The equation describing the amplitude frequency relation is a cubic polynomial equation.
The interference is a cubic polynomial phase function {w_{n2}}(t) = a*{exp left jpi v{t^{3}} + jpi u{t^{2}} + j2pi {omega_{1}} tright)}.
The conclusion is valid because (F omega,tau)) is a cubic polynomial in (omega^{2}) and the fact that (a_{i} ) ((i = 0,1,2)) and (b_{j}) ((j=0,1)) are all continuous functions of τ.
(e) The conclusion is valid because (F omega,tau)) is a cubic polynomial in (omega^{2}) and the fact that (a_{i} ) ((i = 0,1,2)) and (b_{j}) ((j=0,1)) are all continuous functions of τ. □.
where f ( V ) is a cubic polynomial in V which we choose, without loss of generality, to be t ( V ) = V − V 3 / 3. The parameter I ext models the input current the neuron receives; the parameters a, b > 0 and c > 0 describe the kinetics of the recovery variable w.
More precisely, when no danger of confusion is present, we will denote the function (F _{m, n} ^{k}) by F. It is obvious that F is a cubic polynomial on u which may be rewritten as F (u) = Delta t _{k} u ^{3} - (1 + gamma) Delta t _{k} u ^{2} + bigl[ gammaDelta t _{k} + D_{m,n}^{k} bigr] u - C _{m,n} ^{k}.
Define begin{aligned}& Sz_{1} = 1-alpha )^{s}Sz+alpha ^{s}Tz, & vdots & Sz_{n} = 1-alpha )^{s}Sz+alpha+alpha ^{s}Tz_{n-1}, end{aligned} where Sz is injective, Tz is a cubic polynomial and (n=2,3,4,ldots) , then (vert z_{n}vert rightarrow infty ) as (n rightarrow infty).
(21) and (23), replace the factors exp ( 2 i k L ) by exp ( 2 i k 0 L ) therein, and do the double integral by standard contour integral techniques enclosing the poles in the upper half complex plane (for the N = 3 case, for example, the denominator of each transmission amplitude is a cubic polynomial in k, so there are three roots).
The reader will notice that the first part of this model is a cubic polynomial.
Note that the exact displacement field u = (u 1, u 2 ) T is a cubic polynomial.
