example: chemequ description: TITLE : chemical equilibrium of hydrocarbon combustionREFERENCES :This polynomial system describes the equilibrium of the products of hydrocarbon combustion.Keith Meintjes and Alexander P. Morgan: "Chemical equilibrium systems as numerical test problems", ACM Toms, Vol 16, No 2, 143-151, 1990.NOTES :Although the total degree equals 108, there are only 4 real and 12 complex solutions and an infinite number of solutions at infinity.A typographical error has occured in equation (2d), instead of `+ 4Ry5', it should be a `- 4Ry5'.Applying m-homogenization straight to it renders B = 56. Simple linear reduction makes the total degree equal to 48. With m-homogenization, no better upper bound can then be computed.The constants are : R = 10; p = 40; sqrt(p) = 6.3246 1/sqrt(p) = 0.1581 1/p = 0.0250 R5 = 1.930e-01 (2*R5 = 3.8600e-01) R6 = 2.597e-03/sqrt(p) = 4.1062e-04 R7 = 3.448e-03/sqrt(p) = 5.4518e-04 R8 = 1.799e-05/p = 4.4975e-07 R9 = 2.155e-04/sqrt(p) = 3.4074e-05 R10 = 3.846e-05/p = 9.6150e-07 (2*R10 = 1.9230e-06) system: Polynomial variables: y1 > y2 > y3 > y4 > y5 equations: 0 1 2 3 4 length of Janet-like basis: 18 length of Janet basis: 18 length of Gröbner basis 18 Hilbert polynomial: 16 Strategy: degJ highJ lowJ degJL highJL lowJL

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