Download Die Mechanik des Himmels - Vorlesungen Band 2 by Carl Ludwig Charlier PDF

By Carl Ludwig Charlier

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5). 18 1 Asymptotic Estimates On the boundary of the domains for λ ∼ μ−4 , the Newton polygon has one segment determined by the points M2 , M4 , M5 and M6 , which corresponds to the abridged equation μ4 x 6 + 1 − ν 2 λμ4 x 4 − λx 2 − 1 − ν 2 λ2 = 0. 4 Exercises Use Newton polygons to find the first and second terms in the expansions for the roots of the following equations for μ 1. 1. x 3 − 3xμ + μ3 = 0. 2. μ4 x 4 − x 2 + x − μ = 0. 3. μ−3 x 3 + μ−1 x 2 − μ−2 x + 1 = 0. 4. μ5 x 5 − μ2 x 3 + x − μ3 = 0.

M2 , M4 , M5 , M6 ), where the points Mi have the following coordinates: M1 = {0, 0, 1}, M4 = {2, 0, 1}, (see Fig. 7). 3 Newton Polygons 17 Fig. 7 Newton polyhedron Equating the orders of the terms which define the facets we get λ ∼ λ2 ∼ x 2 ∼ λx 2 , λx 2 ∼ x 2 ∼ μ4 x 6 , λ2 ∼ λx 2 ∼ λμ4 x 4 ∼ μ4 x 6 . Hence, for the first and second relations, λ ∼ 1, and for the third, λ ∼ μ−4 , and the entire domain of the parameter λ splits into three subdomains, where the Newton polygon has a similar structure.

M2 , M4 , M5 , M6 ), where the points Mi have the following coordinates: M1 = {0, 0, 1}, M4 = {2, 0, 1}, (see Fig. 7). 3 Newton Polygons 17 Fig. 7 Newton polyhedron Equating the orders of the terms which define the facets we get λ ∼ λ2 ∼ x 2 ∼ λx 2 , λx 2 ∼ x 2 ∼ μ4 x 6 , λ2 ∼ λx 2 ∼ λμ4 x 4 ∼ μ4 x 6 . Hence, for the first and second relations, λ ∼ 1, and for the third, λ ∼ μ−4 , and the entire domain of the parameter λ splits into three subdomains, where the Newton polygon has a similar structure.

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