Sentence examples similar to a function of space and Q z from inspiring English sources

The general form of the corresponding partial differential equation is partial_{t} psi+ sigma_{z}A z partial_{z} psi= Q z,t) psi, (46) with (A z) inmathbb{R}) a function of space and (Q z,t)) the two-by-two gravitational collision matrix associated with Eq. (45).

The GIXD data were reduced first by producing contour plots of intensity as a function of Q xy and Q z.

Note that these steady-state probabilities are functions of α, β, and q z.

The distribution of 17.1 million molecules from the PubChem Compound database as a function of (a) Q x and Q y, (b) Q x and Q z, and (c) Q y and Q z, respectively.

Therefore, q w and q z are the functions of τ, λ w and λ z.

Proof Let us consider a function q given by q ( z ) = z f ′ ( z ) / f ( z ).

Applying the Hadamard factorization of a meromorphic function, we write f ( z ) as follows: f ( z ) = z m P 1 ( z ) P 2 ( z ) e Q ( z ), where P 1 ( z ), P 2 ( z ) are entire functions such that ρ ( P 1 ) = λ ( f ), ρ ( P 1 ) = λ ( 1 / f ) and Q ( z ) is a polynomial such that deg Q ( z ) = q > max { k, λ ( f ), λ ( 1 / f ) } + 1.

These steady-state probabilities π are the functions of p, q w, q z, α, and β.

Since qt is only a function of the coordinates x, y, and z, it integrates out to a volume V, leaving q i = V * q i ′(T).

We now introduce the q-extension of exponential function as follows: e q ( z ) = ∑ n = 0 ∞ z n [ n ] q !

Let us define ϕ ( w ) = β / w 2 and Re ϕ [ q ( z ) ] = Re ( β 1 + z ) > β 2 > 0. Consider the function Q defined by Q ( z ) : = β z q ′ ( z ) q 2 ( z ) = β z 2 ( 1 + z ) 3 2. Further, Re ( z Q ′ ( z ) Q ( z ) ) = 1 − 3 2 Re ( z 1 + z ) ≥ 1 4 > 0. Thus the function Q is starlike, and the result now follows by an application of Lemma 1.3.