The phrase "a relation for which" is correct and usable in written English.
It can be used in contexts where you are defining or describing a specific relationship or connection that meets certain criteria.
Example: "We need to establish a relation for which the variables are directly proportional to each other."
Alternatives: "a relationship that" or "a connection for which".
It is, in fact, a distinct relation for which causal dependence is, at best, a defeasible marker.
One theory that he did not seriously argue against is the view that the C-relation is a primitive transitive, asymmetric relation for which nothing positive may be said concerning it.[11] I will briefly discuss a theory that he entertains in his article "The Relation of Time to Eternity", which was re-published in Mind in 1909 but was written at least two years before.
lineage i. hRd denotes a "more feasible or indifferent" relation, for which the propensity of h is greater or equal to that of d.
The purpose of this article is to establish the existence of multiple positive solutions of the dynamic equation on time scales, subject to the multi-point boundary condition, where is an increasing homeomorphism and satisfies the relation for, which generalizes the usually p-Laplacian operator.
There are two kinds of relation for which there are no transitive laws: intransitive relations and nontransitive relations.
Thus, "…is equal to…" is such a relation, as is "…is greater than…" and "…is less than…" There are two kinds of relation for which there are no transitive laws: intransitive relations and nontransitive relations.
The paper provides a complexity analysis of the algorithm together with some results about its practical performance and describes a class of binary relations for which the algorithm outperforms the most efficient lattice-constructing methods.
Now, we give a relation for the given numbers and polynomials, which are Daehee, degenerate Daehee, partially degenerate Daehee and totally degenerate Daehee numbers and polynomials as follows: sum_{k=0}^{n}binom{n}{k}D_{k}d^_{n-k}= sum_{k=0}^{n}binom {n}{k}d_{k} tilde{d}_{n-k} and sum_{k=0}^{n}binom{n}{k}D_{k}(x d^_{n-k}= sum_{k=0}^{n}binom {n}{k}d_{k}(x) tilde{d}_{n-k}.
