To find the explicit equations of the
right inverse, one has to integrate the set of one-forms, obtained at the last step of the IA, which may be a difficult task.
where [M.sup.R] is the
right inverse of Markov matrix M.
The
right inverse of the Hahn difference operator was proposed by Aldwoah in 2009 [14, 15].
So [mu] is a left inverse for [theta]; in order to show that [mu] is a
right inverse we compute as follows:
Most of the AIC schemes estimate
right inverse [[??].sub.R]([q.sup.-1]) and then it is used as left inverse [[??].sub.L]([q.sup.-1]) by considering left and
right inverse are equal.
The inverse (a left inverse, a
right inverse) operator is given by (2.9).
More precisely, we are interested in the following problem: does P(D) admit a continuous linear
right inverse, i.e., an operator S : [epsilon](K) [right arrow] [epsilon](K) such that P(D) [omicron] S = id [epsilon](K)?
Left and
right inverse eigenpairs problem is a special inverse eigenvalue problem.
It is called a
right inverse property quasigroup (loop) [RIPQ (RIPL)] if and only if it obeys the
right inverse property (RIP) yx*[x.sup.[rho]] = y for all x,y [member of] G.
A quasigroup (Q, -)has the left inverse property,the
right inverse property or the cross inverse property, if for any x [member of] Q there exists an element x-1 such that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII.], respectively.
There is a simple choice for the
right inverse G, namely,