نتایج جستجو برای: legendre curve

تعداد نتایج: 131940  

1995
Ivan T. Ivanov

The Legendre transform and its generalizations, originally found in supersymmetric σ-models, are techniques that can be used to give local constructions of hyperkähler metrics. We give a twistor space interpretation to the generalizations of the Legendre transform construction. The Atiyah-Hitchin metric on the moduli space of two monopoles is used as a detailed example. email: [email protected]...

2016
Payel Das Gnaneshwar Nelakanti

In this paper we analyse the iterated discrete Legendre Galerkin method for Fredholm-Hammerstein integral equation with a smooth kernel. Using a sufficiently accurate numerical quadrature rule, we obtain super-convergence rates for the iterated discrete Legendre Galerkin solutions in both infinity and L-norm. Numerical examples are given to illustrate the theoretical results.

Journal: :Proceedings of the American Mathematical Society 1954

Journal: :Journal of Computational and Applied Mathematics 2000

Journal: :Journal of Computational and Applied Mathematics 1993

Journal: :Rendiconti del Circolo Matematico di Palermo 1897

Journal: :Comptes Rendus Mathematique 2021

The evaluations of determinants with Legendre symbol entries have close relation combinatorics and character sums over finite fields. Recently, Sun [9] posed some conjectures on this topic. In paper, we prove also study variants. For example, show the following result:

Journal: :Des. Codes Cryptography 2015
Ming Su

Background • Legendre Sequence For a prime p > 2 let (s n) be the Legendre sequence defined as s n = 1, n p = −1, 0, otherwise, n ≥ 0, where. p denotes the Legendre symbol. • Sidelnikov Sequence Let q be an odd prime power, g a primitive element of F q , and let η denote the quadratic character of F Background • Legendre Sequence For a prime p > 2 let (s n) be the Legendre sequence defined as s...

2014
Jordan Bell

f((1− t)x1 + tx2) ≤ (1− t)f(x1) + tf(x2), x1, x2 ∈ C, 0 ≤ t ≤ 1, then f : X → R is convex. Proof. Let (x1, α1), (x2, α2) ∈ epi f and 0 ≤ t ≤ 1. The fact that the pairs (xi, αi) belong to epi f means in particular that f(xi) < ∞, and hence that xi ∈ C, as otherwise we would have f(xi) =∞. But (1− t)(x1, α1) + t(x2, α2) = ((1− t)x1 + tx2, (1− t)α1 + tα2), and, as x1, x2 ∈ C, f((1− t)x1 + tx2) ≤ (...

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