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On the uniqueness of algebraic curves passing through $n$-independent nodes | | H. Hakopian
; S. Toroyan
; | | Date: |
18 Oct 2015 | | Abstract: | A set of nodes is called $n$-independent if each its node has a fundamental
polynomial of degree $n.$ We proved in a previous paper [H. Hakopian and S.
Toroyan, On the minimal number of nodes determining uniquelly algebraic curves,
accepted in Proceedings of YSU] that the minimal number of $n$-independent
nodes determining uniquely the curve of degree $kle n$ equals to ${mathcal
K}:=(1/2)(k-1)(2n+4-k)+2.$ Or, more precisely, for any $n$-independent set of
cardinality ${mathcal K}$ there is at most one curve of degree $kle n$
passing through its nodes, while there are $n$-independent node sets of
cardinality ${mathcal K}-1$ through which pass at least two such curves. In
this paper we bring a simple characterization of the latter sets. Namely, we
prove that if two curves of degree $kle n$ pass through the nodes of an
$n$-independent node set ${mathcal X}$ of cardinality ${mathcal K}-1$ then
all the nodes of ${mathcal X}$ but one belong to a (maximal) curve of degree
$k-1.$ | | Source: | arXiv, 1510.5211 | | Services: | Forum | Review | PDF | Favorites | |
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