نتایج جستجو برای: k digamma function
تعداد نتایج: 1538186 فیلتر نتایج به سال:
Let $D$ be a finite and simple digraph with vertex set $V(D)$.A signed total Roman $k$-dominating function (STR$k$DF) on$D$ is a function $f:V(D)rightarrow{-1, 1, 2}$ satisfying the conditionsthat (i) $sum_{xin N^{-}(v)}f(x)ge k$ for each$vin V(D)$, where $N^{-}(v)$ consists of all vertices of $D$ fromwhich arcs go into $v$, and (ii) every vertex $u$ for which$f(u)=-1$ has a...
Let $kge 1$ be an integer, and let $G$ be a finite and simple graph with vertex set $V(G)$.A weak signed Roman $k$-dominating function (WSRkDF) on a graph $G$ is a function$f:V(G)rightarrow{-1,1,2}$ satisfying the conditions that $sum_{xin N[v]}f(x)ge k$ for eachvertex $vin V(G)$, where $N[v]$ is the closed neighborhood of $v$. The weight of a WSRkDF $f$ is$w(f)=sum_{vin V(G)}f(v)$. The weak si...
This research focuses on the Ostrowski–Mercer inequalities, which are presented as variants of Jensen’s inequality for differentiable convex functions. The main findings were effectively composed functions and their properties. results directed by Riemann–Liouville fractional integral operators. Furthermore, using special means, q-digamma modified Bessel functions, some applications acquired ob...
Let $G$ be a graph with vertex set $V(G)$. For any integer $kge 1$, a signed (total) $k$-dominating functionis a function $f: V(G) rightarrow { -1, 1}$ satisfying $sum_{xin N[v]}f(x)ge k$ ($sum_{xin N(v)}f(x)ge k$)for every $vin V(G)$, where $N(v)$ is the neighborhood of $v$ and $N[v]=N(v)cup{v}$. The minimum of the values$sum_{vin V(G)}f(v)$, taken over all signed (total) $k$-dominating functi...
In this work, we study the k-means cost function. The (Euclidean) k-means problem can be described as follows: given a dataset X ⊆ R and a positive integer k, find a set of k centers C ⊆ R such that Φ(C,X) def = ∑ x∈X minc∈C ||x− c|| 2 is minimized. Let ∆k(X) def = minC⊆Rd Φ(C,X) denote the cost of the optimal k-means solution. It is simple to observe that for any dataset X, ∆k(X) decreases as ...
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