋅sin2 (I) = i⋅N⋅2πf IS /c s w⋅sin2 (I), where i 2  = −1, and k, f IS, and c s represent the infrasound wave vector (number), infrasound frequency and infrasound speed, respectively.

The summation in the second term can be rewritten ∑ i = 1 N ∑ j = 1 N M j i ( z j - z i ) = ∑ i = 1 N ∑ j = 1 N M j i z j - ∑ i = 1 N ∑ j = 1 N M j i z i = ∑ i = 1 N ∑ j = 1 N M j i z j - ∑ i = 1 N ∑ j = 1 N M i j z j.

The last term can be rewritten as (6) This is the modified partition function over the alignment of x and x* admitting only those base pairs that can be formed by both structures.

It can be seen immediately that the second term can be rewritten as Here, is the vector connecting the apex (on surface b) and the midpoint on the base (on surface b′) of the Borrmann fan which is supposed for dynamical diffraction in crystal B (see Fig. 3 ▸); is the angle between the diffracting plane and the surface normal.

The dissipative terms can be rewritten in terms of energy dissipations cause to the vehicle and by the payload: E{x}_{vehicle}={m}_vleft( cg{v}_{av}t+frac{1}{2}{v}_{max}^2right)+fraC}_DA{rho}_{airho}{v}ir}{v}_{av}^3t (3). is the component due to vehicle even at zero payload E{x}_{payload}={m}_pleft( cg{v}_{av}t+frac{1}{2}{v}_{max}^2right) (4). is the component due to payload.

As shown in the right-hand side of (11), the original loglikelihood function by dropping constant terms can be rewritten as begin{array}rcl@ L boldsymbol{theta}_{ell}) &=& -sumlimits_{f=,f_{1}}^{f_{N}}frac{left(hat{x}^{[i]}_{ell}left(,fright -alpha_{ell} pleft(,fright -alpha_{ellll}right)right)^{2}}{mathrm{E}left[|n'left(,fright)|^{2}right]} end{array} (17).

The fractional part of the term, which can be rewritten u q 1 + u q, has a minimum value of 0 (which it will obtain when there is no input, when u = 0) and a maximal value of 1 (which it will approach, when the magnitude of the input is much greater than 1).

While for n = 0 (14) reduces to an LS problem, for n > 0, omitting again irrelevant terms, it can be rewritten as d n ( i ) = arg min d n ∈ ℝ L 1 2 d n T R n : N d n + d n T g n ( i ) + λ d n 2 (18).

Condition (5.4) can be rewritten as (5.7).

Then the design equation (Eq. 4) of the reactor in term of the volume Vr can be rewritten in term of the reactor area Ar as follows: frac{{A_{text{r}} }}{{Q_{0} }} = frac{1}{{H_{text{r}} k_{text{d}} }}left[ {frac{{Y(S_{0} - S }}{X} - frac{beta (1 + alpha )}{alpha + beta }} right].