Sentence examples for p e I from inspiring English sources

The information sent from e j to e i can be calculated by I e j - > i = log 2 p ( e i / e j ) P ( e i ).

From Table 1, the probability P(E i /E j ) could be calculated, giving the constraint ∑ i = 1 6 P ( E i / E j ) = 1.

It is clear that ξ i e i T ( t ) P e i ( t ) ≤ V ( t ).

The given constraint is ∑ i = 1 6 P ( E i ) = 1.

Thus, we can conclude P e, i ≐ ρ - L s r i ( 1 - 2 r ′ ).

From the construction of V ( t ), we have V ( t ) ≥ ξ i e i T ( t ) P e i ( t ).

HbNIP2 1 represent typical urea SDPs (H-P-T-A-M-P-G-S-N), and SDPs of V-V-H-P-E-I-I-A-P with the substitution of V for I at SDP2 in comparison to typical boric acid SDPs (T/V-I-H-P-E-I/L-I/L-A/T-A/G/P).

Here, the fermionic field ψ is ψ ( t, x ) = 1 2 π ∫ d p ( b p e − i ω p t e i p x + d p † e i ω p t e − i p x ) (9). with b p ( d p ) fermionic (antifermionic) operators obeying the anticommutation rules { b p, b p ′ † } = δ ( p − p ′ ) and { d p, d p ′ † } = δ ( p − p ′ ).

S e: Energy demand of application i, calculated as: S e = p e − i e; where p e is the energy demand corresponding to sector e and i e. is the amount of fossil fuel renewable sources can not replace.

By setting the first-order term in this equation to 0, we obtain a differential equation for g λ (j): (18) − α (j 0, ω ) 2 g ' λ (j ) = (j − j 0 ) g λ (j ) where, (19) α (j 0, ω ) 2 = − ∑ p p c (j 0, p ) e − i ω p ∑ p ∂ c ∂ j | j 0, p e − i ω p. Thus α is a ratio of discrete Fourier transforms at the frequency ω.

By virtue of Theorem 1, we consider the function g z) defined by g ( z ) = 2 β - p e - i α + 2 ( p cos α - β ) 1 - z ( z ∈ U ). (4.2).